AI Assistant
Transcript
00:02:320Paolo Guiotto: Okay, good morning.
00:14:310Paolo Guiotto: So today, I want to do… The first part.
00:19:640Paolo Guiotto: some problem on convergence, and then we move forward, and we will see some important results concerning convergence.
00:31:740Paolo Guiotto: The two most important results are the law of large numbers and the central limit theorem.
00:39:90Paolo Guiotto: So let's see, this is the program. We start with the exercise 8, 5, 6.
00:47:370Paolo Guiotto: Here we have a sequence of random variable, Xn, random variables with… density.
01:02:790Paolo Guiotto: This is basically what is called the Cauchy distribution, FXN of X equal.
01:09:400Paolo Guiotto: 1 over pi, which is a scaling factor, and divided the 1 plus
01:15:200Paolo Guiotto: N square, X squared, or X real.
01:22:770Paolo Guiotto: Now, we accept that this is a density, this is a standard class, but you can check directly. Now, first question is, is this sequence going to zero in distribution?
01:37:680Paolo Guiotto: Question two, is this sequence X scan going to zero improbability?
01:47:610Paolo Guiotto: And, number 3.
01:51:960Paolo Guiotto: If the exam are independent, Independent.
01:59:110Paolo Guiotto: Then, can you say… can you say that XN goes to zero almost surely?
02:10:90Paolo Guiotto: Okay, so About to defester.
02:15:210Paolo Guiotto: The convergency distribution is the weakest for us.
02:19:590Paolo Guiotto: And it is characterized by a number of equivalent conditions.
02:27:180Paolo Guiotto: So, basically, we have, let me just refresh this…
02:35:870Paolo Guiotto: The list of equivalent characterizations of the convergence in distribution are these, so…
02:44:90Paolo Guiotto: The expectations of, function of the random variable, where functions are continuous bounded, goes to the expectation of,
02:53:810Paolo Guiotto: Pheovaxa.
02:55:990Paolo Guiotto: The cumulative distribution functions converge pointwise to the cumulative distribution function of the limit.
03:04:750Paolo Guiotto: And this is, an X by X, stuff.
03:10:190Paolo Guiotto: or the characteristic function, convergence, again, pointwise… I forgot the… the point-wise convergence of the distribution function, cumulative distribution function.
03:23:240Paolo Guiotto: is for points where the limit CDF is continuous.
03:28:950Paolo Guiotto: While this one is for every value of the variable, the convergence of the characteristic functions. And the last one is the convergence of some probabilities.
03:38:970Paolo Guiotto: Now, which one should we choose?
03:41:780Paolo Guiotto: With these assumptions, so we know the densities, It seems that,
03:50:670Paolo Guiotto: Mmm… well, the first and the last…
03:55:00Paolo Guiotto: looks more or less the same, and they are with the generic functions, phi and the generic intervals. It does not say that we cannot do that, but…
04:05:150Paolo Guiotto: Perhaps if we can compute the CDF, or the characteristic function, it is better to use these two conditions.
04:13:850Paolo Guiotto: For the CDF, we have the density, so…
04:18:00Paolo Guiotto: We could… well, let's see what happens if we compute the CDF, because maybe here, having the density, we can say that D…
04:31:890Paolo Guiotto: CBF.
04:33:770Paolo Guiotto: Off.
04:35:530Paolo Guiotto: XNZ.
04:38:790Paolo Guiotto: F… XN.
04:41:340Paolo Guiotto: of x, let's say, is the integral from minus infinity to X of the
04:47:390Paolo Guiotto: density, F, X, N, let's say, at U, the U.
04:55:470Paolo Guiotto: Yeah, we can integrate this because it's basically an arctangent stuff. If we write this integral minus infinity 2x is 1 over pi, then we have n divided 1 plus n squared X squared, that we could write NX squared dx.
05:15:470Paolo Guiotto: We could give the NX here.
05:19:680Paolo Guiotto: And therefore, or if you want, you can leave the n here and directly write. This is the, derivative with respect to X, apart for the factor, of the arctangent of NX, right?
05:34:420Paolo Guiotto: I'm sorry, this is, you.
05:38:830Paolo Guiotto: arctangent of NU, so we have to do the evaluation of the arctangent of NU between U equal minus infinity and U equal X.
05:52:140Paolo Guiotto: So, this is 1 over pi, then we have an arctangent of NX.
06:01:460Paolo Guiotto: minus the arctangent of NU when U goes to minus infinity. Here, n is natural, no?
06:09:100Paolo Guiotto: So NU goes to minus infinity, and therefore the arctangent goes to minus pi over 2, so plus pi over 2.
06:19:100Paolo Guiotto: Well, this is the FXN.
06:25:30Paolo Guiotto: So, from this, we can now easily compute the point-wise limit even for every X.
06:31:930Paolo Guiotto: limiting N, then… Let's see what happens. The limit when n goes to plus infinity.
06:39:940Paolo Guiotto: of FXN of X.
06:43:700Paolo Guiotto: is equal to… now, you see that, we have just to compute the limit of the tangent of NX.
06:51:930Paolo Guiotto: Now, n is going to plus infinity, and the behavior of this depends on X.
06:58:670Paolo Guiotto: on the case is X positive, X negative, and maybe X equals 0. So we have three situations. X negative. For X negative, NX goes to minus infinity, so we have 1 over pi, arctangent.
07:13:980Paolo Guiotto: Of minus infinity plus pi half.
07:18:60Paolo Guiotto: The arctangent of minus infinity is minus pi half, so all this is zero.
07:25:770Paolo Guiotto: For X equals 0, we have to separate, because when x is 0 and x is constantly equal to 0, so it does not go to infinity. Arctangent of 0 is 0, so we get just 1 over pi times pi half.
07:40:470Paolo Guiotto: Which is 1 half.
07:42:860Paolo Guiotto: And for X positive, now NX goes to plus infinity.
07:49:300Paolo Guiotto: So, the R tangent will go to the R tangent of plus infinity.
07:55:880Paolo Guiotto: Which is equal to pi half.
07:59:870Paolo Guiotto: So this plus another pi half yields pi times 1 over pi 1.
08:07:600Paolo Guiotto: So we can say that,
08:10:260Paolo Guiotto: the limit, the point-wise limit for n going to plus infinity of the CDF of XN at the generic X, this actually converges for every X.
08:22:830Paolo Guiotto: 2. 0 for X negative. 1 half for X0.
08:29:310Paolo Guiotto: and 1 for X positive. This is what happens for the limiter of the CDF.
08:37:299Paolo Guiotto: Now, to say that DXN converges in distribution to zero, now, a condition is… characterization is if and or if the limit
08:53:520Paolo Guiotto: when n goes to plus infinity of the CDF of DXN, at point X is FX, where X is 0, so FX of little x, for every point x.
09:09:940Paolo Guiotto: Such that, well, we can write continuity.
09:15:220Paolo Guiotto: Continuity.
09:18:500Paolo Guiotto: point.
09:20:960Paolo Guiotto: for… FX.
09:24:80Paolo Guiotto: Now, what is the CDF of the variable which is constantly equal to zero? It's a…
09:31:430Paolo Guiotto: The variable is a constant, so we know that this is equal to 0 if x is negative, and it is equal to 1 if x is greater or equal than 0.
09:44:190Paolo Guiotto: So, the CDF of the supposed limit is a function made like this. It is 0, constantly 0, and at 0, it takes value 1 forever.
09:58:880Paolo Guiotto: So the continuity points for this CDF are all the reals except X equals 0, where you have this jump. So that condition here, for every X, continuity point for the CDF, FX means, in this case.
10:18:640Paolo Guiotto: For every X, different from zero.
10:21:950Paolo Guiotto: You see?
10:26:00Paolo Guiotto: Is that clear?
10:30:200Paolo Guiotto: So the characterization through the CDF of the weak convergence says I have weak convergence of the sequence Xn to the variable X if I have pointwise convergence of the CDF
10:46:330Paolo Guiotto: to the CDF of the limit.
10:48:960Paolo Guiotto: at every point X, which is a continuity point for FX, for the limit CDF.
10:56:340Paolo Guiotto: Now, this means that if the limit CDF is continuous, this means for every X. If it is not continuous, if there are discontinuity points.
11:06:270Paolo Guiotto: Well, this is for every x except the discontinuity points. Now, the question is, is this true? So, is it true that the limit when n goes to infinity of FXN of X is FX, of X for every X
11:22:930Paolo Guiotto: different from 0, which is the unique discontinuity point for the FX, and that's true, because if you look at the limit we computed, we have seen the limit exists for every X, so it exists in particular for every X different than 0.
11:37:870Paolo Guiotto: For X negative, we have 0, for x positive, we have 1, which is exactly the value of the limit distribution for X negative and for X positive.
11:47:440Paolo Guiotto: So, we can say which is… True.
11:54:810Paolo Guiotto: So, this implies that the sequence XN converges in distribution to zero.
12:02:720Paolo Guiotto: Now we have question two.
12:07:680Paolo Guiotto: But if you want, we can see, just for the sake of… Hmm…
12:14:980Paolo Guiotto: Just in this case, we could have done also with the characteristic function, just for the sake of,
12:23:150Paolo Guiotto: Of curiosity, alternative, let's say, alternative.
12:27:820Paolo Guiotto: solution.
12:30:190Paolo Guiotto: We compute, Because here it is possible.
12:36:740Paolo Guiotto: We compute the characteristic function of the XN.
12:41:990Paolo Guiotto: Now, in this case, This is formally the expected value of EIXN.
12:52:670Paolo Guiotto: But since we have a density, no, this is also the integral on R of EICX.
13:01:160Paolo Guiotto: Times the density FXNXDX.
13:07:450Paolo Guiotto: Now, the point is that this density, since it is a density, is an L1 function, and therefore this is the ordinary Fourier transform, apart for the
13:18:100Paolo Guiotto: sine on the exponent of E, but that's a detail. So we can say that this is, let's say, let's use the hat for the old L1 Fourier transformer.
13:29:920Paolo Guiotto: It is the creator form of 1 over pi, then we have n divided 1 plus n squared X squared.
13:37:890Paolo Guiotto: So this is the variable, so let's take a point here. Evaluated in the… if we are considering here the traditional L1 Fourier transform, evaluated at minus C.
13:51:310Paolo Guiotto: Now, this is a constant, so it can… the Fourier transform is linear, can be written outside. And this is basically a Cauchy distribution, so one of the distributions for which we have computed the Fourier transform. We can see this if we…
14:08:400Paolo Guiotto: If we, factorize the n squared, so we have n divided n squared, then we have 1, so this becomes a 1 over n squared plus the square of the variable.
14:23:320Paolo Guiotto: So it is the head of this, evaluated at minus…
14:28:210Paolo Guiotto: See? Now, we remind of this, remarkable, Collier transform. We have this 1 over A squared plus the square, so the form of 1 over A squared plus X squared.
14:43:940Paolo Guiotto: evaluated at point C,
14:46:510Paolo Guiotto: If I'm not wrong, this should be pi over A, E minus A modulus of X, if… if I'm not wrong. Okay, so now our A here is 1 over N, now, because A square yields exactly 1 over N squared.
15:07:50Paolo Guiotto: So we can simplify a little bit this, 1 over n, so at the end, we get… so 1 over pi n, the coefficient, then we have pi over A, A is 1 over N.
15:20:220Paolo Guiotto: exponential minus A, which is, A is 1 over N, so 1 over N modulus of… well, be careful, here it's irrelevant, it's minus X for us, okay?
15:35:690Paolo Guiotto: So in any case, after this calculation, we get this cancels this, this cancels this, and therefore, we have E2 minus modulus of Xi divided by N. This is the characteristic function of DXN of Xi.
15:52:390Paolo Guiotto: Now, here we see easily that when we send n to plus infinity.
15:58:290Paolo Guiotto: Also, this point-wise limit is pretty easy to be determined, because clearly, this fraction goes to 0. So this will go to e to 0, and it is equal to 1.
16:12:260Paolo Guiotto: Now, 1 is the characteristic function of a random variable constantly equal to 0, because when you plug into this exponential 0, you get expectation of e to 0, so 1, expectation of 1 is 1. So that's exactly the characteristic function of phos evaluated at generic C, no?
16:34:30Paolo Guiotto: You see? Because phi of 0, C, is the expected value.
16:40:440Paolo Guiotto: of EIC times 0, so it is, expectation of, 1 equal to 1.
16:51:440Paolo Guiotto: And therefore, this happens for… since this happens for every X real, because there is no condition for Xi, we can immediately conclude that Xn converges in distribution to 0.
17:06:700Paolo Guiotto: So… basically…
17:11:79Paolo Guiotto: There is not any advantage in one versus the other, but let's say, in this case, we could have done.
17:17:960Paolo Guiotto: Both these checks.
17:20:760Paolo Guiotto: Now, let's pass to question 2. That concerns the,
17:27:589Paolo Guiotto: the limiting probability. Now, we know what? We know that the limiting probability is stronger than limiting distribution, so
17:38:430Paolo Guiotto: We could have also tried directly, maybe, if there is limit in probability, it will be also the limiting distribution to the same thing.
17:47:410Paolo Guiotto: Okay, let's see what, what should be this. Now, to check that XN goes to 0 in probability.
17:56:880Paolo Guiotto: We need… Check.
18:04:450Paolo Guiotto: if… the probability that the distance between XN and the limit, which is 0 in this case.
18:12:950Paolo Guiotto: greater or equal than some epsilon, this quantity goes to 0 when n goes to plus infinity, and this must be for every epsilon positive. That, at the end, means that probability that modulus of XN is greater or equal than epsilon.
18:30:60Paolo Guiotto: goes to 0 for every… when n goes to plus infinity for every epsilon positive.
18:37:730Paolo Guiotto: So, to discuss this, we have to determine that probability exactly, or… Assess the…
18:44:930Paolo Guiotto: Now, this is the probability that…
18:48:750Paolo Guiotto: modulus of X is greater or equal than epsilon, or if you want, equivalently, there is also probability that modulus Xn is less or equal than epsilon.
18:59:260Paolo Guiotto: This goes to 1 when n goes to plus infinity for every epsilon positive.
19:06:440Paolo Guiotto: Maybe just for convenience, it is, I use this Saxon way.
19:11:400Paolo Guiotto: So this is, since we have a density, no, we can write this probability, that means Xn belongs to the interval from minus epsilon to plus epsilon. So as the integral.
19:24:690Paolo Guiotto: on the range minus epsilon epsilon of the density FXN of, say, still U, DU.
19:36:750Paolo Guiotto: Okay?
19:41:120Paolo Guiotto: Now, this integral can be computed, because it is the same calculation we did above. It is 1 over pi.
19:47:330Paolo Guiotto: So this is, one apart for the factor 1 over pi. We have, that expression n divided 1 plus nu square, which is the derivative with respect to U of the arctangent of NU.
20:05:920Paolo Guiotto: So, when we evaluate the integral, we get the evaluation of the arctangent of NU,
20:17:660Paolo Guiotto: Between the minus epsilon and plus epsilon.
20:22:180Paolo Guiotto: So this yields 1 over pi, then we have arctangent of n epsilon minus the arctangent of… when we put u equal minus epsilon, we get minus NU.
20:38:340Paolo Guiotto: Now, the arctangent is odd, so you can carry the minus outside.
20:46:50Paolo Guiotto: And, therefore we get, 2 over pi arctangent
20:52:730Paolo Guiotto: of n epsilon. But this is exactly what is this probability. So the probability that modulus Xn is less or equal than epsilon is exactly 2 over pi
21:08:10Paolo Guiotto: the arctangent of N epsilon.
21:13:660Paolo Guiotto: And here we can compute the limit. Remind that epsilon here is fixed and positive.
21:22:390Paolo Guiotto: And you have to take the limit when n goes to infinity for epsilon fixed. When n goes to plus infinity, since this is positive, the argument of the arctangent goes to plus infinity, so the arctangent goes to the value pi half, so at the limit, we have 2 over pi times pi over 2.
21:40:610Paolo Guiotto: That means we get exactly one. So we can say now that, yes, it is true that Xn converges in probability to 0.
21:53:130Paolo Guiotto: Now, question three, so if we would… if we could have done directly question,
21:59:740Paolo Guiotto: Question 2, and conclude that, since we have convergence in probability, we have also convergence in distribution.
22:10:800Paolo Guiotto: Now, let's take the question number 3.
22:19:660Paolo Guiotto: it says, suppose that here, you know that DXN are independent.
22:29:820Paolo Guiotto: And it asks to check, to discuss.
22:35:280Paolo Guiotto: whether or not XN goes almost surely, 2-0.
22:45:790Paolo Guiotto: Now, since this is an almost sure convergence, we know that, in general, XN goes almost surely to 0,
22:56:30Paolo Guiotto: If, and only if… The probability that,
23:04:380Paolo Guiotto: the limb subset, you know, of the intersection of N union for n larger than capital N of sets where the distance between Xn and the limit, which is 0 here, is larger than epsilon.
23:20:360Paolo Guiotto: this probability is equal, to 0. So this is basically the probability that XN
23:29:430Paolo Guiotto: This is the event where Xn of omega is, let's say, away… from… The limit? Zero.
23:44:250Paolo Guiotto: Infinitely many.
23:48:10Paolo Guiotto: times.
23:49:570Paolo Guiotto: It is, it is, the event where the distance between Xn and 0 is bounded below by this epsilon positive.
24:00:840Paolo Guiotto: for infinitely many index n, so it is impossible that Xn goes to 0, because that distance should become less than whatever is epsilon.
24:12:360Paolo Guiotto: Okay? So if we prove that for every epsilon positive, this is zero, we have an almost sure point-wise convergence.
24:21:20Paolo Guiotto: Now, to… to do this, we know that, this is a limb soup,
24:27:710Paolo Guiotto: set of these events, modulus Xn greater or equal than epsilon.
24:35:770Paolo Guiotto: And, the Borrel Cantelli is,
24:40:920Paolo Guiotto: a result that can tell what is the probability of this set under certain conditions. So, it's not a true-false, it's not a necessarily sufficient condition, but it is almost, and especially in this context, it will become an if and all if.
25:00:70Paolo Guiotto: So what we do have to estimate is this probability that modulus Xn is greater than epsilon, and the sum of these probabilities.
25:12:680Paolo Guiotto: So we have data by… Boral cantelli.
25:17:880Paolo Guiotto: Dilemma.
25:19:920Paolo Guiotto: we have, that if this sum is finite.
25:28:90Paolo Guiotto: Then, we are sure that this, the probability of the limb subset
25:36:490Paolo Guiotto: of this, let's call shortly EN, events N, this is zero.
25:42:890Paolo Guiotto: And therefore, this will imply XN converges, almost surely.
25:49:770Paolo Guiotto: to zero. If the sum of these probabilities
25:56:600Paolo Guiotto: Is infinite, so this series is divergent.
26:01:570Paolo Guiotto: And this is an alternative, because this is a positive term, serious, so it can be either convergent or divergent. There is no other possibility. So, it's one or the other.
26:13:740Paolo Guiotto: In the second case, however, this information alone cannot imply anything about the probability of the limb soup.
26:23:430Paolo Guiotto: And if we add that these events, again, are independent.
26:30:290Paolo Guiotto: Which means that, in this case, DCN depends on the XN, means that XM independent, so plus is independent.
26:41:170Paolo Guiotto: In this case, we can say that the probability of the limb subset
26:47:990Paolo Guiotto: of DN is equal to 1. So in this case, we would have that it is never
26:54:640Paolo Guiotto: Never convergent, so it is not convergent with probability 1.
27:00:800Paolo Guiotto: So…
27:06:700Paolo Guiotto: is not only… is a strong opposite of this, no? You see, either you have convergence with probability 1,
27:16:370Paolo Guiotto: The, let's say, the…
27:18:280Paolo Guiotto: The negation of this would be you may have convergent with some probability, but this says.
27:26:900Paolo Guiotto: that there is no possibility to have a convergence, so it's never convergent. So all this, turns around the convergence or divergence of this series.
27:38:310Paolo Guiotto: So let's check this. We already computed what is this probability, because it is the probability, more or less, not exactly. It is the probability we computed here. We computed the probability where modus xn is less or equal than absolute, we want greater or equal. Well, that's the complementary. So, we have…
27:59:60Paolo Guiotto: the probability that modulus XN
28:03:200Paolo Guiotto: is less or equal than epsilon is 1 minus the probability modulo XN, well, formally, I should say, greater than epsilon, you know? But since here, these variables, they have a density, the probability that Xn takes,
28:22:180Paolo Guiotto: modulus Xn equal epsilon means Xn equal plus epsilon or minus epsilon. So the probability that Xn takes one single value is zero, because you have a continuous distribution, so I can…
28:34:520Paolo Guiotto: just say, greater or equal, it is the same. So this is 1 minus, and we take that number.
28:41:20Paolo Guiotto: It is 2 over pi a tangent of n epsilon, so 2 over pi.
28:46:780Paolo Guiotto: arctangent of N epsilon.
28:52:250Paolo Guiotto: Okay, so now we know that this quantity actually goes to zero. So this is the minimal requirement to have convergent… convergence, but at the same time, it does not…
29:07:150Paolo Guiotto: It is not income… Incompatible with this.
29:11:440Paolo Guiotto: Because suppose that you have… that this quantity goes to 0 as 1 over n squared, you have that the series of 1 over n squared is fine. But you could go to 0 as 1 over n, for example, and in that case, you would have that the series is infinite.
29:29:400Paolo Guiotto: So the single fact that this quantity goes to zero does not tell what happens to the series of this number. We need a more precise behavior of this, so we need to understand how this quantity goes to zero.
29:49:140Paolo Guiotto: So, we… Need it.
29:53:100Paolo Guiotto: to understand…
29:59:920Paolo Guiotto: Oh… this probability. So, in fact, this is an analytical quantity. 1 minus 2 over pi… R tangent…
30:12:890Paolo Guiotto: of N epsilon.
30:15:700Paolo Guiotto: goes to zero.
30:19:450Paolo Guiotto: So does it go to 0 as 1 over n, as 1 over n squared, or any other quantity? Okay.
30:27:40Paolo Guiotto: Now, how can we discuss this? Now, this has nothing to do with probability, it's, if you want…
30:35:700Paolo Guiotto: A sort of tricky calculus exercise, okay?
30:40:110Paolo Guiotto: Well, I will, give you a couple of possibilities here. Let's see if we can do both. The first one is a trick.
30:51:70Paolo Guiotto: which is related to the arctangent function, it's a very particular thing. The point is that here, the argument of the arctangent is going to plus infinity, right? What do we know about the arctangent at plus infinity? That it goes to pi half
31:06:300Paolo Guiotto: But we cannot use anything more precise than this, because at plus infinity, we cannot talk about the derivative, the tangent, things like this. We do not have a McClure formula and similar facts. Now, when you want to understand a fine behavior, a synthetic behavior of a function, normally you should try to use the… something like the McClure formulas, because
31:28:890Paolo Guiotto: But they are… they work if your argument is around zero, not at plus infinity.
31:34:160Paolo Guiotto: If the argument is around plus 50, you use Taylor formula, okay? But if it is at plus infinity, you cannot do that. But, fortunately here.
31:46:290Paolo Guiotto: We may.
31:49:280Paolo Guiotto: use… D.
31:52:540Paolo Guiotto: Remarkable.
31:57:530Paolo Guiotto: identity. There is a remarkable identity for the Arctangent that says.
32:02:190Paolo Guiotto: If you have a positive number, X, positive.
32:06:10Paolo Guiotto: And you do the sum of the arctangent of X, arctangent of X, plus the arctangent of 1 over x.
32:15:790Paolo Guiotto: You always get pi half.
32:19:240Paolo Guiotto: Well, the idea is, the following. You know that our tangent of X is the angular coefficient
32:28:360Paolo Guiotto: It's the angle whose angular coefficient is X. Now, X, if this line has angular coefficient X,
32:39:390Paolo Guiotto: the perpendicular line has angular coefficient 1 over X. And when you sum the two angles, you get pi half. Now, this is basically the geometric origin of this
32:50:710Paolo Guiotto: This, this relation.
32:55:740Paolo Guiotto: Now, if you use this relation here, so you have that 1 minus 2 over pi
33:01:310Paolo Guiotto: arctangent of N epsilon, so your x is this number, becomes 1 minus 2 over pi.
33:11:450Paolo Guiotto: Then, we write arc tangent of X, this one, equal pi half minus the arctangent of 1 over X.
33:20:670Paolo Guiotto: So this becomes pi half minus the arctangent of 1 over the argument, so 1 over n epsilon. As you see, that now the argument of the arctangent goes to 0.
33:34:720Paolo Guiotto: And at 0, we know fairly well what happens to the arctangent. So we do some little algebraic steps, so we have 1 minus 2 over pi, pi half is 1, plus 2 over pi, the arctangent of 1 over n epsilon.
33:53:970Paolo Guiotto: The ones, go away, and therefore we remain with the 2 over pi arc tangent of 1 over n epsilon, where now this argument goes to 0.
34:04:750Paolo Guiotto: So, our tangent, of X, when X goes to 0, is asymptotic to what?
34:14:670Paolo Guiotto: Well, if you remind that this is simple, it is asymptotic to X itself. Otherwise, you compute the first Maclaurin formula, you get the same conclusion. So we get that this is asymptotic to 2 over pi times 1 over n epsilon. And here we have the
34:34:00Paolo Guiotto: the right information, because… so the series of the probabilities of modulus exam
34:42:409Paolo Guiotto: Is gray, is greater or equal than epsilon.
34:47:560Paolo Guiotto: So what we have done here, was we computed this probability. The probability is,
35:01:780Paolo Guiotto: I did the… Sorry.
35:08:280Paolo Guiotto: this was greater or equal, because we wanted this, no? We want the probability that modulus Xn is greater or equal than epsilon. This is the prob… 1 minus the probability of the complementary, which is X less or equal than epsilon, which is the probability we computed above. I…
35:28:310Paolo Guiotto: I just, flipped this, to, sign.
35:32:390Paolo Guiotto: So now we have that the probability that modulus Xn is greater or equal than epsilon, which is 1 minus 2 over pi there tangent n epsilon. This one is equal to this 2 over pi arctangent of 1 over n epsilon, which is asymptotic to this.
35:50:30Paolo Guiotto: So we can say that this is asymptotic to the serious
35:54:350Paolo Guiotto: So there is a constant 2 over epsilon pi, and then we have a divergent harmonic series, so this is divergent by asymptotic comparison.
36:07:660Paolo Guiotto: And therefore, the series of the probabilities, modulus Xn greater or equal than epsilon is divergent as well.
36:16:900Paolo Guiotto: And, now we can use the Borel-Cantelli. The second, second Borel-Cantelli, lemma says that, since the series of these probabilities is plus infinity, and
36:31:390Paolo Guiotto: We know that DXN are independent, so these events are independent. We get that the probability of the limb subset, limb soup of the sets where modus Xn is larger than epsilon.
36:48:570Paolo Guiotto: It is equal to 1.
36:50:620Paolo Guiotto: So with probability 1, models XN is away from 0 infinitely many times, and this means that Xn can never go to 0.
37:01:810Paolo Guiotto: So, Xena.
37:03:800Paolo Guiotto: Does not go to zero, almost surely.
37:09:120Paolo Guiotto: And this finishes the argument.
37:12:40Paolo Guiotto: Well…
37:13:740Paolo Guiotto: Of course, this argument depends on this remarkable identity that probably, if you don't know, you never will… but maybe you may say, what is the R tangent at infinity, and maybe…
37:25:640Paolo Guiotto: you ask to chat GPT, and it will tell you that there is this identity. Otherwise, we could always say is that, so, alternative…
37:38:750Paolo Guiotto: Argumente, which is slightly more general, because this is very… it's based on a very particular thing.
37:46:230Paolo Guiotto: So, suppose that, let's return to the 1 minus 2 over pi arctangent over n epsilon.
37:57:550Paolo Guiotto: So, I suspect that I know that this goes to zero, and suppose that I want to… I want to bet that this is asymptotic to something like constant divided then to alpha.
38:14:230Paolo Guiotto: And determining this alpha will give the answer, because at the end, we know that this alpha is 1.
38:20:500Paolo Guiotto: And the constant C is 2 over epsilon pi, no? But suppose that I don't know, I want to determine this. How can I do that?
38:28:770Paolo Guiotto: Well, this means that, equivalently, 1 minus 2 over pi n epsilon divided the constant of n to alpha, this quantity goes to 1.
38:41:950Paolo Guiotto: when n, of course, goes to plus infinity.
38:45:370Paolo Guiotto: So basically, you see, I have to compute, I am,
38:50:950Paolo Guiotto: If you want, since I can always, for example, I can introduce an epsilon here.
38:59:120Paolo Guiotto: I'm just doing this… these are simple algebraic tricks to simplify this fraction. I put an epsilon here so I can write epsilon to the alpha here. There is no danger, epsilon is fixed, it's positive.
39:11:730Paolo Guiotto: So, epsilon to alpha is an innocent constant. So, this means that you have to prove that apart for a constant, 1 over C, epsilon to alpha.
39:23:20Paolo Guiotto: Then you have this epsilon n to alpha. Now, this is a double denominator, so it comes up to the numerator. 1 minus 2 over pi, arctangent
39:37:800Paolo Guiotto: of N epsilon, this thing… goes to 1.
39:43:380Paolo Guiotto: So, we are led to consider…
39:47:180Paolo Guiotto: the limit of this expression when n goes to plus infinity, and see whether it is not 1, but a constant different from 0, then we will rescale by dividing by the constant, we get 1.
40:00:960Paolo Guiotto: And since when n goes to plus infinity, I can say, well, let's call epsilon n X. I can say, so we…
40:10:390Paolo Guiotto: We are led… to… the… limit…
40:17:720Paolo Guiotto: for x going to plus infinity of X to alpha psi, 1 minus 2 over pi arctangent of X.
40:29:70Paolo Guiotto: Oh, now we can move this to zero. We do the change of variable, set y equal 1 over X, and this becomes a limit for y going to 0, positive.
40:42:340Paolo Guiotto: of, that X to alpha becomes, what? Becomes 1 over Y2 alpha, so the denominator y to alpha, and here we have 1 minus 2 over pi. The arctangent
40:57:740Paolo Guiotto: of 1 over Y.
41:00:970Paolo Guiotto: Now, if you look at this, when y goes to zero, we already know that the numerator goes to zero, denominator goes to 0, so it is a 0 versus 0 in determinate form, so we can use the below-the-hopital rule, and we have the limit when Y goes to zero.
41:17:290Paolo Guiotto: Plus, denominator becomes alpha Y to alpha minus 1. Numerator, so the 1 disappears in the derivative, minus 2 over pi.
41:28:340Paolo Guiotto: The derivative of the arctangent is 1 over 1 over Y, which is the argument squared, and then we have the derivative of 1 over Y, which is minus 1 over Y squared.
41:43:340Paolo Guiotto: Okay, when Y goes to 0, when Y goes to 0 plus…
41:53:630Paolo Guiotto: So, let's, let's do…
41:56:220Paolo Guiotto: It would have been to work directly with plus infinity, however. Let's do some algebraic restyling here. This becomes Y squared plus 1 divided Y squared, I write at numerator, so Y squared divided 1 plus Y squared.
42:13:550Paolo Guiotto: I can write like that. I can cancel this Y squared with the Y squared in denominator here.
42:20:540Paolo Guiotto: I arrange a bit the sine becomes plus, so I have limit. When y goes to 0 plus of plus 2 over pi, then we have… what remains is 1 over alpha, y to alpha minus 1, 1 plus Y squared.
42:39:350Paolo Guiotto: And this limit can be easily computed, because this 1 plus Y squared goes to 1, y goes to 0.
42:47:80Paolo Guiotto: And all the story depends on this quantity.
42:50:350Paolo Guiotto: Y to alpha minus 1.
42:52:710Paolo Guiotto: So we have three possibilities. If alpha is larger than 1, that exponent alpha minus 1 is positive, Y is going to 0, so
43:03:660Paolo Guiotto: Y to alpha minus 1, we go to 0, positive, so I have 1 over 0, denominator of that fraction, positive, the limit will be plus infinity.
43:15:970Paolo Guiotto: When alpha is equal to 1,
43:18:790Paolo Guiotto: And only for this case, you see that the Y to alpha minus 1 becomes Y to 0, so it's constantly equal to 1, and so the limit is 2 over pi for this case.
43:32:180Paolo Guiotto: While when alpha is less than 1, but positive, here alpha is supposed to be positive, now this exponent is negative, alpha minus 1, y is going to 0, 0 to negative is plus infinity. So the denominator goes to plus infinity, the fraction goes to 0.
43:51:370Paolo Guiotto: And then, from this, you see that there is a unique case where this quantity does not go to 0 or infinity, but it goes to a finite number different from 0, and it is the case alpha equals 1.
44:05:530Paolo Guiotto: So you can say that for alpha equal 1, that limit, the initial limit, is 2 over pi. So, if you put alpha equal 1 in this, it becomes X,
44:18:760Paolo Guiotto: Times 1 minus 2 over pi, arctangent.
44:23:760Paolo Guiotto: of x goes to 2 over pi, so it means that if you divide by 2 over pi, or you multiply by pi half, you go to 1,
44:34:450Paolo Guiotto: And therefore, returning back to our n epsilon, it means that, so, if you put here n epsilon, n epsilon, and you write that at denominator, 1 over 2 over pi, arctangent.
44:51:330Paolo Guiotto: of N epsilon, here, divided by… so, when I have to divide, this becomes a double denominator, and this is a 2 over pi.
45:03:540Paolo Guiotto: and epsilon, this quantity goes to 1. And this says that… 1 minus 2 over pi.
45:12:350Paolo Guiotto: The arctangent of N epsilon goes to 0 asymptotically as 2 over pi divided n epsilon, which is exactly the same
45:25:00Paolo Guiotto: result we obtained here by using the direct… that particular formula. You see? 2 over pi times 1 over n epsilon is 2 over pi divided n epsilon.
45:39:600Paolo Guiotto: Okay, so this is a bit more flat. Now, probably there is some other question. Is there any other question, this? The size?
45:49:80Paolo Guiotto: So we, we discussed the point-wise convergence, yeah, so that's over.
46:00:20Paolo Guiotto: Okay.
46:03:760Paolo Guiotto: I want also to do the next step.
46:07:780Paolo Guiotto: Exerciser, the 857.
46:15:520Paolo Guiotto: Okay, so here we have…
46:22:30Paolo Guiotto: yeah,
46:25:180Paolo Guiotto: I'm sorry, because I've not, totally checked these exercises. Here it is written. UN is a sequence of, random… it is missing something here, that they are independent. So, independent.
46:45:670Paolo Guiotto: Otherwise, you cannot basically compute anything here. Independent,
46:51:160Paolo Guiotto: Random variable, each with distribution which is uniform in the interval 0, 1.
47:01:350Paolo Guiotto: Then it defines XN as the minimum, of these U1 yuan.
47:14:280Paolo Guiotto: Okay, there are two questions. Number one, determine the CVF of XN.
47:27:950Paolo Guiotto: Question 2… discuss… convergence.
47:37:10Paolo Guiotto: Offer.
47:38:820Paolo Guiotto: this sequence N times XN.
47:46:870Paolo Guiotto: Okay, so let's start working on the first Questions.
47:52:330Paolo Guiotto: So here we have to determine the CDF.
47:57:270Paolo Guiotto: And, so, this is a standard, so… look carefully at this example, because it may be asked many times to work with minimum, maximum of a set of variables.
48:11:290Paolo Guiotto: So, the CDF of, XN
48:16:340Paolo Guiotto: Well, here there are two ways. Let me show you. First, let's say, the straightforward way, no?
48:25:430Paolo Guiotto: I say, by definition, this is the probability that XN is less or equal than X.
48:33:700Paolo Guiotto: So this means that I have to consider the probability of this event, the minimum.
48:39:340Paolo Guiotto: of U1, U2, etc, UN is less or equal than X.
48:46:470Paolo Guiotto: Now, this is not easy, but there is a standard idea, so this is the interesting part of the story, which is the following. So, how can you say that this minimum is less or equal than
49:01:00Paolo Guiotto: X. Formally, the minimum of… is less or equal than X if, and only if, at least one of the U is less than X, no? So, I should say, it is U1 less or equal than X. Union
49:18:300Paolo Guiotto: you to… of course, what I'm trying to do is to use what I know. I know the distributions of the UN, so I want to go back to the distributions of UN.
49:28:970Paolo Guiotto: Okay? So I could say that this is, this union.
49:37:480Paolo Guiotto: UN less or equal than X.
49:39:940Paolo Guiotto: Well, in this way, the problem is that that union is not necessarily disjoint.
49:45:620Paolo Guiotto: Because you may have that 3U are less than X.
49:49:410Paolo Guiotto: So, these are not disjoint, and then we can transform this into a disjoint union, because if I say, okay, let's take U1 less or equal than X, then we take the part of U2, which is less than X,
50:05:340Paolo Guiotto: But with U1, which is greater than X, you know?
50:09:560Paolo Guiotto: Because this now becomes a disjoint union. The union of these two is the same of the union of these two. This is something we have done when we have
50:18:590Paolo Guiotto: seen the continuity from below of the measures. Now, out to that, the union has a disjoint union. You take the second set, and you take out the first set, no? So, this formally would be U2 less or equal than X minus
50:36:720Paolo Guiotto: U1 less or equal than X, which is equivalent to say that you take U2 less or equal than X, and U1 greater than X.
50:47:200Paolo Guiotto: This makes, now, this a disjoint union. And then we can continue in that way. So the next one will be U3 less or equal than X, and both U1 and U2 greater than X, and so on.
51:03:280Paolo Guiotto: Okay.
51:05:350Paolo Guiotto: The last one of this union will be…
51:09:320Paolo Guiotto: UN less or equal than x, and all the U1, U2, etc. UN minus 1 are greater than little s.
51:18:480Paolo Guiotto: Now, we have a disjoint union, and we can decompose the probability as the sum of the probabilities. So, FXN
51:27:730Paolo Guiotto: of little x is now the probability of this joint union, so I can say it is the probability that U1 is less or equal than X, plus the probability that U2 is less or equal than X, but U1 is greater than X.
51:46:430Paolo Guiotto: Plus.
51:47:490Paolo Guiotto: the probability that U3 is less than X, and U1 and U2 are greater than X, and so on.
51:57:910Paolo Guiotto: until the last one, the probability that UN is less or equal than X, then U… U1 is greater than X, U2 is greater than X, etc, until UN minus 1 is greater than X.
52:16:400Paolo Guiotto: Okay, now, remind that you are uniformly distributed, so they have the same CDF. So, in particular, this is the CDF
52:27:330Paolo Guiotto: the CDF of U1 at point X. What is this?
52:33:410Paolo Guiotto: Well, this is where the independence enters, so if you miss the independence, you don't know how to move here. So this, by independence, becomes the product of the two probabilities, no? By independence.
52:49:790Paolo Guiotto: becomes, the product of U2 is less or equal than X times the probability that U1 is greater than X.
53:00:880Paolo Guiotto: But then, this is FU2,
53:04:700Paolo Guiotto: of X times 1, I can look at this as 1 minus the probability that U1 is less or equal than X, which is FU1.
53:17:180Paolo Guiotto: Go backs.
53:18:290Paolo Guiotto: And so on. So, the next one will be, you understand easily the mechanism. F3 of X, then we will have a product F1 minus F U1X times 1 minus F U2X.
53:36:170Paolo Guiotto: and so on, plus the last one will be F-U-N.
53:58:650Paolo Guiotto: value of FXNX. Now, we know that all the UN are distributed in the same way, so this means that they have the same CDF, so that FU1, FUN are the same function, okay? So, FU1.
54:16:420Paolo Guiotto: Equal F, U2, equal, etc. F,
54:20:820Paolo Guiotto: UN. Now, what is the, let's call it F of X.
54:26:320Paolo Guiotto: which is the CDF of a uniform random variable. Now, what is the CDF? This is, is, zero,
54:37:810Paolo Guiotto: So the variable is distributed in the interval 01 here, so the CDF is 0 when X is less than the
54:47:280Paolo Guiotto: smallest value, which is 0, because this is the probability that you are less than X. When x is less than 0, that probability is 0. It is 1 when X is larger than the biggest value, which is 1.
54:59:470Paolo Guiotto: And then, in the middle, it is linear from 0 to 1. In this case, the general formula would be X minus A divided B minus A. That's for the interval AB. The interval AB is the interval 0, 1, so it becomes just X for X between 0 and 1.
55:19:170Paolo Guiotto: Okay, so, this means that our FXN of X
55:26:200Paolo Guiotto: is equal to… so FU1 of X, so that function of X, well, we can see that if x is, we can also understand from the definition, if the little x is less than 0,
55:42:240Paolo Guiotto: the probability that the minimum being less than 0 means is 0, because the U are positive, are between 0 and 1, so the minimum cannot be negative, as well as if x is greater than 1, the probability that the minimum of these variables is less than 1 is 1, because the variables are all between 0 and 1, so the result will be
56:06:660Paolo Guiotto: 0 for X less than 0, and 1 for X greater or so equal than 1. And in the middle, for X between 0 and 1,
56:19:670Paolo Guiotto: We have this formula, so…
56:24:180Paolo Guiotto: you see that DF are all equal to X, so I have something like this. FU of X is X, plus, you see, FU of X times 1 minus FU, so it is X,
56:38:70Paolo Guiotto: times 1 minus X plus. The next one is this one, I'm talking about this.
56:45:680Paolo Guiotto: It is FU3X times 1 minus FU1 times 1 minus FU2, but all these F are X, so I have X times 1 minus X times 1 minus X, so X times 1 minus X squared.
57:01:930Paolo Guiotto: And so on.
57:03:620Paolo Guiotto: We continue until the last sum, which is this one, the last term in this sum.
57:09:410Paolo Guiotto: which is X, you have to replace to each FX, no? So X times 1 minus X, n minus 1 times. So it will be plus X1 minus X n minus 1.
57:24:350Paolo Guiotto: So this is… let's focus on this. We can factorize X, and then you see that we have the sum of the powers of quantity 1 minus X for X, for, say, for the exponent from 0 to n minus 1, so sum
57:40:500Paolo Guiotto: for K going from 0 to n minus 1 of 1 minus X to the exponent k, which is a geometric sum.
57:49:400Paolo Guiotto: the value of the sum is known. It is 1 minus 1 minus X to exponent n divided 1 minus X. Of course, we can… we cannot take X equal 1 in this formula, and that's natural, because also
58:04:310Paolo Guiotto: The value of this summer, would be zero for that, for that, apart for the first step.
58:12:860Paolo Guiotto: So, we can say that, finally, DFXN of X is equal to…
58:20:720Paolo Guiotto: 0 for X less than 0 is equal to 1 for X greater or equal than 1, and it is equal to this quantity, well, X times over 1 minus X times 1 minus
58:35:930Paolo Guiotto: 1 minus X to the n for X between 0 and 1.
58:47:270Paolo Guiotto: Okay. Now, I want to illustrate to you an alternative and smart way to do this calculation.
59:01:400Paolo Guiotto: So, we could… we could have, done this way. FXN of X is the probability that
59:16:220Paolo Guiotto: Yeah, the minimum… of these, U1, UN.
59:23:490Paolo Guiotto: is less or equal than X.
59:26:870Paolo Guiotto: Now, as above, let's assume the cases where this calculation is trivial. So, 4X, let's say 4x between 0 and 1, where it's non-trivial, this.
59:40:240Paolo Guiotto: What we could do, we could say that this is 1 minus the probability of the complementary.
59:45:740Paolo Guiotto: The probability of the complementary means the minimum of U1.
59:51:190Paolo Guiotto: UN is greater than X.
59:58:570Paolo Guiotto: Now, if you think about the event, this event.
00:02:890Paolo Guiotto: I think there is something wrong with this calculation, it smells, but let's see what comes with the second way. Now, the minimum is greater than X, means that all the U must be greater than X, no?
00:17:130Paolo Guiotto: Because the minimum is one of them, it's a minimum authentic number of things, so U1 greater than X, U2 greater than X, etc, UN greater than X. Now, we can split this probability with the product, because they are independent, so by independence.
00:36:670Paolo Guiotto: we have that this becomes the product of the probabilities, so product of probability, we'll say UK greater than X for K going from 1 to N.
00:48:610Paolo Guiotto: And this one is now, related to the CDF. It is 1 minus the probability that that UK is less or equal than X.
00:59:380Paolo Guiotto: For which we have the formula, no?
01:01:720Paolo Guiotto: This one is, since X is between 0, 1, this is just X. So at the end, we have a product for k going from 1 to n of 1 minus X,
01:17:290Paolo Guiotto: And since this quantity is independent of K, this is 1 minus X to power n.
01:24:330Paolo Guiotto: At the end.
01:25:680Paolo Guiotto: So, we, we got this much easier, and,
01:32:380Paolo Guiotto: were different, so there is an error in the first part. DFXN of X is 1 minus…
01:39:870Paolo Guiotto: 1 minus X to the end. So we would have that FXN of X, this is for X between 0 and 1. So FXN of X is
01:52:950Paolo Guiotto: 0 for X less than 0 is 1 minus 1 minus X to the n for X between 0 and 1, and 1 for X greater than 1.
02:07:270Paolo Guiotto: Now, the point is, what went wrong for the… with the… definitely this second way is much easier.
02:14:130Paolo Guiotto: less tricky.
02:17:720Paolo Guiotto: But where did the mistake here?
02:24:170Paolo Guiotto: Yeah.
02:27:740Paolo Guiotto: You know what is the mistake? Do you see?
02:31:180Paolo Guiotto: It is here, because the sum for k going from 0 to n minus 1 of, say, Q to K, let's put this.
02:40:910Paolo Guiotto: is 1 minus Q2, this exponent plus 1. So, the N, divided 1 minus Q. But here, what Q is 1 minus X, so I would have
02:52:140Paolo Guiotto: I should have written this, 1 minus 1 minus X. Now it comes the same, because you see that
02:58:730Paolo Guiotto: 1 and 1 disappear, that minus minus X becomes a plus X, that cancels with this X, so at the end, we get, again, 1 minus 1 minus X to the n. So this is the coefficient that shouldn't be there. So now it's the same.
03:17:190Paolo Guiotto: Okay, so now we have the CDF of this random variable, and the second question asks to discuss convergence. It's a bit vague, because it does not tell which sensor of an
03:34:630Paolo Guiotto: XN, so convergence.
03:37:380Paolo Guiotto: of this.
03:40:250Paolo Guiotto: Now… It's, what could be the idea? It's difficult, because, so…
03:51:750Paolo Guiotto: Here, let's… let's see if the weakest convergence fails, or… because if the weakest fails, it is clear, we may say, let's start with the strongest, but then we have the strongest, there is not the strongest. We have two
04:07:360Paolo Guiotto: Not equivalent convergence that we consider stronger, which are the point-wise, the almost sure convergence, and the convergence in mean, no?
04:20:650Paolo Guiotto: But the… you should say, yeah, but they… this converges to 1. Okay, that's the difficulty, because we don't see exactly what should be the limit here.
04:30:940Paolo Guiotto: Or, equivalent, we could try to start with the weakest convergence. If that fails, we cannot have any stronger.
04:38:440Paolo Guiotto: If that sequence converges in distribution, then we may have a candidate for the limit, and we could try to see how better can we improve this convergence. So, let's see what is the convergence of this in distribution.
04:57:120Paolo Guiotto: Now, we need… be careful, because it is not XN, but NXN, okay? So, since I have the CDF of XN,
05:06:500Paolo Guiotto: I'm not going to pass to the limit in the CDF of XN, because that's not the sequence I'm considering. Here, I have to take this one, so say YN equal NXN. However, it's not a big problem, because if you look at the CDF of YN,
05:23:520Paolo Guiotto: So let's say this is the probability that YN is less or equal than
05:30:580Paolo Guiotto: Which is, by definition, Yn is NXN less or equal than Y, and we can clearly divide by N. Here, n is an integer, so it means fxn less than Y over N.
05:43:910Paolo Guiotto: And that's now the CDF we just computed, FXN of Y divided by N.
05:51:460Paolo Guiotto: So we can say that this is…
05:53:690Paolo Guiotto: equal to 0 if Y over N is negative, this means Y negative.
06:00:410Paolo Guiotto: It is 1 if y over n is greater than 1, so means Y greater than nth.
06:08:470Paolo Guiotto: And for Y between 0 and N,
06:12:60Paolo Guiotto: It is 1 minus 1. Remind the formula, FXN of X is 1 minus 1 minus X to the n, so we have to plug this X here, the value Y over N, so Y over N to the n.
06:29:410Paolo Guiotto: So this is the CDF of this sequence, okay? So let's copy here. FYN at Y is equal to 0 for Y negative.
06:43:860Paolo Guiotto: Then, 1 minus 1 minus 1 over n to the n for Y between 0 and N…
06:52:820Paolo Guiotto: and 1 for Y larger than N.
06:57:800Paolo Guiotto: Now, since we have explicitly written this CDF of this.
07:03:950Paolo Guiotto: we could try to see what happens if we do the point-wise limit. Remind that the characteristic property is that this should converges… should be conver- pointwise convergent at every continuity point of the limit.
07:23:560Paolo Guiotto: We don't know yet what is the limit.
07:26:40Paolo Guiotto: So we have to identify what happens here. So let's see, however, what can be said about the point-wise limit of this. Now, clearly, for Y negative, we are doing a limit of 0, we get 0. So…
07:40:130Paolo Guiotto: If Y is negative.
07:43:810Paolo Guiotto: we have that the limit in n of these FYN of Y will be equal to 0.
07:51:910Paolo Guiotto: If Y is positive, what value should I consider for FYM?
07:58:180Paolo Guiotto: Well, you see that since Y is fixed here, and n is going to infinity, clearly, sooner or later, when I have a Y positive, I am in this case, because Y is somewhere at right of 0. So for a small value of N,
08:15:780Paolo Guiotto: I am at left of Y, so Y is larger than N. But, definitely, this N will go to infinity, so it will be here, okay? So my Y is between 0 and N. So, if Y is greater or equal than zero.
08:33:740Paolo Guiotto: for N going to plus infinity, my Y would be between 0 and n. So the CDF
08:43:490Paolo Guiotto: Yn at point Y will be 1 minus 1 minus Y over n to the N.
08:53:29Paolo Guiotto: And what happens to this quantity when we send n to plus infinity?
08:58:950Paolo Guiotto: Maybe you recognize that this is the quantity
09:03:220Paolo Guiotto: More or less, that yields an exponential, so you may remind that 1 plus X over N to the n
09:10:790Paolo Guiotto: This quantity goes to e to the X for every X real.
09:17:60Paolo Guiotto: So we put this with the minus Y, and we go to 1 minus E to minus Y,
09:24:910Paolo Guiotto: And this happens for every Y.
09:28:670Paolo Guiotto: There is no condition for Y. So, in particular.
09:34:210Paolo Guiotto: for every Y positive. So the conclusion is that the limit when n goes to plus infinity of F,
09:49:939Paolo Guiotto: YN. Of Y is… Equal to 0 when Y is negative.
09:56:10Paolo Guiotto: And when Y is positive, it is equal to 1 minus E2 minus Y.
10:03:480Paolo Guiotto: So we got the non-trivial limit.
10:06:990Paolo Guiotto: Now, if you look at this function, f of y.
10:11:290Paolo Guiotto: So you may recognize something, because this is a known
10:15:210Paolo Guiotto: CDF. It is the CDF of an exponential random variable. It is, in any case, if you don't know that… if you don't recognize the specific random variable, you notice that this function, f of y, is made like that. It is 0 for Y negative.
10:33:500Paolo Guiotto: For Y positive, it is 1 minus the E2 minus Y. What is this function? It's a function… so E2 minus Y is the exponential decreasing down to 0, so minus will go up, so 1 minus will be something of this type that goes to 1 when y goes to plus infinity.
10:52:970Paolo Guiotto: So, the function, that f of y, is a function, which is between 0, 1 for every y.
11:01:00Paolo Guiotto: It is 0 at minus infinity, it is 1 at plus infinity. It is increasing. It is right continuous, it is even more, it is continuous. This characteristics makes this a CDF, so there is definitely a random variable with this CDF.
11:20:210Paolo Guiotto: Okay, so this is the CDF of some Y.
11:24:420Paolo Guiotto: As soon as this happens, this means that our YN, so NXN, goes in distribution to that Y. If we want, we can say that Y is an exponential random variable with parameter 1.
11:43:670Paolo Guiotto: Where the CDF is this, and the density, FY of Y, is E2 minus y times indicator of positive Ys.
11:57:210Paolo Guiotto: This is the density of this random variable, which is the derivative, nothing but the derivative of the CDF.
12:05:990Paolo Guiotto: Okay Now, we know that this sequence converges weekly to this Y,
12:14:100Paolo Guiotto: It is now… we should try to see if this convergence is any stronger convergence.
12:25:230Paolo Guiotto: Wow.
12:26:340Paolo Guiotto: Now, is… is…
12:32:120Paolo Guiotto: Of course, this question is somehow open, no? It's not, say, the discussed convergence. It can be, everything, or, or, or…
12:43:360Paolo Guiotto: So…
12:45:10Paolo Guiotto: you could say it converges in distribution, and that's it, not what you can say. We can wonder if, is this sequence X and XN also
12:58:30Paolo Guiotto: convergent… in probability, Which is the next level.
13:05:640Paolo Guiotto: Now, if this happens, since, if this happens.
13:17:980Paolo Guiotto: Then, necessarily, that N, Xn must converge in probability to the
13:24:310Paolo Guiotto: Why? Because this… this… if it converges to Z, it should converges also in distribution to the same Z, okay? So the problem is, is this convergent in probability to that Y? So this means that, equivalently.
13:44:300Paolo Guiotto: the probability.
13:46:760Paolo Guiotto: that the distance between NXN and Y Great or equal than epsilon.
13:53:460Paolo Guiotto: Must go to zero for every epsilon positive.
13:57:340Paolo Guiotto: If this happens, the answer is yes. If this is not verified, the answer is no.
14:04:980Paolo Guiotto: Okay, now, the problem here…
14:07:840Paolo Guiotto: it makes this problem a little bit more complicated than the previous one. For example, the previous one, we had Y equals 0, so this probability depends, once again, by the Xn, so we can't compute explicitly. Here, if we want to compute explicitly this probability, we would need the joint distribution of these two.
14:28:440Paolo Guiotto: And who knows this?
14:30:360Paolo Guiotto: We do not have this.
14:32:850Paolo Guiotto: So how can we assess this probability?
14:35:990Paolo Guiotto: Now, this is, we can do, let's say, in,
14:40:130Paolo Guiotto: direct elementary way, trying to separate X and Y, because we have the single distributions, so we have to use that fact. So how can we do that? Now, this is… this literally means that, N, XN
15:00:30Paolo Guiotto: minus Y, the distance between these two, maybe, yeah, is, either less than minus epsilon, or NXN minus Y is greater than epsilon.
15:16:930Paolo Guiotto: What I want to do is,
15:20:610Paolo Guiotto: to… well, this is a disjoint union, because either you are less than minus epsilon or greater than epsilon, so I can focus on one of the two. Let's focus on, for example, this one.
15:32:410Paolo Guiotto: The probability… so the probability of this is the probability of this, which is the sum of the probabilities.
15:43:900Paolo Guiotto: They are similar.
15:45:920Paolo Guiotto: So let's consider the probability where NXN minus Y Is less than minus epsilon.
15:55:60Paolo Guiotto: Now, I try to split this…
15:58:850Paolo Guiotto: in this way, so I can say that, well, this is, N… Xn less than Y.
16:09:250Paolo Guiotto: Minus steps, you know?
16:12:470Paolo Guiotto: Okay, let's introduce, let's introduce, we can say that this is,
16:24:920Paolo Guiotto: I split this event into Y less than something, that we will decide later, so Y less than alpha.
16:34:340Paolo Guiotto: Or Y greater than alpha.
16:37:980Paolo Guiotto: In the first case, so in this, in this way, so then we have also NXN less than Y minus epsilon, and the same here, NXN less than
16:52:360Paolo Guiotto: And Xn minus Y less than minus epsilon.
16:56:970Paolo Guiotto: So this becomes, because these are two disjoint sets, you know, here you have Y less than alpha, and here Y greater than alpha, this splits into the sum. Probability that Y is less or equal alpha.
17:10:870Paolo Guiotto: N. N.
17:13:10Paolo Guiotto: XN less than Y.
17:15:990Paolo Guiotto: minus epsilon, plus probability that Y is greater than alpha.
17:25:480Paolo Guiotto: And the N. Dixon.
17:30:550Paolo Guiotto: Less than Y minus epsilon.
17:36:460Paolo Guiotto: Okay. Now, reminder that, so…
17:42:910Paolo Guiotto: If we want to prove a convergence in probability, we have to prove that this probability goes to zero.
17:49:720Paolo Guiotto: Normally, either you compute exactly the probability, but that's possible in only few circumstances, or you estimate the probability, saying that this is smaller than something which is smaller than something else that at the end goes to zero.
18:07:890Paolo Guiotto: But if I want to disprove this.
18:11:500Paolo Guiotto: I have to follow a similar but opposite strategy. So, to prove that this does not go to zero, I have to say that this probability is bigger than some other probability, which is bigger than something else that does not go to zero. This is…
18:27:750Paolo Guiotto: the two possibilities. These are the two possibilities. So what I want to follow here is the second one. So I will do something to get a lower bound that shows that this quantity cannot go to zero.
18:40:280Paolo Guiotto: Okay?
18:41:510Paolo Guiotto: So, what I do is the following. Notice that in this event, since Y is less or equal than alpha.
18:53:850Paolo Guiotto: Okay, and NXN is less or equal than Y minus epsilon.
19:22:650Paolo Guiotto: Yeah, I transformed this into…
19:27:980Paolo Guiotto: Because with this way, I have, let's say, with this, written in this way, I could say, if Y is less than alpha, and then Xn must be less than Y minus epsilon, this will be less than alpha minus epsilon.
19:43:900Paolo Guiotto: But this means that this, if I eliminate this, this event is contained into this, so I'm going to…
19:51:790Paolo Guiotto: Assessed from above the probability.
19:57:70Paolo Guiotto: well, I've done the calculation before, I've discovered that if I do this, and I go to the end, I discover that I do not get a good bound, because again, the bound is 1.
20:06:710Paolo Guiotto: Okay? But if I do the opposite, it works. So I want to show an inclusion here, and therefore, I cannot put alpha in place of Y in this case. So what I do is the following.
20:19:910Paolo Guiotto: This was to explain. So, I can see this as an intersection, this and this.
20:27:340Paolo Guiotto: Okay? So, if you are… if an omega is in this event, it means NXL is less than Y minus epsilon.
20:36:780Paolo Guiotto: Right? So I say that this is y less or equal than alpha minus those omega for which NXN is larger than Y minus epsilon.
20:49:340Paolo Guiotto: No? Because if you are there, NXN is less than epsilon, you can also say that you are not in the event where NXN is greater than. So, you write this as a difference. Notice that this event here
21:06:650Paolo Guiotto: Now, here, since Y is less than alpha.
21:11:820Paolo Guiotto: Since NXN must be greater than Y minus epsilon, if NXN is larger than alpha minus epsilon.
21:22:870Paolo Guiotto: This event is, this is in… sorry, this event is contained into this, because if you are larger than
21:34:480Paolo Guiotto: alpha, you are larger than Y, right? Now, so you can say this implies that you are larger than Y minus epsilon. So since this is contained in this, when you subtract, it is the opposite. So this contains
21:48:420Paolo Guiotto: Y less or equal than alpha minus NXN greater than alpha minus epsilon.
21:57:960Paolo Guiotto: You see?
21:59:330Paolo Guiotto: So, the event NXN greater than alpha minus epsilon, if you are greater than alpha minus epsilon, and Y is less than alpha, you are also greater than Y minus epsilon. So, if you are here, you are here. But since here we have a complementary, a minus.
22:16:840Paolo Guiotto: If you are not in this one, you cannot be in this one. So, this means that the right one is smaller.
22:24:600Paolo Guiotto: And therefore, when we do the probability, so the probability that Y is less or equal alpha, NXN,
22:34:260Paolo Guiotto: less or equal than Y minus epsilon. This thing here, is greater or equal than the probability that Y
22:43:410Paolo Guiotto: Is less or equal than alpha.
22:48:850Paolo Guiotto: minus NXN greater than alpha minus epsilon.
22:54:530Paolo Guiotto: Now, we can say that this is greater or equal
22:58:650Paolo Guiotto: This set, this event, is not necessarily contained in this one, so I cannot say the probability of the difference is the difference of the probabilities.
23:07:700Paolo Guiotto: Okay? What I could say, formally, is probability of, say, an event A minus an event B, if they are generic, I could say it is the probability of A minus the probability of what of B is in A.
23:24:530Paolo Guiotto: This, in this case, would be correct, because what of B is not in A, is not in that set. So, formally, this is the correct identity.
23:34:50Paolo Guiotto: But since this one is less than the probability of B,
23:38:890Paolo Guiotto: You see, with the minus in front, this becomes an inequality this way, probability of A greater minus the probability of B. So I can say that the probability of a difference is larger always than the difference of the probabilities. This is the sub-addativity.
23:55:210Paolo Guiotto: So here I get the probability that Y is less or equal than alpha.
24:00:460Paolo Guiotto: minus the probability that N, XN,
24:04:350Paolo Guiotto: Is greater than alpha minus epsilon.
24:09:740Paolo Guiotto: And now, you see, I split this Y and NXN, for which I have the distributions. For this, this is 1 minus E to minus alpha, if I'm not wrong.
24:25:820Paolo Guiotto: Well, we assume that this alpha is positive.
24:28:870Paolo Guiotto: Otherwise this quantity would be zero.
24:32:60Paolo Guiotto: For the second one, the probability that NXN is larger than alpha minus epsilon is 1 minus the probability that NXN is less than or equal than alpha minus epsilon, and for this, we have the formula. This is 1 minus…
24:50:580Paolo Guiotto: Where is it?
24:55:520Paolo Guiotto: 1 minus, the, the exponential.
25:00:150Paolo Guiotto: Thing.
25:02:70Paolo Guiotto: So, 1 minus, 1 minus… this is the Y alpha min of epsilon minus epsilon over n to the n.
25:17:140Paolo Guiotto: Okay.
25:19:250Paolo Guiotto: Now, look, this cancels this, so it remains equal. 1 minus e to negative alpha minus minus plus minus 1 minus alpha minus epsilon over n to the n. Now, what happens to this when n is big?
25:38:870Paolo Guiotto: When n goes to plus infinity, this goes to 1 minus C2 minus alpha.
25:44:400Paolo Guiotto: minus, you see this, goes to E2 minus alpha minus epsilon.
25:51:370Paolo Guiotto: So, this means that,
25:54:150Paolo Guiotto: Basically, we can make this probability… imagine that you take the smallest possible value of epsilon is 0. We can make this probability very close to 1.
26:06:530Paolo Guiotto: Because that exponential, there are… so this is when epsilon goes to 0, 1 minus 2E2 minus alpha. So it means that if you choose alpha large enough, this is…
26:19:990Paolo Guiotto: more or less won this probability. So, in particular, this one won't go to zero.
26:27:190Paolo Guiotto: And if you repeat the same bound on this one, on the second, probability, you get the same, so that probability cannot go to zero. So, from this, it follows that the probability of NXN minus Y
26:44:790Paolo Guiotto: Greater than epsilon.
26:47:740Paolo Guiotto: Cannot go to zero.
26:50:380Paolo Guiotto: So this, this means that this sequence cannot be convergent.
26:58:140Paolo Guiotto: in probability.
27:02:790Paolo Guiotto: Okay, so now it's, a bad way of…
27:06:350Paolo Guiotto: 3 minutes, so I, I do not start the…
27:12:480Paolo Guiotto: It's, we just closed here because…
27:16:70Paolo Guiotto: It would have no sense to, start, a new path now, okay?
27:24:620Paolo Guiotto: Okay.
27:29:460Paolo Guiotto: So… Let's finish here.
27:39:200Paolo Guiotto: Something called it.