So I guess we will start. I don't know. Not seeing Mateo or Noia. OK, so good afternoon. Good morning. Sorry. Good morning to everyone. I need Steve to wake up. So let's start with a brief reminder of what we have seen yesterday. So yesterday I started to talk about random walks and levy flights. And eventually also I should mention probably Brownian motion. So I started with some definitions and mainly discussing this model, a very simple Markov model, where you just have this discrete time random walk.  And this eta is just drawn from some symmetric distribution, which I chose here to be symmetric for the sake of simplicity. And then I started to discuss this Green's function, which is the basic object that you would like to compute here, which is basically the probability or the probability density, in fact, to arrive at x starting from x0 after n steps. OK, so this is this quantity here. I derived this equation for it, which is a recursion, eventually which turns out to be a recursion relation for this quantity, which is this forward or backward Kolmogorov equation. And eventually you can solve it explicitly for any jump distribution p of eta, which is probably something that you knew. And then I moved to the real object that I want to study here, which is the survival probability. And then you will see that it actually plays a very important role when we will study the extreme statistics of this random walk. So yesterday I derived an equation for this survival probability. And I started, if you remember, I started from the constrained Green's function, that means the Green's function, but conditioned to stay positive. And from the backward equation for this g plus, I derived an equation for q x0 n. Now, maybe to start with today, let's re-derive this equation for q x0 of n in a slightly different way, which I hope will be a little bit more enlightening. So I mean enlightening. That's the same idea. Again, the idea is to write an equation for this quantity, and I will write a backward equation for q of x0. So again, it's a backward equation that that's, again, a kind of Kolmogorov equation, if you want. So the idea, again, is to cut. So you start at x0, and you want to compute the probability that you stay positive up to step n. So a typical configuration will be something like that. And so here, you don't care too much. You don't care at all, in fact, about where you end up. So you need to sum over all the possible n points here. Now, to derive an equation for this q of x0 n, it's quite useful to think about this trajectory as consisting of a first step and n minus one step. So you have one step here and n minus one step there. And because you have a Markov process, again, these two segments here, these two time intervals are just independent, simply because this part here doesn't know anything about what happens just right before. Now, if you want to compute this probability that you survive up to step n starting from x0, well, the first way to decompose this probability is just to say, OK, you start at x0. So there is first the probability that I stay positive during the first step. And that will give me some probability. So basically, I will decompose q of x0 n. So that's the first probability that I do a jump which will drive me at, say, x prime 0. And the probability for this first jump is just p of x prime 0 minus xx0. Now, of course, x prime 0 has to be positive, OK, because I need to survive during that first step. So that means that if the first jump is here, then obviously this does not contribute to this survival probability. So of course, here x prime 0 has to be positive. Let's do it here. And then once I arrive here, well, then I need to survive during the n minus 1 remaining steps. And this probability is precisely the survival probability starting at x prime 0. And I have to survive during n minus 1 steps. So that means that I need to do this product. Again, I can do here this product simply because these two parts are just independent, OK? So the probability of the two events are the same. And now you see, I need this x prime 0 here. It can be anything on this positive semi-axis here. So that means that I have to integrate this for x prime 0, and x prime 0 has to be positive. So that means that I need to integrate it from 0 to plus infinity, OK? So that's the central equation here. And that's the backward Kolmogorov equation. So to solve it, you need to add some boundary condition, boundary and initial condition. First condition that you need here is that if you start at any point x0, which is positive, obviously, I mean, when x0 is just positive, if you just don't move, then the probability that you survive is just 1, OK? So that means that it has to be 1. And this actually is true, provided x0 is positive, or maybe 0, OK? So that's still possible here. I could eventually imagine the case where I start exactly for this bit random mode, this is possible. I start at 0, and then I ask what's the probability that I survive up to step n. This will not be possible, in fact. That situation will be different for the continuous-time random motion. I will comment on this a bit later. But for the random mode, this is obviously possible. And then, OK, there is another condition that I already mentioned yesterday, is that if x0 goes to infinity, then for any n, this just also should be 1, OK? So to solve this equation, it turns out that we can solve this equation, or at least one can have a semi-explicit formula for this quantity. You will see how it looks like. But before, I just want to comment on one thing. So this is the survival probability. Now, from the survival probability, one can define what we use, what sometimes is called the first-passage probability. So that's the first-passage probability. So what is that? So that's the first passage at 0, at step n, time step n. So what is that? So basically, you are now looking at something like that. So you look at the probability of the following event. So you start at x0. And so you stay positive, say, up to step n minus 1 here. So that's you do something like that. And then between step n minus 1 and step n, you cross 0, OK? So that's the first time that you cross 0. So that's the probability that you first cross 0 between step n minus 1 and step n, OK? So that's what I want to call f of x0 n. And that's 2, 5, yes, it's here, sorry. So this is obviously related to the survival probability. And this is essentially minus the discrete derivative, right? So if you think a little bit about it, so let's define it properly. So this is this f of x0 n. This is the probability, I just write it in words, to cross the origin immediately after step n minus 1. So again, as I said, if you think a little bit about it, these kind of trajectories corresponds precisely to the difference between q of x0 n minus 1 minus q of x0 n. So if you compare the probability that you survive up to step n and the probability that you survive up to step n minus 1, then the difference are precisely the number of trajectories that have crossed 0 between n minus 1 and n. So that's, again, minus the discrete derivative. Is that clear? You need to think a bit about it, but that should be pretty obvious. So maybe the first thing to do is to try to see what happens for Brownian motion. So that means how can I solve this equation if I make the same Brownian scaling limit that I was doing yesterday? So basically, that means that I will consider the case where the time steps, I mean the time between two steps goes to 0. And also, I need to rescale properly the size of the jump, of course. So I will just, Brownian scaling limit. You remember what we did yesterday. Eventually, I showed you that this integral equation, that moment this was minus infinity, but it doesn't change anything, that this equation here yields back the diffusion equation. So if you redo what we did before in the Brownian scaling limit, this integral equation, this backward equation, becomes the diffusion equation. So remember that. So I did that in detail yesterday, so I will not redo it today. But eventually, what you get is that Q, actually, in the continuum limit satisfies this equation. So I have basically some function Q of x0 t in that case. So Q of x0 n becomes in the continuum limit Q of x0 t. And you get this equation. Yeah, I don't have it, actually. So I guess Erika will bring it. And so again, you agree? I mean, I will not redo it, because this is exactly what I did yesterday. Now, still, there is a difference here. There is some difference because of the boundary conditions. Now, indeed, there is some subtlety here when you look at the boundary condition. Indeed, for the Brownian motion, you cannot impose it to start at 0 and to be strictly positive at time t equals 0 plus.  And in fact, the correct boundary condition that you have to impose, so that's a bit subtle here. But Q of x0 of 0 has to be 0 in this case. So that means that, again, for the Brownian motion, because it's continuous both in space and time, if it's at 0, then it will require 0 an infinite number of times before eventually escaping. The Brownian motion is typically doing something like that. So you cannot actually impose it to be at 0 at time 0 and to be strictly positive at time 0 plus. So this is one of the difference that you get with discrete time random motion. Now, you also see that this equation was, yes, sorry. Yes, excuse me. And in fact, I can actually be even if you start exactly, I mean, it's not only at t equals 0, but it's true actually for all time, of course. So you see, I mean, that this is basically the same equation that, so the equation itself is the same as yesterday. So when I looked at the propagator, I showed you that it satisfies this diffusion equation. And that's why eventually it has this Gaussian form. But it has the Gaussian form that we know, because there was a specific boundary condition that it was basically a delta function at t equals 0. Because for the Green's function, you say, OK, what is the probability that I start at x0 and arrive at x at time t? So that means, what's the probability that I arrive at x given that I was at x0 at time t equals 0? So obviously, at time t equals 0, the Green's function has to be a delta function, x delta of x minus x0. Now here, you see, I mean, the boundary condition is different. And effectively, you will see that the solution for q is quite different. It resembles a little bit a Gaussian, but it's not a Gaussian. You will see. It's basically the interval of a Gaussian. So how do I solve this equation? Well, there are various ways to solve it. But certainly, when you have to solve this kind of partial differential equation, I mean, you have two variables and you don't like this too much. So you find a way to essentially recast it, transform it into a simple differential equation. And the way to do that is to write q or to search for a solution under a scaling form. So here, you know that you have diffusion, scaling. And what you are doing, so when q is a probability, so it's a dimensionless quantity, the Green's function is not a dimensionless quantity, but q is a probability. So it's something which is dimensionless. So that means that I will just look at it. I will look under the form that I'm writing now. So that will be, oh, is it x1 square root of t here? Oh, x1 square root of t here. OK, so I will just look for a solution of this equation simply by assuming that there is a scaling variable and a scaling variable. I mean, you have no choice. I mean, in your problem, the only dimensionless scaling variable that you have is something that will be of the form x0 divided by square root of t. And in fact, if you want to work properly, you need to have, let's look at this. So d has to be here for dimension reason, and 4 is just for commodity. OK, so I just look for a solution under this form. So there is no special, I mean, in principle, there is no special reason, except that you know that here, if you look at your Brownian motion, the Brownian motion itself is scaling variant, right? So that means that you know that basically when you look at the Brownian motion on a time interval 0 square root of t, say it like this. So if I look at typical trajectories, say a Brownian motion up to time t, then basically, that should be equivalent to look at xt over square root of t on the unit time interval. OK, so that means that if you look at this quantity, x of t divided by square root of t, then that should be equivalent in law to the Brownian motion on 0, 1. That's what I'm saying. OK, so that's what I mean by scaling variance. That means that if you look at these whole trajectories on the time interval 0t, then if you rescale all the things by square root of t, then in law, so that means they are not really the same, but in distribution, they will be just the same. That will be similar to the Brownian motion on the unit time interval. Exactly. So that's the reason why I'm looking for such a solution. So in other words, what I'm saying is that basically, looking at the survival probability of the Brownian motion up to time t is similar to look at the survival probability of the process x of t divided by square root of t on the unit time interval. That's basically what I'm saying. Once you have this form, it's just a simple exercise to derive an equation for you.  But now the nice thing is that this is just an ordinary differential equation. There are various ways to solve this problem, but this one is probably the simplest. And of course, I need to satisfy this boundary condition. So that tells me that u of 0 has to be 0. So yeah. So the yellow person is taking a third school time or a third school time. So I didn't get your question. Sorry, when you were doing infinite things? Right? Right. Right. That's a bit of a safe thing. Yeah, yeah. So that's precisely the same. Might you have a semi-infinite? Yeah, but that's basically what happens there, I think. I mean, that's basically what I was mentioning here. OK. Yeah, yeah, this equation is perfectly well-defined. It's perfectly well-defined. And again, you have to, OK, it's another question to relate it to the discrete random walk. I will comment on that. But this equation is perfectly well-defined. And now you can, of course, you can solve it, right?  And eventually, you integrate this equation. So u prime will be a Gaussian, obviously. And then you integrate it, and you find that u of z, you eventually, it's just an error function, OK? So that's 2 over square root of pi integral from 0 to z, exponential of minus u squared du. And this is what we call the error function, OK? So again, this should be relatively simple to see, right? I mean, you can integrate this equation. You can, I mean, it's a simple, it's a first-order equation for u prime, OK? Which you can easily integrate, OK? Yes? I didn't get your question. Oh, what is z? Yeah, z, you see, I mean, u of z is this argument, right? So u is a, u is a, yeah, OK. This is z, right? u, so this is the argument of the function u. So u has a single variable, it depends on a single variable, which I denote that. Is it OK? That means that now you have an explicit expression for q of x naught t, which is this error function of this guy, right? So this error function of x naught divided by square root dt 4. So that's a very important result for the Brownian motion, OK? So you get immediately, I mean, that's fairly, I mean, you get derive many things with that. I will probably comment on this later on. Now, in particular, what you can get is what happens in the large time limit, OK? So what's the probability that you start, that starting at x naught? What's the probability that I survived for a very long time? And if you look at the asymptotic behavior of this function, so you need to understand how it behaves. Essentially, when t is large, that means when z goes to 0, and then what you find is that, you see, it's pretty simple. u of z goes to when z is small. So I want to understand what happens for large t, OK? So when t goes to infinity, that means that I need to understand what happens when this argument is small, OK? Now, if you look at this formula, so I have to understand what happens when z is small. And when z is small, of course, it goes to 0. So the range of integration is very small. And it's basically proportional to z times 2 over root pi times 1, OK? Because exponential minus u square is simply 1 when u goes to 0. So the result is just x naught. So there is a 2 here which will cancel the square root of 4 here. And you just get x divided by square root of pi dt. So there are two things which are important to notice here. The first one is that this goes like 1 over square root of t. So that's probably something that you have already seen before. So the survival probability for the Brownian motion decays as 1 over square root of t. That's the probability of return to the origin. So eventually, it will return to the origin, OK? Because q of x naught goes to 0. And we know this is not true in higher dimensions. In dimension 3, we know that the random work, there is a finite probability that the random work will not return back to the origin. In 1D, with probability 1, it will come back to the origin. And the survival probability decays as 1 over square root of t. That's one thing. The other thing is that it's proportional to x naught. That means that when x naught goes to 0, this also goes to 0. But this we knew because we had sort of input. So that's for the survival probability. Then we can also get, if we want, if you want, you can also get the first passage probability. So what's the probability that I start at x naught? And what's the probability that I cross 0 for the first time at time t? So that's just the minus the derivative of q. I don't know if it's not anymore the black one, but this is just minus dq dx naught. And this derivative, this one is just a Gaussian, OK? So let's write it explicitly. This is just x naught divided by 4 by dt cubed exponential minus x naught squared divided by 4 dt. Now, of course, if you look at the large time behavior of that, it's basically the time derivative minus the time derivative of this guy. And this decays as 1 over t to the power 3 by 2. So that's, OK, I just write it here. Probably, I'm not sure you can see it here. It's probably a bit too far. Can you see it here? OK, so let me just write it. OK, well, that was here. So that's this guy, right? So it's minus this derivative. So asymptotically, for Brownian motion, what you find, so dm for Brownian motion, I would maybe use this later on. So this decays as x naught to the power t to the power 3 by 2. Yeah? Yeah, yeah, sorry, sorry, sorry. Thank you. OK, so there is actually a nice interpretation for that, which one asked about this first passage. So now if you look at typically at the Brownian motion, what is happening? So you can ask this question, right? So you look at the, and we probably, OK, we will encounter this again somehow. So suppose that you are looking at some Brownian motion like this, OK? And then you can ask, you see, I mean, let's look at where it crosses 0.  I mean, you would like to understand that this time intervals there. So if you look at this time intervals here, so you would have, say, your first one here, tau 1, type tau 2, and let's say tau 3. So what it says, I mean, this formula tells you is that these tau i's here, these are random variables. And if you look at basically the distribution of these tau i's is basically this quantity here. And in particular, if you look at the tail of the PDF, the probability distribution of tau i's, then it's something that decays as 1 over tau to the power 3 by 2, which is basically what we are computing here. That's basically what we are computing with that. It's a probability that, since you are here, that that could be here even at tau 0 here. But the distribution of these guys decays as 1 over tau to the power 3 by 2. There is a very nice property in addition for Brownian motion is that these tau i's actually turns out to be independent. They constitute what is called a renewal process. I will probably comment more on that when I will talk about the records. OK, so that means that essentially for the Brownian motion, of course, we have a fairly simple expression for this quantity. Now, let's come back to our initial problem, which was to understand what happens for the discrete time random walk. So for a discrete time random walk, I need to solve the full equation that I showed before, which is this backward equation. And this, of course, is much more complicated. Yes? Well, in which method? Yes, in principle, what you can extend, I mean, this method can be extended. Well, as I said a bit yesterday, if you have some Brownian motion, so for instance here, it was very simple. It's very simple because I don't have any potential, right? But I don't have any drift. But you could imagine that your particle instead of evolving in a free landscape, it evolves in a given potential. And then this can be adapted. You can write, OK, but you can write, basically, the equivalent of this diffusion equation with some potential. And then you can try to solve this Kolmogorov equation. And this can be done in a wide collection of systems. It can be adapted to many situations. But of course, if you leave this, again, I mean, this works very well for Brownian type of diffusion, right? Because if you don't have this Brownian limit, then as we discussed briefly yesterday, then you end up with this fractional type of operators and this becomes a mess. But for all these old systems, all the systems that have a nice Brownian limit, yeah, you can actually extend it quite. Yeah, it's a very powerful method to compute these persistence properties. So you need two things. I mean, yeah, you need two things. In fact, Brownian scaling and also you need some Markov property, right? Because otherwise, you cannot write the equivalent of the, but this Markov for Kolmogorov equation actually is something that we have used a lot. I mean, so we wrote, a few years ago, we wrote a big review on this problem. I think you probably received the reference to it. I could re-do, we give it to you if you want. So now we want to do a much harder thing, which is to solve this equation in the discrete case, discrete setting. So solving this is extremely difficult. And I will probably not show you all the details, but I want to show you the result. And later on, we will see that we can actually use the formula. So let's see how it works. And let's try to analyze it. So this is called Polaxix-Pitzer formula.  So it solves this problem for any jump distribution. Again, you really want to solve this equation here, provided that you have this. Now, again, you are back to this, need to be equal to 1, for x naught, for x naught. And you want to do that for any jump distribution p in principle. So again, this Polaxix-Pitzer formula gives you an explicit expression.  Let me write it, and I will comment. So that's what you want to compute. You would like to have qx naught n equal. Now, of course, unfortunately, the object for which you have an explicit expression is not exactly qx naught of n, but it's something that, so you first take the generating function of this object. So that means that you take this generating function. So if you know this quantity, in principle, you cannot, you can recover q, yes. Yes, yes. So in this case, actually, so you see for the random walk, this is a well-defined problem when you have x naught equal to 0. So you could choose something else, but I want to consider this possibility, which I could not do, in fact, for Brownian motion. But here, you see, I mean, I can't start at 0 exactly and never come back to it. This is definitely possible. Yeah, I wrote, OK, so what I wrote before is that q of 0, 0 for Brownian motion is 0. I insisted on the fact that this kind of situation is not just possible for Brownian motion. But I was saying that this is specific to the Brownian motion. When I say Brownian motion, I mean really that the continuous time, continuous space process, while here, I'm really looking at the discrete time random walk. Now for Brownian motion, what I said is that you cannot be exactly at 0 at t equal 0 and be strictly positive right after. This is just impossible, OK? This is impossible because, again, when Brownian motion costs 0 once, it will recross many times, infinitely many times, right after. You see, I mean, it has typically, if you look at a trajectory of continuous time, Brownian motion, if you just put it at 0, then eventually it will, with probability 1, it will do with this many crossing, OK? So that's, I mean, the origin of that is because we said that the density of 0 is infinite for the Brownian motion. So again, the picture is that when you have Brownian motion, when it crosses 0, in fact, when it crosses any value, at a given time, it will recross infinitely many times right after. So that's why you cannot impose this condition. Well, that means that it's not that it blows up, but you really need to consider it on a finite time interval. And then, with some probability, it would escape from it. But you cannot have, what I say is that it's really a limiting procedure, OK? So for Brownian motion, you cannot have this. This is just impossible for Brownian motion. This is really a problem of limit. This is just impossible. And that's why you cannot impose that for Brownian motion. So again, this is just impossible for Brownian motion. But for random mode, obviously, this is possible. It's a bit subtle, but that's how it is. Again, I think it's fairly easy to understand that for random mode, this is definitely possible. Because you have this quick jump, you just need to have the first jump will immediately drive you far from 0, and then you are done. So that's what I want to show, I want to solve, OK? So again, to have a nice, yeah, we need the blackboard because the formula is quite big. So the object for which you have an explicit formula is, so you first need to take the degenerating function. But in fact, it's not yet enough. Then you take the Laplace transform with respect to x0. You compute, so you integrate over x0, and you take the Laplace transform. So I'm doing two operations. So I'm taking the generating function with respect to time, and the Laplace transform with respect to x0. And this quantity has a fairly, not fairly, completely explicit expression. And let me write it. So you will probably find it a bit, but you have never seen this formula. So it's a bit complicated, but it's quite explicit. Now p hat of k is just a Fourier transform of this. So the very nice thing is that this is true.  I can compute its Fourier transform, and I inject it into this formula, and I immediately get the right hand side. Now, of course, if you want to get back to q of x0n, usually you have to work a little bit. I mean, in principle, you have to invert a Laplace transform. It's usually pretty hard. And then you need to invert this generating function by a Cauchy formula. So that's, in principle, still a bit hard to get information out of it. But still, you can do it. And this requires analysis. And I will show you some application of this formula, what kind of things you can extract without entering too much into details. But still, I mean, trying to show you that it's indeed very, very nice and that you can get some nice information. Yes? Yeah. Yeah. So yeah, you just need to have a density here, and it needs to be symmetric. There are extensions for p of x, which are either discrete.  There are some extensions of it. There are also some extensions when you have a drift. But here, this holds for the simpler case where a p of x is continuous and symmetric. Maybe I should write it. OK, so it's really a beautiful formula. It's a bit complicated. The derivation is quite involved. I will not present it here, but I will present some results that you can extract from it. And the first one, and I think maybe the nicest one, is what is called under the name of the Sparanderson formula. It turns out that, so what is this Sparanderson? Yeah, lambda is not lambda. Lambda is this guy's p. So p, so again, I have this s, which is basically the conjugate which conjugates with respect to n. So it enters only here and there. And p is the Laplace parameter associated to x naught. And indeed, it enters here and there. So now I want to derive this Sparanderson formula. So Sparanderson actually is a single name. Sparanderson was a Swedish mathematician. And he had actually derived before this Polak's experts of formula. So this is a formula from roughly from the 60s, 70s. Sparanderson actually had a series of wonderful results which he obtained in the 50s. And this one that I will present here was obtained in 54. But he had, I mean, he was really a great mathematician. He obtained a lot of nice results for this random work, concrete results. And so there is one which I present here, which, as we will see, has many applications, which is how to extract what I mentioned before, how to extract this quantity here. So I just look at the case where I start exactly at 0 and I ask, what's the probability that I stay positive of 2 step n? That's a fairly natural observable. Now, what I claim is that it's pretty simple to look at. I mean, to obtain what this guy is from this formula. So how does it work? So let's look at the left-hand side. And we would like to have a Q of x naught. So we would like to basically have a way to extract the value at x naught goes to 0. Now, if you look at x naught goes to 0, obviously, this integral of x naught will be dominated by large values of p. And so what I will do is to do the following. I will just first do, on the left-hand side, I will just first do a change of variable. And I will just change p of x naught. I will just set it equal to y. And I will just do this. So I just do that. So d of x naught will become dy over p from 0 to infinity. And now x naught is y over p times n, s to dn. And now I have exponential simply minus y. So I didn't do anything except this change of variable. And now what I do is I take the limit when p goes to infinity. So if I take p goes to infinity, you see that this will go to Q0n, which is the quantity that I am after. So if I look at this quantity when p goes to infinity, it's not really a limit. But essentially, I will get 1 over p times this integral over y. So I will just take this exponential here. I will just take it here. And then I have this sum here from 0 to n, Q of 0n, s to dn. Again, I assume that there is a good limit for this guy. So that will converge to Q of 0n. And I will simply have a 1 over p. Now you see that the integral over y is very simple. It's just 1. And so basically, the left-hand side is just 1 over p times the generating function of the object I am after. Yeah, I mean, I'm not a mathematician. I assume that I can do that. And I assume, of course, that I can do that.  So I mean, if you really want to put the epsilon here, it's very simple, I guess. Actually, you can even show that the absolute convergence. So you are really, everything is under control. So you have this. And so that's the left-hand side. Now if you look at the right-hand side, so that means I want to look at what happens in the large p limit on the right-hand side. So here, this is what it is. This is 1 over p square root of 1 minus s.  And this one behaves like p. So p divided by p square is 1 over p. And when p goes to infinity, this is just 0. So basically, in the large p limit, this is just 1. Very simple. So if I look at the right-hand side, it's very simple, right? I mean, we do this argument. But again, this just is basically minus p divided by p. And the leading term will be p square there. So that's essentially, OK, let's write it like explicitly. What I'm saying is that, OK, so this is just p divided by p square. This is just 1 over p. So what I'm saying is that this just goes to 1, as p goes to plus infinity. So that means that the right-hand side just converges to 1 over p square root 1 minus s. So if you just identify both the left and right-hand side for this beautiful formula, OK, let's keep it on the ground. But eventually, what I showed is that for large p, OK, so this is just 1 over p times this generating function. And here, this is just 1 over p square root of 1 minus s, OK? So this is quite nice because, of course, you see that the 1 over p just cancels. I mean, the left and right side, OK? And you obtain this formula, that the generating function of this probability that I am after, OK, I want to recover this Parandasian formula, is just 1 divided by square root of 1 minus s. So that's quite nice because you see that it does not depend on anything. It does not depend on the jump distribution that you have. That means that, so finally, we end up with what this Parandasian formula is. He derived it in a quite different way because he didn't know about this Spitzer formula. Now, I use ways to derive this, but all of them are quite complicated. In any case, I mean, all these results are quite hard to get. So that's Parandasian, OK? So that's the sum of this generating function. And you will see that we will use that many times in the following. It has many nice consequences. So that's Parandasian. And the nice thing, again, is that this is true for any jump distribution, OK? So for any p of eta, which is symmetric, n continues, OK? There is, of course, a generalization of this Parandasian formula for discrete random walk. Probably I will not discuss this in these lectures, but there is one. So now, in principle, OK, you can use Cauchy's formula to invert this, but you know also the series expansion of this guy on the right-hand side. So you can immediately compute q of 0, n. And eventually, q of 0, n has these very simple forms, which is 1 over 2 to the power 2n, 2n chosen, OK? So you just invert this formula, OK? So here, I did not really invert it. I mean, I just supposed that I know the smallest expansion of this, and this gives you that. Maybe I can just, because I will reuse, OK? Many times, I start, I mean, that's the first time that I use it, but the use of generating function is very common in this business, I mean, in this problem. So maybe I just want to make a remark which concerns what is called under the name of Cauchy's formula. So typically, in many cases, I will not know the quantity a of n, whatever a is, where a of n is. In this case, this is q of 0, n. But usually, I will know something about this generating function. You have such a relation. So if I know a tilde of s, in principle, I can know a of n. And the way I compute a of n is via the Cauchy formula, right? So a of n will be, so that's an integral over the complex plane around 0. And I have to do this ds over 2 pi i s to the power n plus 1 a tilde of s. I guess you have seen this formula. So that's called Cauchy, right? This is due to Cauchy. It's basically a residue formula. So again, here you need to do, so this means that you are integrating over a contour, which is around the origin. And you integrate counterclockwise. That's this symbol. So here, of course, I can do that. But again, I know also that the full expression for this. And so I have this very nice expression, which again, you see, I mean, is true for random work, like if you have jumps, which are Gaussians, but it's also true if you have power loss. It's true if you have heavy types. It's true for any continuous and symmetric distribution. So it's extremely robust result. And now, if you look at what happens when n goes to infinity, well, you just need to use Tling formula to analyze this binomial coefficient. And what you would find is that this actually decays like 1 over square root of pi n. So when n is large, you see that you recover this 1 over square root of n behavior. So it's a bit similar. You remember when I looked at the Brownian motion. Brownian motion has this 1 over square root of T. But it's quite different, in fact, because the result that I have, when I analyze q of x0 T for large T for Brownian motion, it was decaying like 1 over square root of T. But there was x0 in front. Instead here, I have no x0. I just have a 1 if you want. And so that means that the origin of these two formulas actually are quite different. And we recently wrote something about it. I mean, I could give you some reference, but it's quite subtle to understand the link between the both. But I want to emphasize really that this has nothing to do with what I showed you before. So this is not for the Brownian motion. This 1 over square root of n that you have here is not the 1 over square root of T that we had before. These are different observables. And in particular, you see that this 1 over square root of n here holds again for levy flight, for instance, which is, I think, quite remarkable. OK, so these are nice results. And now that we are equipped with this Polak's x-pixel formula on the one hand, Spar-Hundersen formula here, then we can do many things. And in particular, we can analyze in a detailed way the extreme statistics for this random motion. So I will now find or show you some nice applications of this for the extreme value of time series, namely the example of random motion and levy flights. OK, is that fine with this formula? OK, so again, basically everything that you want to compute about first passage, about extreme statistics, about records, is essentially included in this slide. It's quite involved formula, but many information is encoded there. So let's move on then to this extreme statistics now. So let's see how to apply this formalism to extreme statistics. And in particular, I have talked a lot about survival probabilities, but at the moment, I have not shown you any connection with extreme statistics. So I will start with that. So let's start with now really the extreme statistics of random walks and levy flights. So again, I already emphasized that these are sets of strongly correlated variables, simply because if you look at the correlations between the two positions of a random walker, being the sum of random variables, they are actually quite correlated. And that's a very nice example where we can study this problem for strongly correlated systems. So again, I will just set up the problem one more time. So again, we have this. p of eta is symmetric, and it's continuous. OK, p of eta is symmetric and continuous. And so now I want to look at these kind of trajectories. So I have, say, a Brownian motion. OK, start at some point, say here. And it does something like that. So I will look at it on a finite time interval n. And eventually, n will be large, but finite. And what I am interested in is basically the highest value of this Brownian mode, this one or more. So basically, I would like to say something about, say, this guy. So that's the maximum. I will call it capital M index n. And that will be the quantity that I want to compute. So basically, what I really want to compute is this probability. So the distribution of this guy. So I want to compute, I will denote it by f of y and n. OK, this is not exactly the same f as before. I mean, I'm sorry. But this is a new chapter. This is not the first passage. So I want to compute this guy. I would like to say something about this. So the first thing that I want to show you is that we know almost already this guy. Because I can view this cumulative distribution as a survival probability. So there are two ways of seeing that. One is a bit via formulas. And another one is just by making some drawings. So let's do some, let's just first relation to survival probability. How do I do that? Well, I mean, let's do it maybe just. So what does it mean? f, this probability here? Well, the probability that the max is less than y is just the probability that all these guys are just below a certain value y. So that means that this probability here is just the probability that, say, x0 is less. So here, because I'm talking about continuous jump distribution, having Mn strictly or, I mean, that this inequality can be strict or not, maybe, to avoid any confusion. Let's take the strict inequality. It will probably be better like this. So I want to compute this probability, right? Agreed? So maybe I should even start at x1, x2, xn. And I started starting at x0, which, of course, has to be also smaller than y. Well, so now instead of, I will change slightly my variables. And instead of working with xk, I will work with the quantity zk, which is just y minus xk. You have the right to do that. So I take this random walk, and I just make a transformation. It has, of course, a pictorial representation. And what does it mean?  Now I started where? So basically, where did I start it? And I just started at z0 equal to y. But we know this guy, right? This guy is just a survival problem. So xk is a random walk. So zk is obviously also a random walk because it's just y minus xk. And so this probability here is just a survival probability. And that's the probability that I denoted before. This is just q of y. I start at y after n step. That's the survival probability. And that's the same notation as before. OK, so is that clear? Yeah, because here, I just, yeah, OK, sorry. Let's start at, yeah, OK. This is the case indeed. I started at x0 equals 0. Yeah, thank you. Well, there is, so I can't, OK, let maybe I just comment on that. I have not too much emphasize this. So the random walk you see is just translationally invariant. That means that if I want to start, if I start at x0, then everything will be globally shifted by amount of x0. So I can start everywhere I want. I mean, this is just a global translation. So it doesn't really matter where I start. So that's why I usually choose to have x0 equals 0. But OK. In general, I should have y minus x0. So that's my survival probability. Now, of course, this has a fairly simple pictorial way of, it's pretty simple to understand what happens like that. So in the drawing, let me just look at the trajectories that indeed contribute to this probability. So typically, that will be something like that, OK? So let's do it this way. So let's just want to illustrate this thing which I did. But let's do it. Let's draw the correct picture. So I want to look at the trajectories basically that stay below a certain level y, OK? So this is a typical trajectory that indeed contributes to that. And I wanted to compute the probability of such an event. Now I'm doing two things. First thing that I do is that I'm reverting this axis here. So this one, which is up, now will go down. So I just look at, just change the orientation. I'm just doing the same one or more. But I'm just orientating it like this. And I could even do it on the same, maybe because it will be probably more practical, easier. So what I do is that I just first change the orientation of the axis, OK? First thing. And since the germs are symmetric, this doesn't change anything. And then what I'm doing, what I will do is just to shift the origin. And I will shift the origin here, OK? So the origin which was 0 here, I change it. And now I change, I take this here, OK? So this is my new origin. And I have the right, of course, to do that, OK? So when I did that, of course, now this is 0. And this is y. Now you see, I mean, OK, you need to see a little bit on the other side. The probability that I said below this line, while it just translated into this probability that I start at y. But now I cannot cross this line here because this is the maximum. And the probability that you do not cross this line starting from y is precisely the survival probability, OK? That's just what I did here, OK? When you do that, you do exactly what I did, OK? You just reverse the axis. This is the minus sign. And you change the origin. This is this plus y, OK? So this kind of drawing here, I mean, will be this kind of construction is actually extremely common in this random work type of problems. And that's quite nice because that tells you immediately, again, that the cumulative distribution of the maximum is nothing else but the survival probability. And we have a nice formula for the survival probability. This is essentially this Polaxx-Pizza formula. So it's essentially it's done, I mean, in the sense that you have all the formulas at hand and then you need to analyze them and play with them to extract some information. OK, so is that clear to everyone? Yeah? So I want to show you a nice application of this identity here between the cumulative distribution of the maximum and the survival probability starting at y, which has to do with a search process, which is kind of a simplified version of the Shmoloshovsky problem, which is the following. So let me formulate it in a rather simple way. So it's a nice application of that, application of this duality, if you want. So the problem is the following. So basically, you imagine that you have a target which sits at the origin. So it's a one-dimensional problem. It can be a reactant, it can be anything. I mean, it's really a target that you are looking for. And to look for it, basically, you have a set of random walkers that are performing random walks and eventually will touch this target. And once it touches this target, well, the target dies if you want. So that's this problem where you have. So if I look at this, so I have some Brownian motion. So this is time. And I'm looking at some random walkers. So this is time t equals 0. So I have a set of random walkers. So they perform this kind of random walks. So of course, they can cross. They do these kind of things. So they start at some position, say, x1 initial state, x2, x3, x4. And at the origin, I have a target. And this target is immobile, so it doesn't move. So it has a fairly simple straight line in this thing. So that's my target. It's a target or it's a trap. And basically, the idea is that if one of these walkers touches this target, then you can imagine that there is some chemical reaction. For instance, that happens and the target just disappears. There are values interpretation of this thing. So you have this capital N. So this is the origin, 0. And you have capital N random walkers. And they are independent. And what I want to compute is basically, so at some point, the target will be found or will be touched by one of these walkers. And it will die. And of course, this will happen at some random time. And I would like to say something about the probability of this random time. So that means that I will just define PS of t, which is the survival probability of the target. So the probability that it has not been touched by one of the walker up to time t, survival probability of the target. Not of the walkers, but of the target. So I have this N random walkers. And at some point, one of them will touch the target and the target will die. And that will be this time that I want to compute or estimate. And in many cases, you have a large number of random walkers. And then it's typically is very large. And I will assume that initially at time 0, they are randomly uniformly distributed on a given segment of size L. So I will assume that initially at time t equals 0, the walkers are uniformly distributed. And some fixed interval 0, L, I want to consider the limits that I briefly mentioned the other day. When N is large, L is large also. But the density is finite. So L is finite. Sorry, N is large. L is large. So that's a kind of, but the density is fixed. So density 0 is fixed. I briefly mentioned this limit the other day. Yeah, that's true. But this will happen with the base. Essentially, yeah, that's true. In principle, it's possible. But it will happen with probability 0, right? Because they are uniformly distributed. So I mean, it's clear that no one of them will be exactly at 0, right? I mean, if you take a little. In that case, you would say that it immediately dies. So I want to compute that. And you will see that it's very nicely related to the maximum of some random moves. So why is it so? So let's write explicitly what this probability is, right? So the survival probability of the target up to time t. So essentially, you see that if I want to compute this PS of t, well, that's just the probability that none of the random workers have crossed the origin up to time t. So PS of t will be simply this quantity. So let me write it this way. It's just a product from i equal 1 to n of the survival probability, but not the survival probability of the random worker. Is that clear? So that's for a given realization of the x i's, OK? Simply, I mean, if I don't want them to cross, then they are independent. So this is simply this probability. And now I need to average over the initial condition, OK? So I will average over the initial condition. And the average that I do is simply that, OK, you see, I mean, they are just basically, they are just to conclude it, or so it here, OK? So I just average over the initial condition. And this is done with a uniform probability. So I agree that they are distributed over this. All of them are just uniformly distributed between 0 and l. And so that means that they are distributed, I mean, p of x i. OK, let's do it this way, if you want, I feel that you don't like this too much from 0 to l, p of x i, p of x i, OK? So that's the distribution of the p of x i. And p of x i, OK? So in principle, that should go even to plus infinity. And they are uniformly distributed, sorry. So p of x, maybe p is not, so p index x. So that's the initial position of the x. This basically is 1 over l, if x is in the interval 0l and is 0 otherwise, OK? Because they are uniformly distributed between 0 and l. OK, so that's this 1 over l comes in. So eventually, that's just that, right? So that's just, you see, they are all the same. They are independent and identically distributed. So this is, again, the product from i equal 1 to n of this 1 over l integral from 0 to l dx i q of x i t. But again, they are just all the same. So that's, again, is simply this component, OK? So that's just 1 over l integral from 0 to l dx q of x t to the power n. So now I want to take the larger and larger limit. And I will do a similar trick that we used before. You remember probably that to get this larger limit, we would like to have something of the form 1 minus alpha over n. So we just rewrite this in another way. We had already seen this before, right? I will write q under this form as 1 minus 1 minus q. I have the right to do that. So obviously, now if you do the integral over, so I will rewrite this. So if you do the integral dx over l from 0 to l, OK? Then this is just 1. And then here, this is just 3. And I just inject this formula here. So now this is just 1 minus, say, 1 over l that I displaced here. And I have this dx 1 minus q q of x t to the power n. So now I claim that I'm ready to take the larger limit simply because I'm looking at. So what is fixed here is n over l, OK? So I can write 1 over l as rho over n, OK? Yes. You mean here? Yeah, OK. So in here, no, it should not be there. So I am just doing the integral dx over l from 0 to l. So I get 1. And now I'm ready to take the larger limit because this 1 over l here is just rho over n. And rho is fixed. And n goes to infinity. So now I can take the larger limit and I obtain a very nice formula. Is that OK? Would better? OK, well, I guess you have understood what the model is. But I'm saying that now I can take the larger limit. And what you see is that this PS of t now, when l goes to infinity, is just exponential of minus rho times this integral, 0 to l. So now l goes to infinity, OK? So again, I'm taking this limit. So that's just dx 1 over q of xt. So now this guy here, I could have done that. Yeah, OK, I put a t. But OK, this can be a discrete time, right? I mean, I didn't do anything here. What I want to say here, how to show you, convince you, that this is just the average value of the maximum after time t of the maximum of 0 t. It can be discrete or continuous. I mean, that's the same. I mean, the reasoning, all the same. So is that clear? Yeah, it's clear because we have seen that, I mean, OK, I answered a bit quickly. But this q of xt, it's a survival probability. But we have also seen that this is, being a survival, this is actually the probability that the maximum m of t is less than x. OK, we have seen that before. I just showed you this argument. So now I claim that this is true. And this is just the result of an integration by path. So how does it work? Let me write what is m of t. Well, if I want to compute m of t, in principle, we need to do the integral from 0 to infinity dx of, say, x times the PDF of the maximum. The PDF of the maximum is just the derivative of this quantity. I write it, and I hope you will like it. So I claim that this is just that. So is it clear to you? Exactly, that's just the first one. Exactly, so that's the point. So q of xt is a PDF, is a CDF, cumulative distribution. So the PDF is just dq dx, OK? So dq dx is just the PDF, essentially the probability that m of t is equal to x up to dx. And then I just need, if I want to compute the average, I just need to compute the first moment of this distribution, OK? So now you see, I mean, if you do it by parts, if you do an integration by parts, you just immediately obtain that, right? So what I mean by that is that, so how do you do this by parts? So you will take u. So you will basically write that u is x, and u prime is just 1. And then you take dq, sorry, v prime is equal to dq dx. So in principle, that can be q. But in fact, q is not converging when x is large. So actually, you add a constant to it. You have right to do that. And if you do it by parts, you see that there is a first term, which is q times q of xt minus 1. And you want to evaluate it in between x equals 0 and x equals plus infinity. But this will be 0 at x equals 0. This would be 0 at x equal to infinity because of this guy. Then you are left with minus this guy, which is plus integral from 0 to infinity dx 1 minus qx. I mean, I'm doing it a bit quickly, but this should be related. Probably you have already seen this kind of identity, right? So this is 0. And this is precisely that. Is it clear? No, there is no. So this is 0. So you arrive at this very nice result that the survival probability is simply exponential of minus 4 times the average value of the maximum. So that's quite a nice application of this identity. Yes? What would be the result for IID random variables? So what do you mean by random variables? IID in this case? In the x-pronged order. Yeah, exactly, yes. In case of IID, I mean, here I didn't say anything about what the process is really, right? If you give me some, now, I mean, OK, while I use some fact, I mean, which probably is not, I mean, I cannot translate for IID, I already, I mean, the notion of survival probability doesn't really, it's not very clear. OK, one might be able to give a sense to it, but OK, it's not, I think it should be probably not very natural, I would say. I assumed already that I had a sort of stochastic process behind. OK, so now the question is, how do you get this, so this should give us some motivation to compute this maximum, right? So eventually what we have shown is that for large n, so question now is how do you get this m of t? So ps of t is exponential of minus rho m of t. OK, so just a remark here. t can be continuous or real, actually. Discrete, you see, I mean, here I didn't do anything. t can be discrete or continuous, right? So that means I can have, you can use this thing and for for bryan motion, for instance, if you think about it, for bryan motion, but t can be continuous. And in this case, we have already the result. I mean, OK, I've not derived it in detail, but we have the result. m of t will be proportional to square root of t, actually. We will do it in a more, I mean, different way. It can be continuous or discrete. So the question is, how do I get it? How can I compute that? And that's what you want to do now, OK? That's the next question. The question is, how can I get this guy? In principle, we have many tools to, all the tools to get it. And as I said, everything, in principle, is contained in this Polaxx-Pizza formula. So let's try to play a little bit with this formula. I will, I guess, not show you all the details, but at least I want to show you a bit how one can get something out of it. So let's try to see, I want to show you how this information is contained in the Polaxx-Pizza that we have seen. So that will be the first, I mean, that we will see how we can compute computation of this guy. So now I insist on having n here to insist on the fact that I have a discrete. So that means, in this case, t equal to n. And this will use the Polaxx-Pizza. OK, so it's a bit technical, but I want to show you, really, that some information can be extracted from this formula. So how does it work? I mean, it's relatively elementary, but one has to, one has to, yeah. So how do I do that? So you remember that the Polaxx-Pizza formula gave me information about this generating function, so this PS formula. So it gives me something about this guy. So this guy is actually the following, dx0, sum over n, s to the n, q of x0 of n, exponential of minus p x0. So I want to extract some information about the moments, the distribution. So q is a CDF, but I would like to say something about the moments. I mean, that's what I am after. So maybe I should say, so immediately, you see what you would like to have is an information about mn is what precisely in terms of this q. We have already seen it, but what I would like to have is an information about this quantity here. So that would be dxx dq dx. That's what I would like to compute, right? So you see that, again, in this formula, I will do some integration by parts to obtain a nicer expression. So I will consider the integral over x0, and I will do some integration by parts again. So I do it by parts, again, just to have it in a sort of nicer way. So this time, again, I know what I have to do. I'm going to take the derivative of this, q of x0, and that will be q prime of x0. And the other guy, well, I need to integrate it. Exponential minus p of x0, v is just minus 1 over p. Exponential minus p of x0. So now I'm saying that I can rewrite this integral dx0 qx0 minus px0. So what it is, it is simply the following. So this is just, yeah, why did I do that? Let me see. Yes, OK, no, that's fine. That's fine, so OK. So I just have a first term. Then it will be minus 1 over p. So I will get q of x0 and exponential minus p of x0. And I need to have it between 0 and plus infinity. And then I get what? I get plus 1 over p times the integral from 0 to infinity dx0 q prime of x0 exponential of minus p x0. So now I just look at the value at plus infinity. But when x0 goes to plus infinity, this goes to 1 and this goes to 0. And now here I have another term at 0. Still there is some non-trivial value. And this value is just 1 over p q of 0n. And then you get this, right? So you get this. So what is that? It's just 1 over p dx0. This you see, I mean, you can still write it in a slightly different way. Now q prime of x0 is the pdf. So that's just, basically, I could write it as exponential of minus p of mn. And that's a reasonable way of writing that. Is that OK? May have 1 over p in front. That tells me that now I have another way of writing the things. And then from it, I will be able to compute values and moments if I want. So how does it work? So it works in the following way. So I will have done that. So this guy, so politics pizza. So now I have to restore the sum here. So I will have the first term, which is 1 over p sum from n equal to 0 to plus infinity q of 0 n as to the n. But this guy we know, right? This is the Sparander's n. And then I get another term from which I will be able to get the rest of the information. And this is simply exponential of minus p mn. And then of course I take the sum from n equal to 0 to plus infinity now as to the n. That's fine. So now politics pizza tells me something about this whole thing, right? Because we know this has this form. So you see basically why I did that. Now by expanding in powers of p, you will get all the moments of mn. So if I start to expand this in powers of p, then you will see that I can generate all the moments. This is the generating function. So that's why I like it. Because this is just a sum from k equal to 0 to plus infinity of minus 1 to the k, p to the k divided by factorial k of mn to the k. So now we just have to identify on the right-hand side, I will have to identify the powers of p's. And I will be able to read the moments that I am after. In fact, here I'm just looking at the first moment. So that will be pretty simple. But eventually what you get is the following. So we know that this is equal to, from politics pizza, we have this. And we know that this is equal to 1 over p, 1 over square root of 1 minus s times this exponential. So that's, let me write it properly. This is exponential minus p over pi times this integral, dk of log 1 minus s p hat of k divided by p square plus k square dk. And that's why I already need a dk. So you see that the 1 over, that's my formula, nice. So now this guy I know, you remember, I mean that's just the Spar Anderson formula. And I can just simplify it by p. So now this is what I am after. So I will just write it in the following way. So I have the sum from n equal to 0 to plus infinity exponential minus p of mn s to the n. And then this is equal to what? So what I claim, this is what we have computed before. I showed you this very nice Spar Anderson formula. So it comes simply with the 1 over square root of 1 minus s. So you see that at the end of the day, while the formula is not that bad, it's just this exponential of minus p over pi.  So that's just the exponential and minus 1. So that's, of course, a little bit cumbersome. But what is nice is that now in principle I can get any moments that I want. So let me call this big function here phi of p and s. So what I claim is that I can get anything in terms of this. Yes? That's true. Yeah. So here, I mean, I just, so eventually I did, that's true. I mean, to derive the Spar Anderson formula from the Pollack-Tex-Pitzer, I did this trick where I rescaled the variable x0 and eventually take the limit p. That's true. But the final result was a result for this, right? The final result was simply a result for this series. So here, this is another problem. And I just used the formula for this for this quantity. There is no more p there, if you want. Is that fine? So the final really, I used this trick of large p. But now, OK, this p is another p, if you want. Eventually, the result of Spar Anderson is just this identity. And that's what I use here. Now you have this. And let me just finish with that. Now what I'm saying is that I have this identity. And for me, I can derive anything I want. In particular, I want the first moment. And I just, how do I get the first moment? Well, it will be relatively simple because you take this formula here, essentially. And what you do, what you will do is just to take derivative with respect to p. Let me take the derivative with respect to p and take p equal to 0. So if I consider this formula, I can just write, OK, let's do it explicitly, if you want. If you have never seen that. So how do I get the, so I have a nice, I have the generating function of this exponential. So what I'm saying is that I want to do the minus ddp of this formula, from some n equal to 0, exponential of minus p, ddp of that, OK, as to the n. Now if I do ddp, well, it's pretty simple, right? I mean, this is just an integral. So that will be just the sum from n equal to 0. And I will get, basically, mn exponential minus pmn as to the n, agreed? OK, I assume that I can take the derivative under the sums, et cetera, et cetera. But now, if I take p equal to 0, well, then you are done, right? Because if I evaluate this at p equal to 0, that's evaluating this at p equal to 0. And that's just the generating function of what you are after. So that's how it works. And if you would like the higher moments, then you would take higher order derivatives, OK? If I want to have, for instance, mn square, then I would simply need to take the second derivative with respect to p. And essentially, you can get anything you want this way. So that's what you need. And then, of course, due to a Polaxx-Pitzer, you know what this quantity is. OK, so from Polaxx-Pitzer, it's a bit complicated. I will probably not analyze it in detail, except if you ask me to do that. But then it's just minus d d rho of this function. Evaluated at, sorry, it's not a rho, it's a p phi of p s. Evaluated at p equal to 0. And this you can compute, OK? So that's how we use this Polaxx-Pitzer formula here to compute this expectation value. And what is nice, again, is that here you have a formula for any kind of jump distribution. So you can just do the full computation quite easily. Maybe one thing that I can already comment on is that, well, it's obvious that, OK, this formalism is relatively heavy. I mean, it's quite demanding in terms of analytical efforts to get explicit results. But it's also the signature of the fact that when you look at this extreme, I mean, at strongly correlated variables, at some point, of course, you have to pay it. And the price to pay is indeed to play with this pretty involved formalism. But still, I mean, here you see things are pretty explicit. And of course, playing with this formula is then a kind of art.  But at least I think it's important that you know, I mean, that this formula here, and they are quite useful. They can be quite useful in many contexts. OK, so probably last time I will end up this, OK, I will not, if you wish, I mean, I could do the explicit computation of all that. I think I would just give you the main steps. And then we'll go to, I will discuss then, the full distribution of the maximum, I mean, with mainly focusing, in fact, on the Brownian motion. And then, actually, I was thinking of the following. So initially, my plan was like this. My initial plan was to, OK, treat these extreme value statistics for the random walks. And then I wanted to go to discuss the record statistics. First for IID, and then for Brownian motion. And then I wanted at the end, I mean, to discuss some other examples where you can say something about the extreme statistics. So in some sense, coming back to extreme statistics for strongly correlated variables, and I wanted to discuss at the end these questions about RNT. But, OK, after thinking a bit about it, maybe I thought that after finishing discussing the discussions of the extreme statistics for random walks, maybe I could just discuss the random matrix at that moment. I mean, making a little break with Brownian motion and random walks, maybe that will also be a bit of destruction. And then, eventually, I will come back to record statistics, which is sort of new subject somehow, new subject. Partly disconnected to this. And so that would be my plan. So that means that I would like to shift maybe the random, some words about, I will not do a course on random matrix, of course. That would be just to show you some nice application of extreme value questions in RNT. So this is a quite limited subject. And I think it fits well in this discussion of extreme statistics of strongly correlated variables. And I thought maybe something that you would also find useful, I mean, for many of you, I guess, from the matrix.  So I thought it's fair enough or interesting enough to do it after these things. What do you think? Does it sound OK? OK, so I will do that then. So that means that next time I will finish to discuss a few things that we can extract from this Paul-Axex piece of formula. And then I will start a bit on RNT.