 Vse je to, da je tren. Tren je to, da je inšljena ta čest, da je inšljena. No, ne vse. Ne ne. Vse je to, da je inšljena, da je inšljena. Zelo, da je inšljena. To je inšljena. Vse je to, da je inšljena. Tren je to, da je inšljena. Ne, ne, ne. Zato ne lahko se vse izgleda. the students who knows the building of CSAR very well, but probably they've don't such a detailed knowledge about what happens inside, so also PCS students are strongly encouraged to participate because as said we'll give you presentation about scientific activities, ... Which you bude. And also some vese for the labs. Which I think you have never been with, because they are still in the process. And let's see. Thank you. Good morning to everybody. OK, so up to now essentially if I should make a short summary of what we have been doing, I've mainly discussed the questions of extreme statistics or order statistics. That means basically you look at the collection of random variables and you look at the maximum among them. And maybe you can also look at the second maximum, the case maximum. So I first exposed to you the theory of extreme statistics for IID random variables. This was the subject of the three, four first lectures. And then I moved to the case of strongly correlated random variables. And as I told you, I mean there, the situation is much more complicated. And in fact, there are very few systems or classes of systems for which we can say something precise about the statistics of this extreme in such strongly correlated cases. And I treated in details basically two big classes of systems. One is the case of random walks and random motions. And that was what I treated during two lectures and a half, almost. And then the central piece somehow was this Polaksek's Pizza Formula and eventually this Sparanderson theorem, which were eventually the main tools. That means, I mean, most of the questions related to extremes, at the end, boiled down to some questions related to first passage probabilities. And then I discussed the case of extreme statistics of another set of strongly correlated random variables, which is a case of random matrices. I was, of course, a bit faster. But I tried to tell you and show you some natural and nice applications of extreme value questions in that context. And I had sort of as a guideline. I mean, I used this model by May, this ecological model, which discusses the stability with respect to the strengths of interactions. And eventually, yesterday, I just ended up the lectures by mentioning some nice applications of extreme value statistics of random matrices beyond the random matrix theory. And in particular, I mentioned a few applications of this traceridum distribution. So that was the core of these first lectures. They essentially was dealing with extreme value statistics. And during the last two lectures, I want to touch upon another subject, which, of course, is related to extreme value statistics, but which raises new questions, as we will see, and which is the topic of record statistics. So essentially, I mean, the outline will be as follows. I mean, I can just maybe tell you what I plan to do. OK, so today, I would like to give you first a short introduction and some applications, although I will be relatively brief. And then I will, again, use the same kind of structure that I used before. So I will first discuss the case of IID random variables. And you will see that the theory of record statistics is, of course, well-established, but is also quite rich. So that will be essentially what I plan to cover today. And tomorrow, if, at least, that's my plan, I would like to cover the case of record statistics for random variables, because we have all the material to do that now, in particular with the Sparanderson theorem. And you will see that it's actually quite OK. I would like to show you this, because now that we have the tools, this is a very nice playground to sort of apply these tools. OK, so maybe before I can just give you, I will resend it to you, but OK, I just already mentioned it to you. Probably I can give you two relatively recent references. I mean, the one is, OK, this is kind of self-advertising, but we recently wrote a complete review on that with my colleague Claude Godrech from Sackler, Satyama Jundar. And you can find it on the archive. So it actually discusses basically all that. And as you can imagine, I mean, the content of, OK, of course, the review contains much more than what I will discuss, but 0, 0, 5, 8, 6. So this is one. And the other one is actually a short review, I mean, also a nice review by Gregor Vergen, who was a student in Cologne, actually. But we eventually wrote one nice review, which you can find. It's actually on the archive also, but I don't have the, it's in GFISA, 46, 223. Oh, now I realized I forgot something, that you guys cannot see what happens here. OK, so what's written here? I'm sorry, OK. OK, so if you just type, I mean, Gregor Vergen on the archive, he has a very nice review on record statistics and applications. Now I should remember that I cannot cross this line more or less, so I'm sorry. OK, I will send you, again, these references, but in case I forget. OK, so what's the subject about, and what are we talking about here? So let's formulate the problem in a quite simple way. So I will discuss these questions in the context of discrete time series, so I want to think about the collection of random variables, say, x1, x2, xn. And I would like to think about this index, i here, xi, as a time index. And so you look at a given time series. I mean, OK, this is something that we have been looking at quite in detail, but OK, this is not the first time that you would see such a plot. So I would like to plot xi, sorry, as a function of i. And I would like, I will have this kind of data here, OK? So this is, say, i equal 1. Then here I would have a second point here. So these are my values. I just plot them like this. I could have this value here. Then I will have another value there. And suppose that we have another value here. And I will go down with another value. So these are a set of values, OK, so that this is relatively, this is clear enough, OK? So you can also view it as a time, again, a time process like this, if you like. So this is your data. These are your data, sorry. And basically you define a record, say, here. So we are on my records here. So pictorially, so I will consider that the first value is always a record. Now, the record is established at step k, if the value xk is larger or greater than the previous values from x1 to xk minus 1, OK? So this one is obviously a record. This one is not a record. This one is not. This one is another record. This one is not. This one is a record. This one is a record. This one is not. This is not. This one is not a record, OK? So in blue here, I just look at the record. So typically, if I look at step k, xk or record happen at step k, let's write it explicitly here, it's a definition, if and only if xk is greater than x1, xk minus 1, OK? So as maybe a remark already, exactly as I did for the case of extreme value statistics, I will consider the case where the xis are continuous variables, such that there are no ties, there are no degeneracies. So with probability 0, two values here will be the same. I have in mind that there will be continuous random variables. Maybe just as a remark, so here, of course, what I define here, they are sometimes called upper records. So these are. But I could similarly define lower records, which corresponds to basically if xk is smaller than the minimum. OK, that's OK. But to simplify a bit the presentation, I will mostly stick here to the upper record. And in fact, I will just forget about this upper, and I will talk about records that will mean upper records in the following. So there are, of course, what are the questions that you may ask here, and what are the interesting questions. So there are the main questions, say, while there is a first question somehow, which is whether natural, which is, for instance, if you look at your sequence up to a given step n, and if you ask, OK, what's the value of the record on that current record, then obviously, the value of this record here is clearly the value of the maximum among this set of random variables. So in other words, if you look at the records up to step k, and if you ask what is the value of the current record at step k, then this is basically as asking about the maximum of this sequence up to step k. OK, so if you just look at the values of the record, I mean, which, of course, is an interesting question, well, but this question is essentially a question of extreme value statistics, which we have discussed before. So that's back to this. If I stick to this question, then this is just what we discussed before. So I would say that, OK, this guy we know. But, of course, this sequence of records actually raises another set, various sets of other questions, which are quite interesting, and that we will treat here. The first one, which is, of course, not treated by, not included in this, is how many records are there. So if you look at the sequence up to n here, how many such group points are there? So that's how many records. So that's the number of records, if you want, how many records are there up to step n. That's one first question that we will address here. Now, another natural question that you may ask is, typically, how long does it last to break a record? So typically, what's the time that I had to wait between this record and that one? So that's what people, what I will call, the age of a record. So typically, let's use another color. So how long does it take to break a record? And this is something that I will call the age. So how long does it take? So, of course, this list is far from being exhaustive. But these are, I think, the two most interesting questions that one can ask here is the most natural. And that will be the focus of these lectures. So this is basically to try to understand the statistics of both the number of records and of the ages of these records in these two different cases, first in the case of IID random variables, and second in the case of records for random books. Is that clear? OK. So now, of course, one may ask, why should we care about this? I mean, OK, it's a nice problem. But is there any application of it who is interested in that? So, well, of course, one short answer to this question is that records is actually becoming extremely popular. And we are hearing about records in the newspapers every day. The Guinness Book of Records is one of the best-sold book. I mean, each year, I mean, it's completely crazy, actually. But people like records. OK, so that's why should we care, right? So mainly, I mean, why is it, why are they interesting? So again, as I said, OK, they are popular. That's one reason, but we are scientists, and probably it's not a good reason enough to study them. Now, of course, they have many scientific, many applications in the scientific context. Probably one of the area where these questions of records are actually quite important are in the context of climate. OK, so that's one area where people actually are studying actively the sequence of records, if you look at the temperatures, for instance. Or if you look at the many values, indicators related to climate. And in particular, there has been quite a few works during the last years somehow to use records to try to somehow quantify or detect some global warming. OK, so that in the global warming context, there are a series of papers by natural scientists, but also by physicists. And in physics, there are quite a few papers by Joachim Krug in Koloni, in which he has a few papers on that, which you can probably have a look at. Now, this is obviously one nice application. Now, again, in a similar context, I mean, in a different context, but still a little bit out of statistical physics, statistical mechanics, is the context of evolutionary biology. OK, so they are basically the idea is that when you study evolution or mutations in general, in some sense, I mean, any evolution that successfully spreads in a population has to be a record in some sense, in the sense that if it's still alive, that means that it performed better than any other mutations. And in that sense, this must have been a record in some sense. And there have been a lot of work in that context, and in particular in what people call these fitness landscapes and these kind of ideas. And they have been actually quite popular during these last years. Yeah, so what I said, I mean, very vaguely, again, I will not enter into the detail. I can give you some more references, but what I said is that if you look at the evolutions that happened in biological context, say, a mutation, essentially if it survived, it means that it performed better than any other mutations. And in that sense, that's a record, because it was better than the others. So typically, I mean, the number of records could be something like the number of successful mutations if you want, OK, successful. So let's write a bit vague application, of course, there. I mean, I will not enter into the details, but now there are some other problems where people have studied quite a bit the record statistics, in which I already mentioned in the context of finance. I mean, the context of extreme statistics is basically the context of finance, right? So again, in finance, in general, when there is, for instance, I mean, if you consider the case where the price of a stock reaches a sort of records, then it will, in many cases, it will actually creates a burst of activity in the system. And these records are usually watched out, I mean, watched out very carefully. Now, it turns out that, so again, all these applications are relatively still a little bit far from physics. I mean, this one, not that far at the end, but. Now, it turns out that during the last years, 10 or 15 years, people have realized that they have also a very nice, I mean, these questions of records are actually natural in the context of statistical mechanics. So there were various reasons for that. I mean, the one, I mean, first one that I just mentioned, but there is another aspect that I want to look at and to mention here is the fact that records, I mean, can also be used not really for records themselves, but for the kind of dynamics they generate. So by this, I mean, the following. So if you look at how the records, so let me just look at the, so that's this point that I want just to emphasize a little bit. So if you look at the evolution of the records, so as a function of, say, time, so what happens if you look at the current value of the records, so it starts here at some, let me write an explicit series. So suppose that I start here, then I will go down, can go down. So let's look at this simple sequence. So here I have a record here. I have another record here, and I have another record there. So now let me look at the value of the record as a function of time. So the value starts here, then it will not move for a while, and then suddenly it will jump, and then it will stay at rest for a while, and then again it will jump, and again. So if you look at this, you see that it has actually this kind of staircase-like dynamics. And this actually is very interesting, because it turns out that in many system, the dynamics of the object that you are considering is not a continuous one, but it actually is triggered by this kind of staircase dynamics. And in many cases, so, for instance, one example that has been studied quite a bit is the case of imaging that you have an elastic line. Suppose that you have some elastic line in some disordered environment. So I have an elastic line here. Suppose that you have some disorder, so that this disorder will anchor locally your interface. And basically imagine that you apply an external. So this line can be, for instance, a domain wall in a fair magnet between upspins and downspins in anizing model, for instance. Suppose that now you are sort of acting with an external force here. And if you look at the dynamics of this elastic line, well, what will happen is that if you just act with this force, I mean, for some time essentially the line will not move, and then suddenly it will move, and then a part will move. That means that this guy, you will just depin this locally here, and then it will just be pinned by another set of these defects there. So these are defects that pin your line. And what is happening, you see, is that if you look at the center of the motion of the center of mass of your line, then typically it will have this kind of dynamics. So it will not move for a while, and then suddenly there will be some kind of avalanč that will essentially depin locally your interface. So this kind of staircase-line dynamics is quite interesting, because there are not so many ways, natural ways to generate some kind of such kind of dynamics, and these records actually do naturally that. So in this picture, for instance, I mean, the size of the time that you have to wait here, which are random variables, they are typically related to the size of some avalanč. So basically, the time that you have to do, this is related to some characteristic of an avalanč. Some kind of rare events that suddenly has a rather strong impact on your dynamics. So that's certainly one of the, and people have studied that in many contexts in high-tc super conductors, for instance, in spin-glasses, in many, many different disordered systems. Yes. Yeah, yeah. OK, so there are two things. Indeed, so I first consider a sequence of random points, and I look at the records of it. So these are these records. Now what I look, I mean, I am looking at the process which is constructed from this record. So this is this process here. And now for this process, I'm not interested in the records of this process. I'm just interested in the characteristic of it. So typically, what will be the amount of time that I have to wait here, what is the amount of jump that I will have there. But for this new process here in yellow here, I do not talk about records. But what I'm saying is that this one is constructed from the record sequence of some IID random variables, or some random variables, whatever they are. Well, OK, this is an interesting question, but you see that if your process is continuous, then it's pretty hard to define what the records are. So that's why I started from the beginning with the discrete time series, because there I know how to define these records. But that's a good point. I mean, if you talk about Brownian motion, which is continuous, both in time and space, well, there is no obvious way to define what the record is. So this is really well defined for such kind of discrete time process. And eventually, you can define the records of Brownian motion by some subtle continuum limit. But I would like to think about these records just for discrete time series, because otherwise it's a bit complicated to define them. You are right. Is that clear? Other questions? OK. So that's how I became interested, actually, in this problem. If you want to read more about the applications, actually, in particular in these two, these three cases, actually, again, I just mentioned it before, but you can have a look at this review by Gregor Vergan, actually, his records and applications. I forgot the title. I will send it to you anyway. But you can read a bit more about more concrete applications. These are mainly the case of IID random variables that are discussed there, but nevertheless, they are interesting. OK, so this was, more or less, the introduction. Now I want to discuss, I mean, to go a bit, I want to be more quantitative and see what we can compute there. I mean, what can we say about the statistics of these records? And then we start, of course, with the IID case. I mean, as you can imagine, I mean, in general, the statistics of records is quite hard questions. Although the questions are very easy to formulate, but they are usually much more difficult to answer. And it turns out that the case of IID random variables is actually, as you will see, a very nice. I mean, it's very well understood, but the theory is also quite nice. At least I like it, so I hope I will show you how nice it is. How nice and how universal it is. Because you will see that for IID random variables, essentially the statistics of records doesn't depend on anything. I mean, doesn't depend on the kind of random variables that you have. But OK, to understand that, we need to practice it a little bit for IID random variables. Right. So by this, I mean that I will consider x1, xn, again. These are n IID variables. And they have some density, again. So the probability that xi is less than x is, again, dy, p of y. OK, so that's the density that I will consider here. So again, this is continuous. And this is also symmetric, again. I mean, I'm treating the simplest case, density, which is continuous, and symmetric. So these are just the same assumptions that I did for the extreme statistics, to start with, at least. I mean, of course, then you can elaborate a little bit. And OK, so let me fix some notation. I will define as Rn. And I will say that Rn is the number of records to step. I will use step, because I have in mind that this is a time series. So it is time step. It's not a step in the sense of a random mode, but is that clear? So I have a sequence of n, and n for me is a time, because this is the. So this is what I call Rn. And I would like first to discuss the distribution of Rn, basically. So question, what can one say about the statistics of Rn? Maybe distribution is maybe a bit frightening to start with. So let's just say statistics of Rn. But you will see that we can actually come to the full statistics of Rn. I mean, the full distribution. OK, so first question that I want to ask will be, of course, the average of Rn. So to do that, I will introduce something that I've done already before, but I will introduce what I will call an indicator variable, which we play a very important role in the following, which I denote by sigma k. So sigma k is 0 or 1. And sigma k is 1 if xk is a record, and 0 if not. Various ways to treat this problem for the statistics of Rn. But I want to introduce a framework, which is, I think, first, quite natural, and secondly, which can also further be adapted to the case of random works. So let's do this. Let's do that. And then, obviously, I can just write Rn as a sum from k equal 1 to n of sigma k. That's a nice way to think about this Rn as a sum of these 0, 1 random variables, OK? So the full statistics of Rn is, OK, at the moment is still a little bit far, but let's look at the average value of Rn. So typically, you can once say something about the mean number of Rn. So let's compute it. So let's take the average value of this. So what is the average value for the first moment, OK? So Rn, expectation value. Well, this is just the sum from k equal 1 to n of this, OK? This is linear. I have a finite sum, so everything is perfectly well defined. Now, what is that? Well, it has a very nice probabilistic interpretation. This is just the probability to have a record at step k. If you think a little bit about it, think a bit as you would do some experiments. No, this is an average value, OK? So I have this random variable. So where is the randomness in this problem? So I have this x i's. I draw them from a certain distribution, certain density, OK? So I will have a certain collection of random variables. And then I count the record. And then I make an average over this different p of y, OK? So that means that I just repeat the same experiments with the same distributions. So Rn will be random variables. And I want to compute the average of it. Is it clear? Now, my claim is that this average value is just the probability to observe is the probability that xk is a record. Is that clear? This is typically the number of times that sigma k, that xk will be a record. So that's just this probability, OK? Now, how do I compute this probability, OK? Well, let's try to compute it. And for this, let me just introduce a method, which, at first sight, might be a little bit heavy, but which can be adapted to higher, I mean, to other values. So suppose that I have this kind of series here. But I want to compute the probability that so I have k here, and I want to compute the probability that xk, this is xk, and I want to compute the probability that this guy is a record, OK? Now, the first observation, of course, is that this probability does not depend of what happens beyond step k, OK? Because the fact that it is or not a record only depends on the points which are located to the left of this guy. Yes. Oh, yes, it's here. Yeah. Thank you. Please. So, yes, OK, this you have to think a little bit about it. For instance, if you were suppose that you are, OK, suppose that I ask you in a simulation, OK, how do you compute this probability, this one, OK? So forget about this. How would you compute this probability that xk is a record, OK? But so you want to have the probability that this one is a record, OK? So now this, if you want now to compute this sigma k here, so what is sigma k? So sigma k, in other words, to say it, so sigma k is either 0, 1, or 0, OK? Now, it's one with the probability, say, let me call it this probability rk, OK? Let's call rk the probability that xk is a record, OK? Now, sigma k is one with probability. That's what it means, OK? With probability rk, and this is 0 otherwise. You agree that the problem, I mean, either xk is a record, that this happens with probability rk, or it's not a record, and then it will happen with probability 1 minus rk, OK? Now, if you compute this average value, then this will be 1 times rk plus 1 minus rk times 0. Or 0 times 1 minus rk. And this is just rk. Is it convincing enough? So now the question is, how do I compute this guy? So again, as I said, I only need to consider what happens before this guy. Question? OK. So let's consider only these segments, because what happens after I just don't care too much. And let's think, or let's say that, let's suppose that this records at step k arises at some value y. So this one is a record, means what? Means that all the previous guys are actually below this value y, OK? So rk is just the probability that all these guys are smaller than y. And since they are iid, this is just the integral for minus infinity dy dx p of x to the power k minus 1. There are k minus 1 variables on the left. And now, this guy here at step k is equal to y. So this occurs with probability py dy. And now I need to integrate over all the possible values of y. So then I need to integrate from minus infinity to plus infinity. Is that OK? So again, I want, so I specify first the probability of a configuration, such that the value of the record, sorry, is at y. So this will happen with some probability, which is simply the total probability that all these guys here are below y. So this is this one. There are k minus 1 points. And then this guy here, this last guy, is just exactly at y. So this happens with probability py dy. And I need to integrate over all the possible values of y. Because the records may happen with any values between minus infinity and plus infinity in principle. Is that correct? So now, OK, I think that we have understood this. So now you see, if you look a little bit at this integral, well, it's quite tempting to make a change of variable and to use this variable. So let's make a change of variable. And I will just set u is equal to minus infinity to x, sorry, dy p of y. So it's nice, because the Jacobian du is just px dx. So obviously, this integral here. So let's take care about the bounds. So here, I start this lower bounds here. We'll translate. So when x is minus infinity, u is equal to 0. And when x is plus infinity, u is equal to 1. Because the density is normalized. So this is just an integral from 0 to 1. Now py dy is precisely dx. OK, maybe I could have. I hope it's clear enough. Du, and this is just u to the power k minus 1, OK? So that's really nice, because, I mean, first you see that p of x has completely disappeared. There is no p of x. And secondly, of course, this integral is relatively simple to compute. And this is just 1 over k. Sorry, exactly. So what comes out here is that this is universal. Universal provided, for me, here universal is a shortcut to say that it's independent of the distribution p of y, which is continuous and symmetric. Now, of course, this simple result suggests that there is probably another simple way to get it. And there is, indeed, a very simple argument to obtain this 1 over k. So how can I get that? Well, look at the first k random variables, a simple argument. So what I am asking here, I have this x1, xk, iid variables. Now, there is a maximum somewhere between 1 and k. And the question that I am asking when I compute rk is, what's the probability that the maximum is located at xk? But this maximum could equally be at x1, x2, x3 of xk, because they are just identical. So the probability that the maximum occurs at 1 equal the probability that it occurs at step 2 and equals, again, the probability that it happens at step k. And in other words, this probability can only be 1 over k. So the probability that the maximum is at step k is obviously just equal to 1 over k. So that's uniformly. So the probability at this maximum, the law of this maximum is uniform between 1 and k. And therefore, it's just simply 1 over k. So this, of course, did not require this stupid calculation. So why did I do that? Well, I did it because, of course, I mean when you want to compute some more complicated objects, this kind of reasoning would be extremely useful. Of course, in this case, it's very simple. It's almost trivial. So is that clear for everyone? Yeah, this probability. So again, look at it. So I have this IID random variables. Let's make this row in here. So I have x1 here. I may have x2, can have x3, I can have x4, and suppose that here I have xk. OK, suppose that I want. Let's have this one. So I have this sequence here. So this is 1, 2, 3, 4, say 5, k is equal to 5. So I am asking, what is the probability that this one is a record? So this probability is equal to the probability that the maximum of this sequence is located at this stage here. But what I'm saying is that since all these random variables are just identical, the maximum will occur here, there, there, here, or finally there, with some equal probability. So if I ask you, where is the maximum? Well, it will be just uniformly distributed between 1 and 5. It can be anywhere. And the probability that it's here or there is just the same. And therefore it can be only 1 over 5 here, because k is 5. So I have some probability 1 over 5 that the maximum will be here, 1 over 5 that the maximum will be there, et cetera. So in other words, the probability that the maximum occurs at the last steps, such that it is a record, is just 1 over k. Is it clear? So with that result, we can already have a nice result, a nice estimate here, because now are n. So this is the rate at which, so that means that you break the record with probability 1 over k. So that means that with time, you see that this probability is decaying rather fast, I mean, like 1 over k. And as a consequence, if you compute this rn here, then you see that this is just the sum from k equal 1 to n of rk. This is what is written there. And this is just the sum from k equal 1 to n of 1 over k. And so this is just a harmonic series. And you know that for large n, when n is large, this will just behave like log n. And if you want to know more, the first correction, you also know, will be the cellular gamma. And then there will be some corrections of order 1 over n, which I don't care. But the main point here, forget about this gamma e here, because I will not use it anywhere. But the point, really the main point is that you see that for iid random variables, actually the number of records grows extremely slowly, when n is large. So now let's go a bit further. And let's try to compute the second moment, what are the fluctuations of this quantity. Now this is somewhat a bit more difficult. But let's do it. So how would I compute it? So let's go for the second moment, or the variance, if you want. I would like to have the variance of rn. I would like to have the variance. Let's go for it. So I want to compute, basically, sigma r square, which is rn square minus rn square. So this one we know, this will basically be some log square n plus some corrections. Actually, I would plus some corrections. In fact, I will need eventually these terms here, but this will be transparent. But of course, the hard guy to compute is this one. It's a bit harder than what we did before, because you remember that rn is sum of sigma k. So you can actually write it like this. So it's a double sum now, from k equal 1 to n, from k prime, this quantity here, but this one I know. So what I need to compute here is something more complicated, because these are the correlations sigma k sigma k prime. Right? Now, how do I compute that? Well, there is a first way to do that, and to say, OK, start to wave your hands, and say, OK, I'm looking at IID random variables. And it's rather likely that the sigma k is just IID random variables. If you think a little bit about it, it's not a so trivial statement, because naively one have the feeling that there are some correlations. Now, indeed, it's a theorem that the sigma k are indeed IID random variables. This is pretty hard to show. This was actually shown in a very nice paper by Renny, I mean the guy of Erdos Renny, Alfred Renny. Here, I will show it simply for the two-point correlation functions. So I want to show you that you can compute this guy and show that this is indeed equal to average sigma k times average sigma k prime. The full list, OK, you can find a proof of it, I mean, in the review that we did. I mean, we wrote something about it, which is quite nice, I think. I mean, easy to follow. But nevertheless, a bit hard to present right now. So I prefer to show you some computation of these two points just for the two-point correlations and show you how it works. Yeah, but they are not Gaussians, actually. So you need really to, it still works. It's very nice, pretty very nice property, actually. But this is based on the exchangeability of the variables, on the fact that they are actually exchangeable. They are invariant under permutations. So that's right to compute that. So what does it mean? So these two-point correlations, you see, I mean, they have to be, either they are, so when is sigma k sigma k prime non-zero, well, you see, I mean, sigma k sigma k prime is non-zero when both sigma k and sigma k prime are just non-zero. And in that case, this is just equal to 1. So this quantity here, as before, is just the joint probability that xk is equal to the max of x1, xk minus 1. Is equal to, I put it all, I need to have xk. Or say, OK, probably it's better to have it like this. So I want xk to be a record. And I want k prime to be also a record. So to simplify the discussion, I will suppose that k prime is strictly larger than k. If they are equal, then obviously sigma square k is just sigma k. And so then, what I want also, I want that xk prime is a record. So that means that I want xk prime being larger than all the previous values. So I can still write it this way, xk, xk prime minus 1. So I want this kind of configuration. It's a little computation, but it's a good exercise. So we have this. So we start, say, at some value. You have this, you will have that, you have this, something like this. I suppose that this is k, and then this is k prime. So these are both records. You might have, actually, both records in it. Maybe then it's not very, maybe have something like that. In between, you can also have records, in fact. So this is the kind of probability that I want to compute. So this is also a record. But what I want to have is that xk itself is a record. And xk prime is also a record, OK? So first, is that OK with you? So this sigma k, sigma k prime, this correlation function, again, by the same kind of argument that I used before, when I computed rk, the rate when I showed you that, so average value of sigma k is just the probability to have a record at rate k. You should be convinced that this two point correlation function is exactly this probability that these two guys are just records, OK? Because if they are not record, then the value of sigma k is just 0. OK, so let's, yes? Yes, OK. It starts from xk, because, OK, alternatively, I could have the whole series. But since xk itself is a record, that means that xk prime is larger than all the previous values. I could have this, OK? If you don't like it, you can just put that. That's the same, right? Because it's already, since this one is already, then it has to be greater than xk, which itself is greater than all the previous values, so that's exactly, exactly. So again, I mean, another, if you look at this value of sigma, you can basically do what I did before, right? This sigma k, sigma k prime actually can take only two, I mean, two values, 0 or 1. So when it's 0, OK, I mean, this happens with some probability. But it's 1 only when xk and xk prime are records, OK? So I need xk and xk prime to be records. This is the probability of these events. Otherwise, it's 0. And then when I will evaluate this average value, this will be 0 times some probability that I don't want to compute, which will be 1 minus this. This is some probability, q, if you want. And then, so the average value will be 0 times q, so I don't need to compute q. And this will be 1 times the probability that xk and xk prime are records. And this is precisely what I wrote here, OK? The probability that both xk and xk prime are records is precisely that. Is that clear? So let's go for it, and let's compute it. So that's the same kind of computation that I did before to compute the simple result 1 over k. But now, of course, it's slightly more involved. So how does it work? So again, k prime is strictly larger than k. OK, so let's first look at the first sequence here. So for the first sequence, so I want to compute sigma k, suppose that this guy is, say, y1, or yk. And this other guy here will be yk prime. So the same as before. So how does it work? It works like this. So you need to have k minus 1, guys, which will be less than yk. This is for sure. Now, this guy here has to be yk. So this happens with probability p, yk, dyk. Now, what are the possible values of yk? Well, yk actually can be as small as we wish, but it cannot be larger than this guy. Because if this becomes larger than this guy, then this one will not be any more record. And you are lost. So actually, this has to be less than yk prime. Now, what about the next sequence here? So let's see. So here we said there are k minus 1 points in this k minus 1 points here. Now, I need, of course, now that all these points, which are here, they can be as small as they want, but still they need to be smaller than this value, because I want it to be a record. So all these points here need to be smaller than yk prime. So this will happen with some probability, which is basically the integral for minus infinity to yk prime, say dx p over x times, well, I mean, to some power, which is basically the number of points that I have. Agree? Now, how many points do I have? So I have basically k prime minus k minus 1 points. So this will come with some probability, k prime minus k minus 1. Is that OK? And now, of course, this guy eventually will take its value, so that means p yk prime d yk prime. Yes? Prime minus k plus 1. No, because they are, sorry. Yeah, these two guys are excluded, right? The two extremities are excluded. OK, so what you count can't hold the points between this one and this one excluded. So it's k prime minus k plus 1, and you need to get with minus 2, because you don't want to count this one, and you don't want to count this one. Well, you just need to see with some examples, and that OK? And now, it's not completely finished. I still need to integrate over all the possible values of yk prime. So it looks like other complicated. So first, we have to agree on this formula, OK? And now, we want to evaluate it. Turns out, it's very simple. Now, how comes? Is it so simple? So let's first do, we have seen a nice change of variable that we liked before. So let's make this change of variable first, and let's call it v, OK? So v is just this guy, so that means this is just the integral from 0 to yk dx px. Sorry, minus infinity. Yeah, I have, yeah, there is a mistake here. This is minus infinity. It can be as small as I wish. So I have dv is just p of yk dk, dyk, sorry. Precisely, this guy. So let's do it. So I didn't touch yk prime at the moment. Now, here, what do we have? So let's look at the boundaries. So the boundaries, so when yk is minus infinity, this is just 0. And when yk is equal to yk prime, this is just the integral for minus infinity yk prime. Yes? Ah, there is already yk prime. Yeah, thank you. OK, so I just make this change of variable here. Now, here, this is just v to the power k minus 1. So the bound here, upper bound is minus infinity to yk prime. And then the rest I didn't touch. So it's to the power k prime minus k minus 1 p of yk prime. You like it? OK, so now I will do another change of variable, and I will call this new variable u. It appears here. It happens there. And this pyk prime, dyk prime is the Jacobian. So I just, so what I'm saying now is that I'm doing another change of variable. I'm calling u this integral for minus infinity to yk prime, dx p of x, such that du is p of yk prime, d of yk prime. And now something simpler will come out, because, OK, so this I don't touch. I mean, let's first do this. So this is an integral from 0. So minus, so this integral is precisely u. So this is an integral from 0 to u dv v to the power k minus 1. Yeah, yeah, there is another additional dv here, sorry. So this is this integral. OK, maybe not too confused. Let's do it this way. And now what about this integral? So this is just u to the power k prime minus k minus 1. And p of yk prime dyk prime is just du. And now I have to integrate. I mean, now it's the same as before. So when yk prime is minus infinity, then u is 0. And when it's plus infinity, this is 1. So you see that now I have a fairly simple integral. So things are getting better. We started with something quite ugly. We have something quite nice here. Nice, because, of course, we can easily do the integral over v here. So that will be just 1 over k times u to the power k. This u to the power k will combine with this one. And you will get simply 1 over k times an integral over v from 0 to 1. But now, as I said, sorry, an integral over u. And this is just u to the power k prime minus k minus 1 plus k. This is just u to the power k prime minus 1. And what you find is that this is just 1 over kk prime. Yes? It seems, yes. You can do it, but while you see that it becomes a bit, OK, no, no, no, you can do it, you can do it, but it's a bit, yeah, it's a bit, OK. Yeah, instead, what you can do, I mean, and that's how the proof works, is basically that you really study the joint flow of the sigma k. So you write something which is supposed to be the joint flow of the sigma k. It's a kind of, I mean, similar to what you are saying. And then, OK, by manipulations and also some probabilistic arguments, you arrive at the fact that this is just as this factorized form. That's more or less how the proof works. OK, there is one subtility, is that it's not exactly sigma k, that is, it's some function of sigma k, that is, joint flow of some function of sigma k, which you can. So now we are done, because you remember that sigma k is just the average value of sigma k, and one over k prime is just the average value of sigma k prime. So in other words, we are done because you see that we have shown that sigma k, sigma k prime is just sigma k prime. Yes, sorry. Yes, so this holds actually, OK, I suppose here that k was less than k prime strictly. OK, then if I do the reverse, so if I do k prime, I will obtain the same thing if I look at k prime strictly smaller than k. There is no real factor of 2. I mean, it's just that these are two different objects. Yes, ah, yes, so when, of course, when I will do the, so here this is just for fix k and k prime, then when I will do the sum, of course, I will have to consider both, yeah, yeah. I will come to that in a minute, yeah. Now, so this is a computation. Now, of course, you can also compute sigma k square. I mean, if that means the k square k equal to k prime. And this, of course, because sigma k is 0 or 1, this is just sigma k. And this is just one over k. Now we are ready to compute Rn. As you said, now we have to take care. So I will compute the variance immediately, because there will be some nice simplification. I hope, what, it's not that I hope. I know that there are simplifications, so, OK. So let's see here, OK. So, yeah, so sigma k is 0 or 1, OK. So if it's 0, sigma square is 0. And if it's 1, sigma square is also 1. So that's the same, and that's the same as sigma k. This is just one over k. Yes. Counterintuitive? Why? Yeah, the fact that they are, they are not, yeah, yeah, I agree. I mean, I agree with you. I mean, that's, yeah, no, I mean, I agree that this result is actually, yeah, I mean, one of the reasons, again, is that there is some, in fact, the fact that the sigma k's are uncorrelated is not related, is completely independent of the fact that the variables themselves are correlated or not. It only depends on the fact that they are exchangeable. But otherwise, I mean, it's not a matter of correlations between the X-case. That's a good point, yes. It's really a matter of exchangeability, and this is something I think that we don't have a very nice intuition of, I mean, except if you have been playing with these for years. Otherwise, I agree, it's really a non-intuitive result. So let's see how it works. So we have this result. I mean, we have that, right? So we have this, and then we have the sigma k, I suppose, k, which I could write it this way again. And the terms are going from one to n, I mean, since there are many of them, I don't want to write them. OK, so this is rn square, and this is the average of rn times the average of rn. So now what I will do here is I will just disentangle the diagonal terms, k equal k prime, and the all diagonal terms. So I mean, I will just write it as sum over k plus the sum k different from k prime. And here, OK, here I have what I have. So I have 1 over k. I guess I can immediately write these things like that. So this, we know, is just the sum of 1 over k. So this is just this sum that we have known before. This is just the sum from k equal 1 to n of 1 over k. And now this, we also know, actually, because this is just 1 over k, k prime. So I will write it in this way. It's kind of trick, but so I want to write it as the sum from k to k prime unrestricted from k1 over kk prime. But when I do that, I have to subtract the term, the diagonal terms, OK? So I'm doing that. Minus 1 over k to infinity 1 over k squared. So I'm claiming that this is just that. I hope you will agree with it. And this, eventually, this is just the sum over kk prime of 1 over kk prime. Is that fine? Can everyone read what I write here? I realize maybe it's too small. No, you can read it. So now if you can read it, do you agree with it? OK, so now if you agree, I mean then it's done, because you see that the bad guys just disappears, right? This guy comes with a plus, this one with a minus. And I'm done. Yeah, sorry, sorry, sorry, sorry. So at the end, you see that this guy is very simple. It has a very nice analytical structure. This is just the sum of 1 over k minus the sum of over the 1 over k square. Yes? Yes? To the second line. Yes? Well, what is unclear? No, no, OK. So this one, I think, OK, this one is what it is, right? I think this one should be OK. I mean, yeah, so this does not work, yes. I mean, there is no, so the limit k called k prime is indeed singular. But this has to, I mean, OK, I think it's not related. What is, it's not related to that result, which I find a bit counterintuitive. What is counterintuitive to me is that, OK, for k different from k prime is that you can write that, that they are really uncorrelated. And you are pointing out another point, another aspect, which is that the limit k goes to k prime is a bit singular. But this random variable sigma k is also quite singular, right? I mean, sigma k is 0, 1, so, OK, there is no continuity in the behavior of it. You mean to see if I want to do that or this one? OK. No, it's. Yes, there should be a way. But yeah, I mean, the point is that I really assumed that, yeah, for instance, while you see, I mean, look at k prime equal to k in my formula here, you have this one over this probability that pops up, for instance, which is completely crazy. I mean, probabilistically, there is obviously no reason that one over this probability occurs. And I really had to assume, when I did this, while I really disentangled the two blocks, I mean, first from 1 to k, and then essentially from k to k prime. But really, when k is equal to k prime, this piece here doesn't mean anything. So in other words, when k prime is equal to k, this part here has to be understood as essentially as that disappeared. It's really a singular. I mean, yeah, there is no, yeah, OK. Maybe there is a nicer way to see this limit, but I don't have a good feeling of it, at least. OK, so that's our result. And OK, so what I want to say now is that, OK, so you see that it's rather interesting, because the final result is fairly simple. And it's just that this is this one 1 over k minus this 1 over k square. You see, it has a very nice structure. And of course, for large n, we know how it behaves, right? Because for large n, this will again be log n. This one will be a constant. So here we know that this is log n plus gamma e. And this is this nice sum, if you want to write it. And then you will have some terms which are of 1 over n, which you can get easily. I mean, these are harmonic numbers, if you really need them. So what does it mean? It means that you see that rn square, so the variance, scale like log n, so that means that the typical width of the distribution is actually over the square root of log n. So in other words, if you look at the distribution, so now if you really, so that means that, now I need to introduce some nice notation, I suppose. OK, so let me introduce this notation p of mn, which is just the probability that rn is equal to m. So how does it look like? Well, this, if I plot it, so the average value is log n. We have seen it. So if I plot this as a function of m, so it's centered around log n. OK, there is this gamma e. And then it has some shape, which I don't know at the moment. But still, I know something. I know that the width is over the square root. I mean, it's more than order, in fact. This is really square root of log n, OK? The coefficient is 1. So the relative fluctuations, they go to 0. I mean, like 1 over square root of log n, but in a very, I mean, slow way. Now maybe I can, OK, there are two options. Either I show you how one can compute, maybe it's nice. Or maybe I can show you what we can say about the distribution of this quantity. And then next time I will discuss the edges. So next time I will do that, I will discuss a bit the edges. But you need to be prepared, because this is a little, well, OK, it's not very difficult. But yeah, one needs to compute a few things. OK, let's go with it, with this guy, because it's nice. OK, so the main result is that eventually this is a simple Gaussian. So how does it work? And why is it a Gaussian? OK, the easy way to show that is to remind you that Rn is the sum from i equal. So Rn is a sum of random variables. And the sigma i's turns out to be iid. So I told you this. I computed the linear correlations. I showed that the linear correlation, the two-point linear correlations was basically decoupled. But the sigma i's are iid random variables. Sorry, they are independent. But they are not identical, because we have seen already that the average value of sigma i goes like 1 over i. Put compute sigma square i, we have seen is sigma square is also 1 over i. So these are these Bernoulli random variables. So we have to face the sum of independent random variables, which are not identical. But it turns out that nevertheless it goes to some Gaussian, but are not identical. Now you can still show that there is this extension of the central limit theorem, which still converges to the Gaussian under what is called under the name of the Lindemann condition. Lindemann, excuse me, Lindemann condition. And this is still, OK, I will not enter too much in that, but the Lindemann condition in this case tells you that you still converges to the Gaussian. But OK, I want to show you something different, because there is much more structure in this problem. And in fact, it turns out that this problem of records for iid can also be mapped on a problem on random permutation. So I already discussed random permutation yesterday. So I want to show you that the random permutations are also behind this problem. So how do I see that? So to do that, I will do the following. I want to compute this quantity, p of r n m. This is probability that r n is equal to m. And this is what I denoted by p of m n. Now, to see why these guys happens, it's interesting to look at the generating function of p of m n with respect to m. So what I will compute, basically, is the sum from m equal 1 to n. So m, the number of records cannot be greater than n. You remember? Because this is the number of records. So m has to be less than n. It's an integer, and it has to be between 1 and n. And I want to compute this generating function. So let's compute this guy. I would not give it a name, but this is a generating function. We have seen already this kind of object. Now, this generating function, you see, I can also write it. There is another way to write this thing, which is just to write that. So rewriting is z to the power rn. So if you start with that, and if you want to write what this quantity is, then you will realize that this is exactly that. Now, why do I write this in this way? Because this is just the sum of sigma i. And the sigma i's are iid. Sorry, they are independent. They are not identical. They are independent. So because they are independent, this is still equal to z, the product from 1 to n of z to the power sigma i. I just used that the exponential of the sum is the product of exponentials. And this is because now the sigma i's are iid. Sorry, they are independent. And this is just now the product from 1 to n of z to the power sigma i. Now, this I know how to compute now. I know how to compute, because sigma i can only take two values, 0 or 1. And it will be 1 with probability 1 over i, and 0 with probability 1 minus 1 over i. So in other words, z to the power sigma i is just what? So either it's z with probability 1 over i. Sorry, I'm just doing this computation here on this corner of the blackboard. So if sigma i is 1, then z to the power sigma i is 1. And this happens with probability 1 over i, because we have seen that. And then otherwise, if sigma i is 0, then this is just 1. And this occurs with probability 1 minus 1 over i. So I know what this quantity is at the end. This is just a product from 1 to n of what? So it's z times 1 over i plus 1 times 1 minus 1 over i. Is it OK? This you can still rewrite it in a slightly different way. So let's see what's the best way. So I have a 1 here. Let's me go take out this one. And then I would have z minus 1 over i. Nice. OK, so this is just what? This is just the product of, so I can do that, z plus i minus 1 divided by the product over i. So this is just factorial n. And here I have a product, sorry, i equal to 1 to n. It's a kind of generalized factorial. So how does it look like? What is this number? I mean, it's not a number. What is this function? How does it look like? It looks like the following. So I have a generating function. And so what is the generating function is about? So it's, again, this is this guy. So in principle, now, if I know this generating function, then I can extract rn. Because if I know the coefficient of z to the power m, I can extract p of mn. So that's the idea. So it has this structure there. So let's me just, OK, I can still write it this way. So it's z. So let's just write it as it. So for instance, for i equal to 1, this is z. Then from i equal to 2 is just z plus 1. It's a kind of rising factorial. I mean, z plus 1 times z plus 2. So of course it has a nice structure divided by factorial n. Now this, OK, I don't know if you have seen that before. But these functions are relatively well known in combinatorics. And it turns out that it's a polynomial in z. It's a polynomial of degree n, which one expects. Because it was there. And the coefficients actually are known. So let me, I can still rewrite it like this, from m equal to 1 to n, z to the power m. And then I'm almost done with what I want to have. So I have this one over factorial n. I leave it outside for the moment. So I'm saying that this kind of polynomial here is well known. It still can be the coefficient. Let me denote it this way. So these are not combinatorial factors. I mean, these are not the n-shoes n that you know. Usually they are called the stealing numbers. Stealing numbers, there are many of them. There are several kinds. So these are the stealing numbers of the first kind. And there are also the signed and unsigned stealing numbers. So these are the unsigned stealing numbers of the first kind. Just stick it for stealing numbers here. I will not do so much with these numbers. I just want to explain to you what they count. So, OK, once we know that, maybe just to finish it, suppose that we know what these numbers are. Well, you see that you can do them if you take Mathematica or whatever, what you prefer. I mean, you could extract easily. If I give you n and m, you could easily extract these numbers in principle. You have these polynomials here. You just look at the coefficient and you could tell me, OK, what they are. Now, it turns out that they have a very nice interpretation. And I will tell you in a minute what it is. But before that, I just want to tell you that once I'm able to write the things like that, then basically I can extract directly the p of n equal to m. Then this is just this number, by definition. So I just compare this formula here with that formula there. So this p of m n here is supposed to be the coefficient of the term z to the m of this polynomial in z. And so I immediately get it. Now, what are these guys? So it turns out that this nm here, the sterling numbers, they count the number. So they actually play a role in the context of permutations. And they count the number of cycles in which you can decompose permutation of size n. So what does it mean by this? Maybe an additional remark here. You also see that the dependence on p of x has completely disappeared. So the excise, I mean, there could be uniformly distributed. That could be Gaussian numbers. That could be symmetric exponential numbers. This is completely universal, actually. And by the way, this is not an asymptotic result. I mean, already for n equal to 3, this is fully universal, which is not completely intuitive. Once you know that it can be mapped on two permutations, then it's true that this magic is sort of disappearing a little bit. But OK, the connection, the true connection to permutations is hard to show. I will not show it to you. I just want to pinpoint this connection here. That this nm here, this sterling numbers, they have notations because they count something which is relevant. And they count the number of cycles of size m. No, sorry. This is not it. Sorry, sorry. What I want to say, counts the number of permutations which can be decomposed in exactly m cycles. So basically, that tells you that the number of records is associated to number of cycles in a given permutation in exactly m cycles. So maybe, OK, let me remind you what these cycles are. So let's take, for example, let's take a simple permutation. So let's take n equal to 4, for instance. So I have one permutation, which is like this. So I have 1, 2, 3, 4. And one of the permutation that I consider is basically 4, 2, 1, and 3. So as you probably know, when any permutation, so this is one permutation here, sigma, let me call it tau, any permutation can be decomposed in cycles. And this can be done in a unique way up to reordering of these cycles. But what is basically a cycle here? So you start from 1. 1 is sent to 4, 4 is sent to 3, 3 is sent to 1. And you are back to your cycle. So you have a first cycle here, which is 1, 4, 3. So this is one cycle. And then you have another trivial cycle, which is this one, 2 is sent to 2. Now, so here, typically, the number of cycles is 2. And you can do that in a unique way. Again, up to a reordering of two cycles. And then, in general, any permutation can be decomposed into cycles. And an interesting question in the study of this group is how many, I mean, typically, how many permutations can be decomposed in exactly m cycles. And they can be enumerated. And this is what these sterling numbers do count. So I will not say more about it. It turns out, this already suggests that there is a strong link between record statistics and the random permutation problem. For those of you who like these things, I mean, you can read this very nice book by Flajole and Cedric, where they explain this in great detail. They have this fantastic book, which is called Anality Combinatorics. OK, this is mainly math. And then when you have this, OK, people have studied the large, I mean, the large gen asymptotics of these numbers. And for large gen, you can indeed show that this is a Gaussian. OK, so I just end up with that. In fact, you can also get the Gaussian from the analysis of this generating function. It's a little bit complicated. But in the large end limit, this is what you get. You get that this is just m minus log n squared divided by 2 log n. And there is here square root of 2 pi log n, amazing. So I will end up with that. OK, I mean, of course, this last part on the full distribution is a bit more technical. But I wanted to show you this. OK, I like this connection with these permutations. Find it quite nice. It's a bit mathematical, but it's quite nice. But the first part, I mean, I hope that, OK, I convinced you, at least, that the first part, where I computed first, the mean number of records and eventually the variance, you see, I mean, these are relatively simple, I mean, relatively simple computations, a little bit cumbersome, but which, at the end, simplifies in a very nice way. So one should not be afraid to compute them, because with some effort, you will end up with some nice formulas. So that's basically what I wanted to say for the statistics of the number of records. So I guess that what I will do, probably, tomorrow, is that I will discuss, at which degree of detail is this, I don't know yet, I will discuss a bit the statistics of the ages. Maybe I will show you, I mean, what kind of observable that we can get, I think. Yeah, we have all the material now to do that. I've shown you how to do these computations, I mean to do. At least I've shown, I've done for you the computations, I hope that you could follow it. The same kind of computations can be done for the ages, and eventually one has a very nice expressions for the distribution of the ages, and then I will go, I mean, I will show you how to extend these things to the case of random works. In fact, the case of random works is almost a little bit simpler. I mean, since we know already now, the Spar-Understand formula, you will see that Spar-Understand already did all the job for us somehow. So we will use this result and combine them with some, okay, some similar kind of rezoning, and that will be more or less the end of the ratio of them. Okay, thank you very much.