Entropic regularization of optimal transport and applications
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Carlier, Guillaume
Entropic regularization of optimal transport is appealing both from a numerical and theoretical perspective. In this talk we will discuss two applications, one from incompressible fluid dynamics and the other from mean-field games theory.
 with Jean-David Binaloud, Simone DiMarino, Compilzard, and Luca Nenor, which addresses some connection between entropy minimization and minfi games, a certain class of minfi games. So the title is entropy minimization for minfi games. Bonjour, energies. So I want to start with a remark that is entropy minimization is a tractable proxy for optimal transport. So if you want to learn it, so entropy minimization is a tractable proxy for what's in it. So it starts with a optimal transport problem. Let's say on RD, you have a certain cost, C, you look for transport layer, gamma, between given probability measures, Mu and U. Let's say density is nice probability measures. It's in general difficult to solve. It's linear programming in infinite dimension. It's a bit rough. So one way to approximate it is to add a small contribution. And the natural one is entropy, who aspect, let's say, to the Lebesgue measure, as in this case here. And now this can be recast as a relative entropy minimization of gamma in respect to the Gibbs-Carno C over epsilon, e to the minus C divided by epsilon, or more transport methods. So you start with linear programming, and you end up with a sort of projection problem, but not with a square distance, but with a cool backlight block. Epsilon times the integral of gamma along gamma. Epsilon of epsilon times the integral of gamma. It's epsilon times entropy. Right. And here, what I do is I divide everything by epsilon. I put the cost here in the logarithm, and I end up with this. So the problem becomes finding a transport plan, which in cool backlight block distance is as close as possible to the set of invisible plants. So how do you know that you have a gamma sector, gamma and log gamma is not infinity? So how do you know that you can find at least one which is absolutely continuous? Oh, OK. So I make some assumptions on the, OK, imagine that the margin is finite entropy. So I can take the product measure at this finite entropy. OK? Think that, of course, otherwise it's not a good way. It's not a good, it's not very consistent. So assume, for instance, that entropy of mu, entropy of mu is finite, OK, if you want. Otherwise, I can change the reference measure. OK. So why is it better? Because it's a strict complex problem. So that means a unique solution. But there's a density with respect to the bag. Gamma is gamma xy dx dy. And the optimal gamma has a special form. It's e to the minus c of xy divided by epsilon times the tensor product, a of x, b of y. OK? So you can think of a or the exponential of a and the exponential of b as being lab-range multipliers associated to marginal constraints. So it has this form. And now a and b should be such that the marginal constraints are met. So this is just two algebraic equations, which say that a of x times the integral of e to the Gibbs kernel is the first density. And similarly, for the second marginal, you get an equation like this. So the two lab-range multipliers are related by this relation. This is actually, I mentioned this, because this is a system which appears for historical reason in a paper, a famous paper by Schrodinger and the Fertigies. So solving this is just finding two densities, two positive potentials, a and b, such that this condition are met. And you know, if b is fixed, you find the a. If a is fixed, you find the b. And you can iterate. And it turns out that this is an algorithm which somehow amass to iterate some contraction or some fancy metric called the Hilbert metric. So iterating this equation, these two equations in the Schrodinger system converges. This is called the single algorithm, which is very simple to use, very efficient, pretty fast in practice. And of course, as epsilon go to 0, you approximate the initial O. Am I clear so far? So entropy minimization is much easier to solve than linear programming. So now I'd like to give you a message that epsilon equals 1 might be interesting in itself. And I would like to illustrate this in the case of Mifig games. So Mifig games with the variational structures goes like this. So variational Mifig games, you are basically looking at a minimization problem where you function to fix time interval 0, t. You look for a curve of measure in the velocity field which minimizes the kinetic energy plus maybe some running cost plus maybe some terminal cost. You minimize with respect to rho and v. R0 is fixed. It's a given probability measure. Again, think it has density, it has time interval p, it has finite second moments. So rho0 is fixed. And the second equation is this diffusion equation. In D3-Rome, you replace basically the continuity equation. You add a lot of pressure in it. Plus divergence of rho v equals 0. So it's a sort of optimal control problem for this equation. So for the problem equation, and I'd like to convince you in the end of my talk that this can be recast in the end as a problem like this, as an entropy minimization problem. So f of rho t and g of rho t also? Sorry. So this is terminal time. OK, so this is a running cost. So think it's interval of rho square, for instance. And you want to end up at the end of the day at t. There is some potential or some function, which can be 0. OK, so why are we interested in this? Because at least formally, the optimality system conditions always problems. Let me call it 1. For 1, the system of pd is, of course, various, a Fokker-Prandt equation. Since you are minimizing energy, you can figure out that the optimal v should be the gradient of a potential. So we've got a Fokker-Prandt equation for rho, the drift, the gradient of some potential u. And the u, which is some sort of costate, I joined state for this equation. So I'm going to make a viscous and a Newton-Jack-Cogbie equation. So be careful. This is a general state. So it is backward. This is why the minus sign is coming from. And on the right-hand side, there's something which depends on rho, formerly F derivative of capital F. And there's boundary conditions that are important. So rho naught is fixed. So this is forward in time for the evolution of rho. But this is backward in time for this value function t. So your t is j over t, which is a sort of trans-semitic condition associated to the terminal cost in the sense that j is the derivative of j. So we want to solve this system. It has a sort of game theoretic flavor. So of course, you mentioned that all this was introduced by last year in Reynolds. And there's a thousand. And it is just a special case of mythic eggs. Mythic eggs theory has developed a lot. This is a special case. So now the question is, can one, if formulated, let's say, a laschködinger as an entropy immunization problem, which is more tractable, at least from a numerical point. Or maybe it's interesting in itself. And the guess that this is the case actually comes from recent results. And some people in the audience are responsible for two papers, two nice papers, one by Chen, Pavel and George Yu. And the other one by Ivan Fistor-Leonard and Richard Ipani, which consider this is maybe a simple problem as well, but it's related, which is a sort of noisy benign formula, which goes like this. So let me take two probability measures. Again, think that they have densities, which are our finite entropy. And I consider two variational problems. The first one is optimal control of Hock-O'-Plank, if you want. So given new and new given some time t, let me introduce the least cost of Hock-O'-Plank of somehow transporting mutinu. So this is the infinum of the same guy, rho v squared. So rho v forms the equation, same equation. I should have given the name to this in the beginning. 0, it's going to be star. And there are boundary conditions. So rho 0 is mu, rho of t is mu. Could be plus infinity if the measures are not 1bA. And let me introduce another problem, an entropy minimization problem, which I have to define a reference measure on the past case. So the continuous pass of this time interval 0 t, which is a so-called reversible linear measure, which is defined this way. So you take the standard Brownian motion, starting at 0, call it b on 0 t. But to change the initial condition, you launch it at x time of x, and you integrate with respect to the initial condition uniform. So why is it called reversible and venerable? Because the back measure is in binary. So now there's another problem, which I call s t mu mu, which is minimized with respect to probability measure on the past case. Another boundary condition that's the loop of the push forward by the evaluation time 0 of q should be mu. And the terminal measure, so this is a measure on the past case, it induces our genomes for each time. So you prescribe them q t equals mu. And the theorem, which is proven here, so the result of Yvon, Christian, and Richard slightly more general, well, there was a connection between those two guys that the minimum entropy is the same as the Pocopron cost up to the entropy of the initial condition. And now I'd like to make a remark here. Sorry, I'm confused. What is this r? It's a reference measure. So it's a measure. It's not a probability measure. So the b depends on t, or what's the b here? How b is standard pronoun motion, but is there a time? It's a pass. Sorry. Because it's a whole trajectory. You look at one realization of the pronoun motion, but starting at x, and then you integrate with respect to the initial condition. It's uniform. So if I integrate the function, what do I get? So it's 4. OK, that's a good question. So the expectation, the same with respect to this measure, well, it's not an expectation. So phi is a function defined on path. So omega is a pass. d r omega. So this is a point of space, which is infinite dimensional. It is you take one trajectory of the pronoun motion, which you start at x. It depends on the one. Maybe you can take a cylindrical test function, which depends on finitely many time realization of the motion, if you want. And this is random, so you have to integrate. You get the expectation, and then integrate with respect to x. Because it's fine. And now I'd like to make a remark, due to Christiane's name, that you might think that this problem here is complicated, because you look for a measure on a set of paths. This is really a dynamical problem with two layers of dimension, in fact, it simplifies. And the reason why it simplifies is the following. So take q such that the q max equal mu, qt equal mu, q is a probability measure of the path space. You can define a transport plan, which is the joint evaluation and starting and end point of q. So this is a transport plan between mu and mu. And now the relative entropy can be decomposed h2r. And similarly, I can define r0t in the same way. So this is basically the heat gamma that we have here. So it disintegrates. So it's a relative entropy, sorry, a total gamma, plus the integral on the path space of the entropy of the conditional probabilities. So I do not borrow the conditional probability. We've got more x0, xt. So this is the probability on the path space. It induces a conditional probability on paths which event points x0 and xt. And this is a nice property of relative entropy that it decompose this way. But now these are probabilities. So it's always positive, not negative. And this is 0 when this guy coincides with this guy. So in fact, the infinite dimensional path you can forget about. The best strategy is really to optimize, to minimize this with respect to gamma. And then this you can make 0 by equating this conditional probability with this one, which is called the So-called Brouhman Bridge. So in fact, st of mu nu is nothing else than the infimum of our transport plans. So now on rd, now, we're not on trajectories anymore. And the minimum value of relative entropy with respect to r0t, which is very nice. It's e to the minus the square distance divided by 2t, basically. This is a problem we started. This is a problem we started with t equal to sine. We started. How much time do we have? We started a little late. Say six more minutes. Six more minutes. So I'll be quick. Time for the Brownian Bridge. Sorry? Time for the Brownian Bridge. OK. So this is a 2n-point problem. So now this can be generalized to more marginal constraints. Because here, you see in a mid-fielder game problem, we look for a whole trajectory of measure, a whole path of measure, to more marginal constraints. And once you know this nice formula here, and you play a little bit with properties of entropy, what you can prove, in fact, you can discretize a problem in time, which numerically is what you're going to do. And consider more general functional. So let me set some notation. So let's assume that you have a path of measures. So it's a continuous curve valued in the past ocean space. And then you can introduce two least energies. The first one is just the middle. We have to look to the velocity field v of the kinetic energy. So mu now is fixed. Subject to the diffusion equation. So now we have basically a start with heads. Basically, you prescribe the good variables v mu, the momentum variable. So it's like prescribing the divergence of this momentum variable. So this is convex in the initial problem, in fact. And there's another one, which is look for a probability measure on the past space that you prescribe all the time margin, the Qt equals mu t for everything. And now it turns out that they are the same as mu is exactly your mu plus the entropy of the initial condition. And I don't have time to enter the details. And you have district, maybe, time discretization, which consists instead of fixing a wall pass, you fit snapshots at different times. So you can give yourself mu into mu m, mu probability measures on our d. You can define the discretization of this guy. So what is it? It's just a sum of the focal length value, least energy, mu k plus 1. So you just fix the constraint at times kt divided by m. The value of the bastard trajectory should be mu k. So you can do the same here, define as n mu dot mu m as being the infimum h2, we respect, does a more investible. We don't measure, but you don't fix all the marginals, but only those at instance, districts times kt divided by m. And now this is just a solution. Sqrt satisfies star. What is the star? Yeah, it's a dT mu minus r. I see. Yes, right here. So avoid having two unknowns. So maybe I just like to connect the minimization of entropy. So we saw the link with Brouhlin Bridge. Brouhlin Bridge with two legs, two end points. So this problem Sn, it's a bridge with more than two legs. It's nice that some bridges have more than two legs. So the connection, with my properties of entropy, is the following, Sn mu dot mu m. So it uses a lot of the fact that Markovianity properties of the reference measure are. It's just the sum of what we saw before on the time it would be divided by m of mu k, mu k plus 1. That was a neural term which is the sum of the values of mu k, rk, but rk is the effect measure. So this is just entropy. The usual entropy. So if you combine this decomposition of the least entropy and the Bellabou-Boucouin formula, you're recombined. So this is the same as the least energy subjected to focal plane plus these conditions at mu at time kt divided by n should be mu k. It's fixed. This is the same as m mu n plus the initial entropy of the initial connection. So now you see that if you recall the problem of midfield gain while you had to minimize the kinetic energy subjected to focal plane plus something else, but it's the same as minimizing over probability measures. Now the starting point was fixed, which should not be more or less. So this is the same as the least entropy plus the running cost of F of Et, which was equal to t plus terminal cost. And of course we discretize the entire. So this is an example because there is diffusion in this least beginning business where in fact the problem with diffusion, you don't light up silent to zero. So it's really interesting because it's a nice strictly complex problem which is pretty attractive. Thank you very much. APPLAUSE Any questions for you? What is the optimal Q here? That's a good question. You mean here for the whole stuff? Yeah, it's optimal one. So the optimal Q, it's a good point. So first the optimal row is of course the projection with respect to time with the optimal Q. Now the optimal Q, you can obtain it by just a lot of theory. It's a chain, it should be absolutely continuous with respect to the Lebesgue measure. And now it should have a special form because it minimizes entropy as we started with. And in fact it should be given, the change of probability is given by some drift which is related to the Hamilton-Jacques equation we saw in the beginning. I don't know if it's a change of probability on the past phase. But for all these it has the same marginality and utility. Ah, you mean for the wolf, when everything is fixed. Ah, but okay, then you have a multiplier of four million. For each, it should be given by an exponential of something which there is a multiplier for each time margin constraint. So Q should be obtained by R with a certain change of variable with this exponential density. And the phi should solve some dual dorsal limits. So roughly speaking it's also a differential equation. This is formal because it's not clear that you have any irregularity. I don't know what you're doing. That's why here this problem is well posed if you make reasonable assumption on f and j. And of course it takes a complexity itself. Does it answer your question? Is that right? Any other question? Okay, thank you. I hope you had a star studded session this morning. So please reduce, share Robert Ben MacKann with his beard.
 with Jean-David Binaloud,  Simone DiMarino, Compilzard, and Luca Nenor,  which addresses some connection between entropy minimization  and minfi games, a certain class of minfi games.  So the title is entropy minimization  for minfi games.  Bonjour, energies.  So I want to start with a remark that  is entropy minimization is a tractable proxy  for optimal transport.  So if you want to learn it, so entropy minimization  is a tractable proxy for what's in it.  So it starts with a optimal transport problem.  Let's say on RD, you have a certain cost,  C, you look for transport layer, gamma,  between given probability measures, Mu and U.  Let's say density is nice probability measures.  It's in general difficult to solve.  It's linear programming in infinite dimension.  It's a bit rough.  So one way to approximate it is to add a small contribution.  And the natural one is entropy, who  aspect, let's say, to the Lebesgue measure, as in this case  here.  And now this can be recast as a relative entropy minimization  of gamma in respect to the Gibbs-Carno C over epsilon,  e to the minus C divided by epsilon,  or more transport methods.  So you start with linear programming,  and you end up with a sort of projection problem,  but not with a square distance, but with a cool backlight  block.  Epsilon times the integral of gamma along gamma.  Epsilon of epsilon times the integral of gamma.  It's epsilon times entropy.  Right.  And here, what I do is I divide everything by epsilon.  I put the cost here in the logarithm,  and I end up with this.  So the problem becomes finding a transport plan, which  in cool backlight block distance is as close as possible  to the set of invisible plants.  So how do you know that you have a gamma sector,  gamma and log gamma is not infinity?  So how do you know that you can find at least one  which is absolutely continuous?  Oh, OK.  So I make some assumptions on the, OK,  imagine that the margin is finite entropy.  So I can take the product measure at this finite entropy.  OK?  Think that, of course, otherwise it's not a good way.  It's not a good, it's not very consistent.  So assume, for instance, that entropy of mu, entropy of mu  is finite, OK, if you want.  Otherwise, I can change the reference measure.  OK.  So why is it better?  Because it's a strict complex problem.  So that means a unique solution.  But there's a density with respect to the bag.  Gamma is gamma xy dx dy.  And the optimal gamma has a special form.  It's e to the minus c of xy divided by epsilon times  the tensor product, a of x, b of y.  OK?  So you can think of a or the exponential of a  and the exponential of b as being  lab-range multipliers associated  to marginal constraints.  So it has this form.  And now a and b should be such that the marginal constraints  are met.  So this is just two algebraic equations, which  say that a of x times the integral of e to the Gibbs  kernel is the first density.  And similarly, for the second marginal,  you get an equation like this.  So the two lab-range multipliers are related by this relation.  This is actually, I mentioned this,  because this is a system which appears  for historical reason in a paper, a famous paper  by Schrodinger and the Fertigies.  So solving this is just finding two densities,  two positive potentials, a and b,  such that this condition are met.  And you know, if b is fixed, you find the a.  If a is fixed, you find the b.  And you can iterate.  And it turns out that this is an algorithm which somehow  amass to iterate some contraction or some fancy metric  called the Hilbert metric.  So iterating this equation, these two equations  in the Schrodinger system converges.  This is called the single algorithm,  which is very simple to use, very efficient,  pretty fast in practice.  And of course, as epsilon go to 0,  you approximate the initial O.  Am I clear so far?  So entropy minimization is much easier to solve  than linear programming.  So now I'd like to give you a message that epsilon equals  1 might be interesting in itself.  And I would like to illustrate this  in the case of Mifig games.  So Mifig games with the variational structures  goes like this.  So variational Mifig games, you are basically  looking at a minimization problem where you function  to fix time interval 0, t.  You look for a curve of measure in the velocity field which  minimizes the kinetic energy plus maybe some running cost  plus maybe some terminal cost.  You minimize with respect to rho and v.  R0 is fixed.  It's a given probability measure.  Again, think it has density, it has time interval p,  it has finite second moments.  So rho0 is fixed.  And the second equation is this diffusion equation.  In D3-Rome, you replace basically the continuity equation.  You add a lot of pressure in it.  Plus divergence of rho v equals 0.  So it's a sort of optimal control problem  for this equation.  So for the problem equation, and I'd  like to convince you in the end of my talk  that this can be recast in the end as a problem like this,  as an entropy minimization problem.  So f of rho t and g of rho t also?  Sorry.  So this is terminal time.  OK, so this is a running cost.  So think it's interval of rho square, for instance.  And you want to end up at the end of the day at t.  There is some potential or some function,  which can be 0.  OK, so why are we interested in this?  Because at least formally, the optimality system conditions  always problems.  Let me call it 1.  For 1, the system of pd is, of course, various,  a Fokker-Prandt equation.  Since you are minimizing energy, you  can figure out that the optimal v should be  the gradient of a potential.  So we've got a Fokker-Prandt equation for rho, the drift,  the gradient of some potential u.  And the u, which is some sort of costate,  I joined state for this equation.  So I'm going to make a viscous and a Newton-Jack-Cogbie equation.  So be careful.  This is a general state.  So it is backward.  This is why the minus sign is coming from.  And on the right-hand side, there's  something which depends on rho, formerly F derivative  of capital F. And there's boundary conditions  that are important.  So rho naught is fixed.  So this is forward in time for the evolution of rho.  But this is backward in time for this value function t.  So your t is j over t, which is a sort of trans-semitic condition  associated to the terminal cost in the sense  that j is the derivative of j.  So we want to solve this system.  It has a sort of game theoretic flavor.  So of course, you mentioned that all this was introduced  by last year in Reynolds.  And there's a thousand.  And it is just a special case of mythic eggs.  Mythic eggs theory has developed a lot.  This is a special case.  So now the question is, can one, if formulated,  let's say, a laschködinger as an entropy immunization  problem, which is more tractable, at least  from a numerical point.  Or maybe it's interesting in itself.  And the guess that this is the case actually  comes from recent results.  And some people in the audience are responsible for two papers,  two nice papers, one by Chen, Pavel and George Yu.  And the other one by Ivan Fistor-Leonard  and Richard Ipani, which consider this is maybe  a simple problem as well, but it's  related, which is a sort of noisy benign formula, which  goes like this.  So let me take two probability measures.  Again, think that they have densities, which  are our finite entropy.  And I consider two variational problems.  The first one is optimal control of Hock-O'-Plank,  if you want.  So given new and new given some time t,  let me introduce the least cost of Hock-O'-Plank  of somehow transporting mutinu.  So this is the infinum of the same guy, rho v squared.  So rho v forms the equation, same equation.  I should have given the name to this in the beginning.  0, it's going to be star.  And there are boundary conditions.  So rho 0 is mu, rho of t is mu.  Could be plus infinity if the measures are not 1bA.  And let me introduce another problem,  an entropy minimization problem, which  I have to define a reference measure on the past case.  So the continuous pass of this time interval 0 t,  which is a so-called reversible linear measure,  which is defined this way.  So you take the standard Brownian motion,  starting at 0, call it b on 0 t.  But to change the initial condition,  you launch it at x time of x, and you integrate  with respect to the initial condition uniform.  So why is it called reversible and venerable?  Because the back measure is in binary.  So now there's another problem, which I call s t mu mu,  which is minimized with respect to probability  measure on the past case.  Another boundary condition that's  the loop of the push forward by the evaluation time 0 of q  should be mu.  And the terminal measure, so this is a measure on the past  case, it induces our genomes for each time.  So you prescribe them q t equals mu.  And the theorem, which is proven here,  so the result of Yvon, Christian, and Richard  slightly more general, well, there  was a connection between those two guys  that the minimum entropy is the same as the Pocopron cost  up to the entropy of the initial condition.  And now I'd like to make a remark here.  Sorry, I'm confused.  What is this r?  It's a reference measure.  So it's a measure.  It's not a probability measure.  So the b depends on t, or what's the b here?  How b is standard pronoun motion, but is there a time?  It's a pass.  Sorry.  Because it's a whole trajectory.  You look at one realization of the pronoun motion,  but starting at x, and then you integrate with respect  to the initial condition.  It's uniform.  So if I integrate the function, what do I get?  So it's 4.  OK, that's a good question.  So the expectation, the same with respect to this measure,  well, it's not an expectation.  So phi is a function defined on path.  So omega is a pass.  d r omega.  So this is a point of space, which is infinite dimensional.  It is you take one trajectory of the pronoun motion, which  you start at x.  It depends on the one.  Maybe you can take a cylindrical test function, which  depends on finitely many time realization of the motion,  if you want.  And this is random, so you have to integrate.  You get the expectation, and then integrate with respect to x.  Because it's fine.  And now I'd like to make a remark,  due to Christiane's name, that you might think that this  problem here is complicated, because you look for a measure  on a set of paths.  This is really a dynamical problem with two layers of  dimension, in fact, it simplifies.  And the reason why it simplifies is the following.  So take q such that the q max equal mu, qt equal mu, q is a  probability measure of the path space.  You can define a transport plan, which is the joint  evaluation and starting and end point of q.  So this is a transport plan between mu and mu.  And now the relative entropy can be decomposed h2r.  And similarly, I can define r0t in the same way.  So this is basically the heat gamma that we have here.  So it disintegrates.  So it's a relative entropy, sorry, a total gamma, plus the  integral on the path space of the entropy of the conditional  probabilities.  So I do not borrow the conditional probability.  We've got more x0, xt.  So this is the probability on the path space.  It induces a conditional probability on paths which  event points x0 and xt.  And this is a nice property of relative entropy that it  decompose this way.  But now these are probabilities.  So it's always positive, not negative.  And this is 0 when this guy coincides with this guy.  So in fact, the infinite dimensional path you can  forget about.  The best strategy is really to optimize, to minimize this  with respect to gamma.  And then this you can make 0 by equating this conditional  probability with this one, which is called the  So-called Brouhman Bridge.  So in fact, st of mu nu is nothing else than the infimum  of our transport plans.  So now on rd, now, we're not on trajectories anymore.  And the minimum value of relative entropy with respect to  r0t, which is very nice.  It's e to the minus the square distance divided by 2t,  basically.  This is a problem we started.  This is a problem we started with t equal to sine.  We started.  How much time do we have?  We started a little late.  Say six more minutes.  Six more minutes.  So I'll be quick.  Time for the Brownian Bridge.  Sorry?  Time for the Brownian Bridge.  OK.  So this is a 2n-point problem.  So now this can be generalized to more marginal  constraints.  Because here, you see in a mid-fielder game problem, we  look for a whole trajectory of measure, a whole path of  measure, to more marginal constraints.  And once you know this nice formula here, and you play  a little bit with properties of entropy, what you can prove,  in fact, you can discretize a problem in time, which  numerically is what you're going to do.  And consider more general functional.  So let me set some notation.  So let's assume that you have a path of measures.  So it's a continuous curve valued in the  past ocean space.  And then you can introduce two least energies.  The first one is just the middle.  We have to look to the velocity field v of the kinetic  energy.  So mu now is fixed.  Subject to the diffusion equation.  So now we have basically a  start with heads.  Basically, you prescribe the good variables v mu, the  momentum variable.  So it's like prescribing the divergence of this momentum  variable.  So this is convex in the initial problem, in fact.  And there's another one, which is look for a probability  measure on the past space that you prescribe all the time  margin, the Qt equals mu t for everything.  And now it turns out that they are the same as mu is  exactly your mu plus the entropy of the initial  condition.  And I don't have time to enter the details.  And you have district, maybe, time discretization, which  consists instead of fixing a wall pass, you fit  snapshots at different times.  So you can give yourself mu into mu m, mu probability  measures on our d.  You can define the discretization of this guy.  So what is it?  It's just a sum of the focal length value, least energy,  mu k plus 1.  So you just fix the constraint at times kt divided by m.  The value of the bastard trajectory should be mu k.  So you can do the same here, define as n mu dot mu m as being  the infimum h2, we respect, does a more investible.  We don't measure, but you don't fix all the marginals,  but only those at instance, districts times kt divided by  m.  And now this is just a solution.  Sqrt satisfies star.  What is the star?  Yeah, it's a dT mu minus r.  I see.  Yes, right here.  So avoid having two unknowns.  So maybe I just like to connect the minimization of entropy.  So we saw the link with Brouhlin Bridge.  Brouhlin Bridge with two legs, two end points.  So this problem Sn, it's a bridge with more than two legs.  It's nice that some bridges have more than two legs.  So the connection, with my properties of entropy,  is the following, Sn mu dot mu m.  So it uses a lot of the fact that Markovianity properties  of the reference measure are.  It's just the sum of what we saw before on the time it  would be divided by m of mu k, mu k plus 1.  That was a neural term which is the sum of the values of mu k,  rk, but rk is the effect measure.  So this is just entropy.  The usual entropy.  So if you combine this decomposition of the least entropy  and the Bellabou-Boucouin formula, you're recombined.  So this is the same as the least energy subjected to focal plane  plus these conditions at mu at time kt divided by n should be mu k.  It's fixed.  This is the same as m mu n plus the initial entropy of the initial connection.  So now you see that if you recall the problem of midfield gain  while you had to minimize the kinetic energy subjected to focal plane  plus something else, but it's the same as minimizing over probability measures.  Now the starting point was fixed, which should not be more or less.  So this is the same as the least entropy plus the running cost of F of Et,  which was equal to t plus terminal cost.  And of course we discretize the entire.  So this is an example because there is diffusion in this least beginning business  where in fact the problem with diffusion, you don't light up silent to zero.  So it's really interesting because it's a nice strictly complex problem  which is pretty attractive.  Thank you very much.  APPLAUSE  Any questions for you?  What is the optimal Q here?  That's a good question.  You mean here for the whole stuff?  Yeah, it's optimal one.  So the optimal Q, it's a good point.  So first the optimal row is of course the projection with respect to time with the optimal Q.  Now the optimal Q, you can obtain it by just a lot of theory.  It's a chain, it should be absolutely continuous with respect to the Lebesgue measure.  And now it should have a special form because it minimizes entropy as we started with.  And in fact it should be given, the change of probability is given by some drift  which is related to the Hamilton-Jacques equation we saw in the beginning.  I don't know if it's a change of probability on the past phase.  But for all these it has the same marginality and utility.  Ah, you mean for the wolf, when everything is fixed.  Ah, but okay, then you have a multiplier of four million.  For each, it should be given by an exponential of something which there is a multiplier for each time margin constraint.  So Q should be obtained by R with a certain change of variable with this exponential density.  And the phi should solve some dual dorsal limits.  So roughly speaking it's also a differential equation.  This is formal because it's not clear that you have any irregularity.  I don't know what you're doing.  That's why here this problem is well posed if you make reasonable assumption on f and j.  And of course it takes a complexity itself.  Does it answer your question?  Is that right?  Any other question?  Okay, thank you.  I hope you had a star studded session this morning.  So please reduce, share Robert Ben MacKann with his beard.
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