 Come and see either me and Matteo. We have suggestions for improvement. If you have complaints or things like that, go directly with the teaching assistant. So please remember that today there is a poster session. So those of you who have not yet displayed the poster, the poster boards are there. So we clearly see them. We have numbers. So I don't know whether you are pre-assigned a number or otherwise occupy abusively whatever space you like. But please display the poster and be there. Participate actively. This is a good opportunity for you. And then let's see. Enjoy the lecture. OK, thank you. OK, good morning to everybody. So today I will start with some preliminaries on random walks and random motions and levy flights and all these kind of things. The main motivation for that is the following. Basically, we have seen before that basically I've done in the previous lectures, I've done in details the extreme statistics for IID random variables. And I've shown you that we have a fairly complete description of it. And we can actually, we have a lot of tools. A lot of results are available there. And in particular, we have seen the existence of these three universality classes, Gumbel, Frechet, and Weibull. And then I ended my lecture yesterday by saying that these results for IID are actually relatively robust and can be extended to the case of frequently correlated random variables. I've mentioned two examples, first one at the rather heuristic level, which was the case where you have random variables which are exponentially correlated. And I did this kind of real space blocking arguments to convince you that one can still apply the extreme statistics for IID in that case. And then I gave you one of the results that is known in that context, which is the case when you have a Gaussian stationary process. And in that case, if the correlations decay faster than one overall again, then basically you are back to this case. But on the other hand, on the first lecture, I gave you some hints. And I gave you also some concrete examples where strong correlations are actually important, although I didn't describe any models in details. That means that instead of having basically correlations that are decaying, you might have in some cases growing correlations, or at least there are systems that are not described by this realm of independent random variables. And there are value systems which are quite interesting in this context. And basically, during the last years, people have tried to investigate specific systems where one can do some concrete computations of the extreme statistics. The main reason for that is the following is that as I showed you or told you, in this case, we have very powerful results that show you some different finite number of universality classes. Now on the other hand, when you leave this realm of IAD random variables, it's not very clear what these different universality classes are. It's not even clear that there is only a finite number of them. And the idea that people have pursued is basically to try to study some specific examples which are of course relevant for physical applications and for which you can actually obtain exact results, if you want. And among these families, actually, there is only a restricted number, unfortunately, of system for which one can say precise things. But there are two of them. The first one is the context of, I mean, are the models related to random walks? And this is relatively wide. I mean, this also concerns levy flights in general, and why you saw those stochastic processes. Of course, the continuum quantum part would be Brownian motions and the extensions. So that's one first big subject. And some of those stochastic processes actually belong to this list, for which you can obtain exact results. And there is another big subset of systems, if you want, for which exact results for extreme statistics have been obtained is the big subject or the big topic of random matrices. Big in the sense that there are actually many different matrix models and there are many various systems related to them for which people and related models, for which exact results became available relatively recently during the last 20 years, say. And that's, so this is not an exhaustive list. There are some other problems for which you can get exact results. But that's basically the two kinds of families of models that I would like to study. And in fact, I will mainly focus on this one because I think that application-wise, this is certainly quite interesting for many of you. And that's what I want to start to study right now. And then if I have time, and I hope I will have, I will then mention this, but that will be certainly, so that will be a first part, and that will be a second priority if I have time, OK? So since I want to talk about that, I thought that before really going to the extreme statistics of these problems, let me just, I thought it could be useful just to make some preliminary remarks, reminders about random walks, random motion, and levy flights. And let's start with that. So I will recall some well-known, but also some not so well-known properties. So random walks, I just want to give you clear ideas and definitions about what these probabilistic objects are. I'm sure you have seen them, of course. So that will be, I guess, the main topic of today. So let's start with some definitions so that we are sure that we are all talking the same language, at least when we talk about random walks. So what I will call, and what we usually call random walk, is if you want is the simplest discrete time Markov chain, which is the following. So you have some variable x, you start with x0 equal to 0, and then you evolve according to this simple Markovian rule here. So that's under this form here. And now this eta n's here, they are IID random variables. And they will be drawn from a PDF, which I will denote p of eta. Now here in this talk, I mean this talk in these lectures, I should say, I will mainly focus on the case where p of eta is symmetric. I mean, this is just to keep it as simple as possible. But many of the things that I will say do extend to non-symmetric distribution. But let's start because we will have a p of eta is equal to p of minus eta. So essentially the mean of eta n's is just 0. So if I look at the mean value of eta n, it's just 0. I don't say any specific things about the second moment of eta n. And that will just be more specific in the moment. But in principle, that's my definition for a random. So I will mainly focus here also on one-dimensional case. But again, many things can be done in higher d. But OK, one has to start with something. And let's start with the simplest case. So typically, I mean, that's the thing that will happen. So you start at, if you look at your random work, it will typically have this kind of shape. So we have a third job here. I might have a second job there. So after n steps, I will end up somewhere here. And so here I just represent x as a function of n. And I started at x0 equals 0. So there are values random works that can be studied. And let me just mention some few examples. So now you can see that the random work is then completely specified by this distribution here. So each type of distribution will, in principle, correspond to a different kind of random work. First example, and probably the most common one, is the case where these eta n's are just Gaussian random variables. So let's, for instance, just remind one could have this. That's one example. Another example could be, for instance, the exponential distribution. But now, compared to what we have studied before, the distribution is symmetric. So I need to consider this kind of exponential distribution. OK, so it has an exponential tail on both sides. Another case, of course, could be, because we have studied that already a bit before, which would be the case of a uniform distribution. So that could be, for instance, p of eta. OK, I don't want to write it because it's a bit cumbersome. But just draw it. That could be something like that. So 0 outside, and just 1 in the middle, but symmetric. So these are three nice examples. You might also, one case that I will probably discuss a bit less, but still it's present, and it's nice to have it in mind, is the discrete random work. So that's the case where, basically, eta n can just be plus 1 or minus 1. So this belongs to that, I mean, this is also described by this case. And this corresponds to the fact where you would have two direct deltas, 1 in minus 1, and 1 in plus 1. So that's a simple random work on z. But these are well-defined examples. Now, in all these cases, what do they have in common? They have in common that if you look at the moments, but the first moment is 0, but if you look at the second moment, say, eta square, the variance, then in all these cases, so in these cases, 1 to 4, sigma square is well-defined. So if you compute, basically, this guy, the second moment of this distribution is finite. This is quite obvious in all these cases. This integral is perfectly well-defined. And that means that if you look at the large n behavior of this random work, if you look at how it looks like when n will become large, what will happen is that in all these cases, you will observe a diffusive behavior. That means that if you look at the typical scale of xn after n time steps, xn will typically be proportional to square root of n. And in fact, we can even be more precise. It will be proportional to sigma square root of n when n is large. That's a fairly important, quite important remark here. So all these random works, if you want to some extent, if you look at them on the large time limit after a large number of steps, they will roughly look the same. Now let's consider another example here of random works defined by this Markov chain, which differs from all these ones by the fact that the distribution might have, so this is the last example, which will be quite important also in this context. Yes, sorry. Even if the mean is non-zero, yes, yes, yes, of course. OK, I mean in that case, so if you put, so for instance, suppose that the mean of the jumps is non-zero, so I don't know. Let's take this one, but suppose that it's centered to some finite value. So that effectively means that you have an external drift. So it's an external force. So basically, the equation of motion that you would have would be something like xn is equal to xn minus 1 plus c plus eta n. So c would be a finite number, and eta n again would be centered. But in this case, of course, to leading order, xn, again, in these examples, in this case, xn would be proportional to n if you have an external drift or if you have a non-zero mean. So it will be deterministic. The leading term would be deterministic. So that means that your worker would really just follow the drift. But the fluctuations around this drift will be, again, given by square root of n. That's true. Provided, of course, sigma is finite. OK, so let's consider another case, which is the case where p of eta has a fat tail. So if you look at the limit when eta goes to plus or minus infinity, it's symmetric. And let's consider the case where alpha is strictly smaller than 2. Then obviously, in that case, this second moment is not defined, right? Because if you compute, say, sigma square, and you start to do this guy, to do this integral, sorry, eta square times p of eta, then obviously you see that at large values of eta, this behaves like 1 over 1. So this thing here, for large values of eta, you see that it just behaves like eta to the power, basically. So to the power 1 minus alpha. And this obviously, when alpha is strictly smaller than 2, is non-integral. So that tells you that this integral here, so this is for when eta goes to plus infinity. And obviously, as soon as alpha is smaller than 2, this integral is just diverging. It just not defined. So this p of eta, in this case, because alpha. And that actually creates a large, I mean, a substantial difference compared to what happens here. And this type of random works here. They actually called levy flights. No, there is no, so in this case, you don't have the Brownian motion, but still there is a continuum limit. You can still define a continuum limit, but the system is quite different. I will probably comment on that. The trajectories of these levy flights is discontinuous. You have jumps, contrary to what happens for Brownian motion, for which the trajectory is indeed continuous. Levy flights have discontinuous trajectories. And I will probably comment a little bit on that in a while. But one important, at this stage, the thing that I want to say is that in that case, you don't have any more diffusive behavior. But if you look at the typical scale of xn when n is large, that will be of the order n to the power 1 over alpha. So that's quite different compared to the square root of n. And that's typically much larger than n to the 1 half, which is the diffusive behavior. So generally, these levy flights, they display what is called super diffusive behavior. It means that they go faster than square root of n. So now let's try indeed to see and to give a bit more precise definition of this guy and try to see what happens in the limit of long n. Yes. OK, so with this kind of model, you cannot get any, I mean, sub-diffusive behavior. One way to do it is to construct, but I guess I would not construct too much, discuss too much this thing. But what you can do is use this, what is called under the name of continuous, I mean, continuous time random walk. So what you are doing is that you perform a random walk, but between each step, you're waiting sometimes on random time. And this time will be distributed according to some power law. And depending on the power law that you have, you can get sub-diffusive behavior. So you do a first jump, then you wait some time. It can be short. It can be very long. It's also a random variable. And then you perform the jump. So this is called a subordinator. And this can yield a sub-diffusive behavior. I guess I will not touch too much upon that. But it's also interesting, of course. OK, so let's try to understand and describe a little bit better what happens in the continuum limit. And let's focus first on this first case. And these are, of course, they correspond to the Braynian limit. So let's just derive and be a little bit more precise about what it is. So let's consider this first two cases. And what I want to discuss is the case of what is called the Braynian limit. So what I mean by that, again, there is one thing to notice. You remember that before we were looking at the collection of random variables, which are IID. Now here we are considering something a bit different, which is xn, which is this guy. But you immediately see that xn now can be written as the sum of these eta i's. This is quite simple. But of course now, so these eta i's are IID random variables. But now you have a set. So these xn's here are themselves the sum of IID random variables. And obviously they are not independent. That's quite simple to see. I mean, let's just do this simple computation. Well, if you compute the average of xn, I mean, obviously this will be the sum of these eta i's. That's obviously 0. They are all 0, and I have a finite number of them. So it's obviously 0. But now what is more interesting is to compute, say that the two-point correlation. So let's consider this xn prime. And then you see that let's write it explicitly. So xn is just the sum from 1 to n of, and then you will have a second sum over j from 1 to n prime. And that will be eta i eta j. Now these numbers here themselves, these eta i's, are indeed uncorrelated. So if you compute the correlations, then this will be simply a delta ij. So this is 0 if i is different from j. And these are some value. And the value is precisely sigma square. This is what I call sigma square, the second moment. So now you see that you have this second sum too. You have this double sum, but you have here this delta ij. So suppose that n is larger than or greater or equal than n prime. So then you will see that basically this is just the sum from i equal to n to n prime. But this delta ij is just 1, but that reduces this double sum to a single sum. And so you simply have the sum from i equal to n prime of sigma square. And this is just sigma square times n prime. So that tells you that indeed this series is strongly correlated, because the correlations you see, I mean, they are not decaying at all. As a function of n, this is just a constant. And as a function of n prime, this is increasing. So we are far from the examples that you have treated below. Now how does a trajectory typically look like? Well, in this case, if you really look at what happens on the large-scale limit, well, you will see this typical Brianian type of motion. So again, I mean, maybe I should really stress here. That means that this extends and extends and cry. They are really strongly correlated, right? I didn't write it, because probably most of them will not see it. But that's the conclusion for now. So let's go back to that. Typically, what will happen if you really look at the large-scale limit, you start at 0, and you will have this kind of motion, right? I guess you have seen many times, OK? So the trajectories look continuous. And they typically look like that. They are quite, I mean, bit rough. I mean, the path itself looks a bit rough. But it is continuous. Now, how do you construct the Brianian motion limit? Well, what you do is that you define a function x of t, and you will define it this way. So this is x. So you have this capital X there, and you just look at them like this. So you look at it on a time interval, say 0n, OK? So this is just integer values of nt. So t is in between 0 and 1. And then you need to renormalize it, of course, by sigma square root of n. This is the subject. So now I choose to index. Basically, I had an interval which goes from 0 to n. And I prefer to work on an interval 0t, which is between 0 and 1. So I just have this random variable here. I divide it by sigma square root of n because I told you that the typical scale would be over the square root of n. Now, I will take the limit. So the claim is that this guy, in the limit when n goes to infinity, so maybe I should do the other way around, then this will converge to a nice object, which is b of t, and which is the Brownian motion, OK? Now, how it does converge here when n goes to infinity is, I leave this to mathematician. But of course, we can give a precise meaning to that convergence. This convergence is sometimes called what is called non-square theorem. I mean, if there are some mathematicians among you. Now, this b of t here, for us, is indeed this Brownian motion. What it means is that it satisfies a Langevin equation, which I will write in this way, which is b dot t, which is db dt, is equal to eta of t. It starts at 0, b of 0 is 1. And the eta that is here is the Gaussian white noise. So that means that eta as a function of t, if you look at it as a function of t, this would just be Gaussian random variables. And it has 0 mean. Maybe I will write that. So it has 0 mean. So eta of t is 0. And it has delta correlated noise. I will take your question. So this is, I use the same notation. It's a kind of continuous version of this guy. So I just want to use indeed the same notation, because you see that this guy, this disequation, I can again write it is equal to eta n. In the continuum limit, basically this xn minus xn minus 1 will be derivative with respect to n, which is a derivative with respect to t. And this eta of n will translate into this eta of t. So that's the continuum version of this random variable. I will maybe comment on this in a minute. So basically I will then say roughly speaking how you do implement this numerically. And somehow you have to come back to that kind of equation. I will be a little bit more explicit. Yes? Oh, yes. Yes, it's one because I did this normalization here. I divided by sigma square root of n. So if you don't do this sigma, if you don't divide by sigma, so that's one difference indeed between the eta and this eta of t and this eta. And this one has a variance sigma square. This one has variance 1. So basically they are related by 1 over sigma. You're right. Yeah, so that's the point indeed. So somehow that's really the main point is that no matter what you had for the distribution of the eta n's, you end up in the large n limit to this Brownian motion. So that's a crucial point. This convergence here is independent. So that's, of course, independently of p of eta. So yeah, of course, as soon as you have a sigma, with a, OK, but I am in this context here. Brownian limit sigma square finite. Yeah, so yeah, OK, there is one way to see it, which is relatively simple. If you look at xn here on this sum, you see that xn at a given time n is just the sum of IID random variables. This IID has a finite second moment. And that means that if you look at it as a finite n, it will be Gaussian. So the distribution of xn will be square root of n times a Gaussian. Now it's a bit less clear that it's true for the process itself. So that means that that holds. So if you really want to understand this at a finite time, this is simply the central limit theorem that everybody knows. Now if you really want to look at the process, that means, for instance, if you can now justify the fact that these two-point correlations really need converge to the correlations given by the Brownian motion and even higher-order correlations, then of course, this is a highly non-trivial theorem. And this is known under the name of the Donzka theorem. So that's a very important piece of probability theory, which goes beyond the central limit theorem. And of course, I will take it for granted. So that's the case of Brownian limit. Now let me say something a little bit more qualitative about the Levy flights. Right. First, for the Levy, so one remark is that this trajectory is here in the continuum limit. They are actually continuous. I mean, this is more or less written here. Now if you look at the Levy flights in the large n limit, you can also give continuous description of it. I will not do it because it's a little bit more involved. But let me just give you the picture. And in fact, from my point of view, when you have to work with Levy flights, it's usually much more convenient to work with a discrete time or no more limit and then take the large n limit when you can. But this is much more simple. So let just the case of Levy flights and just to point out the difference with this kind of trajectories when n goes to infinity. So you remember that in that case, you have a very fat tails in the distribution of the jumps. And as a matter of facts, what is happening is that for some time it looks very much like Brownian motion and then sometimes you will have some big jumps like that. So that's typically a Levy flight that you would get in the large n limit. So that's quite different. But again, so it admits representation in terms of, I mean, you see that it's a kind of composition of locally some Brownian motion. And then sometimes you have these jumps. But the full description is a bit involved. And my recommendation is that usually in these cases, it's much easier to work with the discrete random one. So that's kind of description of the process now. And these were mainly definitions. Now let's go to introduce some tools to describe these probabilistic objects. So a simple tool to describe these quantities and these objects is called the Green's function. Sometimes this is also called propagator. I want to come back to the discrete time process. And I want to define what I call the free Green's function, which I will denote by j of x, x0, n. So that's the probability density. So x0 is the starting point. So you are considering a trajectory, basically, that starts at x0 and that ends at x, roughly speaking. And so this guy, so this is a density. So now if you look at dx of that, then this is the probability that the worker is within the interval x, x plus dx, given that it started at x0. So this is probably the worker is in x plus dx after n steps, given that it started at x0 at t equals 0. So that's basically you ask this question. You have your random work. Now it can actually start anywhere, not necessarily at 0, if you want. So now I just chose this case, start at x0. And after some time n, it will be in x. And you ask the probability for such an event. Very important, very important object, obviously. Now this is a probability, so it's normalized. And it's normalized such that the integral over x is 1. That's important to understand that. That's always a good rule. I mean, when you are facing some probability, I mean, to ask yourself, how is it normalized? That's usually how and why is it well normalized. OK, so by definition here, the normalization is just such that you just have the integral for minus infinity to plus infinity dx of j x0 n is equal to 1 for all n and all x0. So that just tells you that basically if you start from x0 after step n, well, you have to end up with probability 1 somewhere. Nice. Now the question is, how do you compute that? So there is maybe another, so this is normalization, and there is also an initial condition that I want to stress. You know that at time t equals 0, time n equals 0, sorry. You are at x0, so that means that if I look at this quantity at n equals 0, then this should be a delta function around x0. So if you look at this object, n equals 0, this should be a delta function. At this stage, maybe it's useful to say that from now on, I will consider the case where the distribution of the jumps is continuous. So I exclude from now on the plus minus random walk. It can also be done. Usually there were some difficulties which are due again to this problem that I already mentioned, this kind of degeneracy. So I will just leave it from now. So everything now is for these continuous jumps. So there is no delta function in the densities. The densities are nice, nice, nice functions. Now the question is, how do I compute this j? So to compute this quantity, let me show you quite nice method which uses recursion relation. And that will be some kind of, that will be a method that I will repeatedly use in the following. So that's important to understand it right now. So the idea of constructing recursion relation here is possible thanks to the fact that I have a Markov process. That means that essentially if I look at this process here, suppose that I look at what happens right before it, then essentially this last step is completely independent of what happened before. That's the definition of the Markov chain. And it's also, by construction, it's really written in the equation of evolution, right? xn is equal to xn minus 1 plus eta n. But you don't need to know anything about xn minus 2, xn minus 3, or whatever. So the recursion relation that we are going to write exploits this fact by saying that I will decompose my trajectory first on these n minus 1 first steps and then see what happens during the last step. So in other words, I will say that the probability to arrive at x after n step, that's the probability to arrive at n minus 1 in, say, some value x prime after n minus 1 steps, and then doing a step, just a single step between x prime and x during the last step. So that's the idea of this recursion relation. And that's really just expressing in word if you want the Markov relation. So let's write it explicitly. Here, that simply means that g of xx naught n is something which is just, so as I said, maybe, let's do it this way. So I'm saying that I can start from x naught again and arriving anywhere at x prime in n minus 1 step. And then I will do some jump basically from x prime to x. So the jump will be x minus x prime. And then I need to integrate over all the possible x prime. So that's very simple. And I'm sure you have all seen, or most of you have already seen this kind of formula. And this is called, usually, the Kolmogorov equation. And here, I will actually call it the forward Kolmogorov equation. And that's actually quite important. You will see why, because there is another equation, which is called the backward Kolmogorov equation, which actually is much more useful for us. Yeah, so the only thing is that there, you will have a more complicated object. You will have the propagator, which is much more complicated. But yes, that basically is the same idea. That's true. And in fact, you can also write a path integral for this Brownian motion, of course. That's true. So this is the first observation. Now, the second observation is that mathematically, you see that it's a convolution. I can see it as a convolution, because I have a function of x prime times a function of x minus x prime. I just integrated over x prime. So I will exploit this structure to solve it exactly in a minute. So that's a convolution, because a convolution structure. So now, what I was saying is that there is this forward equation. It's nice, but there is also a backward equation, which in fact is much more powerful, which maybe you haven't seen too much. But the idea is basically the same. But instead of doing at what up, here you see, I mean, we have sort of investigated what happens during the last step. But we can do the same by focusing on what happens during the first step. So I will do basically the same. I will say that, OK, I can do exactly the same kind of argument by saying that the probability to arrive from x naught to x is basically the probability to make just the first step between x naught and x prime 0 and then propagate from x prime 0 to x. And then I will have to sum over this first point here. So that's very simple here. And at the moment, there is nothing special. But you will see in a minute that, or maybe not in a minute, but you will see a bit later, that so that's the same. Let's see whether we got it. So what I'm saying is that now I'm just doing first a step from x naught. OK, let's do the integral afterwards. So I just do a first step from x prime 0 to x 0. And that will happen with some probability p of x prime 0 minus x naught. And then I will propagate from x prime 0 to x. But now in n minus 1 steps. And then what I need to do is basically to integrate over the x prime 0. So they look superficially the same. But in fact, they are quite different. And you will see that later on, what will mainly be interested in are first passage problems. I will define this more precisely later. But I want already to emphasize that is that this approach is very useful when studying first passage. And you will see why when studying first passage. OK, now it's a bit too early to mention it. But I just advertise it. So remember that this backward Kolmogorov equation is actually extremely, extremely interesting. Yeah, exactly. Yeah, so in that case, we are just doing this thing. So I'm just looking at what happens during this first step here. So during this first step here, I will just make some jumps between x0 to x prime 0. OK, so I jump from x0 to x prime 0. So that's just the first step. So that's the probability of this first step. You agree? Oh, I see, sorry, sorry, sorry, sorry. Yeah, yeah, absolutely, yeah. Thank you. Yeah, you're all right. Well, here, since the jumps are symmetric, in fact, these are the same, but you are all right. So this is nice. Now, how do you solve these equations? Well, I already told you how you solve these equations. These are convolutions. So since these are convolutions, I will just use Fourier transform. So I want to compute these objects for any n and any jump distribution. So we want to solve these equations. And in this case, you can solve either of the two forward or backward equations. But you want to solve them. So this is the solution of the Kolmogorov equation via Fourier. So let me introduce my convention for Fourier transform. So I will do some Fourier with respect to x. So I will just write, I will introduce this quantity, j tilde of k, x naught n, which is just exponential a. OK, so that's dx, eikx, jx, x naught, and n. We know, right, that I have these two equations, either this one or that one. They are more at this stage there. For this purpose, there we look the same. So these are convolutions. So basically, if I look at the Fourier transform, this will transform into products of, so the convolution of the Fourier transform of a convolution is the product of the convolution, right? So the product of Fourier transform, excuse me. So let me introduce this. Let me be clear. Let me introduce the Fourier transform of the jumps, OK? So what I know is that if I Fourier transform this, I will just write it. I mean, you can work out the details a bit later. But I suppose that this should be clear for everyone. So if I do the Fourier transform of this or that equation, what I get is basically that j tilde of k, x naught n, is just equal to j tilde of k, x naught, n minus 1, times p tilde of k. That should be clear, right? So now I want to see this as a recursion as a function of n. So it's very simple, right? Because you can just solve it explicitly, right? If I just see this as a function of n only, forget about k and x naught, then you solve the form. a n is equal to a n minus 1 times b. That's pretty simple, right? I mean, this will be only the product of all this and this a n. So in other words, if I want to solve this, you will get it and you can check that this is true. This would be basically up to some constant. This would be p tilde k to the power n. You can show this recursively if you like. So now what I want to do, how do I fix a? I fix a by looking at its value at n equals 0, right? So a is actually k x naught at n equals 0 by definition. But now I need to remember this together with an initial condition. So what is j tilde? So let's compute it. By definition, j tilde of k x naught n equals 0 is just the Fourier transform of the initial condition. So that's minus infinity dx e i k x j of x x naught n equal to 0. And this guy is just a delta function, right? Because at x equal to 0, you start at x naught precisely. So this is just some delta function delta of x minus x naught. And so a eventually is very simple, right? This is just a k x naught. So now I have an explicit expression for this guy. And I can invert or invert this integral, OK? So eventually what you get is now I invert this. OK, let's finish this computation. Let's write it's finished, but let's write it explicitly. So this is k x naught n. There are various ways, of course, to obtain this. I'm not saying that this is the only one. I like it because this is a kind of method that we frequently used in these kind of equations. And so once you have that, OK, now you can invert this Fourier transform, OK? So and you immediately get it, OK? So this will be the integral dk over 2 pi for minus infinity plus infinity exponential of minus kx plus ik x naught then p tilde k to the power n. And p tilde k, I remember you, is just the Fourier transform of the jump distribution, OK? So you have an explicit expression. OK, this is fairly standard, right? I guess many of you have already seen that. If you have not seen it, I mean it's good to see it once. Now question is what happens in the large n limit, of course. So of course, one has to, what happens for large n? Now what happens for large n? You see, when n is very large, it depends a little bit on the value of p tilde k, right? Whether it's smaller or greater than 1. Now, obviously, this p tilde of k is smaller or equal than 1. So p tilde of k, to see that, let's first maybe look at p tilde of 0. Well, p tilde of 0, you see, is just the integral of p of x. Well, just 1, because it's normalized. Jump density is normalized, so you just get 1. And now, if you look at how does this p tilde of k looks like in general, well, you can show that it will be a decreasing function of k. And if you plot it, typically it will look like that as a function of k. So it starts at 1. And then it will decay as a function of k. So that tells you something. That tells you that this integral here, you see, I mean, basically, when k is very large, p tilde of k will be quite small. And if you take this small number to the power n, when n is large, then it will be extremely small. So that tells you that this integral here over k is essentially dominated by the small k values where it's close to 1, because as soon as k is too big, basically, this integral of this integral here will be 0. And so that tells you that this integral here is dominated by small values of k. So in other words, what you want to look at to extract the large n behavior of this Green's function is to look at how this p tilde of k looks like when k is small. Yeah, OK, I just told this to you. I didn't show you. OK, it's not so complicated, though I mean it requires a little bit of work to show that this is a decreasing function of k. But you can, for instance, I think that the best is just to, so for instance, if you take, it's just to look at examples, right? So if you take, for instance, you see that if p of x is a Gaussian, exponential minus k squared, then we know that the Fourier transform of a Gaussian is itself a Gaussian. So that will be p tilde of k. Sorry, if you take p of x exponential minus x squared, then its Fourier transform will be exponential minus k squared. So that will be something that decays, OK? And more generally, it will always look like that. I mean, it's a theorem. If you want to have a nice distribution, yes. So at the moment, everything is so now that's a good question. Up to now, this is exact for any types of jump distributions, OK? Thin or fat types. That's true. But what is different is how this behaves for small k. So the behavior now for small k of p tilde of k will be quite different if you are dealing with fat days or thin types. So now comes the difference, because you want to know the small k behavior of this p tilde of k. So namely, how does it behave here? So for this, well, you need to, I tell you how it is. And then so you need to know something about the small k behavior of p tilde of k. OK, maybe we can just say, let's see, maybe let's do some simple, simple analysis. So let's consider, first, the k square sigma square is finite. The tilde of k, by definition, is this guy. So let's do a small k expansion of it. It's a bit uncontrolled approximation, but still it's the most natural. So if you want to do a small k, well, you will just expand this exponential here. So let's expand it. So aikx is just 1 plus ikx minus half k square x square plus tata. I suppose that I can just integrate term by term. So first term simply gives me the integral of p of x. That's 1. That's what I said before. Now the first term you see is just ik times integral of xpx dx. And this is just the average value of eta. And that's just 0, because the first term is just the first value of eta. And then I have a second term, which is minus half k square times eta square plus tata. Can you see that? Can you see here, or it's already too low? It's a bit too low already, because it's important, so I should take care of it. I just wrote this simple expansion, but still. So what I claim is that this is just the expansion of the characteristic function. I mean, so it should not be that. You see that this guy is 0. And now I'm happy, because in this case I can do that. When sigma is finite, I can actually do that. And because eta square here is well defined. And this is just sigma square. But you immediately see that if I am dealing with fat days, now maybe I can answer more precisely your question. If I have fat days, obviously I cannot do that, because this coefficient here is infinite, so that tells you that basically this expanding here as I am doing here is not allowed there mathematically. Well, here sigma square is finite, so I'm quite happy. So that means that I can write that in just higher order terms. So now let's inject this in that. So now it's not strictly exact, but minus infinity. So I have dk over 2 pi. I have exponential of minus k x minus x naught. And then I get this quantity, 1 minus sigma square over 2 k square to the power n. And I'm saying again, well, I'm dominated by k very small. So I can replace that. I can replace this thing, because k is small, of course. I have in mind that, again, it's dominated for small k. And that's just, I can just replace it by exponential of minus n sigma square over 2 k square. So I'm just re-exponentiating this. I'm just using the same trick as we did last time. Just using that 1 minus alpha to the power n is exponential minus n alpha. So again, this is correct, because k is dominated. I mean, I'm dominated by k by small k. And more precisely, I'm actually dominated by k over the 1 over square root of n. That's why I am allowed to do that. And so you see immediately that now you are most done, because what you have to compute, so let's just write it now, is just the Fourier transform of a Gaussian. And you can, of course, evaluate it. And let's put it with all the correct prefactors. And then when you do this integral over k, again, you have to believe that you are dominated by small k values. I mean, there are more controlled way to do that, which is not that complicated. But I hope that you got the message here. So eventually, what you get is that this quantity here, just go to that. It's just 1 over square root of 2 pi n sigma square. Exponential of minus x minus x naught square divided by 2n sigma square. So that's eventually, so you see immediately that, because you have x square divided by n here, so that means that the typical scale is actually square root of n. So that immediately tells you that x minus x naught is of the order of sigma square root of n. So that's what I told you before. This is the Brownian scaling limit. That's contained in this equation. Because you immediately see that, essentially, for x minus x naught much larger than square root of n, then basically, due to this exponential, this probability will be 0. So all the weight is concentrated around x minus x naught divided by square root of n. So if you start at x naught, then basically, you will end up at x, which is at the distance of the order sigma square root of n from x naught. Is that clear? There is, of course, a finite probability that you end up anywhere else, but this probability is extremely small, and n is large. So what is nice here, again, you see that only the second moment actually matters. It's the same kind of calculation that you have done when you show the central limit theorem. I mean, that's one way at least to show the central limit theorem. And you immediately see that only the second moment actually holds. Now, you see that this Gaussian here, it appears precisely because I have the right to do that here. Now, what about the case of Levi-Fleitz? So what about the case where sigma square is infinite? So there, that's the case of Fattais or Levi-Fleitz, and that's OK. Now, it turns out that if you look at what happens close to k equals 0, then what is happening is that, for instance, so that corresponds to p of eta. So suppose that you have this case like that for the jumps. So suppose that the jumps have a parallel case like this with alpha much smaller than 2. Then if you look at the Fourier transform of this guy, there is a theorem, which is known under the name of the Tauberian theorem, which is one of the Tauberian theorem. Then p tilde of k basically behaves like that. So that's now quite different. It will behave like this. That can take it as a theorem. If you have a large eta behavior like this, then you will get p tilde of k, which is of that form. And a actually can be computed in terms of c and some numbers, which involves some gamma functions. But that's the point. And you see that only when alpha is equal to 2, you recover the exponential minus k square. But otherwise, you have a non-analytic behavior. And that's very important. That's very important because now you will insert this behavior there. And instead of having a k square here, you will get a k to the power of alpha. And that will give you another function. And this function is an alpha stable distribution. So let's do it quickly here. So you will get that g of x, x naught, is just of that form. So I'm just doing the same thing. I'm re-exponentiating the whole thing. And I get exponential minus a n a k to the alpha. If it goes too fast, please stop me. I mean, the idea is not to. So if there is something that is not clear, please tell me. Is that OK? So now this form here, you can actually put it in a nicer scaling form. If you really want to write it in this way. So you can still write it in this way as this kind of scaling form, phi of alpha of x minus x naught n to the power of alpha. So we're just rewriting, change of variable, and rewriting a little bit of things. So do I do that? I mean, I just do the change of variable n k to the power of alpha is equal. n a k to the power of alpha is just q to the power of alpha. And that will immediately introduce this scale there. There is a here. So this a is given here. And this a actually depends on the coefficient, essentially, that you have here. Now what is this phi alpha? So phi alpha is a bit different. But it has a nice four-year representation. So basically, its four-year transform is very simple. Its four-year transform is just exponential minus k alpha. But let's write it. So phi alpha of x, that's this expression. dk over 2 pi exponential of minus k to the alpha minus i kx. So that's exactly, I mean, it's just written there, right? This is just this function. Once you get rid of the n, the a, and you put x minus x not as x, then that's this function that is written there. Agreed? So this function is called an alpha stable distribution. And it's quite different from a Gaussian. Of course, for alpha equal 2, you recover a Gaussian. This is just a Gaussian. That's what we studied before, right? If I set k equal 2, I am back to this formula. And we know that the four-year transform of a Gaussian is a Gaussian. But in general, for alpha much smaller than 2, this is quite different. And in particular, it also has a parallel state. So if you look at phi alpha of x as a function of x when, so it's symmetric, of course, when x goes to plus infinity, then this guy actually behaves like x to the power 1 plus l. So it has itself a parallel state. So that means that if you take a sum of random variables with the fat tails, then basically the sum itself has also a fat tail with the same exponent. And that's due to the fact that I was mentioning yesterday. That's due to the fact that if you look at this sum of large fat tails random variables, then basically the sum will be dominated by one or two guys. And this one or two guys, of course, has this parallel behavior. And that's what you get here. Yeah, basically, yes. Essentially, yes. For any alpha, there is no closed form expression. There are series expansions. Maybe I can just give you the case. There is one case, which is nice, which is the case alpha equal to 1, because we have studied it before. For alpha equal to 1, this is the Cauchy distribution. For alpha equal to 1, you can compute this, obviously. Alpha equal to 1, you can do this integral. And what you would get is that phi 1 of x is just 1 over pi times 1 plus x squared. So that's the Cauchy. I hope you can still read it. This is Cauchy, right? So I don't want to rewrite it if you don't mind. It's just 1 over pi times 1 over 1 plus x squared. It's Cauchy. That's a good question. So stable because why are they stable? Stable because basically, if you take two Gaussians, so if you take two Gaussians, in fact, even with different variances, but if you sum them, the sum is, again, a Gaussian. And that's why actually this appears as a limiting form in the central limit theorem. But there are other functions, other distributions, that satisfy this property. And these other functions are the alpha stables. So if you take two alpha stables, they are again stables. And that's the reason why they are limiting the exactly. They are exactly fixed points of this transform. In fact, you can even really formalize it in a more precise way in terms of fixed point. The same is true also for the extreme statistics, actually. There is a series of recent papers by Eric Bertin, Zoltan Rax, and five years ago, where they do this. Yeah, that's true. OK, I will not comment too much on that, but yes. Yeah, so yes, it could be interesting. I don't know, maybe I can write. Yes, indeed. Maybe I can comment on that a little bit later. In a minute, I will actually come back to that. But yes, that's true. I'm not very comfortable with this fractional compute calculus, but indeed, I mean, they play a role in that. So maybe I will comment on that, because what I want to study is basically I want to come back to this Boolean limit and come back to that. So that's all what I wanted to say at the moment about the limiting forms of the propagators. Later on, I will construct some more complicated objects. But at the moment, this is the only thing that I want to stress. Now you see what I did here. I started with the discrete equation. Let's do the sigma finite. What I did is that I looked at the discrete Kolmogorov equation for J. And I solved it exactly for finite n. And then I obtained the large n limit and obtained this Gaussian behavior. So now I want to present you an alternative description, which is also interesting, which amounts to look at the forward or backward equation, so the Kolmogorov equation, and look at the continuum limit of this equation that we had. Remember that we had this recursion relation. And what I want to show you is that this recursion relation actually turns out to the standard diffusion equation when you take a proper diffusion limit. So let's do that now. And maybe at the end of this, I can come back to your question about fractional calculus. So what I want to do is an alternative derivation of this result, if you want, at least for the Brownian case, by focusing on the Brownian limit and recovering the diffusion equation. Since this morning, actually, I talked a lot about diffusion, but we haven't seen any diffusion equation. But of course, it's there. And you know that this Gaussian here is a solution of a diffusion equation, so I want to show you where this diffusion equation is. So let's do that. I just want to discuss the Brownian limit and the diffusion equation. So maybe it's nice to come back to start directly, actually, with the Brownian somehow. It's probably useful to do that. Let's start with the diffusion equation that I showed you before. So I have v dot of t is zeta of t. Oh, sorry, I called it eta. So let's stick to this notation, eta. Well, eta is a Gaussian. Eta of t, sorry, is Gaussian white noise. And I would like to write the backward or forward, if you want. Yeah, let's focus on the forward. I would like to write the forward equation associated to that. But of course, if I want to do this, I have to discretize my process. So I will just imagine that I have a nice discretized version of it. So first, I need to recover what is that, to recall you what this is. So this is just 0. I mean, 0 mean and delta correlated. But of course, one has to give a meaning to this quantity when you do some discretization. So I want to have a discretized of this equation. And in other words, what I will write is that this b dot of t will just be b of t plus delta t minus. So I just introduce a discretized time if you want delta t. And I just want to use a b t plus delta t minus b of t. And this will just be delta t times my function. OK? So now what I need to do, of course, I need to tell you what this quantity becomes. Now, because in this discretized version, while this actually is well-defined, and it has to be regularized. So in other words, I need to regularize in some way this delta t minus t prime. And the correct way to do it is simply to say that this eta square of t is just 2d divided by delta t. OK, so that's the discretized version of the delta function. And that's obviously, I mean, dimensionally is correct. Right? The delta of t has a dimension of inverse of time. So that's the only thing that I can imagine. And that's the way I discretize this Brownian motion version. OK? So now I want to write the backward Fokker Planck equation or forward Kolmogorov. I told you, yeah, it's better to call it Kolmogorov, in fact. It's almost, it's not a Fokker Planck. So I want to write the Kolmogorov equation for this discretized version, OK? And then take the limit delta t goes to 0. So equation for the discretized Brownian motion. So I will just write the same equation that I wrote before, right? So in other words, I will just write the following. So now it works like that. So I will write g of x that I defined before. And I just look at what happens between t and t plus delta t. Either side. So doing delta t, I will have one jump, OK? Because the jumps happens within a certain time interval delta t. And so that's how it reads, right? So it's basically, I will just write it and comment, j of, I will have x minus. So here you see that you have eta delta t. And then I have p of eta. And I have to sum over eta. You're fine with that? So I want to take the limit delta t goes to 0. So I will just, Taylor, expand this guy here. I'm sure you have seen that many times. Why just, Taylor, expand this guy for delta t goes to 0, OK? So first term is just g of x. So let me write. OK, let's do it this way. First term is g of x. Now I have a second term, which is minus half eta times delta t times the first derivative of j with respect to x. Yes, oh, sorry. Yeah, yeah, I mean, no, no, it's a good. It's just because I was a bit sloppy. No, they are there, of course. So if you don't mind, here I will not write them because it's a bit boring, but they are here, OK? So I don't write, or maybe I can just write j like this. And you have to remind that j, in fact, stands for this guy. So here, sorry. In this line, I use j for j of x, x naught and t. So this is the first term. And then I have a second term. So second term of the Taylor expansion will give me eta square delta t square. And now I get the second derivative, OK? OK, I claim that I can stop here because this will generate higher orders of delta t here. And then I have p of eta d eta. So now let's try to, so I have three terms. But you see that j does not depend on eta. dj dx does not depend on eta. It's the same for the second derivative. So eventually, what I get is relatively simple. I get that g of x, x naught t plus delta t is OK. So I have this first term here, which is just j times the integral over eta of p of eta, which is just 1. Now I have a second term here. Let me write it. This is minus half delta t times j times the first moment, the eta, eta, the minus half, there is no minus half. Thank you. OK, so that's the second term. Yes, I'm getting tired now. Yes, OK? And then the third term, which is here, which is just, now let me take care, 1 half delta t square, second derivative of j. Now I have an integral, which is minus infinity to plus infinity eta square p of eta. So now you see what happens. This one is 1. This is fine. This one is 0 because the mean value of eta is 0. So this term disappears. And I get only the second term. So this term, I mean, superficially is over the delta square, but in fact, you have to remember that this guy is 2d over delta t, so that this term is actually over the delta t side. Now you can also get that the higher order terms, of course, will be of higher order of t. And then eventually you get the equation in discrete time when delta t goes to 0, and we are almost there. So eventually you get this equation that you have j of x, x naught t plus delta t, which is just j of x, x naught t plus 2. So this is just d times delta t times the second derivative. So now when you take the limit when delta t goes to 0, obviously you get your equation. The equation that you get is nothing else, but indeed the equation, the diffusion equation, as expected, as I told you and as you probably were expecting it. This diffusion equation, of course, we know how to solve it because there is an additional initial condition that we have, which is that t equals 0. This has to be a delta function. And if you solve it as you can imagine, you will basically obtain that j of x, x naught d is of this format. It is 1 over 4 pi dt exponential minus x minus x minus x square divided by 4 dt. OK, so that's the answer that you get from this direct continuum approach of the diffusion equation. And that's also, of course, what we got before. As expected, the two approaches are coinciding. And we are quite happy with that. So maybe just one comment to your question. So if you want to do something like that, but I will be brief, if you want to do something like that for Levy flights, I mean, of course, you are facing one problem, which is that eta square will not be defined, so you are not allowed to do that. And in fact, what happens is that you can show that the correct way to do this expansion is, indeed, to introduce a functional derivative of j with respect to x. And that's how it comes. And indeed, essentially, what happens is that you will not have the second derivative here, but the fractional derivative of order alpha on that side. And now you immediately see that in the standard way to solve this equation is to look at Fourier transform. And in Fourier transform, this second derivative here translates to a q square. While if you have the fractional derivative of order alpha, that will be modulus of q to the power alpha. And that gives you the alpha stable. In general, I think it's, at least that's my point of view, it's probably quite difficult to do. It's one way to do, but I usually, as I said, I think that when you have to deal with Levy flights, it's really much nicer to work with in discrete time and look at the larger limit. That's why I showed this first approach. Well, the thing is that in the larger limit, then you really have a good analysis. I mean, you have good tools to analyze what happens in the larger limit, which are basically standard tools of analysis. While here, if you want to start with these fractional operators, it's a mess. I mean, for my point of view, it's very hard not to say, I mean, to say precise things and correct things when you start to deal with this stuff. So this is a recommendation. If you have Levy flights, I think it's better to do it with discrete time. OK, so in the few minutes that are left, I want to introduce to the subject, I mean, to the last part concerning these Brownian and Wandermox and Brownian motions and Levy flights, which has to do with survival and first passage. So I will just probably just give some definitions today. And then the main results, I guess, will come tomorrow, except if I'm very fast. But that's not the idea. OK, so I hope that I gave you a good panorama of what a good overview say of this duality between discrete and continuous, in the case of Wandermox with a finite sigma. And I hope I clarified some ideas about these different objects, which are Wandermox, Levy flights, and Brownian motions. So now for all these stochastic processes, there is a central question, which is relevant for extreme statistics, but which is actually also relevant in many other domains, which has to do with the notion of survival probability, what is usually called first passage problem. So let's try to see what it is. And the first object to define is basically, so up to now we looked at these Green's function, but this was a free Green's function. So I had this random walk. It started at x0, it arrived at x, and I asked what's the probability for such an event, but the random walk could go anywhere in between. Now I'm constraining my random walk, and the way I constrain it is the following. So I just look at this kind of trajectories. So it starts, say, at x0. And I want to ask what's the probability that it arrives, say, at x after n steps. But now I'm putting some constraint. I'm saying that I don't want the random walk to be negative. So that means that I don't want it to cross this axis. That means that I will only consider this kind of random walk like that. So it has to stay positive. So in other words, I don't want to count this kind of trajectories. So suppose that I would have trajectories of that kind. It could be like this. This one I don't want. For the free propagator, that would be OK. But this one is not allowed. Because I'm saying, essentially, I mean, can you imagine that your random walk, as soon as it arrives here, it gets absorbed, and it sort of dies. So it will never reach x. So this one is not allowed. And that means what? So mathematically, the object that I want to construct, maybe let's write it here. So I will write this Green's function as g plus of x, x0, n, which is basically the probability that, again, the walker is in x. So this times dx, sorry, after n steps. So that was like before. But now I want, in addition, that x1 is strictly positive, x2 is strictly positive, et cetera. xn minus 1 is strictly positive, given that I started, well, x, of course, has to be positive also, but given that I started at x0. So that's a quite complicated object if you think about it a little bit. And computing it directly is very hard. Now it turns out that, again, this approach that I showed before, this approach of writing a recursion relation forward and backward equation are quite useful. So one can write this recursion relation. Of course, I mean, these are the same as before. I mean, the idea, at least, are the same. I can exploit the Markov property of the chain of the random walk, if you want, here. And, again, I can write, if I just specify what happens during the last step or during the first step, I will obtain two kinds of equations. And these are this forward and backward equation. So let's write them. I will have the forward. So I will have g plus. Sorry, I prefer to write with g plus like that, because my notes are with that. So I have the forward one, which is as we wrote. So basically, that's the same. So here I arrive in x prime and x naught. And then I will do the last jump between x and x prime. And then I can integrate over x prime. But of course, now x prime has to be positive, because this intermediate point here cannot be negative, because otherwise my walk is dying. So that's a difference, because it's not really a convolution anymore, because here I don't have the integral from minus infinity to plus infinity. So somehow I'm breaking space translation. And if I break space translation, I cannot use any more Fourier transform. And that's really a PT, because things become much more difficult. So this is the forward, but you will see that here the backward is much nicer. And you will see why in a minute. So I can do the same, but for the backward. And that's just g plus of x, x prime 0 in n minus 1. And then I just make the jump between x prime 0 and x 0. But now I integrate over x prime 0 by the same constraint force. OK, so I'm writing it a bit fast, but this is exactly what we wrote before. I mean, the only difference, again, is that the intermediate points, which we are integrating over here, needs to be positive, because our walk is positive. So that's really, this is the important point to notice. It's not minus infinity. Now, these are the Green's function. Now there is actually another quantity, which will be very important in the following, which is called the survival probability. So what is the survival probability? Well, the survival probability starting from x 0 up to step n is basically the probability that starting from x 0, you stay in this half plane up to step n. So that means that what's the probability that starting from x 0? You have not crossed the origin up to time t. So that's the definition of the survival probability. And that will be our central object in the following. So I will denote it by q of x 0 n. And that's really a probability in this case. This is sometimes called persistence probability. This is a true probability in the sense that this is just the probability that x 1 is strictly positive. x 2 is strictly positive. x n is strictly positive given that you started at x 0, which has to be positive. What's the relation with this guy there, with g plus? Well, you see, if you want to compute this probability, you just need to look at the Green's function. That means the probability that you arrive at x. And then you just need to integrate these probabilities over x for any x positive. So the statement is that I look at all the probability that drives me from x 0 to x. So this is this probability that I arrived within x and x plus dx. And then I just integrate over all the possible values of x. So that will be really a central object in the following. And I will just end up with the following remark, which is actually quite important, is that if you look at one of these two equations, there is one which immediately gives you a nice equation for q. So indeed, I mean, if you look at this backward equation, if you just integrate over x, so on the left-hand side, you would obtain simply q of x 0 n. Now, if you look at this side, well, you can also integrate over x because x here is just a dummy variable here. Completely x doesn't only appear on g plus here. So that means that if you integrate over x, then that's just q of x prime 0 n minus 1. Of course, this remark obviously does not hold in this case, because if you integrate over x, well, you integrate over this guy, and you don't know what you are doing. So the nice thing is that if you integrate by integration of the backward equation, and that's why it's so important here, integrating the backward equation. So again, what you obtain is that q of x naught n itself satisfies a backward conmigrant equation. So that means simply here x 0 prime of q x prime 0 n minus 1 p of x prime 0 minus x naught. So that's quite important. So that will be our starting point for last time. And I will show you basically that I will not comment on all the details, of course. But it turns out that essentially this equation can be solved for any p. The solution is a bit convoluted, but there exists a fairly almost explicit expression for q in terms of p. So that will be the subject of last time. Let me just comment on one thing, maybe here, on the boundary conditions. Because of course, the boundary conditions are a bit different for g plus and q. And I just want to mention them to be complete, and then I will be done. So there are various boundary conditions for q. The first one is that at n equals 0, and if you start at x naught here after n equals 0, well, the probability to survive has to be 1, right? I mean, it didn't move. So that means that for n equals 0, the q of x naught has to be 1. That's boundary and initial conditions. The first one, indeed, as I said, is that at step 0, your particle starts, and it has obviously to be 1 for all x naught positive or 0. That's a small caveat. I mean, small, not a caveat, but something that I will maybe discuss, should be subtle. And there is another boundary condition, just let you think about it. But obviously, I mean, if I start from infinity, so that means that x naught, if x naught goes to infinity, then obviously, I mean, I have a probability 1 to survive for any n. It sounds like a bit peculiar boundary condition. So these are my boundary conditions, and we will see how we can solve these things later. Yeah? Well, I mean, I don't know what happens for n goes to infinity. You will see it's quite non-trivial. Maybe since I mentioned first passage, let me just. No, OK, maybe it's too much. OK, I will stop here.