YT YouTube2Text

YouTube2Text → Transcripts → Complex Systems - Jean-Philippe Bouchaud - Lecture 4: Optimisation, HJB & “Chaos”

Complex Systems - Jean-Philippe Bouchaud - Lecture 4: Optimisation, HJB & “Chaos” — Transcript

by EconophysiX Lab · 24,232 words · 4,005 segments · language en · Watch on YouTube

Open in transcriber Summarize with AI Download SRT/VTT

Full transcript

0:01this conference will now be recorded
0:06so good morning everybody
0:09um
0:10so I'm going to continue where I left
0:12the last time but let me give an outline
0:15of what I'm going to talk about today
0:19um so I'm speaking about this relation
0:22between optimization problems and growth
0:25models which is in a sense quite
0:28unexpected and so I'd like to show you
0:30how it works so I'm going to review the
0:34Bellman method that leads to so-called
0:37Hamilton Jacoby Bellman he JB equation
0:42related to path optimizations I'll show
0:45you that in this optimization problems
0:47there are very interesting phenomenon
0:50happening shocks and discontinuities
0:53I'll speak about traffic jam which is
0:56surprisingly uh related to this problem
0:59it's a kind of you know infinite drawer
1:02problem where
1:03problems map themselves onto one another
1:06in an interesting way
1:09I'll mention that the equations that we
1:11get is related to the so-called kpz
1:14equation Canada par easy which has
1:17become one of the uh
1:21totem of uh theoretical physics in the
1:24last 20 or 25 years
1:27um and I'll end with this very
1:29interesting concept of Chaos in
1:32optimization problems so I'm sure that
1:34all of you have heard about chaos but
1:37here chaos takes a slightly different
1:39meaning which
1:41for our general perspective on economics
1:44for example where people think about
1:46human beings as optimizers in some
1:50situations you'll see that minute
1:52changes in the parametrization of the
1:55optimization problem leads to completely
1:58different solutions which you know shows
2:02that this idea of optimization must be
2:05handled with care
2:08okay so if you remember I talked about a
2:11kind of
2:12treasure hunt problem where
2:16a guy on his bicycle say as to
2:20wander around
2:23to catch
2:24bounties on its way
2:27but of course
2:29you know moving the extra mile to get a
2:31good Bounty caused some energy
2:35or there's a cost Associated to moving
2:39away from the straight path and so what
2:41I said is that
2:43the gain Associated to a sudden path is
2:48the integral from 0 to capital T capital
2:50T being
2:52the time at which the game ends
2:55DT
2:58of x and t
3:00and T
3:01so this is the Bounty or the reward
3:05collected along the path along the path
3:09X of t Okay so there's one here bound
3:12here another one here maybe there's one
3:14here that you decided not to pick up
3:18and then there's a cost Associated to
3:21this uh extra mile
3:23that we say which
3:26I'm written as
3:28uh DT V squared
3:33of X of t
3:35and T
3:37and what I said was that V here is the
3:41control is what you decide to try to uh
3:44to achieve on your bike so your
3:47trajectory EXP
3:50is not exactly given by VT this would be
3:54too simple uh but it's it there's an
3:58extra term here which is uh so exp via
4:03vaccine t
4:04Plus
4:06Phi
4:07and this is some some noise
4:11some annoyanced on your way such that as
4:15I said side T side key Prime
4:20cannot be anticipated and so it's it's a
4:23Delta function it's given by 2J
4:27delta T minus D Prime
4:31so there's this random noise along your
4:33path
4:34you're trying to give yourself some
4:37velocity V that's the cost you have to
4:38pay and all in all that's your objective
4:41function
4:43that's called an objective function
4:52and so let me uh say a few more things
4:55about this problem that I already said
4:57last time but just to be absolutely
4:59clear about this so this can be seen
5:01also as a physics problem
5:06you can see that as a theoretical
5:08mechanics problem where
5:11this would be minus the energy
5:15and the B's would be pinning sides
5:18Associated to pinning sites and pinning
5:21sides to track the polymer or the vortex
5:25or the dislocation whatever example you
5:28have you want to have in your mind
5:30and so uh Associated to a certain
5:33location of a pinning site you gain some
5:36energy B but you pay some uh elastic
5:41energy due to uh deforming increasing
5:45the length of of your object okay
5:49so
5:51to add here
5:53also
5:56energy
5:59of pinning problems
6:14and of course in a statistical mechanics
6:16setting
6:17the the T equals zero the zero
6:20temperature limits means that you want
6:22to find the best path
6:24the optimal path
6:27the other thing that I mentioned was
6:29that you can
6:31think of higher dimensional objects not
6:36one-dimensional but two-dimensional
6:37membranes and so on but in this case the
6:41mapping to such a an optimization
6:44problem in the sense of of my
6:48bounty collection or a treasure hunt
6:51makes less sense because in a sense time
6:54would have to be
6:56dimensional on the other hand something
6:58that you can easily generalize to is the
7:02case where X instead of being a
7:04one-dimensional object becomes a d
7:07dimensional object so this doesn't
7:09really
7:11tough much so if you add little arrows
7:15here you can actually have a still a
7:18one-dimensional problem in time but
7:20multi-dimensional problem in space where
7:23you're not confined to a one-dimensional
7:26direction you can actually move around
7:28in two Dimensions or three dimensions
7:30whatever okay
7:33so what I said last time is the
7:35optimization problem is how to best
7:37choose the control
7:40so that's the uh the policy if you want
7:43or the control
7:44and so how to best choose it in order to
7:47achieve the best average gain because of
7:51course this trajectory X of T will
7:55depend on the realization of the noise
7:56which you don't know
7:58and so the best you can do or at least a
8:02reasonable objective function is to find
8:05the best G
8:07in the sense of averaging over pi
8:12okay so this is the setting of the
8:14problem that I wanted to repeat before
8:17moving to the Bellman method
8:27Bellman argument
8:31which you're going to see is a very uh
8:34General argument and also very nice that
8:37allows you to think about this problem
8:40in a in a general setting where this
8:43Belmont argument
8:45leading to the Amazon Hamilton Jacoby
8:48Bellman equation is going to be useful
8:51so the Bellman argument is
8:53let's assume
8:57that the problem is solved
9:04for all
9:07key Prime greater or equal to than 70.
9:12so actually Four
9:15T left or equal and P Prime less than n
9:20t or equal to so what I'm arguing here
9:23is okay imagine that there's a certain P
9:26here
9:27and I know that I solved the problem
9:30already
9:32uh at later times
9:34okay in the sense that I'm able
9:38to
9:39compute to know something that I'm going
9:42to call the gain to go sometimes it's
9:45also the cost to go okay to go
9:49and you'll see why it's called the game
9:51to go in a second so the game to go is a
9:54certain function that I'm going to call
9:56G of x and t
10:00so here I'm resulting back to a scalar a
10:04one-dimensional problem X where X is a
10:07scalar so I'm assuming that I've solved
10:10the problem in the sense that I know
10:12what the optimal gain is going to be
10:16from where I am now at X and at time T
10:20and of course I also know all the G of x
10:23and t Prime for t
10:25for T Prime between T and capital T okay
10:29so I've solved the problem I know
10:31wherever I'm I am
10:33in the X Direction I know everywhere the
10:37G of x and t okay
10:41so in particular
10:43what I want to solve
10:46is for G or the average G of x i which
10:51is by definition
10:52G of x0
10:55P equals zero okay
10:58so that's my my aim I want to compute
11:02this object
11:03where x0 is my initial position
11:08foreign
11:09so that's at the end of the calculation
11:11what I have to compute but in the
11:14meantime I'm assuming that I only know
11:17this function G up to Prime T I mean
11:20coming from the future
11:22so I know
11:25G of x and t here everywhere and now I
11:28want to iterate somehow I want to find a
11:30recursion that allows me to know G of X
11:33at T minus DT okay
11:36so that's going to be the Belmont
11:38argument it's the recursion argument
11:40coming from the future and recasting to
11:43the Past okay
11:45and at the end if I'm able to do this
11:48well I'm going to be able to compute
11:51my objective more optimal objective
11:54function at time 0. so again what I'm
11:59noting here by G is the optimal thing
12:01the best thing I I can do
12:04starting from X at time t
12:08okay so this is the optimal
12:11sorry
12:13is the average value or because I still
12:19have disorder after T right
12:22exactly so G of x and t is D is the best
12:25average value of what I can do from T to
12:28capital T
12:29okay thank you it has the same meaning
12:32as as this one
12:33uh but I'm assuming I know it already
12:36and I'm going to find the recursion
12:38relation spread
12:40okay
12:41so let me move my screen my camera a
12:44little bit
12:50a
12:53so how does this work
12:58so the question is what should I do
13:06I do best
13:09at
13:10a t
13:13minus DT
13:16okay I know what I have to do from T
13:18onwards what should I do at T minus DT
13:21and what is the
13:24um the game to go at x minus each at T
13:28minus DT
13:30so
13:32what I'm going to write is I'm going to
13:35introduce a notation which is U is going
13:38to be V of x
13:41and T minus DT
13:44so it's the velocity of my bike just at
13:47time before t
13:49and I'm going to try to find what's the
13:51best value view possible
13:53okay and what I'm saying is that you
13:57is what minimizes
14:01so admin
14:03over U
14:04of what the little extra cost I have to
14:08pay between t and t plus DT
14:12um and so this is going to be uh
14:17minus Lambda squared minus Lambda over
14:21to U squared
14:24okay this is times DT this is what I
14:28will have to pay in terms of uh kinetic
14:31energy if you want between T and and t
14:34plus DT and between T minus DT and T
14:37then there is the little piece of
14:42of bounty that I'm going to collect so
14:45this is going to be independent of you
14:47this is B of X of t and t e t okay but
14:53this is independent of U so I I cannot
14:56optimize over it
14:59thanks for this view star this is what
15:02what I'm looking for and then
15:04well
15:06I'm going to if I'm at X at time T and I
15:10apply a velocity U to my uh pedaling I'm
15:16going to be at time
15:17T at
15:20so can you still seem
15:27when I'm writing here
15:29maybe not plus u d t plus PSI EG
15:37okay so if I'm at at X
15:43sorry thank you try to write it write
15:47that more properly
15:52so I'm assuming that I'm X at times C
15:54minus DT and at time T I will be by
15:57definition at U
16:00DT away from X but then there's the
16:03annoying noise on top of that plus side
16:06H Okay so
16:12what I'm going to gain in total is
16:17this G function that I already know at
16:20time T and at time T I might as I just
16:24said that I I'm X Plus q d t
16:26plus PSI DT
16:30t
16:33average Rubik's side
16:36okay
16:39so if you want
16:42in this
16:44total gain I'm going to have a piece
16:47that depends only on the future
16:50but that I'm assuming I already know it
16:52is my G of x and t the only thing is
16:55that I don't know where I'm going to end
16:57up
16:58I'm going to end up at X plus u d t plus
17:00PSI DT and PSI is random so
17:04what I'm going to gain further on is
17:07going to be given by a function that I
17:09know but I don't know exactly where to
17:11compute it and that's where I have to
17:13average of x i so this is the big piece
17:16coming after time T and then the extra
17:20piece in this big integral coming from
17:22the little layer between T and T minus
17:26DT
17:27so this little piece here
17:29is given by
17:31minus Lambda squared G squared DT minus
17:34Lambda over 2 U Square DT plus b of x
17:37and t
17:38uh GT okay but this as you see doesn't
17:41depend on you
17:45okay
17:46so I I'm cutting the optimization
17:49problem in pieces I'm assuming I know
17:51what to do Beyond Time T and now I'm
17:55only concerned with the optimization
17:57between T minus DT
17:59and T okay
18:02so
18:04let's expand
18:09sorry
18:11oh yes
18:12you star is it an admin or in a Max
18:22you're absolutely right I was uh in my
18:25head I was thinking of the problem in
18:26terms of energy where all the signs are
18:29flipped but it's obviously the r Max
18:32thank you we want to maximize the gain
18:34and not minimize it we want to minimize
18:37the energy but maximize the date yes
18:40thank you
18:43okay so
18:45um let me again move my
18:52camera
18:54all right
18:58so now I'm going to use the a Taylor
19:01expansion
19:03to write that x g of X Plus
19:08UDT
19:11plus PSI DT
19:14and key
19:18average of PSI
19:20this is
19:21equal to so the first term is G of x
19:26and T
19:27a then there's a term
19:31the first term in the Taylor expansion
19:33is Plus
19:34U
19:36plus PSI
19:39average with PSI
19:41ET
19:42EG
19:44the X
19:46and then
19:48there I can't stop at all the DT because
19:51of of the PSI term you remember that
19:55I told you in a very early on that when
19:59you write this thing
20:01it's a little bit ill-defined from a
20:03mathematical point of view because uh
20:05the Delta function is infinite well T
20:08equals T Prime so you remember what I
20:11told you it means that PSI is actually
20:13of order one of
20:15of all this of all the one over square
20:18root of DT
20:19okay
20:20and and therefore
20:23this side here is is formally infinite
20:27when DT goes to zero but more precisely
20:30is of all the one over square root of BT
20:32and so I have if I wanted to work at all
20:36the DT I have to keep the term PSI
20:39squared DT squared because it's going to
20:42be also of all the DT
20:44so let me write it plus one half
20:48uh so in principle I should keep U plus
20:52PSI squared so let me do it I squared DT
20:56squared
20:57e to G DX squared
21:01okay
21:03but the only term of all the DT in all
21:05this when I average over five
21:12is the PSI squared term
21:15so let's let's do the averaging
21:18um
21:20the average value of PSI is zero so this
21:23goes away
21:25u d t squared this is of all the DT
21:27squared so I don't have to keep it
21:29uh the Double product is also zero which
21:33is the average value of size zero and so
21:35the last the next the only term that
21:37remains here is one half
21:40of uh 2 J
21:44um
21:45DT
21:48okay so it's a little hand waving but
21:51this comes from the fact that Pi is of
21:54all the whatever square root of DT and
21:56you can can make that rigorous with the
21:59the formalism of stochastic differential
22:02equation
22:04okay so all in all what do I have
22:07I have that U star
22:14is the ARG Max
22:17so the value of U that maximizes
22:21minus Lambda over to
22:24uh
22:25U Square
22:28so I'm dropping the terms of all the GT
22:30that I'm going to factorize and then the
22:32only term that depends on you here is
22:36this one plus hue
22:39DG DX
22:43MPT Plus
22:46something that doesn't depend on you you
22:49see that my my term here which is j e 2
22:53G over DX squared doesn't depend on U
22:55and the the Bounty term doesn't depend
22:58on you in either
23:00so the only two terms that depend on you
23:02are this one coming from uh the direct
23:05cost of paddling faster and this term
23:08which changes the arrival point at time
23:13t plus DT
23:15okay but you see that this is now a
23:18trivial problem I have as a function of
23:20U this is a quadratic problem
23:24and so uh the maximization is a function
23:28of U is
23:30um is given by the a linear equation
23:33which comes from taking the derivative
23:36of this function with respect to U
23:38and so what I get
23:40is
23:43U star
23:44equal
23:48um one over Lambda
23:52EG
23:56okay
23:58this comes really you know if you take
24:00the derivative of that with respect to U
24:02you get minus Lambda U and this is the G
24:05DX so the solution is done
24:08so it's very easy if you know G
24:10everywhere you know the optimal control
24:17the optimal control
24:19is the derivative of
24:22the cost to go or the gain to go as a
24:25function of X with respect to X okay
24:29so that's what I got for the optimal
24:32control and now what I can do is compute
24:37key
24:39of x
24:40at time T minus DT
24:43of X I find T minus DT is simply
24:48G of X at sine t
24:51plus whatever I've I've had to add to it
24:55through this extra little piece using
24:57the optimal control okay
25:00because as you remember what I told you
25:02is that g is obtained as the optimal the
25:05the optimal gain you can have from now
25:08on and so it's assumes that you use the
25:11optimal uh policy
25:14so what I have here is G of X and C
25:16minus DT is G of x and t
25:19plus a bunch of times
25:22that come from from here so for example
25:27there's this tough time here which is
25:30minus Lambda over 2
25:33U squared DT so this is
25:37one over Lambda
25:39egx
25:42squared
25:43PT
25:45okay
25:49then there's a this sum here plus u d g
25:53DX so Plus
25:571 over Lambda
25:59egdx
26:02squared
26:04so you see that these two terms were
26:05actually combined together
26:08and that's the what happens usually when
26:10you have a quadratic optimization
26:12problem at optimal add Optimum the two
26:16terms are of the same order of magnitude
26:18so that's what we're finding here
26:21then there's this
26:24extra term coming from noise
26:27so Plus
26:29J okay wait I have to be careful
26:34I still have a little room
26:37I am I'm paranoid with this
26:42question of space so plus J
26:48sorry it has a DT
26:50okay the same DT as this one
26:53and then plus J
26:56uh D2
26:58T DX squared DT
27:02this is this term here
27:05and finally
27:06the Bounty term Plus
27:11DT
27:13D of x
27:15and T minus DT
27:30okay so to order DT I therefore get a
27:35differential equation for G
27:40so G of x e minus DT is uh G of X and T
27:45minus DT DG DT so what I get is minus DG
27:51DT
27:53equal
27:54so now I cancel the DTs everywhere I
27:58regroup these two terms together and so
28:01what I get is one over two Lambda
28:04EG EX
28:07squared
28:10Plus
28:12J
28:14d to g
28:16DX squared
28:18Plus
28:20D of x and t
28:23okay
28:28and this is called the Hamilton
28:34Jacobi
28:37Bellman equation
28:44actually it has many names
28:48but in this context that's what it's
28:51called can you still see here
28:55there is a little piece out of the
28:58screen yeah
29:09right
29:10where I stopped
29:16okay
29:18so you see
29:20what you have to do is to solve a kind
29:23of non-linear diffusion equation there's
29:26a diffusion term here
29:28there's a non-linear term yet DG DX
29:31Square which comes from the optimization
29:33problem in u
29:35[Music]
29:36there's a bounty term you have x and t
29:38which I haven't specified
29:40and something you could say is wait
29:43usually the diffusion equation is DG DT
29:46equal plus
29:48something plus the diffusion constant DG
29:51e to G DX squared
29:53but here I have a minus and you know the
29:56diffusion equation with a with a
29:58negative diffusion constant is something
30:01very sick but here remember that what I
30:04have to solve is a backwards equation I
30:07I don't solve this equation forward in
30:10time I I I solve it backwards in time
30:13okay so what I have to specify is what's
30:16the final condition okay and so
30:20this equation should be solved backwards
30:23in time plus with
30:27a final condition
30:33which is a certain G of X and capital p
30:38that I give myself
30:40okay I'm going to discuss that in a
30:43second but so there's no problem with
30:46the minus sign because I'm actually
30:47iterating this equation backwards in
30:49time okay and
30:51as I said once I've computed this this
30:55the solution of this equation up to T
30:57equals zero then I have the solution of
30:59my
31:00initial problem so I'm going to give you
31:03an example where we do this explicitly
31:05but let me first uh say a few words
31:09about
31:11um the Hamilton Jacoby Berman equation
31:14and the final condition
31:17so um
31:20first of all
31:23the final condition
31:44is this okay
31:58so the final condition depends on the
32:00problem
32:04so in the Bounty rate
32:07it may be that the final condition is
32:11that the the organization of the
32:13organizer of the game has asked you to
32:16come back to your initial position at
32:18the end of the game okay so so you know
32:22you could have a a final condition which
32:24is that
32:26if you end up at the end of the game not
32:28on your final condition which at your
32:31initial position sorry which is at zero
32:33then you're penalized
32:36so for example you could choose that g
32:39of x
32:40and capital t is minus some penalty
32:45x minus X zero squared
32:49for example
32:57so this means that you know you should
32:59pick up your bounties along the way but
33:02you should be mindful of the fact that
33:04you should come back to your initial
33:05position at the final
33:07uh at the end of the game so this is one
33:11possibility but of course you can invent
33:13whatever possibilities you want and I'll
33:16I'll give you an example uh in a second
33:19which we're going to solve explicitly
33:21for
33:23so the second remark
33:26is that
33:28the hjb equation as written
33:33is also called
33:39the kpz equation
33:45when D of x and t
33:50Is Random
33:56so kpz again is for cardal party design
33:59and I've actually I've I've sent you
34:03yesterday evening uh some notes on with
34:07references and there's the whole review
34:10paper on the kpz equation if you're
34:12interested but
34:14um I'm going to say a little more about
34:15this equation here today but essentially
34:18up to now I haven't specified at all
34:21what the what this field is okay this P
34:24of x and t it can be anything it can be
34:27deterministic
34:28but it can also be random
34:30so for example in the pinning problem
34:33it's usually considered to be random
34:35because the the the impurities that pin
34:38the polymer are located at random uh
34:41sites in space and so we can consider
34:44the spinning field to be a random field
34:48and when B of x and t is random then
34:52this equation here the Amazon Jacobi
34:54Bellman is called the kpz equation
34:59and actually just the in passing the kpz
35:03equation was not introduced that far
35:06fast in the context of thinning but
35:09rather as a model for surface growth so
35:12this equation here
35:14can also be seen as a as an equation for
35:17randomly growing surfaces but that's
35:20another story that I won't get into but
35:23just again to show you that there are
35:25many many problems that are accounted
35:28for by this Hamilton Jacobi Bellman
35:31equation and it's really nice to have
35:33uh all these mapping
35:37okay but
35:39as you remember I'm in a chapter called
35:42uh models of random growth
35:45and what I promised although I know you
35:48don't see my outline is to show you the
35:51relation between random growth and
35:54optimization
35:55so that's what I'm going to do now and
35:58actually it's not going to be very
36:00difficult
36:01because now I have my Amazon Jacobi
36:05Bellman equation
36:16which I'm going to
36:19keep
36:20in the screen for you to see
36:24okay you see it
36:33so what I'm claiming now is that using a
36:37simple change of function
36:40I I will get back to the growth model
36:43that I've told you about
36:45uh in the last weeks so I'm introducing
36:48now a new function
36:51which I'm going to call Zed for
36:55an obvious reasons so I'm defining G of
36:57x and t to be 2 Lambda J
37:01log
37:04of Z of x from t
37:07okay
37:08where you remember Lambda is
37:11uh Associated to the kinetic energy term
37:14and J is the variance of the noise
37:18and if I do this then I won't you know
37:22do the calculation but it's really easy
37:24to do and you can do it for yourself so
37:27if I inject this definition in the AGB
37:30equation I get an equation for v which
37:34reads minus easy
37:37EP
37:39equals J
37:41e to Z
37:43DX squared
37:47Plus
37:49ETA
37:49of x and t
37:52Z
37:53of course z is a function of x and t so
37:56it's useful if I keep this dependent
38:01where ETA
38:03of x and t
38:06is equal
38:07to
38:09B of x and t
38:12divided by 2 Lambda J
38:16okay
38:20and so if you remember because you still
38:23have your notes and I
38:26I have erased my screen but this is the
38:29equation that I've given the name to
38:31last time that I've called a
38:33and if you remember this is an equation
38:36describing random growth with diffusion
38:40so this is an equation describing a
38:43population
38:44that lives in the one-dimensional space
38:47but again you as I've said you could add
38:50a narrow to X and make this a d
38:54dimensional problem if you wish anyway
38:56this is describing
38:58uh except there's a minus sign here but
39:01if I change the the sign of time uh it's
39:06describing the evolution of a population
39:09that diffuses with this term here but
39:13also grows randomly or decays randomly
39:17because of this random growth term that
39:20I've discussed okay
39:21and so here we have a problem an
39:25optimization problem
39:26which actually maps onto a growth
39:30problem
39:31and this is really interesting because
39:33it means that
39:35you can either see this problem as
39:37trying to find the optimal path
39:40in an optimization problem or see it as
39:44a kind of darwinian evolution of species
39:49that diffuse and randomly grow or Decay
39:53okay and if you solve one problem you in
39:57a sense solve the other and so I'm going
39:59to go back to that later on to explore a
40:03little further this analogy between
40:04finding an optimal path
40:07and solving for population growth okay
40:12so that's the surprise if you want that
40:16we've gone through uh a method that was
40:20introduced in the 50s to solve
40:22optimization problem and I think it's
40:24really important to for you to have seen
40:26it somewhere in your curriculum because
40:28it uh it's it's not often uh mentioned
40:32in physics courses but it's it's one of
40:35the pillars of uh many engineering
40:39applications in particular
40:41in financial engineering but also
40:44there's a lot of application of the
40:45Amazon Jacobi Bellman equation or the
40:48Bellman method which doesn't necessarily
40:51lead to Amazon Jacob Rebellion equation
40:53but sometimes it's a different problem
40:55for example a discrete problem where you
40:58have a different type of equation at the
41:00end but anyway this Bellman argument is
41:03Central to many models in economics
41:05because in economics as I've told you at
41:08the beginning people are assumed to be
41:11utility maximum sizes and so
41:15agents in an economic world have to are
41:18faced with optimization problems and
41:21often they are assumed to be able to
41:24solve these optimization problems using
41:26a Bellman argument okay
41:31so I'm going to speak later about
41:34this analogy between natural selection
41:37if you want and optimization
41:40but for those of you who are uh uh
41:45familiar with the statistical method
41:47statistical physics method I just want
41:50to add one more comment
41:52and for those of you who don't really
41:58um are first in these topics don't worry
42:01it's not going to affect the rest of the
42:02lecture you can also see a
42:07is is also the transfer Matrix solution
42:16to the pin problem to the pinning
42:18problem
42:25so what I mean by this is that
42:30as I said you have this objective
42:32function in the in the context of
42:34optimization it's also the energy uh or
42:37minus the energy in a pinning problem
42:40and so you can see this as a statistical
42:42mechanics problem try to compute uh the
42:45boltzmann weight the partition function
42:48and the transfer Matrix approach to such
42:52problems is to find a recursion relation
42:55on the partition function and Z of x and
42:59t is exactly the partition function for
43:01all works ending at X at time T and then
43:06you can find the recursion relation
43:07which is directly given by this equation
43:10okay so this is another way yet to see
43:14where this equation comes from which is
43:17maybe a little closer to the initial
43:19problem than the uh the random growth
43:23interpretation okay
43:25Excuse me yes so um of course your
43:30notations uh remind I mean are
43:32reminiscent maybe of uh what would be
43:36partition functions and so on so I was
43:38wondering is there a particular reason
43:39for which the uh transfer Matrix
43:42approach would be related by a logarithm
43:44to uh to this uh again to go function
43:51no no it's not a coincidence because the
43:54indeed if you think in terms of
43:57statistical mechanics you you know that
43:59uh Zed
44:01is a partition function and log Z is a
44:04free energy
44:05so in a sense G of x and t can be seen
44:08as a kind of free energy to go if you
44:11want at non-zero temperature
44:14so right
44:16it is not it's not a confident
44:20okay so now let me show you how this the
44:26this this version of the HEB equation
44:29which is equivalent I mean I can write
44:32it in terms of G or in terms of V now
44:35I'm going to solve uh this equation here
44:38in a particular case and it's it's a
44:41very cute graphical method
44:44and I hope you're going to enjoy it
44:48um I'm particularly fond of this
44:50example
44:53but please interrupt if you
45:03if you feel that I'm not clear
45:06okay
45:07so the problem I'm going to try to solve
45:10is a problem where
45:14uh
45:16Zed
45:18of X and capital t is given
45:24so it's it's given by a certain function
45:26uh f of x and t
45:31specified
45:37okay
45:40um where and I'm going to assume that f
45:43of x and t
45:46can be written using this mapping here
45:50I'm going to introduce another
45:51annotation which is related to the
45:55previous one which is exponential of
45:59uh D of x and t
46:03divided by
46:06um two Lambda J
46:10okay
46:12so there's a there's a in my problem
46:15there's a specific bounty
46:18um at the end of the game
46:21and I'm going to uh assume that V of x
46:24and t is a is a generic function
46:33that's given so your your your bounties
46:36at the end are you know you know what
46:38they are but on the other hand B of x
46:41and t everywhere else
46:45is zero
46:47okay
46:49so ETA of x and t everywhere else is
46:51zero
46:53so this is a simplified problem where
46:57the only bounties that exist
47:01are
47:03at the Final Destination
47:06so you have bounties
47:08distributed randomly here
47:11but on the other hand every everywhere
47:13else there's nothing to gain okay
47:16so clearly
47:18you know you can have an intuition of
47:21what's going to happen what you should
47:22do is to try to Target a good Bounty at
47:25the end and bike straight between the
47:27two
47:28these are the trajectories that are
47:31going to emerge from the solution of the
47:33problem okay because you don't want to
47:37you know air around or
47:40uh
47:41go in in random directions you go
47:44straight but you try to Target the best
47:48um the the best Bounty without having to
47:51bike too fast in in between now and the
47:54end of the game so that's the physical
47:55interpretation
47:57but let's see how it works
47:59in the in the present context
48:04so what I have to solve here
48:07is essentially because data of x and t
48:11will be zero everywhere except at the
48:14end but this is included in the in the
48:17final condition
48:19um I have to solve a backward diffusion
48:22equation okay
48:24and so what well I guess everybody knows
48:28what's the solution of the diffusion
48:30equation is is the gaussian so let me
48:33write the solution
48:36and you tell me if you agree with it
48:40so Z of x and t is
48:43the integral of d y over let me write it
48:47and then I'm going to comment it
48:494 Pi J
48:52capital T minus multi
48:55exponential of minus x minus y
48:58squared
48:59over
49:014 J
49:04capital T minus multi
49:06times Z of Y
49:09capital T
49:15so that's my claim
49:18and I think you should recognize what it
49:21is so first of all note that without the
49:25ETA term or actually even with yet the
49:27term but without yet the the diffusion
49:30equation is linear
49:31so the solution can be built as a linear
49:35superposition of special Solutions
49:39and here what you see is that what I've
49:42built here is a superposition of many
49:45solutions Each of which having an
49:48initial condition that's localized at
49:50some point Y at time t so what I'm
49:53saying here is that the population on X
49:56at time T is what I whatever I had on
50:00site x y at time capital T propagated
50:03with the diffusion propagator up to
50:07small X at times small T okay
50:10so in particular if I started with
50:14say a Delta function so maybe that's
50:16going to be the thing that you may that
50:18will make you recognize what I'm talking
50:20about so imagine that instead of Y and T
50:22is a Delta function Delta of Y okay
50:27so I'm I'm I'm Coalition all the
50:30populations on side 0 then the integral
50:34over y disappears and what I get is z of
50:38x and t is exponential of minus x
50:40squared over 2 for 4J T minus capital T
50:44minus multi so you get the the gaussian
50:47that spreads out from an initial
50:49condition that's localized on y equals
50:51zero and so the only thing I'm doing
50:53here is superimposing
50:56different initial conditions that all
50:58are going to propagate according to a
51:01gaussian up to time t
51:04note also that here I'm propagating as I
51:07said backwards in time so the effective
51:10time or diffusion is not t but capital T
51:14minus multi
51:16okay
51:18so is everybody okay with that
51:19representation of
51:22the solution to this equation the other
51:24other way to do it if you if you're not
51:27convinced is to plug
51:28this into uh the equation and check that
51:34it is indeed a solution with the correct
51:37sound boundary
51:40uh
51:42condition
51:45okay
51:47so now I'm going to replace Z of Y and T
51:52I'm sorry I shouldn't have introduced F
51:54that's really useless so I'm replacing Z
51:57of Y and T by
51:58exponential
52:02of B
52:04of Y and T
52:06divided by 2 Lambda J
52:09okay
52:12and I'm going to try to tell you what is
52:15z of x and t obtained as this integral
52:29so bear with me it's going to be
52:31slightly tricky but not that much but
52:35the the end result is going to be
52:37interesting so I hope you you'll like it
52:41so in general this without even
52:45specifying B of Y and T I can't do much
52:48but what I'm going to uh
52:51look into is the special case When J
52:56goes to zero
52:59so I'm assuming that the noise
53:02in my uh on my treasure hunt
53:07is small
53:10okay
53:13and so now what you have
53:15is a a
53:20an integral
53:22that looks like integral d y
53:26of exponential
53:281 over J
53:31times the southern function that I'm
53:33going to write
53:35uh a
53:36of Y given X
53:40and then there's the square roots
53:43here but I don't really care about it
53:46okay
53:50and so you know very well what happens
53:52when you have this type of integral
53:55because you have the exponential of
53:56something extremely large
53:58one over J is very large because J goes
54:01to zero
54:02then this integral is going to be
54:04dominated by the Y's that maximize a of
54:09Y given X
54:11okay so this is the usual argument so
54:14it's called the LaPlace method if you
54:16want
54:20and it relies on the fact that because
54:23of the exponential it amplifies any uh
54:27small variation of the function a of Y
54:29and X
54:30and exponential of 1 over J A of Y and X
54:33becomes extremely peaked around the
54:36Maxima of a of Y given X so I guess that
54:40you must have encountered this method
54:43many times already so I'm not going to
54:45spend too much time so what I'm going to
54:48look for is I'm going to
54:51look for y star
54:54which is the ARG Max
55:00again of a of Y given X and and I'm
55:03going to write it explicitly so this is
55:06D of Y
55:09and capital T divided by uh
55:18Lambda
55:21minus
55:23x minus y squared divided by
55:30[Music]
55:33four sorry
55:35is the fine the factor 2 here I'm
55:39missing
55:42four
55:45seven T minus multi
55:49so it's the same expression as this one
55:51where I factor out
55:53uh J okay
55:57so that's that's my aim in life now I
56:00have to find for each X I have to find a
56:03y star that maximizes
56:05the sum of these two terms
56:09and
56:10what does it look like in terms of
56:12problems that you know about
56:15well you see that if I interpret B of X
56:18as minus the pinning energy minus an
56:21energy
56:22x minus y squared is a spring energy
56:27that you need to pay in order to
56:33deform the trajectory so what I what I
56:36mean by this is that in this very
56:38simplified pinning problem
56:40the only pinning is in the end okay and
56:43what you have to do is to find the best
56:45pinning side the best thinning site
56:48knowing that you're going to pay some
56:50spring energy some elastic energy which
56:53is the difference between your initial
56:55point and your final Point squared
56:59so this is what it looks like this is a
57:01the elastic energy you have to pay to go
57:04from X to Y in a Time uh capital T minus
57:09multi
57:10so if you want this is this is another
57:12way to think about this problem it's a
57:14it's an elastic spring that's pinned at
57:17one end
57:18and that you you move the other end so
57:20one end is y it's pinned by this term
57:22and the other end is X and you move it
57:25around okay
57:27so so let's let's try to address now
57:30this General problem
57:33foreign
58:06so let me plot as a function of Y
58:09a generic b of Y and T
58:15divided by Lambda
58:18so this is going to be a certain
58:20function that I'm I'm going to draw like
58:23this so I'm assuming it's it's a random
58:26function for example so it has
58:28Peaks and troughs so this is bad for
58:32your game this is good for your game
58:34and now what I want to do is solve this
58:37problem so let's let's think of it in a
58:41slightly different way
58:43I'm going to introduce
58:46a value that I'm going to call a a star
58:54a sorry for the moment and I'm going to
58:57to find to try to find the value of the
58:59maximum value of a such that
59:04a plus
59:06x minus 1 my minus y squared over 4
59:11T minus mole t
59:14is equal to
59:16ETA sorry to be
59:19of Y and T
59:21divided by 2 Lambda
59:24so
59:25here I'm I'm just repeating the
59:27definition of a but I'm going to look
59:29for the value of a such that this
59:32equation has a non-zero solution in y
59:36okay
59:37and such that a is maximum
59:40so
59:41a must be maximum I'm going to call it a
59:44star such maximum value of a
59:52such that
59:54their ex the solution is is not empty
1:00:00there is a solution
1:00:07okay
1:00:09so I have on the left hand side the
1:00:12parabola
1:00:13as a function of Y and on the right hand
1:00:17side a certain function that I've drawn
1:00:19here and I'm looking for the value of a
1:00:22such that these two terms are equal so
1:00:25there must be a solution to this
1:00:26equation but on top of that I want to
1:00:28find the arc Max here so I need to find
1:00:32the highest value of a such that this
1:00:35equation has a solution
1:00:37so let me try a very high value of a
1:00:41let me put it here so that's my trial
1:00:44value of a
1:00:48I'm assuming that X is here for example
1:00:50this is the value of x
1:00:52and this function this the left hand
1:00:54side it's a plus the parabola as a
1:00:57function of Y
1:00:58so it's going to look like this
1:01:01okay
1:01:06so you see here there's no intersection
1:01:08between the green line and the white
1:01:11line so this equation has no solution so
1:01:14I picked the value of a that's too high
1:01:17so what I'm going to try to do is to
1:01:19lower the value of a
1:01:22in such a way
1:01:24that I find the solution
1:01:26okay
1:01:29and you see that there's going to be a
1:01:32solution maybe that's going to be
1:01:36VC okay
1:01:39yeah
1:01:47this is the value of a I'm looking for
1:01:51foreign
1:01:55where
1:01:59the parallel touches
1:02:01uh the the white line
1:02:05and it's the highest one
1:02:07okay
1:02:08so for a given x what I have to do
1:02:11is
1:02:13um a little for those of you know Atomic
1:02:15Force microscopy imagine that this is a
1:02:19a the atomic Force
1:02:22um
1:02:23sensor and you have to lower it until
1:02:27you touch the rough surface okay or you
1:02:30can think of this as a finger and you
1:02:32you you
1:02:34you lower your finger until you touch
1:02:37for the first time this random object
1:02:40which is
1:02:41uh the the white line okay
1:02:45so this is a graphical
1:02:46uh representation of the solution that
1:02:50we're looking for
1:02:52and so you see that in this case
1:02:55why
1:02:56star
1:02:58is this point here is the is the touch
1:03:00point and in in the case I've just
1:03:03chosen the touch point is very close to
1:03:062X okay
1:03:08but now choose another value of uh of x
1:03:15let me do it in color
1:03:22now I'm choosing a value of x that maybe
1:03:25is here
1:03:27and you see that my Parabola is going to
1:03:32be maybe like this
1:03:36I haven't chosen the same
1:03:40curvature so this is a little bit
1:03:42misleading let me try to do my
1:03:45drawing correctly keeping the same
1:03:48curvature as this one so or maybe this
1:03:52thing to look
1:03:57well okay not too bad
1:04:02so this is the this is the value of a
1:04:04for this particular X
1:04:10and the value the and the Y star now is
1:04:12here
1:04:15so this is A1 a star one
1:04:18responding to some y star one this is a
1:04:21star two corresponding to a certain way
1:04:23y star 2. and so on so for each value of
1:04:27x you should do this construction find
1:04:30the value of a that is maximum while
1:04:33allowing a solution of to this equation
1:04:35uh to exist for this equation to exist
1:04:38and then you have your y star and your a
1:04:42star
1:04:44so once you've done that
1:04:47well you've you've completed your
1:04:50calculation
1:04:52because in this case now what you have
1:04:56is that
1:04:58once you know y star you also know a
1:05:01star
1:05:07so a star
1:05:09is equal to
1:05:12um
1:05:15B of Y star
1:05:18capital T divided by two Lambda minus
1:05:22X star x minus y star
1:05:25squared divided by four capital T minus
1:05:29minus multi
1:05:31and once you know y star and a star you
1:05:36know the value of this integral because
1:05:37it's dominated by the value of y star
1:05:40that you just found and the value of the
1:05:42integral is
1:05:44approximately equal to exponential of 1
1:05:47over j a star
1:05:50so let me write it here so what I get is
1:05:54that Z of x
1:05:57and T
1:05:58in this LaPlace approximation
1:06:02is given by 1 over square root of
1:06:07something that I don't really care about
1:06:10times exponential of minus x minus y
1:06:15star
1:06:17squared divided by
1:06:204 J
1:06:22P minus multi
1:06:25Plus
1:06:27B of Y star and capital T divided by 2
1:06:32Lambda J
1:06:35okay
1:06:37and now if I want to go back to
1:06:45the cost to go because there's a
1:06:48relation between Z and the cost to go g
1:06:51for the gain to go
1:07:01you remember the relation G
1:07:03was 2 Lambda J
1:07:06log Z
1:07:09so g g of x and t
1:07:14is
1:07:15equal to
1:07:17approximately equal to B
1:07:21of Y star and capital T
1:07:26uh over
1:07:29to Lambda
1:07:31minus
1:07:33Lambda
1:07:36x minus y squared
1:07:40over
1:07:42to
1:07:47T minus t
1:07:50and I think there's no two here right
1:07:54right okay so what I'm saying is that
1:07:56once you get
1:07:58the touch point Y star and the altitude
1:08:02a star you have solved your problem in
1:08:05principle
1:08:06you know that
1:08:08your cost to go or your partition
1:08:10function or your whatever you want to
1:08:12interpret it is given by the sum of two
1:08:16terms that are computed at this
1:08:18particular value
1:08:19of the touch Point okay so this is a
1:08:23very general solution but now I want to
1:08:25show you how
1:08:28um
1:08:30how is how it behaves as the function of
1:08:33x
1:08:34this uh this function
1:08:37G
1:08:39so first of all
1:08:45I'm going to
1:08:47first of all plot y Star as a function
1:08:50of x
1:08:52so what you see here you see that
1:08:56at this value of x I had a solution that
1:08:59was close to X
1:09:00but now let's start moving slowly x to
1:09:05the right
1:09:06then this Parabola here is going to move
1:09:08and you see that at one point
1:09:11there's going to be a special situation
1:09:13where the parabola is going to touch the
1:09:17white curve at exactly two points
1:09:19simultaneously not one point anymore but
1:09:22two point
1:09:23and then as as you start as you continue
1:09:27moving
1:09:28forward so let me try to indicate what
1:09:32it looks like just at the touch point
1:09:36okay so there's a situation where it's
1:09:38going to look like this so my drawing is
1:09:40not very good but you imagine that as I
1:09:42push the purple uh Parabola to the to
1:09:46the right there's a point where I'm
1:09:48going to touch exactly simultaneously at
1:09:50two different points white power one and
1:09:53one star two and then when I continue
1:09:56carrying on I'm going to lose the first
1:09:59the first touch point and continue with
1:10:01the second touch point
1:10:03so what it means is that if you
1:10:06carefully think about this graphical
1:10:07construction here what you'll find is
1:10:10that as a function of X the solution
1:10:13looks like this
1:10:19so for a while you're going to
1:10:23uh
1:10:24remains close to a fast solution
1:10:27why star one and then suddenly you're
1:10:31going to jump from a solution that was
1:10:34close to this point to a solution that's
1:10:36close to this point
1:10:37and that means the jump in the value of
1:10:41y star
1:10:43and then you're going to continue around
1:10:45the new touch point
1:10:48until suddenly you have to again move to
1:10:51another touch point
1:10:53and this this jump here happens exactly
1:10:57when there are two solutions
1:11:03to the touch Point problem
1:11:13okay
1:11:14so why did I do all this why did I take
1:11:17you this to this long journey of trying
1:11:20to construct this solution graphically
1:11:22well it was really to reach that point
1:11:25so if you haven't followed in details
1:11:28what I've done maybe you can think about
1:11:30it or listen to to me again uh later on
1:11:34but the main important point to
1:11:36understand here is this graphical
1:11:38solution sometimes as as two touch
1:11:42points and when it has two touch points
1:11:45is the moment where the solution is
1:11:46going to come to just continuously shift
1:11:50from One Touch point to another touch
1:11:53point
1:11:54so what does it mean in terms of our
1:11:56optimization problem
1:11:59it means that
1:12:04let's go back to uh
1:12:06the optimization problem
1:12:16and here I think I need to move my
1:12:21screen
1:12:40so remember my
1:12:43treasure hunt problem is a function of C
1:12:47so I have my bounce keys that are
1:12:49localized only
1:12:51at the end point
1:12:53and what I'm saying is that as a
1:12:55function of my initial point
1:12:57so as I move my initial point up which
1:13:01is going to correspond to moving X here
1:13:03to the right
1:13:05then for a while I'm going to Target
1:13:08this side here so I'm going to to do
1:13:12this
1:13:13this is going to be my optimal path
1:13:16so this pinning site or this bounty is
1:13:21sufficiently interesting that I
1:13:23disregard this one which was closer to
1:13:26my initial position so you know in a
1:13:29sense I had to bike less hard to reach
1:13:31that point but it was not as interesting
1:13:34so I decided to go to that one
1:13:36and then for a while I'm going to keep
1:13:39that as a Target
1:13:44okay
1:13:47so all these initial points
1:13:50are going to aim for the same Target
1:13:52which is that bounty
1:13:54but then as I move still
1:13:58further upwards
1:14:02at a certain point and this is a
1:14:04discontinuous
1:14:06transition When J goes to zero then I'm
1:14:11going to aim for a new Target okay
1:14:17and I'm going to keep the same Target
1:14:19for a while
1:14:22and so on okay so if you want this was
1:14:25my y star one
1:14:28and this is why star two
1:14:31and so on
1:14:34so that's that's what the succession of
1:14:36jumps means it means that
1:14:38you're going to focus on one given end
1:14:41point for a while but then if you're too
1:14:43far away from it you're going to
1:14:45suddenly jump to another Target point
1:14:49so what what's interesting about this uh
1:14:54um
1:14:55this scenario here is that you have a
1:14:58well-defined optimization problem
1:15:01but the solution to this optimization
1:15:02problem develops shot this is called a
1:15:06shock in in uh in the context of fluid
1:15:09mechanics and so even if your initial
1:15:13problem is perfectly continuous so you
1:15:17have a perfectly continuous function B
1:15:21of Y and T the solution to the
1:15:23optimization problem create this
1:15:25continuity so you know whether you're
1:15:29the slightly to the left of the shock or
1:15:33slightly to the right of the shark
1:15:35the solution to the optimization problem
1:15:37is going to be completely different
1:15:40so this suggests that
1:15:42in general
1:15:44these optimization problems are in a
1:15:48sense very fragile it means that if you
1:15:50change a little bit the problem that
1:15:52you're trying to optimize you might find
1:15:54a completely different solution
1:15:57and this is the idea of of chaos that
1:16:00I've alluded to and I'm going to say
1:16:02more about this so anyways so what I've
1:16:04done in this up up to now is to speak
1:16:08about the Amazon Jacobi Bellman equation
1:16:12and introduce you to this interesting
1:16:14notion of stocks in the context of
1:16:18um
1:16:19of optimization
1:16:26so maybe if this is the only thing that
1:16:28you uh
1:16:31remember from this lecture is this graph
1:16:34that I've just drawn the fact that
1:16:37you can't very suddenly change
1:16:40the solution to an optimization problem
1:16:49so what I want to do now is to introduce
1:16:51you to
1:16:55um
1:16:57another
1:16:59way of of uh discussing exactly the same
1:17:02problem
1:17:08which will drive us into traffic jams
1:17:11Drive being an appropriate
1:17:14term here
1:17:17okay so let me come back to uh the
1:17:20Amazon Jacobi Bellman equation
1:17:26so minus
1:17:28I'm going to rewrite it here minus DG DT
1:17:31equals
1:17:331 over 2 Lambda
1:17:36PG DX
1:17:39squared
1:17:43plus J
1:17:45to G DX Square
1:17:48plus b
1:17:53okay now I'm not going to do it
1:17:57completely in details for you but what
1:18:02I'm claiming is that if I introduce
1:18:06um a velocity field
1:18:08which is one over Lambda
1:18:13ejdx
1:18:16okay
1:18:19and
1:18:20so essentially I take the derivative of
1:18:23this sub Amazon Jacoby Bellman equation
1:18:25with respect to X
1:18:28so that's the first thing I do and at
1:18:29the end I change T into minus t and X
1:18:33into minus X
1:18:36then the equation you get is
1:18:41dvdt
1:18:43plus v DV DX
1:18:48equals J D to V
1:18:51DX squared
1:18:54Plus
1:18:581 over Lambda
1:19:00e b yes
1:19:04okay
1:19:07so I haven't gone through the detail of
1:19:09this derivation but it's very easy you
1:19:12see you take the derivative of this
1:19:13equation with respect to X so for
1:19:15example you have by doing this you have
1:19:18a DG DX d2g DX squared and this is a v
1:19:23dvdx
1:19:25and so on and and so what you find after
1:19:28doing this is something that's called
1:19:30the Navy stocks Burgers equation
1:19:43so this is an equation for
1:19:45that you must have uh encountered in
1:19:48fluid mechanics this is the the standard
1:19:51description of the Velocity flow in the
1:19:54fluid
1:19:55driven by some term which is related to
1:20:00to be here so what you see is that yes
1:20:03another language comes around which is
1:20:06the language of good dynamic
1:20:08and that's why I talked about these
1:20:10discontinuities as shocks here because
1:20:13exactly as you have shocks in fluids
1:20:18discontinuities into it you you
1:20:21have for the same reason this continuity
1:20:24is in the optimization problem
1:20:27okay so fluid mechanics now
1:20:33I'm going to show you another
1:20:36interpretation of this navigation in the
1:20:40context of of traffic flow modeling
1:20:47which is going to lead
1:20:50naturally to the same
1:20:53Navy stocks or Burgers equation
1:20:59and so we'll have another way to think
1:21:01about optimization problems in terms of
1:21:06flow or traffic jams
1:21:11so
1:21:12let me call this the mystery of traffic
1:21:14jam
1:21:30so you you probably are aware of the
1:21:32fact that traffic jams they seem to
1:21:34appear randomly or sometimes for a
1:21:38reason of course there might be actual
1:21:40blockage or along the the road but
1:21:44how often what happens is that there's a
1:21:47there's a Slowdown of the traffic
1:21:50and then when the velocity of your car
1:21:52picks up again you realize that nothing
1:21:54special was blocking the traffic there
1:21:56it was just a kind of self-induced
1:21:58effect where people slow down and uh but
1:22:02for no particular reason
1:22:04and what I want to show you is that if
1:22:06you adopt a very natural uh model for
1:22:10traffic jam for traffic flow you will
1:22:13end up with shock Solutions and the
1:22:16shock Solutions are associated in this
1:22:19picture to uh to traffic jams
1:22:23so what I'm going to describe to you is
1:22:26it's called it's called a
1:22:28phenomenological approach
1:22:40that's how physicists call it and
1:22:43economists call this type of equation
1:22:45reduce formed equation
1:22:56but the logic is the same is to find the
1:22:59simplest possible equation describing uh
1:23:03the the phenomenon you want to describe
1:23:06okay
1:23:09so first thing I'm going to do is to
1:23:11introduce
1:23:13again
1:23:15a one-dimensional road X
1:23:18and along that road I will going to I'm
1:23:21going to have the density field row
1:23:25which tells you how many cars you have
1:23:27around
1:23:29position X along the road okay so here
1:23:33I've
1:23:34I've drawn a bump so it means that
1:23:36there's a lot of cars around here less
1:23:39so on both sides and I want to know how
1:23:43this profile is going to evolve this
1:23:46time
1:23:48so the first thing you should write as a
1:23:50as a synthesis is a conservation
1:23:53equation
1:23:54so you know that cars will not disappear
1:23:57spontaneously so I'm assuming that
1:23:59there's no entries or exits in the in
1:24:02the road so that's another problem that
1:24:04can be accounted for if you want but I'm
1:24:07going to assume that you know cars once
1:24:10they're on the motorway they can't get
1:24:11out and so the first thing I'm going to
1:24:14write is um
1:24:17the conservation law which which tells
1:24:20me that 0dt is minus d by the X
1:24:25U times rho
1:24:29where U is the velocity
1:24:34of the column sorry you can't see that
1:24:36we don't see yeah ah sorry thank you for
1:24:40telling me
1:24:53should be space
1:24:58foreign
1:25:11that for each density there's a typical
1:25:15velocity U of rho
1:25:18that describes how fast cars can move
1:25:21for a given density okay and of course
1:25:25you know what's kind of intuitive is
1:25:28that as a function of rho
1:25:32you a pro
1:25:34you you assume that it's going to be
1:25:37given by something maybe like this
1:25:41I mean this is the simplest thing you
1:25:43can think of right so when rho equals
1:25:46zero
1:25:47you go to maximum speed you Max
1:25:51the 130 kilometers per hour on French
1:25:55motorways and then as the density of car
1:25:59increases you have to be more and more
1:26:01careful and there's a maximum density
1:26:04when cars are
1:26:07touching each other where you you
1:26:10actually have to
1:26:11come to a hole because otherwise uh
1:26:14you're hitting the car in front of you I
1:26:16mean it's not comfortable to to drive
1:26:18when the the velocity the density of
1:26:20cars is high and so this is the simplest
1:26:23uh way to think about it
1:26:26but what I'm going to argue is that it's
1:26:28not enough
1:26:30it's not enough in the sense that
1:26:31there's another effect that you should
1:26:34take into account
1:26:35is that it's not only the local density
1:26:38that determines your velocity but also
1:26:41what you see in front of you so if you
1:26:43see in front of you that the density is
1:26:45going down that it clears up you're
1:26:48going to increase your speed and on the
1:26:50on the contrary if you see in front of
1:26:52you that the density of cars is
1:26:54increasing you're going to slow down
1:26:57so a phenomenological way to capture
1:27:01these two effects is to write that U of
1:27:04rho
1:27:05is going to be given by
1:27:09umax
1:27:111 minus rho over row Max
1:27:15okay this is uh this is this shape here
1:27:18but then I'm going to add here
1:27:22minus
1:27:24new
1:27:25over row
1:27:270 to the x
1:27:32so
1:27:33zero one over row 0 DX is the relative
1:27:37change of density in front of you
1:27:40and let me explain the sign minus U is a
1:27:43coefficient that is a phenomenological
1:27:45coefficient
1:27:47um that tells you how sensitive you are
1:27:49to what's going on in front of you
1:27:51and uh the minus sign comes from the
1:27:54argument I just gave that is if you see
1:27:57the the density in front of you
1:27:59increasing you know if you see a lot of
1:28:01people ahead of you you're going to slow
1:28:03down so if zero directs is positive uh
1:28:06you slow down U is going to uh decrease
1:28:10so there's a minus sign here and vice
1:28:12versa okay
1:28:13and then of course
1:28:15you know if you want to be generic you
1:28:18should also take into account that there
1:28:20are things that you don't control maybe
1:28:23there's a you know
1:28:26some unexpected thing happening in the
1:28:29fields the around you and so in general
1:28:34one can also add a noise term here
1:28:38so the velocity nominally should be this
1:28:41but because there's there are things
1:28:43happening you may add some some time
1:28:46dependent noise in in this equation
1:28:50Okay so
1:28:52yes yes just a question so you say that
1:28:55your speed depends on the density of
1:28:57people before you but would it change
1:28:59something if instead of taking the
1:29:02social density of people you take the
1:29:04velocity of the people in front of you
1:29:07because very often it breaks before then
1:29:10you will break yes you you could indeed
1:29:13account for that but
1:29:14there might be other terms you're right
1:29:17but what I'm saying is that first of all
1:29:19it's much easier to see the density of
1:29:22what's in front of you than to estimate
1:29:24the velocity of what's in front of you
1:29:26so the first the the basic thing you're
1:29:29sensitive to is whether in front of you
1:29:31there's room or there's no room velocity
1:29:34comes as a
1:29:36as a second correction if you want so
1:29:38I'm neglecting as I said I'm taking a
1:29:41phenomenological approach so I'm
1:29:43dropping terms that I consider to be
1:29:45less important but you're right in
1:29:46principle there could also be a gradient
1:29:49of velocity that affects your own
1:29:51velocity okay thanks
1:29:54okay so now again time is flying
1:29:58extremely quickly so um I won't uh go in
1:30:02the details of the derivation but if you
1:30:04put this shape here in the continuity
1:30:08equation in the in the conservation
1:30:11equation
1:30:12so you plug U of row in in here you do a
1:30:16little bit of manipulation you change
1:30:20um uh you put yourself in the in a in
1:30:25the moving frame
1:30:26um okay so this is not very important
1:30:29but at the end what you find is an
1:30:32equation
1:30:34for rho that's 0 DT
1:30:38Plus
1:30:39rho zero DX
1:30:42where X now is in the is in the moving
1:30:45frame but it doesn't really matter
1:30:47equals the certain coefficient that I
1:30:51can call J
1:30:52which is a mixture of of these things
1:30:55that I've introduced
1:30:56uh D2 row DX squared
1:31:00plus noise
1:31:04okay
1:31:09so you find for row an equation that is
1:31:12that has exactly the same structure as
1:31:15the Nagi Stokes equation that I've
1:31:17written here okay so if you think of rho
1:31:21as a velocity description of traffic
1:31:24flow is uh the Navy stogsberg is
1:31:28equation for the density not for the
1:31:30velocity for the density although the
1:31:32velocity is related to the density uh as
1:31:35I've written here okay
1:31:37so what is nice is that
1:31:40what I've explained to you is that
1:31:43um
1:31:44in the in in the Amazon Jacobi Bellman
1:31:49equations they are generically shocks
1:31:51that that appear
1:31:53okay here I've shown the existence of
1:31:56clocks in the context a very simplified
1:31:58problem which is that b was only
1:32:01non-zero at the end but the existence of
1:32:04flux is the generic features of these
1:32:06equations
1:32:07it essentially comes from nonlinear
1:32:10non-linearity that you see here or the
1:32:12nonlinearity that you see here we know
1:32:15that these non-linearities generate
1:32:16shocks and therefore we know that this
1:32:19equation will also
1:32:20generate shocks okay
1:32:23and so this is uh
1:32:26exactly what we need to understand that
1:32:29as a function of x
1:32:38okay
1:32:40okay
1:32:58so what it means this this equation and
1:33:00the analogy would have would have uh
1:33:02that I've explored is that as a function
1:33:05of X you expect the density field rho of
1:33:07x and t
1:33:10to evolve from a smooth density field
1:33:12maybe you know initially it's going to
1:33:14be like this
1:33:16that's at T equals zero
1:33:20and then after some time
1:33:23you let this evolve
1:33:25and after some time
1:33:27you're going to generate
1:33:29the shock
1:33:32so instead of remaining smooth
1:33:34this may evolve as something like this
1:33:45good so it means that the velocity
1:33:49now is going to
1:33:52if you transfer transfer this row into
1:33:55the velocity U of rho
1:34:06I'm going to plug U of x and t here
1:34:10then what you see is that
1:34:13from uh this uh this thumb here okay now
1:34:18you you have to massage a little bit the
1:34:19argument because
1:34:21once you have a discontinuity in in row
1:34:24the derivative is the a little bit
1:34:26ill-defined but essentially what it what
1:34:29it shows is that the velocity is going
1:34:31to decrease
1:34:33and then suddenly
1:34:34uh
1:34:36increase again
1:34:39and decrease and increase again okay and
1:34:43so
1:34:44this point of low velocity is the is a
1:34:47kind of traffic jam
1:34:51and what you see is that the existence
1:34:53of stocks means that suddenly uh the
1:34:56velocity is going to pick up again
1:34:58for no particular reason
1:35:00you see here that the evolution of this
1:35:04um this initially smooth profile into a
1:35:07shock profile doesn't require anything
1:35:10special happening on the road it's just
1:35:13a consequence of the nonlinearity of the
1:35:16equation itself so this is very
1:35:18interesting because from the point of
1:35:20view of economic theory you know agents
1:35:24should be rational and should adapt
1:35:27their speed in in such a way that the
1:35:30whole flow is as continuous as possible
1:35:32but in a sense you know the
1:35:34non-linearities kick in and prevent
1:35:37these smooth solutions for uh persisting
1:35:39forever and so you have unwanted
1:35:42uh situation like shocks or Jam that
1:35:47arise from the non-linearity of the
1:35:49problem
1:35:53okay so here I've described this the
1:35:56traffic jam using this phenomenological
1:35:58argument but if you're interested and
1:36:01you know if you like
1:36:03simple models I I just
1:36:07tell you very quickly that you can also
1:36:10model these traffic jams with lattice
1:36:15discrete model where you have particles
1:36:20that can only hop to the lock to the
1:36:22right
1:36:24and the rule of the game is that between
1:36:27t and t plus DT there's a finite
1:36:29probability to hop
1:36:32and the only reason you cannot hop is if
1:36:34there's some somebody in front of you so
1:36:36this is possible but for example
1:36:40if you have one particle here and one
1:36:42popsicle is there
1:36:44then this hop is not possible
1:36:48and so what you recognize
1:36:51is in spirit but in a very schematic
1:36:54manner the the ingredients that I've put
1:36:57here
1:36:58in particular this zero DX which means
1:37:01that if in front of you there's a
1:37:04already a particle then you don't move
1:37:07ahead and this is called
1:37:10uh
1:37:11the totally asymmetric
1:37:14exclusion Problem ASAP that you can look
1:37:17into in the literature if you want but
1:37:19what's what's interesting if you want is
1:37:22that
1:37:22the coarse grain description the the
1:37:25large scale long time behavior of such a
1:37:29discrete model
1:37:31exactly maps onto uh the buggers
1:37:34equation the kpz burgers equation
1:37:39okay so that's what I wanted to tell you
1:37:41about traffic jams now let me finish
1:37:45uh
1:37:47I fear that I won't have time which is
1:37:50too bad because
1:37:52um
1:37:54okay
1:37:55now maybe I'm going to finish on this um
1:37:59analogy between optimization and the
1:38:03population growth I promised you that I
1:38:06would go back to that
1:38:08but it will take me more than five
1:38:11minutes so uh I will I will do it uh
1:38:14early next week
1:38:16before moving to a new chapter
1:38:19so I'm going to stop here and
1:38:21and ask for questions if you have some
1:38:30excuse me
1:38:31I was wondering why this equation is
1:38:33called Hamilton Jacobi Bellman equation
1:38:36because I mean we know Hamilton the
1:38:39equivalent equation from classical
1:38:40mechanics and I didn't really see the
1:38:42connection
1:38:45well you know it's it's
1:38:47um if you have
1:38:50um so In classical Mechanics for example
1:38:52you have um an action
1:38:55okay and when you when you find the it's
1:39:00the same is the same uh logic if you
1:39:04when you look for the path
1:39:07uh minimizing the the action
1:39:10then you find exactly the same equations
1:39:13for
1:39:15both I mean the equation describing this
1:39:19minimization of the action is the Amazon
1:39:21Jacobi equation
1:39:25so maybe something I haven't said is
1:39:28that you can an alternative way of
1:39:30thinking about all this
1:39:33uh so I've erased my uh
1:39:37my treasure hunt problem but you can
1:39:40also formulate the treasure hunt problem
1:39:42in terms of a path integral
1:39:45so maybe that's something that you might
1:39:48find interesting
1:39:49which is related to the transfer Matrix
1:39:52approach that I've described
1:40:01I think now you have to move the camera
1:40:03a little bit yes I will sorry
1:40:17so you remember I told you that there's
1:40:19a
1:40:20again which is the integral from 0 to T
1:40:23DT of B of X of T and T
1:40:29and minus the cost that's integral from
1:40:330 to T
1:40:34DT of V squared okay
1:40:39then if you want to see this as a
1:40:41statistical mechanics problem you you
1:40:43will say that the partition function Z
1:40:46is the integral over all paths
1:40:52so the path is is an actual path it's a
1:40:56this is the trajectory you choose to
1:40:58follow of exponential of minus beta
1:41:03the energy but I said that the energy is
1:41:06minus the gain in the analogy so you're
1:41:09going to have exponential Plus
1:41:12um integral
1:41:14from 0 to T
1:41:17e t b of x and t
1:41:21of X of T and T
1:41:24um
1:41:25minus
1:41:26Lambda over 2 integral D Squared
1:41:30okay okay
1:41:32and this is the action of a classical
1:41:36mechanic problem
1:41:38where B of x and t is the local
1:41:42potential and V is the velocity
1:41:45okay
1:41:46and so if you if you're looking at the
1:41:49solution at large beta
1:41:52or if you think of this problem as a
1:41:54kind of quantum mechanics problem and
1:41:55this is this would be a Feynman
1:41:57representation of the problem then you
1:42:00look at uh
1:42:03the stationary phase approximation in
1:42:06semi-classical in the semi-classical
1:42:08limit or you in the statistical
1:42:10mechanics language if you look for the
1:42:13optimum path the optimal path and then
1:42:16what you see is that you're going to
1:42:18take the functional derivative that is a
1:42:21function of X and you'll find the the
1:42:23Amazon Jack B equation
1:42:26okay nice
1:42:30thank you
1:42:33but again the Bellman method and the
1:42:35Amazon Jacoby Bellman equation are not
1:42:39necessarily linked you can use the
1:42:41Bellman argument of trying to think
1:42:43forward and trying to say well I'm going
1:42:46to assume that my problem is for up to
1:42:48some time T and I'm going to one step
1:42:52before and Solve It Again by recursion
1:42:55this is something that doesn't require
1:42:57continuous Prime and continuous space
1:42:59it's a it's a very general method that
1:43:02can apply to discrete problems as well
1:43:04so that's why you know Amazon Jacoby
1:43:07Bellman relates to the kind of problems
1:43:10that I've been alluding to today but the
1:43:13Bellman method is much more General
1:43:18okay let's see
1:43:24sorry
1:43:26yes to to solve the a equation you you
1:43:30gave the solution which was a classical
1:43:33Evolution solution the final condition
1:43:36times the hit kernel
1:43:39and so it was a solution of a classical
1:43:43diffusion equation without taking into
1:43:45account the non-linear it is in the
1:43:47equation a
1:43:48so well they in the solution
1:43:54what's what's interesting interesting is
1:43:56that the G equation is the Amazon Jacob
1:43:59emailman equation which is non-linear
1:44:04but then by taking logged uh G equal log
1:44:09V or Z equal exponential of G then you
1:44:12find a linear equation
1:44:15linear diffusion equation
1:44:21this is often called the Vino hop
1:44:24transformation
1:44:25but so the equation for G is non-linear
1:44:29but the equation for Z is linear
1:44:32and you see that what is maybe I I I
1:44:36understand from from your question is
1:44:38why do we get shocks in a linear
1:44:41equation
1:44:42well you see that how it appears it's
1:44:45it's really
1:44:47how the initial condition propagates in
1:44:50time that creates the shock but once
1:44:54you're at the level of the linear
1:44:55equation there's no non-linearity
1:44:57anymore
1:45:02yes yes thank you and and it's it's
1:45:05really this this uh optimization uh
1:45:09solution here where you as I explained
1:45:11you find you make a saddle point uh not
1:45:13subtle point the LaPlace method to
1:45:16estimate
1:45:17um the yeah maybe I should have said
1:45:21something about this so the stocks only
1:45:23exist When J go to zero
1:45:26and in the fluid mechanics language this
1:45:29means that the viscosity goes to zero
1:45:32zero viscosity limits
1:45:39but if J is small but not zero the
1:45:43stocks are going to be smoothed out and
1:45:47this is well known also in fluid
1:45:49mechanics where instead of having a
1:45:51velocity field that is actually
1:45:53discontinued it's surrounded if you zoom
1:45:57on the shock itself there's a structure
1:46:00that that is continued actually so when
1:46:02J is very small but non-zero instead of
1:46:05having
1:46:06you know
1:46:08real shocks like this then if you zoom
1:46:11in it's actually uh
1:46:13around the like this a little bit and
1:46:16the width of this surrounding is
1:46:19proportional to J
1:46:24so the the the language of shocks is a
1:46:27little bit of um
1:46:29of uh abusive language because for J not
1:46:32equal to zero
1:46:35these are not really discontinuities but
1:46:37very abrupt changes
1:46:49so today was a little technical I'm
1:46:51sorry for the Maybe by long or
1:46:54not very clear explanation I hope that
1:46:56it was clear enough that you can
1:46:58reconstruct it but I wanted to speak
1:47:00about this because I think that this
1:47:02these graphical methods you'll see I'm
1:47:05going to use graphical methods again
1:47:07later in the lecture and I think that
1:47:10you know to think about a problem using
1:47:12graphical tools is very important to
1:47:15form intuition so in this case when you
1:47:19do the this graphical construction you
1:47:21see something you see very clearly the
1:47:23existence of shocks which come when when
1:47:26when this geometrical problem uh
1:47:29acquires two solutions
1:47:32um so that's why I really wanted to tell
1:47:34you about this this method
1:47:44anything else
1:47:48okay well I'll leave the chat to
1:47:51Valentina then
1:47:54and again don't hesitate to um
1:47:57send mails or interact if you need uh
1:48:01more explanation
1:48:03I have a nice week and until next week
1:48:05then
1:48:08thank you bye-bye this conference will
1:48:10now be recorded there we go
1:48:14okay so wake up everybody to the third
1:48:18today especially to those who joined the
1:48:21course more recently so the today of
1:48:25today is going to be a little bit more
1:48:26technical because it is about issues of
1:48:29stochastic calculus
1:48:35and I guess we will not cover everything
1:48:38that is in the file that I sent you in
1:48:41particular we will not cover I think the
1:48:45derivation of the soccer plank equation
1:48:48for a multiplicative noise
1:48:51so I will give you the solutions for
1:48:53that and then we will go back to this in
1:48:55the today seven in a few weeks
1:48:58but I will just give you an example uh
1:49:00at the end of this of this today or what
1:49:03this different prescriptions for
1:49:06stochastic eyecolus imply for the soccer
1:49:09blank equations and in particular we
1:49:11will discuss the difference in the
1:49:13stationary solutions that we get for
1:49:16these equations depending on whether we
1:49:18choose ether versus Stratton which
1:49:21prescriptions for stochastic calculus
1:49:24and so let me just uh just to introduce
1:49:28uh what is the topic of today let me
1:49:30just recall something that has been
1:49:33already discussed in the previous
1:49:34lectures which is launch event equations
1:49:38so we saw several examples already of
1:49:41this type of equations so I wrote it in
1:49:44here in the most general form for let's
1:49:47say a one-dimensional problem
1:49:50so this is a stochastic differential
1:49:52equations where you have this
1:49:54stochastics because you have the noise
1:49:56term of course which in general can be
1:49:59multiplied by some genetic function that
1:50:02is itself a function of your stochastic
1:50:04process and when this function is
1:50:07non-trivial so when this is not a
1:50:08constant then we talk about
1:50:10multiplicative multiplicative noise
1:50:13and an example of multiplicative noise
1:50:15was given in lecture two or three where
1:50:18it was discussed uh this problem of uh
1:50:22independent multiplicative growth and if
1:50:25you remember the equation was looking
1:50:27like this so it's very similar to what
1:50:30you saw in the homework number three
1:50:33where indeed the noise is multiplicative
1:50:36and it was also uh this equation was
1:50:39solved through a change of variables
1:50:41using stratanovic that was essentially
1:50:45the logarithm so if you remember we took
1:50:48the log of the die divided by the
1:50:50average overall the z i and we called
1:50:54this variable U and if you do this
1:50:57change of variables then you go back to
1:50:59one equation which in which the noise is
1:51:01is additive and no longer multiplicative
1:51:06and uh somehow this uh the reason why
1:51:09I'm putting this in the most general
1:51:11form is that the things that we are
1:51:13going to discuss today so the difference
1:51:14between
1:51:15stratanovic and Ito prescriptions for
1:51:19stochastic calculus are particularly
1:51:21important whenever we have this type of
1:51:23structure uh in the noise
1:51:26and we will see that important
1:51:28differences will arise when we look uh
1:51:31when we go from the launch of an
1:51:33equation to the so-called focal Planck
1:51:35equations which is an equation in which
1:51:38essentially what you do is to average
1:51:40overall these different trajectories for
1:51:43your process which correspond to
1:51:45different realizations of your noise and
1:51:48you write down an equation for the
1:51:50probability that your process is at a
1:51:54given point or configuration X at time T
1:51:57given that you started from some
1:52:00position x 0 at time t equal to zero
1:52:05uh sorry I think uh you have to move the
1:52:08camera a little bit yes exactly
1:52:15so I guess what's on the left we don't
1:52:18need it anymore
1:52:20okay so so today I I would like to start
1:52:23with a little bit of uh Theory so uh I
1:52:26will discuss a little bit uh what is the
1:52:29distribution of this noise what is the
1:52:31limit in which issues arise and how do
1:52:35we solve this issue by choosing uh in a
1:52:38sense prescriptions for our stochastic
1:52:40calculus and I will give a summary of
1:52:43what are the main things that one has to
1:52:45remember when uh doing calculations
1:52:48either with The Ether prescription or
1:52:50with the shatanovic prescriptions and
1:52:53then we go and do some points of the
1:52:57exercises in particular exercise one two
1:53:00and the last point of exercise three I
1:53:03think
1:53:05okay so let me start with some general
1:53:07things about the noise
1:53:12so the noise as we already saw is a
1:53:15stochastic process which usually is
1:53:19chosen to be uh to be gaussian so it's
1:53:22it's a process which is labeled by the
1:53:25time that is a continuous variable and
1:53:27it is gaussian in the sense that we
1:53:30assume that in order to characterize
1:53:33distribution we only need to specify
1:53:35what are the first two moments
1:53:38uh for this process so usually we take
1:53:42the average which is equal to zero for
1:53:45any time
1:53:48and we can think about and this was
1:53:50let's say the starting point of uh
1:53:54think about genetic processes where you
1:53:57have some correlations uh in time of
1:54:00these random variables so
1:54:03the correlation or
1:54:06covariances will be given by a genetic
1:54:10function let me call it g of p and p
1:54:12Prime
1:54:13and we can assume that this function is
1:54:16for instance an exponential so there
1:54:18will be a scale which I call Sigma so
1:54:23this is essentially a generalized
1:54:24variance so that's why I have this scale
1:54:26Sigma and then I put an exponential so
1:54:29let me write it like this one over two
1:54:32uh Tau C and then I put e to the minus
1:54:36P minus t Prime divided by 2 Tau C
1:54:41and what this means is that there is a
1:54:44typical time scales which is the
1:54:46correlation time Tau C that controls uh
1:54:50how much indeed the noise is correlated
1:54:52over time so this is a function
1:54:54which
1:54:56if I put D minus t Prime and this is G
1:55:02it decays exponentially over a certain
1:55:05correlation length or correlation time
1:55:07uh Tau C and what this means essentially
1:55:11is that as soon as you go beyond this
1:55:15time scales then you can more or less
1:55:16assume that the noise at your two times
1:55:20are Russian correlated whereas if you
1:55:22look at time differences which are
1:55:24smaller or of the order of Tau C you
1:55:28feel that this correlation is there
1:55:32okay now let me
1:55:34um
1:55:35yes I think there isn't
1:55:39um the two in the exponential is not in
1:55:42the trendy here right
1:55:45I mean in the TD uh we don't we divide
1:55:49On Me by 2C and not to see
1:55:53okay
1:55:54um
1:55:55so we are on the same page yeah I I want
1:55:58it to be normalized so I think it should
1:56:00be
1:56:03um if you put a two it should be
1:56:05everywhere
1:56:07thank you
1:56:08[Music]
1:56:09um
1:56:11or perhaps I want to put a sigma Square
1:56:14over two yeah so so in what I'm gonna do
1:56:17in the following let me take let me put
1:56:20uh
1:56:22maybe the two in both so that if I
1:56:24integrate
1:56:25overall positive times I get one one
1:56:29times Sigma Square
1:56:31okay and then I will check uh with it
1:56:34today that everything is consistent but
1:56:37that's a good point let me write down
1:56:39uh okay anyway what uh what is one thing
1:56:44that uh we should remember uh already
1:56:47from here
1:56:49that was trashed uh today as well
1:56:52uh so this thing is that when you ask
1:56:55what is the typical value so how big is
1:56:58roughly the noise so you can look at
1:57:00correlations that equals time and the
1:57:03typical value of the noise will go like
1:57:05the square root of this equal time
1:57:07correlation so the idea is that roughly
1:57:10Theta typical
1:57:12has a scaling at each time which is of
1:57:15the order of Sigma divided by the square
1:57:17root of Tau C
1:57:21yes you're right so if I do this I
1:57:25shouldn't it's Sigma or Sigma over
1:57:27square root of two uh depending on
1:57:30whether you put the two or not so I will
1:57:32make sure that in the final version of
1:57:34the today all the factors of 2 are in
1:57:36the right place
1:57:37but the important thing that one has to
1:57:39remember is this factor of 1 over square
1:57:41root of Tau C and why is this important
1:57:44now well it is important because what we
1:57:47want to do in Practical applications
1:57:50is to take
1:57:52the limit of white noise
1:57:56where essentially this correlation time
1:57:58goes to zero so now let me try to see
1:58:02where I can write
1:58:03nope
1:58:11okay
1:58:14foreign
1:58:19and this is the limit of white noise of
1:58:22course
1:58:28so the White Noise limit corresponds to
1:58:31taking
1:58:32LC going to zero
1:58:34and what happens to this function
1:58:36whenever it is properly normalized is
1:58:39that when this correlation time goes to
1:58:41zero the function tends to to a Delta
1:58:45function so the noise indeed
1:58:48uh becomes uh uncorrelated so you have
1:58:53the G of t
1:58:54T Prime in this limit goes to Sigma
1:58:58Square Delta
1:59:00of T minus C Prime
1:59:04okay now this is the limit that one
1:59:07would like to take but of course this
1:59:09limit is uh is a little bit uh
1:59:11problematic so one has to be careful
1:59:14when considering white noise and there
1:59:18are two sources that makes this limit
1:59:20let's say Not Innocent uh so the first
1:59:24thing is that as you see from what I
1:59:26just wrote
1:59:28That You Don't See but
1:59:30from this
1:59:33let me try to be Zoom
1:59:35sorry about this I have to
1:59:40okay let me rewrite it so the first
1:59:42problem is that as I said
1:59:44the typical value of the noise goes like
1:59:46one over square root of Tau C so when
1:59:48you take tausi going to zero this is
1:59:51this function is essentially becoming a
1:59:54very Singularity it's diverging it is
1:59:57not continuous
1:59:58and uh and this is a problem in
2:00:01particular when you want to look at
2:00:04differential equations and differential
2:00:07calculus and the idea so I will shape a
2:00:11little bit uh
2:00:13roughly in here and then we go into more
2:00:15detail but the basic idea is that when
2:00:18you do differential calculus
2:00:26there is another time scale which is
2:00:28going to zero that and you use the limit
2:00:30of this time scale is going to zero when
2:00:32you write derivatives for instance and
2:00:34this is the uh the DP
2:00:37so if you go back to the discretized
2:00:40version of the launchman equation which
2:00:43I wrote
2:00:44before
2:00:46okay look I will rewrite everything in
2:00:49here so if you think about larger one
2:00:51you discretize it then what the equation
2:00:54is telling you is that
2:00:56sorry maybe you can move the
2:00:58the camera a bit up
2:01:00so that yeah because the screen is uh
2:01:07if I just zoom do you still see
2:01:10because it would be nice if we have both
2:01:13Blackboard
2:01:20but then I'm afraid that it's too small
2:01:31go up up up
2:01:38this works
2:01:43thank you
2:01:44okay
2:01:49so the idea is more or less the
2:01:51following so as you've seen here
2:01:52what I can do when I have this
2:01:54differential equation is essentially to
2:01:56discretize at time so I take slices with
2:02:00a width which is of the order of this
2:02:03Delta tin here and then I can write this
2:02:06as a discrete difference equation so the
2:02:09value of my function at time t plus DT
2:02:12is the value at time t class I have the
2:02:15function f which I have to compute at
2:02:18some point which is within the interval
2:02:21t t plus BT and I will go back to uh to
2:02:24the choice of this point uh in a minute
2:02:26times DT and then I have uh the same for
2:02:30the function G and the product of my
2:02:32noise times times DT
2:02:35and what I'm saying is that essentially
2:02:37to write a language I need to take DT to
2:02:400 but at the same time to have
2:02:42uncorrelated noise I want to take a
2:02:44tausi uh going to zero and this means
2:02:48that I have to specify
2:02:50how does this product somehow behave
2:02:53whenever I look at these two limits
2:02:56together
2:02:57so the main issue is that basically you
2:03:00have two limits going to zero and these
2:03:01two limits do not commute with each
2:03:04other so we have whenever we want to
2:03:07discuss this type of differential
2:03:10calculus we always have to specify
2:03:12somehow how we take these two limits and
2:03:15this is what gives rise to the different
2:03:17prescriptions for uh for stochastic
2:03:20calculus so very roughly we will discuss
2:03:22today two of these prescriptions one is
2:03:25the so-called Ito and the other one is
2:03:27satanovic
2:03:29and uh in a nutshell they correspond to
2:03:31the following ideas so whenever you
2:03:33think about uh stratanovic what you are
2:03:36assuming is that your DT in here goes to
2:03:40zero before or faster than the time
2:03:44scale style C which controls the
2:03:47correlation of your noise
2:03:49so what this is telling you is that if
2:03:51this goes to zero faster than at the
2:03:54scale of delta T your noise function is
2:03:57somehow still continues so you can
2:04:00divide this by delta T and you can take
2:04:02the limit DT going to zero and so you
2:04:05would expect that you recover in this
2:04:07way the usual let's say rules for
2:04:10stochastic calculus and and the
2:04:12differential equations
2:04:15but at the same time since uh and then
2:04:18after doing this you take Tau C going to
2:04:21zero but you should remember that since
2:04:23Tau C goes to zeros lower at the scale
2:04:26of BT you will have still some
2:04:30correlations uh between your random
2:04:33variables that give you the noise and
2:04:36this correlation will matter uh in a way
2:04:38that I will specify in a minute so the
2:04:41bottom line is that if you uh which you
2:04:44assume the Tau C is the time scale which
2:04:46goes to zero
2:04:48afterwards and therefore you should
2:04:50expect that the rule of of calculus are
2:04:54the usual one that we have in absence of
2:04:56noise but there are some correlations
2:04:58which will show up when you look at the
2:05:01time scales of the order BT
2:05:03on the other hand what you can do is to
2:05:06assume that this DT goes to zero as fast
2:05:09uh or or even slower than Tau C
2:05:13and in this limit you are really uh in a
2:05:16situation in which your noise is
2:05:18uncorrelated so it is truly uncorrelated
2:05:21except at equal time so even if you zoom
2:05:24at this scale of DT you will have that
2:05:27correlations will go to zero
2:05:30but you're gonna have to be very careful
2:05:32about the rule of stochastic calculus
2:05:35and there will be corrections to the
2:05:37usual uh chain rules for example which
2:05:39comes precisely from this order of
2:05:43limits that you are assuming so this is
2:05:46the idea now let me uh give you a little
2:05:48summary or table uh to discuss
2:05:52concretely
2:05:54what this means when you have to do
2:05:55calculations
2:05:57and then we go to the exercises
2:06:00so first of all
2:06:02okay
2:06:12so I'm sure that you will have this
2:06:14discussed in some other courses in
2:06:16particular if you will follow
2:06:18the advanced statistical Physics course
2:06:21this will be discussed towards the end
2:06:24so what I'm what I would like to do
2:06:25today is to give some
2:06:28let's say operative
2:06:30summary
2:06:33which is a condensate of the main things
2:06:35the main difference is that one has to
2:06:39remember
2:06:40okay so we have
2:06:43stratanovic
2:06:46on one side and then we have
2:06:48Ito
2:06:56and as I as I said one of which morally
2:06:59corresponds to taking
2:07:01DT going to zero before
2:07:05C
2:07:07go into zero and it will corresponds to
2:07:09the opposite or if you want
2:07:12to see both go to zero
2:07:15with some scaling
2:07:19okay
2:07:21and what are the main differences that
2:07:24you find
2:07:25so as I mentioned there should be a
2:07:27difference when you look at correlations
2:07:30in particular at correlations at equal
2:07:33times or let's say overtime skills which
2:07:36are of the order of this PT that you
2:07:38want to send to zero
2:07:40so this is
2:07:42equal
2:07:45time
2:07:47correlations
2:07:53so what happens so suppose that we have
2:07:55now a function f of our stochastic
2:07:58process and we want to compute for
2:08:00instance the correlation of this
2:08:02function
2:08:03with the noise
2:08:05variable
2:08:06exactly at the same time t
2:08:09and based on what we just said what you
2:08:12should expect is that if you take the
2:08:14Ito prescription
2:08:15you're sending taosi uh to zero very
2:08:18fast and so what you should expect is
2:08:20that all correlations uh somehow uh
2:08:24vanish and what this implies is that
2:08:26this type of correlation function this
2:08:28type of expectation this is of course an
2:08:30expectation over the noise and its
2:08:33distribution has to be zero and why does
2:08:36it have to be zero well the idea is that
2:08:38you're looking at here at the noise at
2:08:40the time t
2:08:42and you are Computing here the value of
2:08:44the function x x time t
2:08:48and now if you look at the discretized
2:08:51version of our launch of an equation
2:08:53then you can write X at time t as a
2:08:57function of x x time T minus PT
2:09:01and this function will contain so let me
2:09:03write it very fast so you have that X at
2:09:07time t
2:09:08is X at time P minus and my discretized
2:09:12time delta T
2:09:13plus you have the value of the function
2:09:17within the interval which now goes from
2:09:19T minus BT to T
2:09:21times VT and then you have maybe the
2:09:24function G but
2:09:25you have somehow the noise at the
2:09:27previous time
2:09:29uh T minus DT
2:09:32and so the value of the function at the
2:09:34time T is fixed by or depends on the
2:09:37value of the noise as T minus PT and
2:09:40these two are uncorrelated so the value
2:09:42is T minus OCT and the value at time T
2:09:44of the noise are uncorrelated and this
2:09:47is why what you get is that this
2:09:49expectation value in The Ether
2:09:50prescription is is exactly equal to zero
2:09:55now of course if you do a stratanovic
2:09:57you know that this is not true because
2:09:59you have that this tausi goes to zeros
2:10:02lower so you have some correlations
2:10:04which show up
2:10:05and the way they show up and this is
2:10:07what we will show in the exercise number
2:10:11one
2:10:12is as follows so when you compute equal
2:10:15time correlation with the noise
2:10:17these are non-zero but are equal to
2:10:20Sigma Square over two
2:10:29times the expectation so I hope that you
2:10:31see this this is the expectation over
2:10:33the noise of the first derivative of the
2:10:36function that you have on the left hand
2:10:38side evaluated at X of t
2:10:42and I think that this factor of two
2:10:44comes because I didn't put a factor of 2
2:10:46in the denominator of G so uh the person
2:10:49who raised the comment was right so we
2:10:52should actually stick to the notation of
2:10:54the today and there shouldn't be a
2:10:55factor of two uh in the denominator so
2:10:57thanks for uh pointing this out
2:11:01okay so this is the first yes
2:11:04uh the bar over the x t what the is it
2:11:07is it the average or no I I will comment
2:11:12about this in a minute but the idea is
2:11:14that when you discretize for instance
2:11:16with the Euler discretization scheme or
2:11:19differential equation and you somehow
2:11:22have to choose
2:11:24where to evaluate so you have your time
2:11:26you take slices
2:11:29of with delta T
2:11:32and now let's say that I have t and t
2:11:34plus DT
2:11:36and in the discretized equation I have
2:11:39to choose where do so this function f is
2:11:41smooth on the scale of BT and what I
2:11:45have to do is to choose somehow where to
2:11:47compute it within the interval t and t
2:11:50plus DT
2:11:52and uh as we will see so as soon as we
2:11:55have terms which are not multiplied by
2:11:57the noise you can more or less choose it
2:12:00and choose this point arbitrarily so you
2:12:02have to choose let's say a point within
2:12:05this interval and it doesn't matter
2:12:06which one you choose because any
2:12:08variation in the point that you choose
2:12:11will lead to corrections to this
2:12:13discrete version of the equation which
2:12:15are of order delta T Square so when you
2:12:18divide by DT and take uh delta T going
2:12:21to zero they will disappear
2:12:23but this is not true when you look at
2:12:25terms like this so when you look at
2:12:27terms which are multiplied by the noise
2:12:29you really have to choose and decide
2:12:32which is the point within the interval
2:12:35at which you are Computing this function
2:12:36and different choices will correspond to
2:12:39different stochastic processes and this
2:12:41is what I will comment in a minute
2:12:44okay but let's say in general what I
2:12:47mean by this x bar is not necessarily
2:12:50the middle point but it is one point
2:12:51within let's say this small interval
2:12:54density
2:12:58Okay so
2:13:00Let Me Now go to the second difference
2:13:03so as I said that we will show this
2:13:07um
2:13:08in exercise one
2:13:11and of course when you write this
2:13:13expectation value what I mean is that I
2:13:15compute the expectation with respect to
2:13:18the noise distribution the distribution
2:13:20will depend on Tau C and then I will
2:13:23take after Computing the expectation I
2:13:25will take Tau C going to zero eventually
2:13:29okay these are correlations now let's go
2:13:31to uh uh differential calculus and there
2:13:35is uh in particular one thing that one
2:13:37has to remember
2:13:39that is about the chain rule
2:13:43uh for differential calculus So based on
2:13:46what we said before if you use this
2:13:49autonomics prescription
2:13:51then anytime you divide by DT and take
2:13:55DT going to zero you're assuming that
2:13:57Tau C is finite and so you have a
2:14:00function ETA which is somehow continuous
2:14:03over this small time interval and so you
2:14:06know that you have no singularities that
2:14:08appear in your discrete differential
2:14:10equations and so you can take safely the
2:14:13limit DT going to zero and recover for
2:14:16instance the the usual derivative and
2:14:19what this means is that your chain rules
2:14:22that we are used to will be valid in
2:14:25this set on which prescription so just
2:14:28to be concrete this means that if I want
2:14:31to compute the derivative
2:14:33over time of this function f
2:14:36which let's say for generality may
2:14:39depend both on my stochastic process and
2:14:42also it can depend independently on the
2:14:45variable T then what this is this is you
2:14:49have a partial derivative over time
2:14:51if you have an independent dependence of
2:14:54on T and then you have to do the
2:14:57derivative of a
2:14:59over the x times the x of T over DT
2:15:04and this is now a total derivative
2:15:05because X depends onion so this is the
2:15:08usual uh chain rule that we have in
2:15:11normal calculus
2:15:13but this chain rule breaks down if you
2:15:16do this game of uh taking the two limits
2:15:19together
2:15:20and it breaks down in the sense that if
2:15:22you do this you have to add an extra
2:15:24term which is a so-called Ito term
2:15:27and the Ito term looks like this so you
2:15:29have the same
2:15:33as before so now I'm dropping the
2:15:35dependencies
2:15:36yes yes already teas and you have the
2:15:38usual
2:15:39DF over DX times DX over DT
2:15:45and then you need to add this extra term
2:15:48which is
2:15:50Sigma Square over 2.
2:15:54now if I have a generic range of an
2:15:56equation with multiplicative function in
2:15:59front of the noise here I also have a
2:16:01factor of
2:16:02g-square
2:16:03which was the constant that was
2:16:05appearing sorry the function which was
2:16:07appearing in front of the noise in
2:16:09algebra and then I have the second
2:16:12derivative
2:16:13of my function with respect to X and
2:16:16this is the so-called Eco term
2:16:23okay
2:16:26and this arises because and again this I
2:16:30will comment more in the following but
2:16:32the idea is that in your discrete
2:16:34version of Lounge van you have G times
2:16:38ETA times delta T and if you take the
2:16:42limiter delta T going to zero and Tau C
2:16:45going to zero let's say together then
2:16:48you see that that extra term is of the
2:16:50order of square root of DT so when you
2:16:54want to write the differential equation
2:16:56to linear order in DT you also have to
2:16:59take the square of that term which will
2:17:02give you a contribution of the order of
2:17:04delta T and this is why this term is is
2:17:07appearing in here
2:17:08but let's say as a rule of thumb what
2:17:10one has to remember is that in beta
2:17:12prescription equal times correlations
2:17:14are easy but you have to remember this
2:17:17extra term uh when you do derivatives
2:17:20whereas in strattonovic the chain rule
2:17:23is is the same one so differential
2:17:24calculus is as you expect it to be but
2:17:27you have to remember that you have some
2:17:29correlations which show up in this in
2:17:33this particular form
2:17:35sorry
2:17:36yes uh the the equation for the chain
2:17:41rules are valid on average or for really
2:17:46the function f
2:17:49or
2:17:53no no here you you really take the
2:17:55function uh for one particular
2:17:58trajectory
2:17:59uh depending on the noise
2:18:05uh okay now I want to uh go back uh for
2:18:09a minute uh to that issue of choosing uh
2:18:12the point within the interval DT
2:18:15and the reason why I want to do this is
2:18:17that that is particularly important when
2:18:20you try to derive when you derives the
2:18:23soccer plank equation starting from the
2:18:25launch of an equation because the usual
2:18:28procedure is to you take the lunge of an
2:18:30equation you discretize and then
2:18:32depending on how you choose this uh this
2:18:36intermediate point in your interval you
2:18:38will get different focal Planck's
2:18:40equations in general
2:18:41so as I mentioned we we will do this
2:18:44explicitly in the today number seven but
2:18:46now let me give you just a summary and
2:18:48then you can look at the derivation by
2:18:50yourself
2:18:52um in the homework
2:18:54uh and maybe let me continue
2:18:58to be stable
2:19:01um
2:19:02in the second Blackboard
2:19:13okay so now we discuss this critization
2:19:18foreign
2:19:21this is important of course if you want
2:19:24to solve luncheon equations on the
2:19:26computer the only way to do this is to
2:19:29discretize and then
2:19:31solve the equation for a finite DT
2:19:35okay
2:19:37and let me
2:19:40so as I wrote before
2:19:43let me rewrite
2:19:47the discretized equation in general
2:19:51so what do we want to do
2:19:54so suppose that you want for instance to
2:19:56simulate your launch event on a computer
2:19:58then as I said you choose a
2:20:00discretization step that is
2:20:02and then you try to do what I wrote
2:20:04before so you try to solve for your
2:20:07process at time t plus DT knowing the
2:20:10value of your process at time T and what
2:20:12you have to do is
2:20:15compute this function somewhere and
2:20:18multiply by the DTs and then you have
2:20:21the same for multiplicative noise
2:20:24and you have your noise at time p
2:20:27okay
2:20:29and now in a discrete I deciding we
2:20:31still want to capture this uncorrelated
2:20:33noise so what we the discretized version
2:20:36let's say of the Delta function we look
2:20:39as follows so we assume
2:20:43that the correlation between the noise
2:20:46evaluated at two steps Beyond or t Prime
2:20:49and T Prime
2:20:51is equal to Sigma squared
2:20:55times a chronic or Delta now because we
2:20:58are discrete so this is the chronicle
2:21:00delta T T Prime
2:21:02and then so your time here to discretize
2:21:05a Delta function you have to remember
2:21:06that there's a function as uh the
2:21:09dimension if you want of 1 over time if
2:21:13you are looking at a data function in
2:21:14time and so in the discretized setting
2:21:17the way you implement it is by dividing
2:21:21let's say your correlational function by
2:21:23the small interval BT
2:21:26that you're looking at which gives you
2:21:28back uh this uh this thing that the
2:21:32typical value of the noise is a order of
2:21:34one over square root of something that
2:21:36you want to send to zero
2:21:38now I hope that the comment I'm gonna do
2:21:40now is not too confusing but uh somehow
2:21:44remember what I said about taking the
2:21:46order of limits so once we write the
2:21:49equation in this way somehow what we are
2:21:52already assuming is that Tau C is going
2:21:55to zero meaning that we can really write
2:21:58down this correlational function as a
2:22:00Delta function in the discrete setting
2:22:03and your delta T inside is remaining
2:22:07finite at least when I write this finite
2:22:11difference equation
2:22:12so it looks like I'm already
2:22:14implementing in here the uh the ETO
2:22:16prescription
2:22:17but there is a way to recover within
2:22:20this discretized setting the
2:22:22stratonovich rule and this comes uh by
2:22:26choosing appropriately
2:22:28the point in which I evaluate the
2:22:31function G in this right hand side and I
2:22:35want to choose it in such a way that
2:22:37when I when I do the ether choice I
2:22:40recover the vehicle time correlations
2:22:42are zero as I saw as I wrote up there in
2:22:45the Continuum whereas I want a choice of
2:22:48this point such that when I go to the
2:22:50continuous limit I get an equal time
2:22:52correlation which is no zero
2:22:55and the way this problem is is uh solved
2:22:57is the following so if you do it though
2:23:02then that's easy you can just choose
2:23:06the point where to compute this
2:23:08functions f and g as
2:23:10the
2:23:11let's say
2:23:13marginal point of your interval DT
2:23:16so remember we have t t plus DT
2:23:19so whenever you do it you choose to
2:23:23compute those functions exactly at this
2:23:26time T which is the beginning of your
2:23:28interval and then you get an equation
2:23:30which is very easy to to solve because
2:23:32whatever you have on the right hand side
2:23:35just depends on the value of your
2:23:37function at a given time T and so
2:23:39recursively you get the value of the
2:23:41function at the next time simply by
2:23:43Computing uh directly at the right hand
2:23:46side
2:23:47now when you want to do uh some of which
2:23:49you have to do something else meaning
2:23:51that you have to choose now
2:23:54this point to be what maybe this
2:23:58notation suggests which is the Middle
2:24:00Point
2:24:02in the interval
2:24:06so this means that you are Somehow Here
2:24:10when you do stratanovic whereas this is
2:24:13Ito
2:24:14and then this is a treat so you can
2:24:16solve this explicitly you you discretize
2:24:19you assume this choice
2:24:20first of all you can notice that then
2:24:23solving your equation numerically is a
2:24:25little bit more complicated because you
2:24:27have X of t plus BT which appears both
2:24:30on the right hand side and on the left
2:24:32hand side so you have an non-linear
2:24:34equation that you have to solve for x of
2:24:38t plus DT but then you can really check
2:24:41going through the discretization that if
2:24:44you look now at equal time correlation
2:24:46functions with the noise you get that
2:24:49the value of those correlation is
2:24:50non-zero exactly as we want it to be in
2:24:53the continuous limit so maybe we will do
2:24:55this at the end if we have time
2:24:58but uh some of this is a step that is
2:25:01necessary if we want to do to go through
2:25:03the solution to exercise C because it is
2:25:07uh necessary when you want to derive
2:25:10your soccer plank equation from uh from
2:25:13the launch of an equation
2:25:14so you will see that this goes through
2:25:16indeed uh discretizing your stochastic
2:25:19process
2:25:20and these two different
2:25:22uh prescriptions for this this
2:25:26critization will give rise to two
2:25:29different forms of your focal Planck
2:25:31equations
2:25:33uh that are the following so let me
2:25:35start from
2:25:37um no let me start from Ethan
2:25:41and let me Define a quantity d
2:25:45X you can still see
2:25:48so I Define the X to be Sigma Square
2:25:51over 2
2:25:54times
2:25:55G square of x
2:25:57okay again this is what appears in front
2:26:00of the noise in my lunchable equation
2:26:03and with this notation what you get if
2:26:06you look at focal Planck equation is in
2:26:08The Ether prescription is the following
2:26:11so this is an equation for
2:26:14as I said before the probability that
2:26:17your process takes value X at a given
2:26:20time t
2:26:21it is a differential equation so you
2:26:23have DP over DT that will be equal to
2:26:25minus
2:26:27the derivative over X
2:26:31of f times p
2:26:34f using this is what appears in the
2:26:37language equation
2:26:39Plus
2:26:41I hope that I can squeeze everything in
2:26:43here plus the second derivative
2:26:46over X of
2:26:49D which I just defined which depends on
2:26:52x times p
2:26:55okay
2:26:58so this is poker Planck in detail
2:27:01prescription
2:27:02now what happens when you look at the
2:27:04stratonovich prescription well what
2:27:06happens is that you will get a term
2:27:09which comes from the noise because of
2:27:11course this B here as you recognize uh
2:27:14comes from taking variances of the noise
2:27:17which has a slightly different form so
2:27:20which looks like this
2:27:23I still have DP over DT I have minus
2:27:27the let's say term which depends on that
2:27:31is is unchanged
2:27:33but instead here I have to take D over
2:27:36DX
2:27:37a fourth of square root of B
2:27:41times D over DX of square root of B
2:27:45times p
2:27:48so there is a structure that is a little
2:27:50bit more symmetric you split B into a
2:27:53square root and you have one term which
2:27:56is inside the derivative and one which
2:27:57is outside
2:27:59and and having two different equations
2:28:01corresponds to two different uh
2:28:04processes so this comes from here this
2:28:07comes from here and as we will see at
2:28:09the with the exercise number three this
2:28:12processes are really different uh so for
2:28:15instance if you ask what is the
2:28:16stationary state that you get in the
2:28:18long time limit the stationary state
2:28:20will look different depending on whether
2:28:22you choose to describe your process with
2:28:25Ito or with a Statin of each rules in
2:28:29general of course there is one case
2:28:30where nothing changes so can anybody
2:28:33guess
2:28:34uh what is this case
2:28:37where these two equations becomes the
2:28:39same
2:28:43depends on X exactly so when you have a
2:28:48noise which is uh not multiplicative or
2:28:51if you if you want when G is just a
2:28:53constant that you can absorb into the
2:28:55variance of the noise then these two
2:28:57equations become exactly the same so the
2:29:00bottom line is that you have to be
2:29:01careful about you your choice of
2:29:04discretization and prescription of
2:29:07calculus basically whenever you have
2:29:09multiplicative noise when you have
2:29:10additive noise you're at the level of
2:29:13the focal flank nothing changes and you
2:29:16can
2:29:16you get an equation which is the same
2:29:20okay so uh this is a little bit of a
2:29:24crash
2:29:26course or summary and what we are going
2:29:29to do uh now as I say this first of all
2:29:32with exercise one we want to show
2:29:34explicitly this expression for the
2:29:37correlation function with satanovic then
2:29:40exercise two so the idea is to go back
2:29:42to
2:29:43um to this problem of multiplicative
2:29:46growth that was discussed in the lecture
2:29:48and if you remember there it was
2:29:50discussed in the Stratton of each
2:29:52prescription so I just wanted to sketch
2:29:54what changes if you instead look at The
2:29:57Ether prescription and things changes
2:29:59change because in that case the noise
2:30:01was multiplicative so G was just linear
2:30:05in x and then the in the last part we
2:30:09will see we will compute for that
2:30:11particular problem the stationary State
2:30:14involves prescription to see that uh
2:30:17that it changes
2:30:19if we have time
2:30:22okay so let's start
2:30:26with the first exercise
2:30:29which is a little bit
2:30:32of a functional calculus
2:30:38which I think is good to see at least
2:30:40once
2:30:42foreign
2:30:52the expression of what we want to prove
2:30:54but okay you have it in the paper
2:30:57in the textile day today
2:30:59so the idea is to show that correlation
2:31:02but before
2:31:04we are
2:31:06going to show an intermediate
2:31:09expression that is the equation 22.
2:31:15uh let's say so this is exercise one
2:31:19Point number one
2:31:20so the intermediate thing that we want
2:31:22to show is that if we take a function
2:31:26of the stochastic process
2:31:30and its correlation with the noise we
2:31:33can write this as
2:31:35an integral
2:31:37over some time t
2:31:40then I have this function G which is the
2:31:42correlation
2:31:44that I introduced before
2:31:46evaluated at TNT Prime
2:31:49and then I have the expectation value of
2:31:51the functional derivative
2:31:54of my function s with respect to the
2:31:56noise
2:31:57at the time T Prime that I am
2:31:59integrating on
2:32:02okay
2:32:04and remember that we take in here Tau C
2:32:08let's say finite for the moment and then
2:32:11we will take the limit at the end we
2:32:13will take the limit as Tau C going to
2:32:15zero
2:32:17okay now first of all what is the
2:32:18expectation value in here so the
2:32:21expectation value is an expectation with
2:32:24respect to this stochastic process so it
2:32:26is a functional integral
2:32:28so if
2:32:29[Music]
2:32:30um
2:32:31if anybody uh if there is somebody who
2:32:34has not seen functional integrals I
2:32:36think this is actually very easy there
2:32:39is um is a continuous limit of discrete
2:32:42calculus when you have many variables
2:32:45so in in the casing which you have many
2:32:46variables what you will have to do is to
2:32:49write integrades where you integrate
2:32:50over all the possible values of your
2:32:53variables and then you take the
2:32:54continuous limit and what this will give
2:32:56you
2:32:57is is a measure that is usually
2:32:59indicated like this so if you discretize
2:33:03into your time stats Tien
2:33:07let's say that Tien is
2:33:09delta T times n
2:33:12then this the functional measure the
2:33:15sorry the measure that you would get in
2:33:17this discrete setting is just the
2:33:18product Over N of BTN
2:33:21divided by a square root of 2 pi usually
2:33:24this is what people put for
2:33:27normalization and then if you take the
2:33:29limit of BP going to zero you define
2:33:32your functional measure to be let's say
2:33:34the continuous limit of this infinite
2:33:37product and this is what I'm denoting
2:33:38with this notation in here so if anybody
2:33:41has questions this is something that we
2:33:43can discuss in the question and answer
2:33:45file there are also some questions from
2:33:47the previous year on this point so we
2:33:50can go back to that
2:33:51but anyway you you have to take an
2:33:54average so you integrate overall your
2:33:55process some probability of your full
2:34:00process so for all times and this
2:34:02probability is what is uh is a gaussian
2:34:05because we saw that and we said that the
2:34:09process is gaussian so this P of ETA is
2:34:12what it will be one over some
2:34:14normalization
2:34:16of uh a gaussian factor which is e to
2:34:21the minus one as
2:34:23integral over two times DT DT Prime
2:34:28of
2:34:29the variables is of T and then you have
2:34:33here C to the minus one so the inverse
2:34:36of the correlation function at times TNT
2:34:40Prime Times
2:34:41ETA T Prime so I hope this is not true
2:34:44that you can see this
2:34:49but what is this so I'm just writing the
2:34:51continuous version of and of course the
2:34:53two integers go let's say from minus
2:34:55infinity to Infinity
2:34:57so this is just a continuous version of
2:34:59of a multivariate gaussian if you want
2:35:01where you know that whenever you have a
2:35:03gaussian what you have at the
2:35:05exponential is the inverse of the
2:35:08covariance Matrix which is exactly what
2:35:09I'm writing in here in in functional
2:35:12form
2:35:14so the fact that you
2:35:16have shape of the distribution
2:35:18like this you
2:35:18[Music]
2:35:20because
2:35:21now you see that what I have in here is
2:35:23a factor of ETA
2:35:25and what I have in the measure is e to
2:35:28the ETA Square essentially so what is
2:35:31one way to bring down one factor of beta
2:35:34well this is to take the derivative of
2:35:37my probability measure with respect to
2:35:40Eta itself
2:35:41which is now a functional derivative but
2:35:43that's not too difficult so uh let me
2:35:47compute
2:35:50what is the derivative of my gaussian
2:35:53weight with respect to
2:35:55Eta evaluated at some time let's say U
2:36:00so if I do this derivative what do I get
2:36:02I have one over Z which is untouched and
2:36:05then I will bring down uh let's say I
2:36:07derive with respect to D to this and I
2:36:09have C to the minus 1 times this then I
2:36:12derived with respect to this so I would
2:36:13have ETA times this but using that this
2:36:18um
2:36:18covariance function is symmetric is a
2:36:21yes symmetric in TNT Prime and so it
2:36:24will be the inverse I can collect these
2:36:27two terms so if you do this carefully we
2:36:31get just minus
2:36:33the integral over one of the times the
2:36:36CDT Prime
2:36:38C to the minus 1 and now evaluated a few
2:36:41P Prime Times e type time T Prime
2:36:49times the factor of P itself
2:36:53which I compactly write in the following
2:36:56way
2:36:58okay so now this is almost uh ETA except
2:37:01that we have this G this C that we don't
2:37:04like so what is a trick to single out
2:37:07ETA well what we can do is to multiply
2:37:09what we have in here uh by C to the uh
2:37:13sorry here we have the inverse of C so
2:37:15we want to multiply this from the left
2:37:18by uh by C itself so if you discretize
2:37:23you would have a matrix multiplication
2:37:25here we have to do this with function
2:37:28and
2:37:29the matrix multiplication corresponds to
2:37:32so in a matrix multiplication you have a
2:37:34sum over internal indices and in the
2:37:37continuous limit the sums becomes just
2:37:40an integrand so what I'm saying is that
2:37:42if I do the following so suppose that I
2:37:45integrate
2:37:46over X
2:37:48a function a c of u x times
2:37:53e to the minus 1 X
2:37:58so this is like a matrix times it's
2:38:00inverse the internal index now becomes X
2:38:03and I'm summing over it and what I
2:38:06should get out of this is the identity
2:38:09Matrix if I am looking at the discrete
2:38:12formalism and in the continuous limit
2:38:14what this will give me is just a Delta
2:38:17of U minus t
2:38:19okay so I want to do this in here so
2:38:23what I do is I multiply by C of T and U
2:38:27and then I integrate over U
2:38:30and so if I do this
2:38:35this is a little bit lengthy but it is
2:38:38not difficult so
2:38:50let me do it fast
2:38:52uh
2:38:55so what I get is the following so
2:38:59I take
2:39:01the U
2:39:03P of let's say t u times my derivative
2:39:07of T at
2:39:11you
2:39:14functional derivative
2:39:16and then I look at the right hand side I
2:39:18contact and see with C to the minus one
2:39:21this gives me a Delta of TNT Prime so
2:39:23I'm left with minus
2:39:26ETA of t
2:39:27times
2:39:29my gaussian factor which is unchanged
2:39:33okay and this is a little bit what we
2:39:35have up there maybe you don't see up
2:39:38there no indeed
2:39:44uh yes
2:39:46so you see that up there we have the
2:39:49expectation value of f times C time this
2:39:51is what is the integral of f times ETA
2:39:54times p
2:39:55so what I can do is to write
2:39:58this expectation value
2:40:05ETA of t
2:40:07using this expression in here as minus
2:40:11now I hope that I don't do mistake but
2:40:13this is minus I can take out the
2:40:16integral over U
2:40:17so I have C of EU
2:40:21and then I have my functional integral
2:40:23which
2:40:25is the expectation
2:40:27times uh
2:40:30well the function is
2:40:35times this functional derivative
2:40:39of d
2:40:46and how to get
2:40:49and this is now computed
2:40:52at the time U yes
2:40:56and now to get out the expression that I
2:41:00have in there what I have to do is
2:41:02essentially in integration by parts of
2:41:05this expression in here so here I have
2:41:06the derivative over p and I want to
2:41:08bring the derivative
2:41:09uh
2:41:10in front of f and this
2:41:14integration by parts so we'll have
2:41:16boundary term which go to zero basically
2:41:18because this is a distribution that has
2:41:21to Decay at Infinity
2:41:24in some senses so uh those boundary
2:41:27terms cancels this minus cancels because
2:41:30of the formula of integration by parts
2:41:32and the thing I end up with
2:41:36is
2:41:39this times
2:41:42uh then I bring the derivative in here
2:41:44and what I'm left with is just an
2:41:47expectation over the noise of
2:41:49the functional derivative of f
2:41:52with respect to
2:41:55ETA
2:41:57which is uh apart from you becoming C
2:42:01Prime is exactly what we wanted to show
2:42:05okay
2:42:06and now that we have this what we have
2:42:10to do is
2:42:11to compute this expectation in here
2:42:16so
2:42:19let me go on on this side of the
2:42:21Blackboard
2:42:26and to compute
2:42:28that expectation we first have to
2:42:30compute the functional derivative
2:42:33and to compute the functional derivative
2:42:34we use
2:42:36what we know about our stochastic
2:42:38process namely that it satisfies
2:42:41a larger one equation and now I take it
2:42:44with additive noise
2:42:46for Simplicity
2:42:50uh okay so where are we
2:42:59we are here
2:43:03okay so uh what is this uh functional
2:43:07derivative
2:43:11well first of all we are using
2:43:13stratonovich so derivatives work in the
2:43:18way that we are used to so when I want
2:43:20to compute
2:43:21BS of x t with respect to D ETA
2:43:26a few
2:43:28this will be so I will have first a
2:43:31derivative let me call it f Prime
2:43:35X of T and then I have to take the
2:43:38functional derivative of x of t
2:43:41with respect to
2:43:43okay
2:43:47and this is something that we can
2:43:48compute uh using larger one equation and
2:43:52the reasoning is as follows so if you
2:43:55see in here
2:43:57we want to integrate this object against
2:44:00this correlation function and this
2:44:03correlation function has this
2:44:04exponential form it became exponentially
2:44:08over this Tau C and eventually we want
2:44:11to take Tau C going to zero so what this
2:44:14is telling you is that what will matter
2:44:16in this integral are only times t or
2:44:20actually times U which are sufficiently
2:44:23close to T so that those are the values
2:44:25of times where this correlation function
2:44:28will be will be essential in all zero so
2:44:31what we can do so you see that we have T
2:44:34and U in here this term in here selects
2:44:37times U which are close to T and then we
2:44:40have to compute the derivative of x with
2:44:42respect to the noise at time U and we
2:44:45can assume that the two no the two times
2:44:47tnu are closed because of this argument
2:44:50in here and therefore what we can do is
2:44:52to expand this derivative to linear
2:44:56order in T minus U which is uh and see
2:45:00which terms in this expansion Will
2:45:02Survive once we take the limit Tau C
2:45:05going to zero
2:45:07so what is the the way to do this well
2:45:10we take our language equation and we try
2:45:12to expand it uh to linear order so let's
2:45:16say that we have a Time U
2:45:19and we have a Time T and they are closed
2:45:22and then I write X at time t as
2:45:26as what so let's let me call
2:45:29let me assume that we have a t 0 and U
2:45:32is in between
2:45:35so my launch of money is now of the form
2:45:37f of x plus ETA so if I integrate it
2:45:40assuming that this time difference is
2:45:42small I will have
2:45:44x x time t 0
2:45:46plus the function
2:45:49at some point within the interval but
2:45:51now I'm looking at an interval which I
2:45:54assume to be small this function will be
2:45:56smooth inside the interval so I can
2:45:58approximate this
2:46:00the value of this function as
2:46:02the value of s at the point T so
2:46:06whatever choice I make within of points
2:46:09within this interval I will get
2:46:10differences which are order of T minus t
2:46:140 to the power of 2 so I can neglect
2:46:16them
2:46:17at this level so let me do this and then
2:46:21we see if we agree
2:46:23so I have this and then I have the
2:46:26linear noise now for the noise I cannot
2:46:28assume this argument of continuity
2:46:30because the noise will eventually become
2:46:32a function which is known smooth
2:46:35but I can simply integrate it formally
2:46:38from P0 to T
2:46:40and I get something like this
2:46:44okay
2:46:46and now what I do well I take the
2:46:48functional derivative with respect to
2:46:50the noise at some time U which is close
2:46:53to T
2:46:55and I get the DX over
2:46:59the Italian this is uh is not in there
2:47:02here I have a dependence so I will have
2:47:06F Prime of X of t
2:47:09times the derivative
2:47:12of X of T in the U and from here I have
2:47:18a factor of 1 basically because I
2:47:20integrate I derive inside the integrals
2:47:23so I have a factor of one
2:47:25plus well even here if you want I have
2:47:29orders of T minus t 0 to the power 2 so
2:47:32they will still be there uh also in this
2:47:35expression
2:47:36uh did I forget yes here I forget
2:47:39T minus P0
2:47:43plus one
2:47:44plus order of this time difference
2:47:47Square
2:47:49and let me add something just to be
2:47:52precise which is acetal
2:47:56T minus U because of course this
2:47:58derivative with respect to e times time
2:48:01U will be no zero Only If U is within
2:48:03this interval so it has to be smaller
2:48:06than t
2:48:07okay so now this is what I want you see
2:48:09the disappears also on the right hand
2:48:11side multiplied by DT and I want an
2:48:14expansion of the order of DT so what I
2:48:16can do is I recursively plug in here
2:48:22the expression that I that I that I get
2:48:25out of this equation so this will be
2:48:28one plus something which is a water PT
2:48:31right one comes from here and then I
2:48:34have something which is already DT
2:48:36and so uh if I do this
2:48:40what is the only term
2:48:43which I can keep out of these so can I
2:48:48let me do it directly so if I replace
2:48:50this by one plus order BT and I keep
2:48:53only terms which are ordered VT then it
2:48:55is enough for me
2:48:57put the one in here
2:48:59and everything else will go into this
2:49:02correction
2:49:06s
2:49:09very good and now this is what uh what I
2:49:12want to plug uh in here and remember
2:49:15that all of this we have to take the
2:49:17expectation value with respect to the
2:49:19noise so in the end
2:49:22how much space do I have here
2:49:26until here
2:49:29okay so in the end what I will get
2:49:32plugging everything in this expression
2:49:34in here is
2:49:36is the following is the integral over d
2:49:40u
2:49:41of C
2:49:43and maybe what I call C now before was G
2:49:46sorry
2:49:49it is the correlation of the noise is
2:49:51the same function uh C of t u and then
2:49:55if I plug the expectation value of this
2:49:58and what do I have
2:50:01I have
2:50:06V and remember I also have this Factor
2:50:09here so I have F Prime times one
2:50:12which is just F Prime of X of t
2:50:18Plus
2:50:20plus this is still inside the
2:50:22expectation
2:50:23plus what I have here so this is again
2:50:26as Prime
2:50:28of X of t f Prime
2:50:32of X of T times DT
2:50:36okay
2:50:38plus higher orders
2:50:42okay
2:50:45and now I have more or less everything
2:50:48that I need
2:50:50because now what I can do is I plug the
2:50:54expression for the correlation of my
2:50:56noise so this will be
2:50:58Sigma Square over Tau C without the two
2:51:03e to the minus
2:51:06T minus U
2:51:08divided by
2:51:1020.
2:51:12and there is this Theta function with
2:51:14tells me actually that T minus U is
2:51:17positive
2:51:18this object here will not depend on you
2:51:21so I can take it out of the integral and
2:51:23I integrate over the um
2:51:27over the correlation function now and
2:51:29this will give me a constant factor
2:51:31which does not depend on Tau C so this
2:51:33you can do it I have all of the
2:51:36solutions are written so let me not do
2:51:38it here to save a little bit of time but
2:51:41uh let's say the point is that you take
2:51:43this out and you can integrate over C
2:51:46and for the second term you can take out
2:51:49the expectation value of these two
2:51:51derivatives and then you have to
2:51:52integrate C of t u times this DT which
2:51:58is of the order of T minus U and if you
2:52:00do this second integral you will see
2:52:02that what comes out is a is a term which
2:52:05is of the order of Tau C so when you
2:52:08take the limit Tau C going to zero this
2:52:11part of the expectation value will go to
2:52:13zero
2:52:16let's see when
2:52:18now C goes to zero and you're left with
2:52:21uh with the expression that we wanted to
2:52:24show so with uh let's say b
2:52:28um a constant which will be Sigma Square
2:52:30over two times the expectation value of
2:52:33the first derivative of the function
2:52:36which is exactly uh the rule for equal
2:52:38time correlations in the certain which
2:52:41prescription
2:52:42so I didn't do the all of the
2:52:44calculations in detail but let's say
2:52:46what remains to be done are integrals of
2:52:49this exponential functions and these are
2:52:51uh quite easy to do so I think I hope
2:52:56the idea is clear otherwise
2:52:58just stop me and I can do them more in
2:53:01detail
2:53:03now to to conclude with this
2:53:13let me just do the point number three
2:53:18uh maybe address one question when you
2:53:21use the language equation you don't have
2:53:24this function T anymore
2:53:26you don't have the
2:53:28function G anymore that was multiplied
2:53:30by the noise
2:53:31yes and here I'm assuming that that we
2:53:34are using additive noise
2:53:37okay in this direction
2:53:43you're right
2:53:45I hope it's specified in the exercise
2:53:48otherwise tell me and I will correct it
2:53:57and now I just want to do so let's say
2:53:59the 0.3 of this exercise just because it
2:54:02is
2:54:02[Music]
2:54:03um
2:54:03it is a simple way to to recap the
2:54:07different prescriptions so in that case
2:54:09uh what it tells you is look at now a
2:54:12large of an equation where you don't
2:54:14even have the function f so you have DX
2:54:16of t
2:54:17over DT this is just Pure Noise
2:54:22and so
2:54:24as we saw before X of T will be the
2:54:26integral now let me take a t 0. T over D
2:54:31Tau
2:54:32of Tau
2:54:34and uh what we want to do just to check
2:54:38uh what we said so far is to compute for
2:54:41instance the expectation value of
2:54:43x square of t
2:54:46with the two different prescriptions so
2:54:48with uh both uh ether and stratonovich
2:54:52so now we can do two things so the first
2:54:54one is
2:54:55let me assume that I'm using stratanovic
2:54:59and let me plug in here the expression
2:55:02for uh for ETA sorry for X so this will
2:55:07give me
2:55:09I will have two factors of X so two
2:55:11integrals integral over Theta 1 beta2
2:55:17and then the expectation value of itself
2:55:20Tau 1
2:55:23ETA Tau two
2:55:24which is what we
2:55:27defined to be C or G I don't know let's
2:55:31say G no C like before this is C of Tau
2:55:361 Tau two so this is the exponential a
2:55:40function that that we wrote before so I
2:55:42can plug the exponential I can do this
2:55:44uh two Integrations and if you do this
2:55:46again exercise in uh in Computing
2:55:51integrals of the exponentials what you
2:55:52get out of this should be
2:55:55a factor of Sigma Square Times t
2:55:58so this you can do of course Computing
2:56:02this expectation value for a finite Tau
2:56:04C and then taking calci going to zero
2:56:08but now let's just do to to fix the
2:56:10ideas let's try to Red arrive this
2:56:13result uh playing a little bit with Ito
2:56:18versus strathonovich
2:56:19and the way to do this is let me not
2:56:22compute the expectation of x squared but
2:56:25let me compute the expectational
2:56:28the derivative of x square and then I
2:56:30will integrate it
2:56:31over time
2:56:34and now to compute the expectation of
2:56:36the derivative of x square I have to
2:56:37decide which are the prescriptions that
2:56:40I use because the derivative changes
2:56:43so in the Ito case
2:56:50which is yes
2:56:53in the case of stratanovic the
2:56:54derivative is simple it's the usual one
2:56:57so this will be twice the expectation of
2:57:00x v t x of T sorry times
2:57:04x dot of T so dot is the time derivative
2:57:09and now what is x dot well if I derive
2:57:12this I just have a factor of ETA so what
2:57:14this gives me is
2:57:17the equal time correlation
2:57:19of X of t with the noise at the same
2:57:23time t
2:57:24and using the rule of correlations that
2:57:28we just derived this is what this is
2:57:30Sigma Square over 2 times the
2:57:33expectation of the derivative of the
2:57:35function f now my function capital F is
2:57:38just X so its derivative is one so you
2:57:41see that what I get out of this is just
2:57:43a factor of Sigma Square
2:57:46which is good because this is the
2:57:47derivative of what I wanted to compute
2:57:49and now it's constant so I can integrate
2:57:51over time and I get
2:57:53this resulting here
2:57:56now how do you get it with Ito well with
2:57:59Ito
2:58:00what you have to remember is that the
2:58:03chain rule is not that simple so you
2:58:05have the same factor which is 2 X of T
2:58:09times
2:58:10e times t
2:58:12and then you have an additional factor
2:58:15which is the second derivative of your
2:58:19veto term and the term is Sigma Square
2:58:22divided by two the two cancels this and
2:58:25then I have the second derivative of uh
2:58:28of the function of which I want to
2:58:30compute uh the chain Rule and the second
2:58:33derivative is again equal to one is
2:58:36equal to two because I have to rederive
2:58:37this by X so I have a factor of 2 so the
2:58:40two uh cancers so this is let me write
2:58:44it
2:58:45by Ito I have this and then I have the
2:58:48expectation of two which is the second
2:58:50derivative and then I use the fact that
2:58:53this is a correlation at equal time so
2:58:55in the Ito prescription this is zero and
2:58:58this is just the same factor of Sigma
2:59:01Square as for for satanovic so for this
2:59:05simple expectation value and for this
2:59:08simple type of process
2:59:11which indeed is additive nothing changes
2:59:14you can choose which prescription to use
2:59:17and as soon as as long as you are
2:59:19consistent then you get the same result
2:59:23okay now we uh do not have much time I
2:59:27just want to maybe say something about
2:59:30exercise too
2:59:35I hope this was not too fast
2:59:38but exercise 2 will be fast because it
2:59:40is basically stuff that was discussed
2:59:42already
2:59:43in the lecture number three last time
2:59:47and it was also discussed in the
2:59:50homework if you had time to look at it
2:59:54foreign
3:00:07concretely when you have
3:00:09multiplicative processes
3:00:12I'm sorry what is this mean value of 2
3:00:17uh you asked me to move the screen sorry
3:00:19if I asked you to repeat but the
3:00:21microphone is a really bad mine but mine
3:00:23yes sorry what is the mean value of two
3:00:26just in the upper board yes no sorry
3:00:30ah I am muted myself okay and I just
3:00:35meant uh yes it's um I was writing down
3:00:39pedantically
3:00:40the EPO term so if you have a function f
3:00:44you see here yes if you have a function
3:00:47f remember that the term so the term
3:00:50that you have to add to your chain rule
3:00:52was of the form Sigma Square divided by
3:00:55two
3:00:56eventually you have
3:00:59d square of x if you have a
3:01:01multiplicative Factor but this was not
3:01:03the case in here and then you have to
3:01:05compute the
3:01:06expectation value of
3:01:09um the first derivative
3:01:12of the function
3:01:14right
3:01:15and in here the function of which
3:01:18[Music]
3:01:19so the function
3:01:21uh that sorry I did I forget
3:01:25one derivative in here
3:01:29wait a minute
3:01:35uh okay
3:01:39I think it's the second derivative
3:01:41yes indeed is it
3:01:45but then was I consistent in my notes
3:01:47before
3:01:51uh no wait
3:01:55uh just said no that should be a prime
3:02:01so what do I have to do so I want to
3:02:03derive
3:02:07um all right
3:02:09[Music]
3:02:13my function X as
3:02:17of X of t
3:02:19is x square of t
3:02:23no sorry uh yes is x square of T right
3:02:28so what we have to do
3:02:32yes
3:02:35sorry I was giving you the formula for
3:02:37the correlation in the industry sorry go
3:02:40back to the chain rule so if you um what
3:02:44was the chain rule so the idea was DF
3:02:46over DT
3:02:48over the X times
3:02:52the X over DT
3:02:54plus Sigma Square over 2 eventually G of
3:02:58x
3:02:59d square of X of X over The x square
3:03:03right and this is what I call the ETO
3:03:05term
3:03:08okay
3:03:10and so now my function f is x squared so
3:03:14you see that the uh the second G is
3:03:17equal to one and the second derivative
3:03:19that I get in here is just a factor of
3:03:21two so what I'm writing in here
3:03:26should be
3:03:28so I have an expectation
3:03:30outside
3:03:32of the derivative
3:03:34of x square the derivative of x squared
3:03:37is the usual term plus the equal
3:03:39correction and the equal correction is
3:03:41just a constant
3:03:42so when I take the expectation this goes
3:03:44to zero because I have equal time and
3:03:46this will give me a square
3:03:49so before I was just bringing the
3:03:50expectation inside and splitting it in
3:03:53this way
3:03:55this is
3:03:56you know of course okay
3:04:00uh okay
3:04:03thanks
3:04:05for this
3:04:07so okay let me just maybe comment and
3:04:11then we we stop and the point number
3:04:13three we do it uh next time or
3:04:17um or I will summarize it very briefly
3:04:19but the idea of uh the point number two
3:04:22is to look at the process similar to
3:04:26uh you remember in the lecture uh we had
3:04:29the DC over DP so now I call it s
3:04:32because I have only one uh let's say
3:04:35city of which I want to describe the
3:04:37growth
3:04:38and my equation is the S over DT equal
3:04:43to
3:04:44mu so this was was called m in the
3:04:48lecture s of C plus multiplicative noise
3:04:51which is now just linear so it's s of P
3:04:55itself
3:04:56times the noise
3:04:59okay
3:05:01and so so in the lecture this was an
3:05:04equation
3:05:07lecture three this was of the form D is
3:05:09that I
3:05:11be T equals to basically the same thing
3:05:14now we take only one value of I and I
3:05:17just wanted to point it out that
3:05:18whatever was discussed in there was in
3:05:21the strattonovic prescription
3:05:23and uh what do you find if you try to
3:05:26solve this equation with the Ito
3:05:29prescription
3:05:30well you have to do something similar
3:05:32than what we did in the lecture namely
3:05:34you have to do the change of variables
3:05:35and introduce for instance a variable
3:05:39Phi of T which is the log of s of t
3:05:46but now the change of variable is Not
3:05:48Innocent so you have to understand how
3:05:51to use it when instead of photonovich
3:05:54you think about ether
3:05:57and what's the way to uh to see what
3:05:59happens well what I can do is to compute
3:06:02the time Evolution now of this quantity
3:06:05site
3:06:07uh
3:06:09so let me compute the x i of T over DT
3:06:14and let me use the Ito rule that I just
3:06:17wrote up there to compute this
3:06:19derivative so what would this give me so
3:06:21I don't have explicit dependence on T so
3:06:25I just had to derive with respect to the
3:06:27argument
3:06:28so I have the usual term which I get
3:06:30which I get deriving with respect to S
3:06:32so this will give me 1 over
3:06:35s of t
3:06:37times
3:06:38and d s of T over DT
3:06:44which is what behaves in this way
3:06:47and then I have the Ito term that I
3:06:49wouldn't have in the Stratton which
3:06:50prescription and The Ether term is as we
3:06:53saw Sigma Square over 2. now I have a
3:06:57function G which is no zero so what is
3:06:59my G here is just s
3:07:03so I have a factor of
3:07:05s squared
3:07:07in here
3:07:08and then I have the second derivative
3:07:10and the second derivative is minus one
3:07:13over s Square so this is minus these
3:07:16divided by S square of t
3:07:19and this is a term that we didn't have
3:07:21in the lecture because this is precisely
3:07:23The veto term
3:07:26and now what we can do is we substitute
3:07:29now our lungement equation in here and
3:07:32what we get out of this is
3:07:35curses
3:07:38uh then we have uh this
3:07:42minus s were
3:07:45over two
3:07:47so we have this constant
3:07:49and then we have the term from the noise
3:07:51so
3:07:54the simple noise ETA of t
3:07:56so you see that by making this change of
3:07:58variable so this is useful as it was for
3:08:00stratonovich in the sense that you
3:08:02mapped the problem into a larger one
3:08:05equation with some linear noise but with
3:08:08a drift in here that is changed because
3:08:12you are looking at The Ether
3:08:13prescription so if you were doing which
3:08:15you wouldn't have this extra correction
3:08:17to uh to the drift
3:08:20but apart from that the formalism is the
3:08:22same so now you can integrate this
3:08:24equation
3:08:25that you have in here so you know that X
3:08:28of T will be uh so very reasonable let
3:08:32me write it here
3:08:34I hope you see it so X of T is what is
3:08:36this constant
3:08:38which I call
3:08:39okay
3:08:42mu minus Sigma Square over 2 times T I'm
3:08:46just integrating from 0 to t plus the
3:08:49integral of the noise
3:08:51from 0 to T in detail of my itself now
3:08:56and this is uh essentially a Brownian
3:08:59process so you can think about the white
3:09:02noise as being the derivative of a
3:09:04Brownian process so when you integrate
3:09:06over it this is what mathematicians
3:09:10would call a inner process at time t
3:09:12with a drift that is given by this term
3:09:16in here and so if you re-exponentiate
3:09:19and you go back to your original
3:09:20variable you have that this will be
3:09:23essentially a log normal exactly as it
3:09:25was discussed in the lecture so it's a
3:09:27variable whose logarithm is is
3:09:30essentially gaussian
3:09:32or it's generalization in terms of
3:09:35stochastic process
3:09:37and this is the the what was called in
3:09:41the lecture the statistical monster
3:09:42which has funny properties uh about uh
3:09:46it's having essentially a distribution
3:09:49which looks like a power law in the tail
3:09:51but having moments that uh that we can
3:09:54compute uh explicitly and let me just
3:09:58give you a hint on how to compute the
3:10:00moment and that's uh the end of this
3:10:03exercise too
3:10:06so what you can use
3:10:08foreign just to do it fast is
3:10:14is the formalism of generation functions
3:10:17for uh for a gaussian so what I'm saying
3:10:20here is that at any time T this variable
3:10:24PSI is is a gaussian
3:10:27with uh with an average that is non-zero
3:10:30that is this mu minus Sigma Square over
3:10:32two times T and with fluctuations which
3:10:35have a variance that is Sigma squared
3:10:38okay so when I want to compute the
3:10:41moment of this s of T what I have to do
3:10:45is I want to compute
3:10:47the expectation value of
3:10:50s to the power n at a certain given time
3:10:52t
3:10:53and this is what this is the expectation
3:10:56value of e to the N times
3:10:58the log of s of T which is
3:11:02the variable which I know to bigger
3:11:03option
3:11:07and this is nothing but the generation
3:11:09function so remember that when you look
3:11:11at generation functions
3:11:13of a variable random variable X computed
3:11:16at a given value U
3:11:19let's say this is what this is the
3:11:22expectation value of e to the U
3:11:24h x
3:11:27you have to choose between plus and
3:11:29minus but Limitless plus and for the
3:11:31gaussian this is
3:11:33e to the U
3:11:36H your average that I call here M so let
3:11:40me say that the average of X is equal to
3:11:43n and the covariance
3:11:46the variance in this case is equal to s
3:11:49for my random variable in gaussian and
3:11:52variable so from here I have that this
3:11:54would be
3:11:56of the following form
3:12:00uh okay and uh and the idea is that what
3:12:04you want to compute in here is exactly
3:12:06the same thing so this is a gaussian
3:12:07random variable so if you replace what I
3:12:10call U here with n you get immediately
3:12:14from here the expression for uh for the
3:12:17higher moments of your variable s of T
3:12:20which was something that was already
3:12:21discussed in in the previous lecture
3:12:25number three
3:12:27okay now I think we have to stop it's
3:12:31about time so I will give you the
3:12:33solution for the exercise three
3:12:37so the idea of the exercise C is as I
3:12:39said you derive copper Plank and then
3:12:41you try to look for stationary solutions
3:12:43to uh to the focker plant equations so
3:12:46stationary Solutions are solutions which
3:12:49do not depend on time
3:12:52so
3:12:54my equation is of the form DP over DT
3:12:58was minus B over DX F times p and then
3:13:03depending on your prescription you add
3:13:04here
3:13:06different forms so let me use either you
3:13:10have here
3:13:12this expression in here so what you have
3:13:15to do to compute the stationary value is
3:13:17to set this to zero and you will see
3:13:20that you depending on what is the form
3:13:22of this term in here for the simple case
3:13:25of the exercise two where we choose V of
3:13:28x to be just
3:13:30Sigma Square over 2 times x
3:13:32and the form of a stationary solution
3:13:35would be different so you will get power
3:13:38rows out of these equations but the
3:13:41exponent of the power law is different
3:13:43depending on whether you choose it or
3:13:45set on which which tells you again that
3:13:48when you have multiplicative noise you
3:13:50have to specify always and before what
3:13:53is your choice and be consistent with it
3:13:55and we will really discuss this uh I
3:13:58think in into the seven which is another
3:14:00example of plank so that's a good
3:14:03point
3:14:04will be a good point to go back to this
3:14:06scene
3:14:08okay are there questions
3:14:14about this
3:14:22if not let me stop the registration

About this transcript

This page contains the full transcript of Complex Systems - Jean-Philippe Bouchaud - Lecture 4: Optimisation, HJB & “Chaos” by EconophysiX Lab, generated from the public captions YouTube serves with the video. The transcript has 24,232 words across 4,005 segments, with the original timestamps preserved so you can click any line to jump to that moment in the embedded player.

What you can do with it

Use the transcript to take notes, quote the speaker, build a study guide, generate a summary with ChatGPT or Claude via the YouTube Summary tool, or export it as a timed subtitle file with YouTube to SRT. You can also re-open it in the transcriber to translate the transcript into 100+ languages.

Free YouTube transcript tool

YouTube2Text is a free YouTube transcript generator — no signup, no daily limit. Paste any YouTube link and get the full transcript instantly, with timestamps, click-to-jump, translation to 100+ languages, AI prompts for ChatGPT, Claude, and Gemini, and exports to TXT, SRT, VTT, or Markdown.