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Complex Systems - Jean-Philippe Bouchaud - Lecture 5: Optimal Liquidation (Valentina Ros) — Transcript

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0:01this conference will now be recorded
0:05okay so before we start uh let me just
0:11a warn you of the fact that we forgot to
0:13record the previous lecture but there
0:16will be lecture notes that I will upload
0:18uh to our ens folder so at least the
0:22material will be there available for
0:24everybody to read and I saw that there
0:27are some questions on both homeworks and
0:30lecture notes in the question and answer
0:31five so now we go through them between
0:34today and tomorrow
0:35so hopefully that will also be addressed
0:39uh very soon
0:41okay and now let's come to the topics of
0:45the uh today of today which is related
0:47to this problem of optimal control that
0:51was discussed uh in the last lecture so
0:54uh what we're going to do is go through
0:56this exercise so we will do the main
0:59part and then there are some
1:00calculations that I will leave
1:02an exercise and if there is time I will
1:05go back to uh to the final part of today
1:09number three uh just to discuss a little
1:11bit how to solve for the stationary
1:13state of soccer plank equations for
1:17multiplication noise so that's uh quite
1:20an easy thing but it's good to see it at
1:23least once uh as a Blackboard
1:27before let's start with with the today's
1:32the text
1:35mail
1:36and the idea is to discuss a simple
1:39application of this Hamilton Jacobi
1:42Bellman equation
1:43and the problem that we have in mind in
1:45here is you can think of it as a problem
1:48of optimal liquidation which means that
1:52there is a summaries or some agents that
1:55has for instance x times zero a given
1:58amount
1:59q0 of some tradable instrument as we
2:03call it here so this can be whatever you
2:05can even think about some stocks uh that
2:09you have at time t equal to zero and you
2:12would like to sell all of this uh
2:14quantity of uh of stocks or of this
2:17instrument and you want to do it before
2:19a given final time T so this is the
2:22finite time Horizon and the question is
2:25what is the optimal way to do this so
2:27what is the speed at which you should
2:30sell should you try to sell everything
2:32very fast at the beginning or should you
2:35somehow have a strategy that is more
2:37flat over this time interval or or let's
2:41say in general what is the best way to
2:42do this
2:43and we will try to discuss two
2:46Frameworks so the first part is about
2:48let's say one single agent that has to
2:52face this uh optimization problem and
2:55the second part is uh about uh so in the
2:58second part we will introduce some
3:00interactions that we will treat at the
3:03mutual level so the idea is that you
3:05have one agent that wants to sell its
3:07quantity of each amount of this uh of
3:11these stocks let's say but he or she is
3:14not alone and there are many others with
3:16whom there is some interaction
3:19okay so let's start with uh with part
3:22one
3:23and let me summarize a little bit
3:26uh what are the quantities uh let me
3:29first uh let me
3:31um tell you that the way we will treat
3:34this problem with interaction uh Falls
3:37within the framework of the so-called
3:39meaningful gains and there is some
3:40reference in the text of the today that
3:42you may look at if you're interested
3:44into this subject of maybe games
3:47okay but first let's start with uh with
3:50a single agent
3:53foreign
3:58so there are some random variables which
4:00describe the state of your systems that
4:03I will encode into a vector Y which
4:06depends on time
4:07and these random variables are Q
4:10Depending on time are some X of T and
4:14some f of t
4:17so this Q is the amount of instrument
4:21that you want to sell in this finite
4:23time interval so I will call it to be
4:25inventory
4:28it's time t
4:30x is essentially the amount of money
4:33that you make out of this selling so I
4:35will call it the wealth just to
4:39uh just to have a name for it and X is a
4:43variable which does not describe
4:44directly the state of the agent but if
4:47you want to describe the state of uh of
4:50the market in some sense so this is the
4:52price Associated to uh to the object
4:55that the agent wants to sell
4:59and of course this quantity are a
5:01dynamical quantities so they satisfy
5:03certain equation of motion in time
5:06so in particular
5:08Q of T will vary over time because of
5:11course it will decrease whenever you are
5:13selling this quantity and it will
5:16decrease proportionally to some
5:18velocities that I call VT
5:21and the velocity is precisely the
5:24control so it is the function of time
5:27that you on which you have some power so
5:30you can choose how fast you want to sell
5:32this quantity over time so let me call
5:35it as we did in the lecture
5:39the control
5:42so you see that when you're selling this
5:45quantity is diminishing and so your
5:46velocity is is negative in this
5:49definition
5:51then how much money do you make in an
5:55interval DT by selling this quantity
5:56well this is this has to be proportional
6:00to minus VP
6:02times uh what what you would put in here
6:06a priority would be
6:08somehow the price of of the quantity
6:11that you are selling
6:13uh that that is given by S of T but then
6:17the idea is that when you do this type
6:19of financial operation let's say there
6:21is also a cost which is associated to
6:23the fact that you are selling or uh or
6:26buying and in the model that we will
6:28consider today the cost we will take it
6:30to be simply proportional to the
6:32velocity at which you are selling these
6:35quantities so I will introduce
6:37in here
6:39an additional factor of K times the
6:43velocity and this is telling you that
6:46essentially when you are selling so this
6:48velocity is negative the effective price
6:51at which you sell your your good is uh
6:56is let's say the price of the good
6:57itself minus the cost Associated to the
7:00operation which is uh which is given by
7:03this kvt
7:04and then you have an equation for the
7:07third dynamical variable in here which
7:09is the price
7:11and uh and this will be given by some
7:15quantities that I call MU of t
7:18and this function mu is uh is where the
7:22interaction will enter when we will
7:24discuss it uh in in part two but for the
7:27moment let me write it generically like
7:29this
7:30plus what we can do is to add some noise
7:33so some volatility that tells you that
7:35there are fluctuations uh in the price
7:37that due to uh two different reasons and
7:41that you encode generically into into
7:43some noise term so just to give you some
7:46terminology this quantity in here that
7:50decreases uh the price is uh often
7:54called in this literature market price
7:58and this quantity here which controls
8:01how the price is varying with respect to
8:04the to
8:05say some external input which may
8:09contain uh also the state of of other
8:12agents is what is usually called so let
8:15me try to put an arrow like this it's
8:17called Market impact
8:19and can you put up the Blackboard a
8:22little bit because it's out of the sure
8:25this way
8:29perfect thanks
8:32okay that's good
8:35okay so this is what we are going to
8:36discuss in part one now let me just uh
8:39mention what shall one do uh to
8:42generalize this to the case of many
8:45agents so I will be a little bit floppy
8:47in here but
8:49let's say that the idea is that
8:53if you have many agents you can label
8:55them with some index a so you will have
8:58one of these State vectors
9:01the Y of T will go to some y of T
9:05labeled by a and a identifies each of
9:09your agents so let me
9:11assume that they go from one to n
9:13and the interaction between them as I
9:16said will enter through this quantity mu
9:20in here and what we are going to
9:23consider in the following is the case in
9:26the in which the MU
9:27and let's see that enters into the
9:30equations for the agent a
9:33is the average of the velocities of all
9:38of the other agents in the in the market
9:41let's say so I will think about
9:43something like this this will be one
9:46some one over and some overall other
9:49agents of
9:52their velocity
9:54and what we will we will do in part two
9:56is to go to some sort of continuous
9:58limit and replace let's say this average
10:01with with A continuous average but I
10:04recommend more about this in the
10:07following
10:08okay so to conclude uh the introduction
10:10to the model what I have to do is to
10:13specify the objective function or the
10:16cost function which is actually a
10:18functional which depends on all of the
10:21trajectory over time of this Vector y
10:25so let me call it g
10:27of
10:29e0 and Y 0 so t 0 and 1 y 0 are the
10:33initial time and the initial condition
10:35and this function is given by the sum of
10:38two terms
10:39so there is a first term which is the
10:42game
10:43that is actually just a function of the
10:46final State the state at time capital T
10:50of your system that I will write in a
10:53minute and then you have some so this is
10:55the game
10:57that you want to maximize
11:00and you have also some cost which
11:02contributes to this objective function
11:04which is the integral from the initial
11:07time to the final time
11:10of
11:12your inventory to the uh to the power
11:16two so this is something that penalizes
11:18you whenever you still have uh the
11:21quantity of of your good uh to sell and
11:25you have not yet uh sold uh all of it
11:29so this will be a cost Associated to the
11:33fact that you have still a finite
11:35inventory at any time uh s here
11:39and uh how do we write the gain well
11:43so again this is an example of a model
11:48that we take here and in here we choose
11:51the gain to be to be what well you have
11:54the final wealth uh that you get by
11:56selling your quantity at the final time
11:58capital T that of course you want to
12:01maximize
12:02and then if you are left with some
12:05amount of these quantities that you have
12:07not yet sold then there is some value
12:10Associated to it that in principle would
12:13be the product so the amount of this
12:16quantity times the price at that
12:19particular uh final time but again you
12:22may have uh or it is realistic to assume
12:26that the price is not really so a little
12:28bit similarly to what you have in here
12:30uh the final price is somehow diminished
12:33by the fact that maybe you want to
12:36liquidate at finite time T so you really
12:38want to sell whatever you're left with
12:40at the final time and when you want to
12:44sell it so fast at the end there might
12:46be a penalty so let's say a cost an
12:49execution price this is how it is hold
12:52associated with selling at the finance
12:55time which again diminishes the uh the
12:57effective price
12:59at the time capital T
13:02okay so what we're gonna do is to try to
13:05look at this example and derive some
13:08Hamilton Jacobi Bellman equation for
13:11this particular model
13:15and before let me check that you see
13:19okay
13:22so let's start with the first part and
13:24before let me just make a comment so
13:26this is part one
13:33so the first point is is just a comment
13:35if you want to uh if what we can do is
13:39to look a little bit at the model and
13:41try to figure out how many parameters we
13:43have which are quite many so we have two
13:46parameters which enter into the equation
13:48for the price which are Theta that
13:51measures uh among the strength of this
13:54Market impact and we have the variance
13:57Sigma Associated to the noise
14:00and then you have three other parameters
14:02so this Kappa here in the equation for
14:05the wife
14:07and uh whereas we have the Phi in in the
14:12in the cost function there and we have
14:15capital A which controls some of the
14:19cost Associated to selling everything at
14:22the final time
14:23and what I just want to point out so we
14:25will go back to analyzing a little bit
14:28the effects of these parameters and once
14:31we have the solution to the problem but
14:32I just want to uh pose a moment to to
14:35figure out uh what should be at the
14:38effects of all of this parameter in
14:40particular of the last three
14:42and the idea is that more or less the
14:45parameters Phi and the parameter capital
14:48a Are quantities that uh won't you or
14:52push you to sell faster in a way right
14:55because Phi is associated to the uh cost
15:00that you pay at any time by having a
15:03finite inventory so it's due to the fact
15:06that you have not sold everything at
15:08that time and so there is a cost
15:10associated with so when this high is
15:13large you expect that you want to choose
15:15a velocity that is somehow fast enough
15:17so that you have all of this quantity
15:21fast enough and more or less the same
15:23role is played by this uh capitalane
15:25here
15:26uh whereas K as you see here penalizes
15:29uh those choices in which you uh sell
15:33very very fast because if you sell very
15:35very fast you are diminishing the
15:37effective price for your good and
15:40therefore you are diminishing the total
15:42wealth that you are making
15:44so this is just to point out that Phi A
15:47and K are a little bit in competition so
15:51this would want you to sell fast
15:55and this would want you to sell
15:58not so fast let's say
16:03so the problem is not uh totally trivial
16:07okay so having said this what we can do
16:10now is to try to derive uh Hamilton
16:13Jacobi Batman equation along the same
16:16lines
16:17okay uh
16:20of lecture four
16:25so now
16:29but why is it cut
16:31my video
16:34okay
16:38okay so let's
16:40try to derive this Hamilton Jacobi
16:43Batman equation
16:46so to do this I will introduce uh let's
16:49say just for notation let me call small
16:53y a particular value for the vector for
16:56the random variable Capital wiser so
16:59this will be the vector given by small
17:02Q's Molex
17:03and small s
17:06so these are my state variables
17:08and let me introduce as it was done in
17:12lecture four the so-called cost to go
17:18okay
17:19which is a function V of time and of a
17:25particular configuration of of my system
17:27at the time T and in general so what we
17:30are gonna do in this first part as I
17:32said is to work at fixed value of this
17:35function I'm one of the most B so we are
17:38not assuming that this comes from
17:40interaction but just that somebody gives
17:41you uh this function and then you try to
17:44solve the optimization problems for
17:46these dynamical system so this is why
17:48there will always be a dependence on
17:51this fixed function mu of T in whatever
17:53I'm going to do in this first part
17:56and what is the definition of this cost
17:58well well this is the soup over
18:00the velocity
18:02t
18:04at any later time of the expectation
18:07value of my cost function up there
18:12from uh
18:14so as you see in the way that I wrote
18:17so this is an expectation value over the
18:19noise it's an average over the noise and
18:21I wrote the cost function so in
18:23Principle as I say this is a function of
18:26the whole uh let's say trajectory for
18:28your vector Y at any time but I'm
18:32indicating in here only the dependence
18:34on the initial condition so this is the
18:36cost that you get starting at time t 0
18:39from a given configuration let's say y0
18:43and therefore what I have in here is uh
18:46is the following so I assume that I have
18:49some time T that is larger than zero but
18:52smaller than capital T and I assume that
18:55I know uh what is the optimal strategy
18:58to follow from that time T
19:00onwards and this defines precisely my uh
19:04my cost to go
19:06so just to repeat
19:09a little bit the sketch
19:13of the lecture
19:15the idea is as follows so you have time
19:19you have your final time capital T
19:23and then you assume that you know how to
19:27solve the problem at any time so this is
19:29zero at any time late larger than this T
19:33here so in this region you assume that
19:35you know
19:38what is the optimal strategy to follow
19:40so what is the cost to go and you know
19:41this for any possible
19:44initial value at a time a small T so in
19:48in the lecture of course the problem was
19:50formulated in terms of a one-dimensional
19:53function here we have a vector so let me
19:55draw it like this
19:56uh
19:58so of course this is higher dimensional
20:00but the idea is that whatever is the
20:03particular value of the vector y s time
20:05T you know what you have to do from time
20:08T onwards so you know what is the
20:10velocity that optimizes uh that gives
20:13you the optimal strategy up to the final
20:16time
20:17and the question is once I know this for
20:20any particular initial point it's time T
20:23can I derive an equation that tells me
20:27what I should do in the little time
20:30interval which is between T minus DT and
20:33N time t
20:36and that is that to choose the optimal
20:39strategy in this user-time interval I
20:41have to usually if you remember from the
20:44lecture I always have to compensate
20:46between two things
20:47so first of all I told you that we know
20:49how to solve the problem for larger
20:51times
20:52so we know the value of the cost to go
20:55for any particular choice of why and
20:58therefore for instance we know what is
21:00the best point where to start from that
21:02would optimize our gain at the final
21:05time t
21:06but then we assume that at time T minus
21:09DT we are in some particular initial
21:12condition some particular point and it
21:15might be very costly to go uh from that
21:17particular point to the initial
21:19condition at time T which optimizes your
21:22your total gain because there is some
21:24cost Associated for instance to moving
21:27much into this small interval DT so you
21:30always have to compensate between the
21:32extra costs associated with movement in
21:36this small step and again that you would
21:38have from the final point that you reach
21:41with this movement up to the final time
21:44and this is what gives you an
21:46optimization equation which is which is
21:50not totally trivial
21:52so let's see this and uh just a reminder
21:55so in the in the lecture the cost which
21:58was Associated to this small step was
22:00essentially uh due to the elastic term
22:03in here things will be a little bit
22:04different but let me show this
22:07concretely
22:09okay so to derive this uh Hamilton
22:11Jacobi Bellman I will do
22:15introduce
22:17just for the sake of notation so I will
22:20introduce
22:23uh I will call the Velocity in this at
22:27this time T minus DT which is what we
22:29want to optimize over I will just call
22:32it U to simplify the notation
22:35and then I will assume and I will cause
22:37the noise
22:38in this interval T minus DT simply by
22:42ETA
22:44um writing
22:46and we can see what you're writing
22:48you're right but I don't know why okay
22:51wait
22:54ah now I see because I'm not moving
22:58the camera
23:01okay
23:06now you see it
23:09foreign yes okay
23:13so I was just introducing I mean I just
23:15want to drop this dependence on T minus
23:18DT and I will assume that if I start at
23:24a point why
23:26no let me write it like this so if I am
23:28at a point Y at time T minus DT
23:33at time T I will be at a given point
23:38y plus b y
23:44and this b y will depend on what well it
23:47will depend both on the velocities that
23:50I choose in this small interval DT and
23:53and it will also depend on the noise uh
23:56in this small intervality and so I will
23:58try to solve an optimization problem for
24:00this velocity in the interval DT
24:02averaging over the noise in this small
24:05interval DP so I'm just repeating what
24:07was done in the lecture
24:10and so to derive this equation what we
24:13have to do so does it fit in here
24:17so what I can do
24:21is to try to write an expression for the
24:25cost to go
24:27at the previous time P minus BT
24:31so V at T minus DT assuming that I am at
24:35the point Y at P minus V T So Y is a
24:38variable so I will use it uh
24:42here I use it for time T but now let me
24:46assume that y identifies my position at
24:48the previous time and let me drop the
24:51dependence on mu p in the following
24:55so what would be this equal to well
24:58exactly as we had in the lecture so as I
25:01said I want to optimize over my velocity
25:03in the legal interval you
25:06the uh total cost so the total cost will
25:09have two contributions
25:11there is one contribution that comes
25:13from the extra cost Associated to this
25:16small step that I'm doing and this is
25:19proportional
25:20to this integral in here
25:24that's given that the time intervalism
25:27is very small I can approximate as minus
25:31Phi Q squared
25:32DT
25:35where Q is one of the components of this
25:37Vector y the first one
25:40and then the remaining cost is encoded
25:43in uh in the function that I assume that
25:46I already know so it is encoded in the
25:48cost to go from time T except that I
25:52have to compute this cost to go and to
25:55the point that I reach starting from
25:57point Y at the previous time and
25:59applying the given velocity U so what
26:02this means is that in here I will have
26:04my function V at time D but computed at
26:09y plus d y
26:12and Y plus d y is is noisy so it depends
26:15as I said both on this small U that I'm
26:18optimizing over but it also depends on
26:20the noise which acts in the small
26:22interval so what I have to do is to
26:24average uh that function there with
26:27respect to this noise in the small time
26:30interval
26:32okay and now exactly as we did in the
26:36lecture the the way to proceed is to
26:38give an expansion a failure expansion
26:41for this function here and then average
26:43over the noise
26:48so let me do it in here
26:54what do you see so we need
26:57this which I think you still see okay
27:17but now I have to received a little bit
27:42okay
27:46so let's try to expand and we want to
27:49expand to linear order in DT so this is
27:52V of Y plus v y
27:56uh and in order to uh to do the
27:58expansion we should remember what are
28:00the equations so remember that Y is
28:03the vector of all our three variables
28:05which satisfy some equations which are
28:07just erased
28:09but you have them written so let me do
28:12this uh Taylor expansion here so the
28:15first term will be just real
28:17d y of course
28:20and then I have a term which is the
28:22partial derivative of my function V with
28:25respect to the first uh sorry I forgot
28:27the DT
28:31I hope that you see it there is a small
28:33bit in here
28:36Plus
28:37DV over DQ and here I would like to put
28:41Q dot times DT right
28:45so the derivative of Q over time times
28:47uh the DT and the derivative of Q over
28:50time we have it from the equation so
28:52this is just given by U times DT where U
28:54is the velocity in the small interval DT
28:59then we have to take the derivative with
29:00respect to the variable X
29:03foreign
29:05x dot if you want so the time derivative
29:08of x and this was minus
29:10the velocity times the price
29:14plus k u
29:18okay
29:19and then there is the derivative with
29:22respect to S will be V over DS
29:25and the equation for S dots sorry I keep
29:28forgetting the DT
29:33there is always a DT so the equation for
29:35s dot was what so there was the Theta
29:38term
29:39we will see
29:41times DT let me put it at the end so
29:44there was Theta mu of t
29:51uh this one you're right
29:55exactly
29:57thanks
29:59so here we have Theta mu and then there
30:02was a noise term so let me write it like
30:04this so that was Sigma times
30:08each of the DT
30:12and so let me put a detain here
30:16okay
30:18I'm becoming messy sorry
30:21um and here we have so remember this is
30:24White Noise I I didn't specify but I'm
30:26assuming that this is quite nice so
30:28remember what we discussed last last
30:30time namely that this noise in the in
30:33the limit of uh correlation function
30:35which goes to a Delta function ease of
30:37the order of 1 over square root of DT
30:41and therefore the product it of T times
30:44DT is of the order of square root of BT
30:47and this tells us that if we want all of
30:50the contributions which are awarded DC
30:52we also need to take the term which
30:55comes from the square of this uh factor
30:58in here and this will come with the
31:00second derivative
31:02of V with respect to the variable s
31:07right because then this will be
31:09multiplied essentially by uh the square
31:13of this equation of motion for S and let
31:16me just keep the contribution that is
31:18awarded DT which is
31:20of the form
31:22so let me write it like this
31:26ETA Square Times DT square plus higher
31:29order terms that we will neglect when
31:31deriving the uh
31:33when going to the continuous limit
31:36so now what I do is I take the average
31:39of this expression with respect to the
31:41noise
31:46okay both of the left and the right hand
31:49sides
31:50so this was what I called eat them
31:52before it's the noise which acts in the
31:55little time intervals and you see that
31:57once I take the average so this is
31:59independent this is independent this is
32:00independent the average of these uh
32:05is equal to zero because the noise has
32:07zero average whereas the average of this
32:10quantity in here is precisely equal to
32:16DT so it will give me the uh the
32:20contribution that I'm looking for to uh
32:22to linear organity so this is exactly as
32:25as it was in the lecture
32:27and once I have this what shall I do
32:30well what I can do is
32:32to
32:34um
32:34where is it
32:36to go back to the expression in there
32:38and substitute uh to the right hand side
32:41what I just derived so what I'm
32:44left with is V of T minus b t
32:48y
32:50let me bring this to the other side so
32:52that I have minus
32:54V of t y
32:57and then on the right hand side I have
32:59the sup
33:01over you
33:03of the first term which is untouched
33:07so now I have c q squared and the DPI
33:11will collect it at the end actually let
33:14me divide everything by DT
33:16directly
33:17so these terms come from uh the cost in
33:21the leader interval DT and then I have
33:23to plug my expansion so I have
33:25DV over DQ
33:28times U
33:30minus DV over DX
33:34you
33:36S Plus k u
33:39okay and then I add
33:42DV over the S Theta
33:45mu of t
33:47Plus
33:49Sigma Square over 2 this square of V so
33:53the contribution of the noise
33:55and that's it
33:57okay
33:59and of course if you if I now take the
34:01limit of uh DT going to to 0 what I get
34:05in the left hand side is nothing but
34:09minus the time derivatives
34:12of my cost to go with respect to teams
34:17okay so this is our form for today of
34:21the uh Hamilton
34:24Jacobi Bellman equation
34:27which is an equation where your time
34:29derivative has the minus sign in front
34:32as you remember from the lecture and
34:35were so this is no longer as we are used
34:37to we are used to initial value problems
34:40where we specify the value of the
34:42function at time zero and then we want
34:43to solve forward in times whereas what
34:46we have in here is more a boundary value
34:49problem where we specify the value of
34:51the function at time capital T and then
34:53we want to solve backwards in time and
34:55get the solution at time zero so what is
34:58the boundary value what is the value of
35:01the function at
35:03capital T
35:05well this we know it from the expression
35:08for the uh for the function G that
35:11unfortunately I erased but if you
35:13remember the function G had a cost which
35:16was Associated precisely to the value of
35:18this function at the final time plus
35:20sorry s again which is associated to
35:23that plus a cost and the cost is an
35:25integral which goes from T up to capital
35:28T so the integral Advantage is when I
35:30compute it at the lower Edge equal to
35:33capital T so this is just to say that
35:35the boundary value in here is just given
35:38by X
35:39flash Q
35:42s minus A2
35:46which is the value of the function G
35:49evaluated at a vector y so this is the
35:51boundary condition
36:02okay
36:04and now what we have to do is to try to
36:06solve this equation for the cost to go V
36:10and we try to do this and I don't
36:13remember if this is 0.2
36:16probably yes
36:19um
36:23yes so we try to do it with an answer so
36:27we we just
36:28[Music]
36:29um
36:30try to get to to guess a little bit what
36:33is the shape of this function B at any
36:35time and the guess comes from the
36:38structure that you see at the final time
36:40so as you see the your boundary
36:41condition is of the form
36:43X Plus QX
36:46minus A2 squared
36:50so what we can assume or try to guess
36:52this is an asset is that at any time T
36:55our function V has in some sense the
36:58stature so there is this linear term
37:00plus some genetic function of Q so we do
37:03the following answer which as we see
37:06we will see we simplify the calculation
37:09and now I hope that you see it
37:12so we assume that our V of t y can be
37:16written as
37:18X Plus Qs
37:21plus some generic function of time and
37:24only of q that I call Omega
37:27of d q
37:30and these answers is good because it
37:33allows us as I will show in a minute to
37:36reformulate the problem as more or less
37:40a problem in terms of uh of Q okay so
37:44let's see this uh how this works
37:48so what I have to do is just to replace
37:50this answer into my equation and see
37:54I think simplify a bit
37:57okay
38:02this is just algebra
38:15so on the left hand side I will just
38:18have minus the derivative of this
38:21forming a function with respect to time
38:23which is the only thing that depends on
38:24time
38:25and on the right well I have terms which
38:29do not depend on the velocity use so let
38:31me write them
38:33separately so there is this High then
38:35there is
38:37uh
38:39the Theta term
38:41DV over Dash
38:45your steep
38:47and perhaps already in here so you see
38:51if we make Visions that's the only
38:53dependence on S is linear so DV over
38:55Dash gives me just a factor of Q so let
38:58me write it here
39:01okay
39:02plus there should be another term up
39:05there which does not depend on you that
39:06is this second derivative but because we
39:09are making designs that which is linear
39:11we see that the second derivative
39:13vanishes so I don't have any
39:15contribution in this particular example
39:18which is proportional to uh to Sigma
39:20square and let me comment about this in
39:22a minute but before
39:25let's write what remains so what remains
39:27is the soup over you
39:30of what
39:32a few times the derivative with respect
39:34to q that gives me
39:36S Plus the derivative of w with respect
39:40to Q's we have a factor of U
39:42t w over DQ
39:45and then we'll have a factor of U times
39:47s
39:48but the factor of U times x comes us
39:51with the contribution the first
39:53contribution from the derivative over X
39:56which is simply equal to one so you see
39:59that the factor U times s will cancel
40:00and what I'm left with in here is minus
40:04okay
40:07U squared
40:10and that's it
40:13okay
40:16so our equation is now quite uh simple
40:19or simpler and we can directly solve for
40:24uh the optimal value of U so if we
40:26maximize this quadratic function what we
40:29get out of this is that U optimal
40:33will just be given by 1 over 2K
40:37times
40:39V Omega over the Q
40:45and if we
40:47now substitute this value of U optimal
40:50into this expression
40:52we see that the contribution in here is
40:54just
40:55one over four K times
40:58the Omega over DQ
41:01to the power 2.
41:02yes
41:05okay and this is the final form within
41:07our answer of the Hamilton Jacobi byman
41:10equation and let me add two comments
41:14which are actually point three of the
41:16exercise I think
41:21so the first comment is that this
41:25equation is a little bit different with
41:26respect to the one that was given in the
41:28lecture and the difference is that there
41:31is no diffusion term so the diffusion
41:33term would correspond in here to a
41:35second derivative of our function
41:38Omega and here the second derivative
41:40disappears because so in general the
41:43second derivative comes from the noise
41:45so from higher order contributions
41:48due to the noise as we saw in here
41:50and we are killing this contribution in
41:53here because the noise is only coupled
41:55to the price viable and we are making an
41:58assets which is linear in the price
42:00variables so in this particular example
42:02what we find is that our optimal
42:05strategy
42:09which will be the velocity
42:14does not depend
42:18on what
42:21people calling in the economics language
42:23the volatility so the fluctuations of of
42:26the price
42:30foreign so at least one among the many
42:34parameters that we have simplifies uh
42:36once we make and goes away once we do
42:39this onsets and the second comment is
42:41that there is no diffusion term so this
42:43is an equation which is uh non-linear
42:46because you have a derivative square but
42:48which is first order in the derivative
42:50so there is no
42:52diffusion term
42:57so it's a little bit simpler with
42:58respect to the one given in the lecture
43:01and indeed we can solve for for it
43:04explicitly
43:06as I will show right now
43:11okay
43:24so now what we can do is try to solve
43:26directly for this function Omega of Q
43:30and this is a calculation that can be
43:32done explicitly I will not do all of the
43:34steps but
43:38I will just highlight
43:40some things which are perhaps more
43:42interesting
43:45and so the first thing that helps you
43:49when you try to solve it
43:55is again so what we can do is try to so
43:59if you look at the structure of the
44:00equation I don't know if this is still
44:02visible yes
44:05so what one can do in general is try to
44:08write down so you have any function
44:11which depends on two parameters times
44:13and Q you can try to write down some
44:16uh let's say power expansion in your
44:19variable queue
44:20okay
44:21which in general has some coefficients
44:24that will depend
44:26on time and then you have q to the n
44:32so this is announced that's again that
44:34you can make in general but now if you
44:36look at the structure of our equations
44:37you realize that things are actually
44:40very simple so you can truncate this
44:43expansion and you will get only a few
44:46coefficients that you have to solve for
44:48and the reason is that so you see on the
44:50right hand side I have only terms which
44:53depend on at most on Q Square
44:55and then I have the derivative of this
44:58function with respect to Q to the power
45:012. and so you see that if I truncate
45:04this expansion to the power Q Square
45:06what I get on the right hand side are
45:09only terms which are at most of order Q
45:11squared and on the right hand side I
45:14will always have terms which are at most
45:16of the order at which I truncate and so
45:18I can close the equation just uh
45:21stopping to say n equal to
45:25so what this means is that
45:28I will write this in the following form
45:30so there will be an h0 of t
45:33plus H1 of T Q and then just for
45:37Simplicity let me put a minus in front
45:41of the quadratic term
45:46Q squared and this is in us so if I plug
45:49now this expression into my equations so
45:53you see on the left hand side I will
45:55have just the derivatives of the
45:56coefficients with respect to time on the
45:59right hand side I will have several
46:00terms and then I can equate all of the
46:03terms which correspond to the same
46:05powers
46:06of Q and if I do this I will get
46:08equations for this coefficients uh
46:11depending on T so I'll give you just the
46:13form of the resulting equation this is
46:16easy to uh to derive
46:20so the equation for h0 is of the
46:23following form
46:25the derivative of h0 will be equal to
46:28minus
46:29H1 of V over 2K
46:33then I have a
46:36the equation for H1
46:41which is minus Theta times this function
46:44mu of t
46:47plus 1 over 2 K
46:50H1 of t h two of t
46:55this comes from the square on the right
46:57hand side
46:59if you do the calculation and then I
47:01have the third equation for H2
47:04foreign
47:10Plus
47:121 over 2K
47:15H2
47:16Square
47:17depending on t
47:20okay so this is a system of equation now
47:23of course I also write a boundary uh
47:25term which gives me conditions for these
47:29functions at the time capital T so
47:32remember that due to our answer
47:36Omega of capital T Q
47:39will be simply equal to minus A2 square
47:42if you go back to the equations that we
47:44wrote before and so this means that at
47:47time
47:48capital t h 0
47:51and H1
47:55will be equal to 0 whereas H2
48:00is equal to minus 2A
48:07no plus 2A because I have a minus here
48:11foreign
48:13okay and now given this what one has to
48:16do is to solve uh somehow this system of
48:18equations for the coefficients so what I
48:21want to do in the next uh half an hour
48:24maybe are just two things
48:26so the first thing is that if you look
48:29at the factor of this equation you see
48:31that equation three
48:32is closed in the sense that it only
48:35depends on H2 so we can solve it
48:37directly and I just want to show how to
48:40solve this equation by a separation of
48:42variables because this is something that
48:44comes up uh came up several times in the
48:47lecture so we can just do this exercise
48:50once together in here and then what you
48:54would have to do is to solve for h0 and
48:56H1 and these are two coupled equations
48:58where the variables appear in both of
49:01them and most importantly you have now
49:04this function U of T so so far we said
49:06okay let's assume that mu of T is a
49:09fixed function and we try to look for a
49:12solution at fixed value of mu of T but
49:14now you see that to solve this uh this
49:17system of equations it would be nice to
49:19specify what is this function mu of T
49:22and this brings us to this idea of uh
49:25interacting systems and meaningful games
49:27so I will give the idea for that and
49:29then perhaps leave most of the
49:32calculations to uh to you as an exercise
49:36okay so this is the plan so let's start
49:38by solving equation three so this is a
49:40parenthesis if you want
49:43um
49:45just to discuss together this idea of
49:49separation of variables which
49:51comes up when you want to solve for the
49:53stationary state of Planck for
49:55instance and it was also given I think
49:57today in the lecture uh several times so
50:02let's do an uh the tour
50:08for example actually
50:18okay you still see it but
50:22that you have to do this
50:26okay
50:27so there is as follows we have a
50:29differential equation on the left hand
50:30side you have dh2 over DT so let's
50:33rewrite it in differential form so I can
50:37write that dh2 is equal to what I have
50:42on the right hand side so minus 2 5
50:45Plus
50:47H to square over 2K
50:51times PT
50:53foreign
50:58of 2K
51:00and so here I have minus 4 K Phi
51:04okay
51:06and now you see that I have here a
51:08product of something which depends only
51:11on H2 times the differential in DT so
51:14what I can do is to bring this product
51:16to the other side
51:18and get to the other side so that I get
51:22a left hand side which depends only on
51:24H2
51:26the H squared minus 4
51:30K Phi
51:32and the right hand side depends only on
51:34t
51:36through the DT
51:38okay
51:40and once I have this what can I do well
51:43I can integrate both sides
51:46so I integrate the right hand side with
51:49respect to T from a given time P0
51:53up to a Time
51:55uh
51:56what
51:58p in general
51:59and the left hand side I integrated with
52:03respect to the corresponding value or
52:05between the corresponding value of the
52:07function H2
52:08so from H2
52:10of t 0 to
52:12H2 of t
52:15so I hope you see it maybe it's small
52:17okay
52:19now what is the right hand side well
52:21okay let me first do the left hand side
52:26so the left hand side is an integral
52:28that I know how to do so
52:31it's
52:32anarchical tangent hyperbolic
52:36this you can check
52:40uh okay we need the equation again but
52:44never mind
52:52so you can check that the following
52:54holds that the integral
52:57in the x of x square minus a where a is
53:03some positive constant
53:05so for a larger than 0 this is given by
53:08minus
53:091 over square root of a times
53:15the hyperbolic
53:18the inverse of the hyperbolic tangent
53:20evaluated at x divided
53:23by square root of a so this is just an
53:26example it allows me to compute the last
53:2910 sides so in particular it tells me
53:32that
53:36the hyperbolic Arc sound of
53:39my function H2 of t
53:43divided by square root of 4K C which is
53:47what plays the role of a
53:49is equal to minus well now let me do it
53:53correctly I think there is a Phi over k
53:56e
53:59Plus
54:01e so C is a constant which incorporates
54:03if you want P0 and also what you would
54:07get so what we will get from the left
54:09hand side is the value of the integrand
54:12sorry of the integral at H2 minus the
54:15value at H2 of t 0 which is a constant
54:17so I can absorb everything into this
54:20constant T and what I get out of this
54:22equation is that
54:24by inverting the hyperbolic arctan I
54:28will get that this is
54:29simply given by square root of
54:334 K Phi
54:35so I go a little bit fast but that's
54:38just because it's
54:39it's not so relevant
54:42just to give you an idea pi over KT
54:45Flash
54:47my constant C
54:49that is now
54:51it's the same constant okay and once I
54:55have this how do I fix C well remember
54:58that we had an information about the
55:00boundary value of this function so we
55:02know that H2
55:04at time capital T has to be equal to 2
55:08times a
55:10and therefore what I have to do is to
55:12compute this quantity at Capital time T
55:14and then solve for C and I will get an
55:18expression for C which depends on on
55:21Capital a okay so in particular
55:24you get the C is
55:27square root of C over K times capital T
55:30Plus
55:36a over square root of K Phi okay so this
55:39you can check and afterwards it is just
55:42algebra and this gives you the solution
55:45for
55:47the coefficient H2
55:50and somehow the only thing that I wanted
55:52to uh to to do is to discuss an example
55:55for this separation of variables which
55:58as I said comes up several times
56:01uh in this type of problems
56:04okay now we have a H2 and so we have
56:07solved the equation three here so now
56:09let's go to uh this problem of trying to
56:12solve equation one and equation two uh
56:15for one particular choice of mu of T
56:18which comes from the interactions
56:20and this is part two
56:24of the exercise
56:30it's a square yes
56:32so the question very much is that there
56:35is a squaring here thanks that I forgot
56:39people see it
56:41I hope so
56:45okay
56:48now let's go to part two
56:57and let's try to be a bit more precise
57:00about the idea that I sketched before
57:03that now we want to choose uh this New
57:06York team
57:07to be somehow the average of all of the
57:10velocities of all of the other investors
57:13which are in the market
57:16and we want to do it with the reasoning
57:19that is very close to main field so this
57:21is let me summarize
57:23the point of
57:26uh
57:28of 0.5 of the exercise
57:32and various the following so suppose
57:34that now we have many investors
57:36okay labeled by uh as I say by a so you
57:41will have
57:42this Vector a
57:45uh which labels my investors you will
57:47have a solution for the optimal velocity
57:50uh for each of these investors and you
57:54want to Define mu as an average over all
57:57of the others velocities
58:00that you get from the corresponding
58:02equations
58:03so how do you do this in a meanful
58:05scheme well let me sketch it very
58:08briefly so the idea is that you start by
58:10choosing one
58:15representative
58:20agent
58:24let me label
58:26her by a one specific value of a so this
58:31is a little bit like when you do easy
58:32mean field you choose one side that you
58:35assume to be representative of all of
58:37the sides
58:38and then once you have chosen this agent
58:41you assume
58:43that you know
58:44uh
58:49that you know
58:53this New York tea
58:55that is given by the behavior of all of
58:58the other agents so in sometimes you
58:59know you assume that you know what all
59:02of the other agents are doing and you
59:03solve the problem for your
59:05representative agent given these
59:08particular value of Mufti which is
59:10precisely what we did in part one so if
59:13you do this you get the results
59:16that we get so you get the optimal
59:18velocity now for our particular agent a
59:23given value of mu T we can write it as
59:281 over 2K
59:30the derivative of this function Omega
59:32with respect to q and this is a function
59:34of p q
59:36at the given u t
59:39so you assume you know what everybody
59:41else is doing you use this information
59:43to solve for the velocity of your
59:46particular agent so if you want
59:48thinking about easing you assume that
59:51you know the magnetization of all of the
59:52other sites this tells you what is the
59:55effective field which acts on the
59:57particular side that you are looking at
59:59you solve for the magnetization of that
1:00:02particular site at fixed value of the
1:00:05field and then you have to impose
1:00:06self-consistency
1:00:08and so this is the step which is missing
1:00:11here that is
1:00:14impose
1:00:16self-consistency meaning that
1:00:18the results that you get is consistent
1:00:22with your assumption on mu and namely
1:00:24with the assumption that mu is the
1:00:26average of this velocities so let me
1:00:28write it down and then I will specify it
1:00:30so the self-consistent equation in this
1:00:33setting will correspond to saying that
1:00:35mu of t
1:00:36is what is an average
1:00:40over if you want all of the
1:00:43representative sites of the solution for
1:00:47this particular site
1:00:50that you get
1:00:52that you got
1:00:56a fixed value of new p
1:01:00so you see that this is I hope you see
1:01:02it now I will tell you what I mean
1:01:05precisely by this average but before let
1:01:08me stress so this is really a
1:01:09self-consistent equation so it is an
1:01:11equation in which it's an equation for
1:01:13Mu of T in which mu of T depends appears
1:01:17on both sides yes out of her screen ah
1:01:21yes you're right
1:01:23usual problem
1:01:25thanks a lot guys for
1:01:38okay
1:01:41so you have Mufti both on the right and
1:01:42then on the left hand side and you have
1:01:45to solve for this function you'll see
1:01:47knowing what is the velocity at fixed
1:01:50profile from your team
1:01:53so let me tell you how should we intend
1:01:55this average with respect to all of the
1:01:58other agents
1:01:59so the way we can interpret it
1:02:04is as follows
1:02:16uh now I can write here
1:02:20yes
1:02:27so what is the average
1:02:31well to make this average precise we can
1:02:34think in the following way so as I said
1:02:36we have many many agents and all of
1:02:38those have a particular initial value of
1:02:41their
1:02:42inventory
1:02:45q0
1:02:46labeled by by the agents
1:02:49and these initial values we may assume
1:02:52that they are distributed so that there
1:02:54is a probability
1:02:55P0 of
1:02:58of this distribution of initial uh
1:03:02values for the inventors inventories
1:03:05so what you want to sell or buy
1:03:08and then we know that each of this agent
1:03:12has its own velocity that determines the
1:03:15time evolution of this quantity Q so if
1:03:18you look
1:03:19at the differential equation for Q and
1:03:22we integrate it then you know that Q T
1:03:24of a
1:03:26is q0 of a plus
1:03:29the integrals from t0 to T in the S of
1:03:34your velocity
1:03:36BS of a which depends
1:03:40as we saw solving the optimal problem it
1:03:42depends on your trajectory and in
1:03:45General on your teeth
1:03:47times DT that I'm integrating over uh in
1:03:51here
1:03:53okay
1:03:56so this is the optimal solution
1:04:05and then you see that if I have a
1:04:06distribution for the initial value the
1:04:08initial value of my inventory will
1:04:11determine uh what is the optimal
1:04:14solution together with all the other
1:04:16parameters and this equation will tell
1:04:18me that I will have also a distribution
1:04:20at any time T So if this is a random
1:04:23variable Q at any time will be itself
1:04:25random variable with a given
1:04:27distribution
1:04:29pity of QT and so in some sense when I
1:04:32say that we want to average over uh all
1:04:36of these the investors so what I mean
1:04:38with this average
1:04:40that again you don't see with this
1:04:43averaging here is essentially an average
1:04:44with respect to uh to the state of the
1:04:48system at any time assuming that we have
1:04:50a given distribution of uh of the
1:04:53initial uh inventory for all of our
1:04:57agents
1:04:59okay so this is how we should interpret
1:05:01the self-consistent equation and now
1:05:03let's see just very briefly
1:05:07um
1:05:08how to
1:05:10finish and solve or at least let me
1:05:14sketch
1:05:15uh how the solution can be obtained
1:05:20for this system of equations so let me
1:05:23start from here
1:05:27so to
1:05:29to go on
1:05:30we now need essentially another equation
1:05:33for this mu of T to solve our system of
1:05:37equations
1:05:38and it is convenient to introduce
1:05:42the quantity that I call E of t
1:05:45which is the average in the senses
1:05:47digest described
1:05:50of
1:05:51Q of P
1:05:54so it's just the average with respect to
1:05:56this distribution
1:06:00and why is this nice to introduce well
1:06:03the idea is that if I take the
1:06:04derivative
1:06:07of this language so let me say what is
1:06:10this average so let's look at this
1:06:12equation in here and take the average so
1:06:14if I do this this will tell me that I
1:06:18get a factor which is the average at
1:06:19time 0. and then I have the integral I
1:06:22bring the average inside the integral
1:06:26so now t 0 I must mean it to be zero
1:06:29and I will have the integral over DF
1:06:32of the average of all my velocities
1:06:35optimal velocity which depends
1:06:38on Qs and on mu
1:06:41okay
1:06:45and this average is precisely what I
1:06:47defined as to be equal to Mu of T
1:06:50through my self-consistent equation so
1:06:53this means that if I take the time
1:06:54derivative of this quantity that I just
1:06:57introduced this
1:06:59so the time derivative of the constant
1:07:01is zero and the time derivative will
1:07:03just give me the integrands so it will
1:07:06give me the average
1:07:07of
1:07:09my velocity at the time t
1:07:13given mu of T that through the
1:07:15self-consistent equations they want to
1:07:17impose is just equal to Mu of t so if I
1:07:21solve for this quantity is taking the
1:07:23derivative I have this function mu of T
1:07:26which is what appears in my system of
1:07:29equations
1:07:30so now to complete the system of
1:07:31equation we just have to derive an
1:07:34equation for this e of t
1:07:37in here
1:07:41okay
1:07:44and before yeah
1:07:50yeah and then we are almost done
1:08:00okay
1:08:04so once I'm here let me also or let's
1:08:07say once I am here let me look at this
1:08:09equation that I just wrote and remember
1:08:12that I know the solution for the optimal
1:08:14uh velocity at fixed value of mu of T so
1:08:19I found before that VT so here there
1:08:23should be a not uh
1:08:26a subscript that tells me that I am
1:08:29Computing this at the solution of uh for
1:08:33for the optimization problem so let me
1:08:35write it here V opt
1:08:38as a function of Q
1:08:41and if we're given U of T was just given
1:08:44by
1:08:451 over 2K
1:08:48the derivative
1:08:53of the function Omega that I introduced
1:08:57to the power 2.
1:08:59and using the answer for the function
1:09:02Omega this is just 1 over 2K
1:09:06times so the derivative of Omega is H1
1:09:10of t
1:09:14minus
1:09:16H2 of t
1:09:18the two countries with a one-half Q to
1:09:21the power two
1:09:23you see it okay
1:09:27here I'm just using the answer that I
1:09:30made for the expansion of Omega
1:09:33and therefore if I now plug this
1:09:35expression into this equation for E
1:09:38Prime I get a fourth equation
1:09:40that I call equation four
1:09:43which tells me that e Prime of t
1:09:49is also equal to the expectation value
1:09:52of this quantity here with respect to Q
1:09:55so now I'm
1:09:58I'm let me write small q so this is a
1:10:02solution for any possible value of small
1:10:04q now if I compute the optimal
1:10:06trajectory as a fixed uh the optimal
1:10:09velocity get a fixed trajectory for my
1:10:11random variable q and I take the
1:10:14expectation value what I will get so
1:10:16this is just from here I will get that
1:10:18this is
1:10:191 over 2 K
1:10:23there is no Square here right all right
1:10:27okay
1:10:29H1 of T minus h2c times the expectation
1:10:35value of Q which is nothing but e of T5
1:10:44so from all using all of these
1:10:46identities and my solution for V let's
1:10:48fix mu I get this fourth equation that
1:10:51allows me to relate the quantity that I
1:10:53just introduced to H1 and H2 for which I
1:10:57have the two other equations up there
1:11:00and so now what's the strategy to solve
1:11:03the system of equations I will just
1:11:05sketch it so the first thing that you
1:11:08can do is you take another derivative
1:11:11with respect to time of equations four
1:11:15so now I want to close to get a closed
1:11:17equation for E of t
1:11:19so what I can do is I take
1:11:22the second derivative of my equation 4
1:11:25with respect to time
1:11:27and what I have is 1 over 2K
1:11:31H1 Prime
1:11:34with respect to D minus H2 of T Prime
1:11:38this is not a problem because we solved
1:11:40the problem for H2 so we know what is
1:11:42the explicit form
1:11:44of this derivative times e
1:11:49minus 1 over 2 K
1:11:53h two of T times e Prime of t
1:11:57okay
1:11:59so now we have a true H2 that as I said
1:12:02we know we have e that is what we want
1:12:04and we have this
1:12:06H1 Prime
1:12:09that uh that the the say depends
1:12:13indirectly on uh on the function e why
1:12:18because if you look at the equation two
1:12:19this gives you precisely H1 Prime as
1:12:23minus beta mu and mu is what mu is e
1:12:27Prime of t
1:12:29plus 1 over 2K H1 H2 so you will see
1:12:33that this Factor if I now replace
1:12:36I place equation 2 in here
1:12:39okay
1:12:41what I see is that and I use that mu of
1:12:44T is e Prime of T I will get out of this
1:12:47so this you can check
1:12:49I will cancel terms like this I think
1:12:51and I will get a second order equation
1:12:55for E which is of the following form
1:13:01so I'm going a bit fast in here but just
1:13:03because this is
1:13:05just algebra
1:13:08I have H2 Prime of T minus H2 square of
1:13:14T over 2K
1:13:18equal to zero
1:13:24okay so this is what you should get
1:13:29and the quantity in parenthesis you can
1:13:31further simplify it by using equation
1:13:34three so you know that this will be
1:13:37equal to minus 2 5.
1:13:41okay
1:13:43so this quantity in parenthesis well
1:13:45okay you see the quantity parenthesis
1:13:47here is minus 2 5.
1:13:49so now you have a simple second order
1:13:52differential equation for e
1:13:54and you can solve it
1:13:59so you know that when you have equations
1:14:01of this form the solution is always
1:14:03given by
1:14:05linear combinations of exponentials
1:14:12so again I will just give you the
1:14:15general form and then you can fill in
1:14:18the details
1:14:20foreign
1:14:25second derivative of t
1:14:28plus T by Prime of B
1:14:31and then we have minus 2 Phi
1:14:34e of t
1:14:36equal to zero
1:14:39and if you
1:14:41do all of the calculation you will find
1:14:43that you can write
1:14:45the solution so y will be an exponential
1:14:48well okay because this is a linear
1:14:50equation in the derivatives so if you
1:14:52make an answer which is of the
1:14:54exponential form e to the alpha T you
1:14:57will be able to rewrite this as an
1:15:00equation for the coefficient Alpha and
1:15:02by solving these equations you find that
1:15:08you can write it as
1:15:10in the following form
1:15:19or what you can do is you take these
1:15:21answers you plug it into the equation
1:15:22and this gives you a second order
1:15:24equation for this
1:15:26omegas
1:15:27that each of the following form
1:15:34so which will depend on the parameters
1:15:36you see in this equation so 5K and Theta
1:15:41and if you do it you will find this
1:15:44dependency here
1:15:48okay
1:15:49so the equation will allow you to fix
1:15:51the values of Omega plus and Omega minus
1:15:54and then you see you have an extra
1:15:55constant C and you have to fix it
1:15:58knowing the boundary value for uh for uh
1:16:02for your quantity e that you can deduce
1:16:06for example
1:16:08from uh you can deduce it from the
1:16:11boundary values of H1 and nh2
1:16:16uh that you have from before
1:16:20okay so uh I don't want to do uh now do
1:16:25all of the details uh of the solution of
1:16:27the system of equations because we also
1:16:28do not have time but what is the idea so
1:16:31the idea is that through this game of
1:16:33introducing e of T and solving for E of
1:16:36T you now have everything that you need
1:16:39to go back to what I call before
1:16:41equation one two and three so equation
1:16:44three we solved the equation uh one or
1:16:47if you want just look at
1:16:50it's in this form here
1:16:53so from here you know that if you know
1:16:55H2 and you know e of T and its
1:16:58derivatives then you can solve for H1
1:17:00and if you know H1 you can solve for the
1:17:04equation of for h0 which was just that
1:17:07the derivative of h0 is minus H1 Square
1:17:10over 2K and so you have sold all of your
1:17:14equations and what you can do and this
1:17:16is a little bit an exercise for you is
1:17:19to
1:17:20try to write them down this equation and
1:17:23try to plot them as a function of the
1:17:25various parameters of the problems that
1:17:28that we have
1:17:30and of course I'm not going to do it at
1:17:33the Blackboard but I just want to
1:17:35make perhaps
1:17:37one final comment that is related to
1:17:40part three
1:17:42so part three are just questions about
1:17:46the dependence on the parameters
1:17:51so I will just make two comments so let
1:17:54me keep
1:17:55Omega in here
1:18:00or maybe one comment so one comment is
1:18:03uh the following
1:18:06so let's go back to equation four up
1:18:09there
1:18:10which you see
1:18:12yes so equation four
1:18:17okay so let me conclude with the final
1:18:19comment which
1:18:20is one of the questions
1:18:233.7
1:18:27foreign
1:18:37so equation four is of the form
1:18:40so you see that you have
1:18:43e Prime and e but let me maybe just take
1:18:50let me look at the line which is above
1:18:55equation four which gives me the
1:18:58um the optimal velocity
1:19:01so the idea is the following so I can
1:19:03write
1:19:04that the optimal velocity which is e
1:19:07Prime of T because of the
1:19:10self-consistent equation
1:19:11is given
1:19:13by what so using principle given
1:19:19no sorry this I can write in the
1:19:21following way
1:19:29okay
1:19:33and where does it come from well it
1:19:36comes from
1:19:38essentially this
1:19:42so I have this equation up here and then
1:19:45I can rewrite
1:19:47H1 as a function of e and if I rewrite
1:19:51H1 as a function of e by solving
1:19:54equations four for H1 I get this
1:19:57expression for uh for the optimal
1:19:59velocity
1:20:01and it is useful to write it in in this
1:20:03way just to tell you the final comments
1:20:06and the final comment is that you have
1:20:07two contributions to the optimal
1:20:10velocity so you have one contribution
1:20:12that goes like e Prime T and remember
1:20:15that e Prime T is
1:20:19what we call U of B
1:20:22so it is an average of all of the
1:20:25velocities of the other players in the
1:20:28game
1:20:28so what this is telling you is something
1:20:30that people in this literature call
1:20:33follow
1:20:36the trading flow
1:20:44so this is the term that is basically
1:20:46telling you that if everybody is selling
1:20:48very fast
1:20:50each of the other agents will have a
1:20:52velocity that is large and negative and
1:20:55so your e Prime contribution in here
1:20:57will be the average of all of these
1:20:59velocities that will be large and
1:21:00negative and so you also want to have a
1:21:02velocity that is large and negative so
1:21:04you want to match with what all of the
1:21:08other people are doing uh on average and
1:21:10this is what uh
1:21:12people in the literature call indeed
1:21:14following the flow with an extra
1:21:17contribution though which comes from uh
1:21:20from this difference in here
1:21:22and this difference in here it tells you
1:21:24so typically you can show that this
1:21:25quantity is positive
1:21:28and therefore it's telling you that yes
1:21:30you may have a large negative
1:21:32contribution uh from the the flow of the
1:21:36other people but then you can compensate
1:21:38a little bit with this this trend by
1:21:41this term and this term is proportional
1:21:44to let's say Q so the amount of
1:21:47inventory that you still have at the
1:21:49given time c and e of T is what is the
1:21:53average of this of the inventory of all
1:21:55of the other Traders so what this is
1:21:57telling you is that if you have an
1:21:59amount of inventory that is smaller than
1:22:02the average so you have already so sold
1:22:05more than what the average has sold then
1:22:10you get a contribution which is positive
1:22:12which compensates a little bit uh the
1:22:15negative contribution of the velocities
1:22:16of all of the others so you somehow have
1:22:19to adjust to what the average is doing
1:22:21and this is the essence of
1:22:24um in field meaning that if everybody is
1:22:26selling fast you want to sell as they
1:22:28are are to satisfy also the
1:22:30self-consistent equation and at the same
1:22:32time you want that your average
1:22:34inventory matches somehow the average
1:22:36for all the others so if at the time you
1:22:39have an inventory which is smaller you
1:22:41slow down a little bit you're selling so
1:22:43as to compensate and somehow be
1:22:47consistent with uh with what the system
1:22:50is doing on average
1:22:52so this is uh slow down
1:22:57if if Q is smaller than
1:23:01than the average at a given time t
1:23:05okay so that's basically uh What uh
1:23:09say and the other comments that you find
1:23:13in section three are again about
1:23:15analyzing a little bit uh the structure
1:23:18of the solution that you see in
1:23:21particular as you see
1:23:23um well okay you can go through it by
1:23:27yourself one observation that you can
1:23:28make is that in the behavior of e of T
1:23:31you see that the parameters enter in
1:23:35here like Theta K and Phi but the
1:23:37parameter a which was related to the
1:23:41cost of liquidation at the final time
1:23:44doesn't enter in here so this tells you
1:23:46that maybe you can somehow to this level
1:23:49you can forget about that extra
1:23:52parameter and what you can really play
1:23:53with to get to different strategies are
1:23:56the three ones that appear in here but
1:24:00if you can check by yourself
1:24:03and okay I think this is enough and for
1:24:06the plank that's really very fast
1:24:09but maybe we do it uh next time so there
1:24:12is the next day
1:24:14to the number five is
1:24:17um
1:24:18or or there are two weeks that I
1:24:21allocated for the same today and that is
1:24:23just to have a little bit of pose to
1:24:25also catch up with what was left behind
1:24:28and I think that is next week and uh and
1:24:30the following one after the holidays so
1:24:33this is to say that we will have time to
1:24:34uh to discuss soccer plank again
1:24:37and for the moments we can stop here I
1:24:39think so are there questions
1:24:45nope
1:24:51okay
1:24:53okay then I stopped
1:24:56the record

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