AI Assistant
Transcript
00:06:990Paolo Guiotto: Good morning, guys.
00:08:770Paolo Guiotto: So… Right.
00:11:790Paolo Guiotto: to start with some equal sizes. I checked that,
00:17:400Paolo Guiotto: The exercises on, on the notes are not…
00:22:210Paolo Guiotto: Particularly easy, but however, so let's, at least those I gave you.
00:30:670Paolo Guiotto: So I left you to do this exercise 8, 5, 3…
00:37:30Paolo Guiotto: This is not difficult. It says the XN are independent.
00:45:360Paolo Guiotto: And it says that, well, first of all, there is this notation, which is not the standard notation.
00:52:870Paolo Guiotto: It was an old notation, B011 minus 1 over N. What does it mean? This is a Bernoulli random variable that takes value 0, 1,
01:03:420Paolo Guiotto: So it means that the probability that Xn is equal to 0 is that value, 1 minus 1 over n, and therefore the probability that Xn is equal to 1 will be the
01:17:370Paolo Guiotto: complementary, so 1 over N.
01:22:900Paolo Guiotto: So, the problem consists in, discussing, discuss… LP, almost sure… Convergences.
01:40:220Paolo Guiotto: And also, convergence in probability.
01:46:270Paolo Guiotto: And in distribution.
01:51:480Paolo Guiotto: Well, we have not yet introduced these two lasts.
01:54:650Paolo Guiotto: So let's, start the focusing on the…
01:58:230Paolo Guiotto: First. Now, for the LP convergence, we needed to identify a limit
02:07:60Paolo Guiotto: So, what should be a reasonable limit? We know that the probability that XN is 0, Xn takes just two values, 0, 1, no? So, Xn of omega
02:19:30Paolo Guiotto: Is equal to zero with, probability.
02:25:470Paolo Guiotto: 1 minus 1 over n, so for n large, this is very close to 1, while XN is equal to 1 with a residual probability 1 over n. So, when n is large, this is very small. So this makes reasonable the guess that maybe XN
02:42:730Paolo Guiotto: Go to zero, because it's the most probable value, no?
02:48:610Paolo Guiotto: Now, the problem is, is this true in, any LP norm? Since we know that the LP norm are ordered, and the L1 norm is the weakest, so we can say that, let's discuss first,
03:11:50Paolo Guiotto: the L1 convergence, so… XN goes to… 0 in L1.
03:19:900Paolo Guiotto: This happens if and only if the expected value of the modulus of the difference, Xn minus the limit, which is 0, goes to zero. Now, this is the expectation of modulus of XN,
03:35:360Paolo Guiotto: And since XN takes value 0, 1, it is positive, in particular.
03:40:750Paolo Guiotto: So this is the expected value of XN.
03:45:100Paolo Guiotto: Which can be easily computed, because the Xn takes just two values, so we have 0 times the probability that Xn is equal to 0, plus 1 times the probability that Xn is equal to 1.
04:02:80Paolo Guiotto: In this case, the expectation reduces to this, because… actually, that's a simple function, takes two values.
04:11:420Paolo Guiotto: So, this is, of course, equal to 0.
04:14:580Paolo Guiotto: And, since this is 1 over N, the total is 1 over n, so this is the…
04:22:130Paolo Guiotto: expected value of modulus Xn minus 0, therefore you see that it goes to zero, and we can conclude that, yes, Xn goes in L1, so in probability, we say in mean to, 0.
04:40:50Paolo Guiotto: About the almost sure convergence… Or… equivalently, or convergence… with probability… Wow.
04:58:690Paolo Guiotto: We know that the LP convergence is not stronger than almost sure convergence, so we cannot say that this first fact implies the almost sure convergence, but we know
05:14:860Paolo Guiotto: We know that, Since, the sequence XN
05:23:270Paolo Guiotto: converges in L1 to 0, there exists at least a subsequence, X and K, of the sequence XN.
05:34:230Paolo Guiotto: Such that, X and K goes almost surely.
05:40:580Paolo Guiotto: to zero, to the same limit. So this says that, in particular, if XN, if the sequence XN has some almost sure limit, this must be necessarily equal to zero.
05:56:400Paolo Guiotto: So… the question is, is XN, the full sequence, going almost surely to zero?
06:07:220Paolo Guiotto: Now, we have seen yesterday that this is equivalent of saying that the probability that the limit of DXN
06:20:40Paolo Guiotto: Limited n is equal to 0 is 1.
06:24:870Paolo Guiotto: And, let's just quickly refresh. This means that, writing the definition of limit for every epsilon, the existing initial index, etc.
06:35:840Paolo Guiotto: This means that… sorry, it's the intersection of epsilon, because there is the forever, there exists the initial index is a union, then we have an intersection on
06:46:460Paolo Guiotto: n greater equal than capital N, such that modulus of Xn minus the limit 0 is less or equal than epsilon.
06:56:600Paolo Guiotto: So, if you are in this set, it means that exactly this thing. The limit of the sequence XN is 0, so we want that the probability of this be equal to 1.
07:09:950Paolo Guiotto: Or, equivalently, the probability of the complementary, so the union of epsilon, the section of N, the union for n larger than capital N of sets where modulus Xn is larger than epsilon, this be equal to 0.
07:30:260Paolo Guiotto: Now, since this is a union of sets, well.
07:34:280Paolo Guiotto: We can always rewrite this as a countable union if needed, but this union has probability 0 if and only if these sets have probability 0. So, if and only if the probability that… of the intersection in n of the union for N larger equal than capital N
07:52:680Paolo Guiotto: of modulus Xn greater equal… greater than epsilon, greater than or equal. It's the same as 0 for this, for every epsilon positive.
08:03:630Paolo Guiotto: Now, this set here…
08:05:720Paolo Guiotto: intersection of N union for little n greater or equal than capital n of these sets, say, let's call them the sets EN.
08:15:710Paolo Guiotto: is the set is a particular type of set. This set is the intersection in capital and union.
08:22:840Paolo Guiotto: for n greater than capital n of the yen is the set of,
08:30:500Paolo Guiotto: of, here we are in omega, so a set of little omegas, in the sample space, such that they belong to, infinitely many EN, okay? Omega belongs…
08:50:330Paolo Guiotto: 2… infinitely…
08:56:750Paolo Guiotto: Manny.
08:59:640Paolo Guiotto: Yeah, so there are infinitely many indexes, N, for which omega belongs to
09:06:390Paolo Guiotto: to the corresponding sets, yeah. It doesn't mean that omega must be in all the yen, or it must be in all the yen, starting from some initial index capital L. It must be in infinitely many, no?
09:20:730Paolo Guiotto: In other… the interpretation is, this is the bad set, this is the bad event, the event of omega for which the sequence does not go to zero, because infinitely many times, it remains away, not that epsilon, from zero, no?
09:38:910Paolo Guiotto: If many times you are away from zero, you never go to zero. This is the point.
09:44:780Paolo Guiotto: Now, this set is called the Lim Soup.
09:49:820Paolo Guiotto: set of the yen.
09:52:910Paolo Guiotto: There is a remarkable fact here, which is the Boracan Deli. Let me refresh what's happening here.
10:05:980Paolo Guiotto: So, the Borencantali says that, in general, if the series of the probabilities of these events, EN is convergent, so the sum is finite, you can say that the probability of the limb subset
10:24:810Paolo Guiotto: is equal to zero.
10:28:880Paolo Guiotto: And there is a second statement that contains the case when the sum is infinite. If the sum of the probabilities is infinite.
10:45:90Paolo Guiotto: And here, we need to have some more information, in fact, and the sets E and R.
10:53:570Paolo Guiotto: Independent.
10:57:330Paolo Guiotto: Then, the theorem says that in this case, the probability of the limb soup Is equal to 1.
11:05:530Paolo Guiotto: So if we are in the case where the limb soup is the limb soup of this set, that means the sequence does not go to zero.
11:14:930Paolo Guiotto: In the first case, we would have that the probability that you don't go to 0 is 0, so you must go to 0 almost surely. In the second case, we would have that the probability you don't go to 0 is 1, so the sequences never converge, okay?
11:31:770Paolo Guiotto: So to do this, we need to assess the probability of EN, that's easy, in our case.
11:42:750Paolo Guiotto: The probability of N is the probability of the set where modulus Xn is greater than some fixed epsilon positive.
11:54:630Paolo Guiotto: Now, remind that X is just 01 valued, so clearly, when XN takes value 0, that inequality is not verified, 0 greater than epsil. So this reduces just to XN
12:10:250Paolo Guiotto: greater than epsilon.
12:12:810Paolo Guiotto: Now, since XN takes values 0 and 1, if epsilon is too big, this is 0 again, so we can say that when epsilon is greater or equal than 1, we get 0. But when epsilon is less than 1, but positive.
12:28:540Paolo Guiotto: So the set of omegas when XN is larger than epsilon is the set where Xn is equal to 1. It corresponds to this.
12:39:160Paolo Guiotto: And the probability of this is 1 over N.
12:43:150Paolo Guiotto: So we can say that.
12:45:160Paolo Guiotto: Remind that the goal is to show that… discuss whether is it true or not, that for every epsilon positive, that probability is zero.
12:56:990Paolo Guiotto: We can say that if epsilon is greater or equal than 1, that's true.
13:02:560Paolo Guiotto: If epsilon is greater or equal than 1, then the sum of the probabilities of the sets N,
13:11:500Paolo Guiotto: would be equal, to, would be a sum of zeros, so it would be zero. But if epsilon is less than 1, then the sum of the probabilities of the EN
13:24:620Paolo Guiotto: would be equal to the sum of 1 over n, which is plus infinities.
13:30:430Paolo Guiotto: Okay, so in this case, we are… in this second situation. Now.
13:39:750Paolo Guiotto: What about the events, Ian? So we need to know if they are independent.
13:48:220Paolo Guiotto: Yes, in the problem, it was given that the variables are independent, so the events generated by them are independent. So, since
13:59:320Paolo Guiotto: since the exam.
14:02:990Paolo Guiotto: R.
14:04:130Paolo Guiotto: independent.
14:07:80Paolo Guiotto: the events, EN, which are generated by these XNs, so they are models of XN greater than epsilon.
14:15:300Paolo Guiotto: R.
14:16:580Paolo Guiotto: independent.
14:20:230Paolo Guiotto: And therefore, you see, we are in the condition where the second, or I can tell, statement applies, so by second…
14:31:500Paolo Guiotto: But I can tell you, we have that the probability of that limb subset
14:39:70Paolo Guiotto: N is equal to 1 for this epsilon, for every epsilon positive and less for less than 1.
14:50:550Paolo Guiotto: So, in particular, it is not true that that probability is zero for every epsilon positive, so this condition here, let's give a name, star.
15:01:190Paolo Guiotto: So In particular… Stop.
15:09:500Paolo Guiotto: is Pulse.
15:12:840Paolo Guiotto: for epsilon between 0 and 1. And since to have that, the limit is 0, you must have that property verified for every positive epsilon. This means that the sequence XN
15:28:150Paolo Guiotto: does not converges… does not converge almost surely to zero, but then it is never convergent, because this is the unique possibility. Actually, so Xn…
15:41:930Paolo Guiotto: is not… convergent, almost sure.
15:49:340Paolo Guiotto: And this ends the… Exercise.
15:54:450Paolo Guiotto: Now, before we attack the next one.
15:59:600Paolo Guiotto: I want to do something, I found some exercises around, so I wrote the assignment that seems to be
16:09:150Paolo Guiotto: Because this one, it's a bit tricky. What's happening?
16:21:880Paolo Guiotto: This is… My problem
16:30:480Paolo Guiotto: A little bit.
16:41:300Paolo Guiotto: That's it. We are busy.
16:45:270Paolo Guiotto: So, let's do this.
16:50:990Paolo Guiotto: Let's hope that this… so let's see a couple of other exercises. This is very easy. It's on L1 convergence. So, exercise…
17:01:740Paolo Guiotto: It says, you have a sequence XN that converges in L1 to X, then prove that the expectation of XN
17:13:30Paolo Guiotto: goes to the expectation of X. This is a numerical sequence, no? The sequence of the expectations. And it asks, is the vice versa 2?
17:28:730Paolo Guiotto: So, let's say, question one, we have to verify that if the sequence XN converges in L1,
17:39:820Paolo Guiotto: to X, then the sequence of the expectations converges to the expectation of X. Well, we can just take the modulus of the difference between the expectation of Xn and the expectation of X,
17:55:420Paolo Guiotto: And say that this is, because of the linearity of the expectation, this is the expectation of XN minus X.
18:04:490Paolo Guiotto: And then by the triangular inequality, we could carry the modulus inside, so triangular… Inequality.
18:14:290Paolo Guiotto: So this becomes less frequent than the expectation of modulus Xn minus X. But that's exactly, by definition, the L1 norm of XN minus X.
18:28:790Paolo Guiotto: And this is the quantity that goes to zero because of this assumption, no? So, we concluded that, since this goes to zero, and this one is less or equal than this quantity, also that quantity will go to zero, and therefore.
18:45:950Paolo Guiotto: expectation, what is inside the expectation of XN minus expectation of X, will go to 0, and this is the conclusion.
18:58:190Paolo Guiotto: So, the first point was this. Now, the second point asks, is it true that if you have that the expected value of XN that goes to the expected value of X, this means that XN goes, in L1 sense, to X?
19:16:240Paolo Guiotto: So what about the vice versa?
19:19:910Paolo Guiotto: So either we approve, or we, we look for a counterexample.
19:32:280Paolo Guiotto: So…
19:37:520Paolo Guiotto: What do you think?
19:40:520Paolo Guiotto: So, just to simplify things, let's do it in such a way that the expected value of X be 0. So, in fact, if you replace X equals 0, we would say, is it true that if the expected value of XN
19:55:900Paolo Guiotto: Goes to zero, then… XN goes to 0 in L1.
20:05:190Paolo Guiotto: Is that true?
20:06:970Paolo Guiotto: Well, XN goes to zero, and L1 means that the expected value of modulus Xn minus the limit 0, so modulus XN, this goes to 0.
20:19:680Paolo Guiotto: So, in other words, the question is, is it true that if the expectation of Xn goes to 0, also the expectation of modulus of XN goes to 0?
20:37:990Paolo Guiotto: Do you think anything? So, a good, model
20:42:680Paolo Guiotto: Here we can do with discrete probability, but, for example, you can take, as model, the interval 0, 1,
20:52:390Paolo Guiotto: with the usual sigma algebra, so F is the class, the Barrett class, of R in 01.
21:05:200Paolo Guiotto: And, as probability dealer begged measure.
21:08:340Paolo Guiotto: Now, you just need to have the expectations of the XN going to zero. So, for example, being equal to zero.
21:15:660Paolo Guiotto: No? So, something like this.
21:20:170Paolo Guiotto: If I have on 01, which is my sample space, I divide in two parts. This is plus 1.
21:28:490Paolo Guiotto: And this is minus 1.
21:30:510Paolo Guiotto: Take this sequence.
21:32:570Paolo Guiotto: You can say constantly, XN constantly equal to this one. So, to 1 times the indicator of 0, 1 half, formed analytically, minus 1 times the indicator of 1 half 1.
21:49:140Paolo Guiotto: So if you want to… it doesn't work for this.
21:52:670Paolo Guiotto: Now, clearly, when you do the expectation of this XN,
21:58:470Paolo Guiotto: Since this is the integral, on 01 of that function, Xn of omega in D omega, the omega is just the ordinary LeBague measure, this will be equal to 0.
22:12:70Paolo Guiotto: But if you do the expected value of absolute value of XN, now the function is constantly equal to 1, so you get 1.
22:21:730Paolo Guiotto: And doesn't go to zero.
22:24:380Paolo Guiotto: Okay? So… This is… False.
22:33:780Paolo Guiotto: Okay, as a second exercise, I found… well… All this requires the convergence But let's take this one.
22:45:940Paolo Guiotto: That has to deal. It's very similar to the exercise we have done
22:50:870Paolo Guiotto: In the opening today. So we have a sequence XN of independent random variables.
23:03:130Paolo Guiotto: They are Bernoulli, but, not 0.1. This says that the probability that Xn is equal to n squared is equal 1 over N, while the probability that Xn is equal to 0, it is equal… no, sorry, yeah, 1 minus 1 over N.
23:23:840Paolo Guiotto: So this for N greater or equal than 1.
23:29:710Paolo Guiotto: Now…
23:30:690Paolo Guiotto: what can be said about, let's start here from the almost sure convergence. So, is XN almost surely convergent to
23:44:660Paolo Guiotto: Well, if we have to take to a candidate, this is similar to the previous one. This XN takes two values, 0 and n squared. So, when n is big, n squared, we have 0 and a big value, no?
23:57:120Paolo Guiotto: But the big value is taken with a small probability, so it seems reasonable to think that this goes to zero. So, is this going to zero? Most surely? The second question is, is that…
24:11:210Paolo Guiotto: going to zero in L1, or LP.
24:16:440Paolo Guiotto: So we say L1 because, because LP, or let's say here, XN is going to zero also in LP.
24:29:120Paolo Guiotto: will be greater than 1. And question 3… yeah, here we have this, which is something new.
24:38:850Paolo Guiotto: is the series of the Xena, so this is the infinite sum for n going from 1 to infinity, is…
24:47:10Paolo Guiotto: This area is a serious convergent.
24:52:780Paolo Guiotto: It's, you know, convergent, almost surely.
24:58:740Paolo Guiotto: So this is the series of the variables.
25:03:160Paolo Guiotto: the variables are positive, and, so we, we wonder if this is convergent. You see that it seems,
25:13:640Paolo Guiotto: well, to… you know, this XN are 0 or n squared, no? So…
25:19:610Paolo Guiotto: basically, to be convergent, this thing, you cannot have infinitely many N squared in this number, otherwise the sum will blow up, no? So it's, sort of, could be attacked in this way.
25:33:580Paolo Guiotto: So the… to converge, you must have that XN is, definitely equal to zero.
25:40:800Paolo Guiotto: Why is that? However, we will return on this later. So let's take question one.
25:47:270Paolo Guiotto: So, again, we can say that XN is convergent almost surely, to zero. If and if we do not repeat the story, the probability of the intersection
26:01:660Paolo Guiotto: for capital… sorry, yeah, the section of the union for n greater capital N of modulus Xn greater than epsilon, this is a probability 0 event.
26:16:830Paolo Guiotto: So now, to compute this, we apply the Burel can tellilemma, assessing these probabilities. Now, the probability that the modulus of XN is greater than epsilon, that's,
26:31:240Paolo Guiotto: pretty much the same of the previous exercise, so XN is positive.
26:37:700Paolo Guiotto: is greater than or equal than zero, so this is the probability where XN is larger than epsilon.
26:45:00Paolo Guiotto: Now, we can say that this is, what? Well, definitely, it's not the probability where xn is 0, because it cannot be greater than epsilon, and since XN is N squared.
26:58:880Paolo Guiotto: the other possible value, we say that if epsilon is fixed there.
27:05:160Paolo Guiotto: as soon as n is big enough, this is the probability that X is in equal to n squared. So we can say that this is the probability where X n is equal to N squared, let's say, when n
27:20:270Paolo Guiotto: square is larger than epsilon. Otherwise, when n squared is smaller than epsilon, this will be for a finite number of n, this will be equal to,
27:34:230Paolo Guiotto: No.
27:37:400Paolo Guiotto: Yeah, this will be equal to zero. No, because Xi is less than epsilon in this case.
27:46:300Paolo Guiotto: Now, this means that we have, the first one is the probability where Xn is equal to N squared is equal to 1 over n, when n is larger than root of epsilon, whatever it is, or 0, if n is less than root of epsilon.
28:04:960Paolo Guiotto: Okay, so now, what do we do with this? We have to decide, again, with the… because of the Borel-Cantelli lemma.
28:16:150Paolo Guiotto: what happens to the series of these probabilities? So, yeah, we have that the series, so 4 epsilon, positive, fixed.
28:28:70Paolo Guiotto: Because it is important, this, now to keep in mind that here, epsilon is fixed, okay? It's out of this probability, so it's fixed. We can say that the sum of n of the probabilities of sets modulus Xn larger than epsilon
28:47:440Paolo Guiotto: Now, for certain N, this will be 0, no, but when n, as soon as n is larger than this number, this becomes the sum for, say, n larger, wherever I want to be.
29:02:490Paolo Guiotto: extremely precise, I should take the first integer after this one, so the integer part of root of epsilon plus 1 to infinity. Now, for this n, the value of the probability is 1 over n. It doesn't matter if I miss the first
29:20:170Paolo Guiotto: Finite number of terms, but this is the tail of a divergent series, so this is equal to plus infinities.
29:27:980Paolo Guiotto: And therefore, we can say that,
29:34:200Paolo Guiotto: We are in the case, again, in the case where the sum of the probabilities is infinite.
29:41:320Paolo Guiotto: Now, here we also know that the variables XN are independent, so these events are independent, and therefore the probability of the lint soup is equal to 1.
29:50:580Paolo Guiotto: Okay, so, and since…
29:56:860Paolo Guiotto: the XN.
29:59:470Paolo Guiotto: independent.
30:04:670Paolo Guiotto: Then, because of the second Borel-Cantell Dilemma.
30:13:720Paolo Guiotto: The probability of the limb soup
30:18:60Paolo Guiotto: of these sets modulus Xn greater than epsilon is actually equal to 1, and therefore this means that XN is never convergent to 0.
30:30:550Paolo Guiotto: Okay?
30:32:280Paolo Guiotto: Almost surely.
30:37:270Paolo Guiotto: So, the answer to the first question is negative. Let's look at the second question.
30:44:240Paolo Guiotto: Is Xn going to zero in L1? Here, it's similar to the previous discussion, so we have the expected value of the dis… so this is the distance between XN and 0 in L1 normal.
30:58:990Paolo Guiotto: L1 normal of the difference.
31:02:510Paolo Guiotto: And this is, at the end of the expected value of XN that coincides, since XN is positive, with the expected value of XN.
31:13:330Paolo Guiotto: Now, this is… this variable is 0.
31:16:980Paolo Guiotto: When, is 0 when,
31:23:340Paolo Guiotto: It is 0 or n squared, so we will have, again, 0 times the probability that Xn is equal to 0.
31:31:20Paolo Guiotto: plus n squared times the probability that XN is equal to, N squared… I'm sorry, not. This is 1 over N.
31:43:620Paolo Guiotto: No, I'm sorry.
31:45:30Paolo Guiotto: The variable is n squared, yeah, n squared times the probability that Xn is equal to n squared, but this is equal to 1 over n.
31:55:60Paolo Guiotto: So at the end, we have that this is equal to N, it goes to plus infinity, so we can say that this Xn does not converge in L1 to 0. And actually, it's not convergent in L1, because, and…
32:13:560Paolo Guiotto: Actually… It… cannot… be convergent in L1 to any X.
32:29:940Paolo Guiotto: Because, you see that, actually, what we computed is the norm, because we proved that the L1 norm of XN is equal to N, so in particular, the sequence XN is unbounded
32:49:560Paolo Guiotto: in L1, and the…
32:53:20Paolo Guiotto: To be convergent, a sequence must be bounded. This is not sufficient, but it's a necessary condition.
33:00:620Paolo Guiotto: We cannot have any other LP convergence, we cannot.
33:08:610Paolo Guiotto: Hi, but… It's a pretty… Cannot have any… The… LP convergence, because…
33:29:320Paolo Guiotto: We know that if XN would converge to, in LP norm to some X for P greater than 1, this is stronger than L1 convergence, so the Xn would converge in L1 to X, but this is impossible.
33:51:620Paolo Guiotto: Now, probabi, this problem wasn't too…
33:59:840Paolo Guiotto: Interesting.
34:04:120Paolo Guiotto: Because, now, about the question 3, the convergence of the series, actually, we could,
34:10:800Paolo Guiotto: We could,
34:14:150Paolo Guiotto: say this. Now, if you look at this series, the first point has proved that XN is never going to zero.
34:24:900Paolo Guiotto: Okay? Almost never. So, we can just, give a short answer. Since, by… Q1.
34:37:620Paolo Guiotto: XN.
34:39:659Paolo Guiotto: Does not go to zero.
34:42:320Paolo Guiotto: for almost, every omega in omegas.
34:49:130Paolo Guiotto: Well, you know that the fact that the general term goes to zero is an accessory, but not sufficient condition to have convergence for a series. So, and, since…
35:02:60Paolo Guiotto: For a genetic series.
35:11:730Paolo Guiotto: some… let's say this is a fact that regards the numerical series, sum of AN, to be convergent, to be convergent.
35:27:90Paolo Guiotto: We must… have… That the general term goes to zero.
35:36:220Paolo Guiotto: So we can say that, our series, the sum of the XN, Cannot be convergent.
35:49:330Paolo Guiotto: And that's it.
35:53:670Paolo Guiotto: Well, do the same, repeat the same calculation, do.
36:00:900Paolo Guiotto: Dead.
36:02:550Paolo Guiotto: Same, but with these probabilities. Probability that Xn is equal to E to N is equal to E2 minus N.
36:13:370Paolo Guiotto: And the probability that Xn is equal to 0 is equal to 1 minus E2 minus N.
36:22:810Paolo Guiotto: Do the same exercise.
36:25:830Paolo Guiotto: with this, and see if anything changed. So you have just to say to… Repeat these, calculations.
36:36:860Paolo Guiotto: Okay, now, let's take this, prop… well, it's time we introduce also the, let's… let's, do a…
36:48:460Paolo Guiotto: a bracket with the exercises, or maybe we can now This one…
36:56:50Paolo Guiotto: Well, let's do the 854, which is the last one where we have only the convergence, the almost sure convergence. So, this is exercise…
37:07:540Paolo Guiotto: 8, 5, 4…
37:11:820Paolo Guiotto: It says that we have a sequence epsilon n of numbers positive, such that the series of the epsilon n is finite.
37:21:260Paolo Guiotto: It is not better specified.
37:24:760Paolo Guiotto: And we know that,
37:31:560Paolo Guiotto: We know that.
37:33:740Paolo Guiotto: I hope that it's not missing a condition here.
37:50:830Paolo Guiotto: And do we need independence?
37:56:260Paolo Guiotto: No. We should…
38:02:860Paolo Guiotto: Please research this size here.
38:05:550Paolo Guiotto: 854…
38:16:380Paolo Guiotto: Now, we shouldn't… we shouldn't need independence.
38:20:860Paolo Guiotto: Apparently. Okay, so it says that we have random variables XN, are such that…
38:35:800Paolo Guiotto: the probability that models of XN is greater than
38:42:490Paolo Guiotto: or equal epsilon n, this is less.
38:46:900Paolo Guiotto: Or equal, then, epsune.
38:50:840Paolo Guiotto: So the probability that Xn is… now, these epsilon n are positive numbers, since their sum is finite, they must go to 0, so when n goes to infinity, these numbers
39:03:00Paolo Guiotto: Are going to zero.
39:05:80Paolo Guiotto: So this is saying the probability that Xn is far from 0, a little bit, is small, is very small when n is large.
39:14:580Paolo Guiotto: Now, this, says that show… that this series… some.
39:22:760Paolo Guiotto: of XN… Ease.
39:26:950Paolo Guiotto: Absolutely.
39:29:00Paolo Guiotto: convergent.
39:31:230Paolo Guiotto: We did… probability.
39:37:130Paolo Guiotto: 1.
39:43:120Paolo Guiotto: So, now, what does it mean, absolutely convergent? Means that the series of the absolute values is convergent. So, sum of Xn is…
39:54:450Paolo Guiotto: Absolutely.
39:56:910Paolo Guiotto: convergent.
40:01:960Paolo Guiotto: if… the series of the absolute value of the Xn.
40:07:380Paolo Guiotto: Is convergent.
40:13:210Paolo Guiotto: So, the goal becomes this,
40:16:650Paolo Guiotto: To prove that the probability that the sum of the modules of XN is finite, so this means the series is convergent, this probability is 1.
40:30:610Paolo Guiotto: Now, since we have these informations that the series of the epsilon n is convergent.
40:40:200Paolo Guiotto: We may say the fo- we may say the following. If… I noticed that if Modulus of XN,
40:50:110Paolo Guiotto: is less or equal than epsilon n for every n
40:55:30Paolo Guiotto: larger than some initial N, so if there exists an initial index such that the modulus of Xn is less or equal than epsilon n for every n, then I could say that the series of the modulus Xn
41:13:140Paolo Guiotto: will be convergent, because I could split the sum into the sum for N going from,
41:19:720Paolo Guiotto: well, I don't know where this index starts, but let's say that it starts from 1.
41:25:460Paolo Guiotto: to, say, capital N minus 1 of the modulus Xn, and this is a finite sum, whatever it is, is a finite value. And then I have a second sum, that's an infinite sum, sum for n going from capital N to plus infinity of modulus Xn.
41:43:190Paolo Guiotto: But if there is an N for which this bound holds
41:49:820Paolo Guiotto: So, if modulus Xn is bounded by the epsilon n, I could say for n greater than capital n, this will be bounded by the epsilon n, so my total sum would be bounded by sum for n going from 1 to capital n minus 1 of modulus Xn, which is a finite value.
42:10:630Paolo Guiotto: plus the sum for n going from capital N to plus infinity of the epsilon n.
42:15:970Paolo Guiotto: And both these are finite, because the first one is a finite sum, Find it.
42:22:600Paolo Guiotto: Sam, huh?
42:24:260Paolo Guiotto: And the second one is, convergent.
42:28:100Paolo Guiotto: serious.
42:30:480Paolo Guiotto: Because we, suppose, we assume that the sum of the epsilon n, all the epsilon n, is fine.
42:39:180Paolo Guiotto: Okay, so if I prove that.
42:41:980Paolo Guiotto: The sequence of absolute value of Xn is bounded from a certain initial capital N
42:48:620Paolo Guiotto: by these numbers, epsilon n, I am done. Now, let's write what does mean this property. Now, we have to remind that
42:58:700Paolo Guiotto: Xn depends on omega, so this means, let's see what is the set of omegas for which this is true. The set of omegas in the sample space for which there exists an initial N
43:12:510Paolo Guiotto: Such that the modulus of Xn is less or equal than epsilon n for every n greater or equal than capital N. It's a similar set of… it has the same kind of form.
43:29:710Paolo Guiotto: Of the set, we consider to prove that.
43:32:950Paolo Guiotto: The sequence is… a sequence is, pointwise convergent here.
43:38:740Paolo Guiotto: You see, where we add the pointwise limit, this is more or less the same kind of structure, but it's not the same end set, and it has not the same meaning. But the final point is that we will add to a limb subset.
43:54:350Paolo Guiotto: So, in fact, this is what?
43:57:600Paolo Guiotto: There is a union, so there is… there exists an N, means there exists a union, you are in a union of sets where modulus Xn is less or equal than epsilon n for every n larger than this capital N.
44:14:340Paolo Guiotto: And here we have an intersection, so union of intersection for n greater or equal than capital n of sets where modus Xn less or equal than epsilon n.
44:27:530Paolo Guiotto: So the goal is to prove that the probability of this set is 1, because if this probability is 1, it means that for almost every omega, that property is verified, so the CS in particular will be convergent.
44:42:890Paolo Guiotto: So… if we… prove.
44:50:760Paolo Guiotto: That.
44:52:920Paolo Guiotto: the probability of this event, union on capital N, intersection for N greater than capital N,
45:00:980Paolo Guiotto: of modulus Xn less or equal than epsilon n.
45:05:650Paolo Guiotto: This probability is 1.
45:08:570Paolo Guiotto: It means that this property here Star.
45:13:390Paolo Guiotto: That implies convergence, okay, is verified for almost every omega, so star… is true.
45:24:950Paolo Guiotto: 4… almost every omega in the sample space. And therefore, the series of the XN converges
45:37:310Paolo Guiotto: For almost every omega, or almost surely.
45:41:960Paolo Guiotto: So this is now the plan. To prove that, that probability is 1.
45:46:420Paolo Guiotto: Or, equivalently, this is not the limb soup set, because the limb soup is just the contrary, the opposite of this, no? It is the intersection of the unions. So we put in that form, this means that we have to take the complementary.
46:02:20Paolo Guiotto: This is the intersection of N of the union for n greater or equal than capital N of the complementaries of these sets, modulus Xn greater or equal than epsil, or greater, strictly greater.
46:14:650Paolo Guiotto: then epsilon n… This probability must be zero.
46:20:200Paolo Guiotto: But now, what you see here, this is a limb soup set
46:24:970Paolo Guiotto: of this modulus Xn greater than epsilon n. And so these are the events EN.
46:34:30Paolo Guiotto: to which we, applied the boreal cantelli. We noticed that, we… Notice…
46:43:830Paolo Guiotto: that, the probability of Vienna
46:48:360Paolo Guiotto: That's, this. There is some information in the assumption. It is this one. It is the probability where modulus Xn is greater or equal than epsilon n, so modulus of XN greater
47:03:830Paolo Guiotto: Well, it is here greater epsilon n, but if I put the equal at worst, I enlarge the set, so I enlarge the probability. This one is less or equal than epsilon n by…
47:19:450Paolo Guiotto: Assumption.
47:21:780Paolo Guiotto: And so, the sum of the probabilities of the sets EN will be less or equal than the sum of the epsilon n, and this is find it again by
47:36:40Paolo Guiotto: assumption. So, we are in the case of the first Boracantelli lemma.
47:45:330Paolo Guiotto: We have the sum of the probabilities. It's finite, the probability of the limit subset is zero. We don't need independence for this case, so we get the conclusion, basically. So,
47:58:750Paolo Guiotto: First… Or I can tell you. Lemma.
48:05:250Paolo Guiotto: The probability of the limb soup.
48:09:920Paolo Guiotto: E n is equal to 0.
48:12:730Paolo Guiotto: And this is exactly this property here. Let's give name to star.
48:18:770Paolo Guiotto: That is… 2-star.
48:24:80Paolo Guiotto: Holds, huh?
48:26:370Paolo Guiotto: And therefore, this means the conclusion, the series of exam, is absolutely convergent.
48:36:850Paolo Guiotto: We'd… probability one.
48:48:550Paolo Guiotto: Okay.
48:51:100Paolo Guiotto: Let's introduce… Another, way… another way to converge. This is the definition of convergence in probability.
49:04:520Paolo Guiotto: So, we say that, so, let omega F… B… be a probability space.
49:16:180Paolo Guiotto: We have a sequence XN of random variables, so just, for the moment, measurable functions.
49:23:950Paolo Guiotto: We say that, the sequence, XN.
49:31:40Paolo Guiotto: It's converges.
49:33:230Paolo Guiotto: in probability.
49:42:70Paolo Guiotto: to X, which is another random variable.
49:47:270Paolo Guiotto: And we use this notation.
49:49:980Paolo Guiotto: We say that XN, we put a P sign under the… over the arrow.
49:58:730Paolo Guiotto: to Axon.
50:00:30Paolo Guiotto: If… the probability that XN is away from X of a little epsilon.
50:12:00Paolo Guiotto: This is going to zero when n goes to plus infinity. This for every epsilon positive.
50:21:630Paolo Guiotto: equivalently, It is sometimes useful to know
50:27:300Paolo Guiotto: that we can also write this in the complementary. If we do the complementary, what happens?
50:33:490Paolo Guiotto: The complementary means that, you know, that when you do the probability of the complementary, this becomes 1 minus the probability, okay, of the direct set.
50:43:720Paolo Guiotto: So this can be recasted as follows. You take the probability when the distance between Xn and X is less or equal than epsilon. It is not exactly the complementary because of the equal… to be the complementary, I should write less than epsilon, but it's the same thing, no? Now, this probability is 1 minus the previous one.
51:04:800Paolo Guiotto: If the previous one goes to 0, this one must go to 1.
51:08:100Paolo Guiotto: Okay? So this goes to 1 when n goes to infinity, For every epsilon particle.
51:16:870Paolo Guiotto: It seems the convergence, point-wise, because It does, more or less.
51:24:200Paolo Guiotto: flag, or in fact, there is a clear connection, you will see that almost should convergence. It's stronger than this one.
51:32:370Paolo Guiotto: The idea is that the probability that you are away from the limit is very small, and… but it's going to zero, and it goes to each line. So it seems the pointwise convergence, but it is not, as we will see.
51:50:500Paolo Guiotto: So, in fact, let's see a few relations with the other convergences. The first one, which is… well, let's… let's just start with the almost sure convergence. So, we have that, almost sure convergence.
52:13:610Paolo Guiotto: is stronger.
52:20:380Paolo Guiotto: Stan.
52:23:850Paolo Guiotto: probability convergence. So this means that whenever you have that the sequence converges almost surely to X,
52:33:550Paolo Guiotto: Then, it converges also in probability.
52:39:550Paolo Guiotto: To the same limit.
52:42:910Paolo Guiotto: Well, let's see the little proofer, it's just a combination of the definitions, okay?
52:54:10Paolo Guiotto: Now… I want to prove that, let's take the first characterization. So, suppose,
53:04:230Paolo Guiotto: the XN goes almost surely to X.
53:09:780Paolo Guiotto: We have repeated, also this morning, this means that, The probability that,
53:17:840Paolo Guiotto: You know, you don't go to AXA, it means you stay away from AXA,
53:23:780Paolo Guiotto: So the distance between XN and X is positive, let's say.
53:28:490Paolo Guiotto: uniformly controlled infinitely many times in n, so this is the union for n larger than capital n intersection with N. So this set here has probability 0. So this is equivalent, we refresh the
53:44:780Paolo Guiotto: Just this morning.
53:46:950Paolo Guiotto: It was all this discussion here.
53:50:430Paolo Guiotto: No?
53:52:160Paolo Guiotto: So we say that this was specialized to the limit 0, but the music is the same, and just replace XN by XN minus the limit X, and you get this. So, the characterization is this one. This probability of this limb subset must be equal to zero.
54:13:130Paolo Guiotto: Now, what this says, with respect to these probabilities?
54:19:100Paolo Guiotto: the probability that Xn is a wave that Xn, that not just DXN infinitely many. This says that
54:28:390Paolo Guiotto: Let's say this is the set of omegas such that Xn minus X is greater or equal than epsilon infinitely for infinitely many
54:41:890Paolo Guiotto: N.
54:42:910Paolo Guiotto: Okay? While this thing is the event of omega where that specific XN is away from X, at least epsilon.
54:53:970Paolo Guiotto: So, it smells as if this one is a little bit less than this one, because you are away from X in Philly many times, not just one.
55:03:970Paolo Guiotto: and there you are, away from X at distance optional, just for the XN variable. But let's make this precise. Well, we can say that, so…
55:17:270Paolo Guiotto: If we take the probability that Xn is away from X of this quantity epsilon.
55:27:60Paolo Guiotto: Well, you can say that if you are away here, it means that you will be in the union, well, let's use a… let's change… I have to change an index, otherwise…
55:42:570Paolo Guiotto: I do a mess. I am in the union for k greater or equal than n of the set where modulus XK minus X is larger than epsilon.
55:55:290Paolo Guiotto: So this is contained into that set.
55:59:750Paolo Guiotto: Now,
56:03:550Paolo Guiotto: So, I can say if this is contained in that, the probability will be smaller. So, this will be less or equal than the probability that you are in this union for K less or equal than N.
56:20:00Paolo Guiotto: modulus XK minus X greater or equal than epsilon.
56:45:520Paolo Guiotto: this…
56:49:880Paolo Guiotto: Okay, yes. Okay, so we have this. Now, look at the right-hand side, these sets here.
56:59:130Paolo Guiotto: Now, we have to take the limit, so we have to send them to infinity, no? We want to compute the limit
57:05:360Paolo Guiotto: So from this, it follows that the limit when n goes to plus infinity of the probability where Xn is
57:14:390Paolo Guiotto: away from X of this quantity epsilon, this is the limit in n of these probabilities. The probability of the union for k larger than n modules XK minus X greater or equal than epsilon, okay?
57:32:20Paolo Guiotto: Now, you have a limit of a tribe of probabilities.
57:37:510Paolo Guiotto: when you move this index set. What happens to this set when you increase it?
57:44:740Paolo Guiotto: Since you are doing a union, this is a tail union, no? You take the union over K that starts from N, so k equals n, n plus 1, n plus 2, N plus 3, and so on.
57:55:970Paolo Guiotto: So when I increase N, I basically reduce the sets of this union, because when you flip from N to N plus 1,
58:05:300Paolo Guiotto: The union for k greater or equal than n plus 1 has just one set less than the union for k greater or equal than n. What is missing is the set modules XN minus X greater or equal than Epson.
58:17:930Paolo Guiotto: So this means that this sequence of sets, so if you call this, I don't know, F, N,
58:25:470Paolo Guiotto: is the union for k greater or equal than n of modulus X, k minus X greater or equal than epsilon. This is a decreasing family of sets.
58:39:490Paolo Guiotto: And here we have a limit of probabilities of decreasing… a decreasing sequence of sets, so we invoke the continuity from above.
58:51:240Paolo Guiotto: that for a probability measure, holds always, so without any restriction, and this turns out to be what? The limit of probabilities is the probability of the limit set, but what is the limit set?
59:05:140Paolo Guiotto: is the intersection of all these sets Fn, so it is the intersection of N of the sets FN.
59:12:770Paolo Guiotto: So, going back, this means F, probability of the intersection of N of the unions for k greater or equal than n of modulus XK minus X greater or equal than epsilon.
59:28:190Paolo Guiotto: Right?
59:30:550Paolo Guiotto: So, what I'm saying is that the limit of that probabilities is equal to the probability of this set. And what is the probability of this set?
59:41:30Paolo Guiotto: But just because we have pointwise convergence, that limit is zero.
59:45:670Paolo Guiotto: So, that's zero. And this proves that… what? This proves that the limit of these probabilities is zero. So, this means the convergence in probability. So we have that.
00:00:190Paolo Guiotto: Xn converges in probability to X, and this ends the pro. Okay, so let's review, because maybe,
00:11:630Paolo Guiotto: It has not been entirely clear. So, we have this sequence, by assumption, is almost surely convergent, so this is the hypothesis. We want to prove that it is also convergent in probability. So, we have to show that the probability that XN is away from X for this quantity epsilon.
00:31:560Paolo Guiotto: It goes to 0 when n goes to infinity.
00:34:310Paolo Guiotto: I want to relate this with the fact that this XN goes to X almost truly, means that the probability that Xn is away from X infinitely many times.
00:49:220Paolo Guiotto: This is what this, feels… this is, you know, which is the characterization of almost Scholtenberg.
00:56:760Paolo Guiotto: Okay, so, we start from the probability that XN is away from X for this one, yeah, so…
01:05:540Paolo Guiotto: we should say that this is one of these sets of when K range from N to plus infinity. So, this set is definitely pertaining to the union of all these sets.
01:19:120Paolo Guiotto: So, the probability of that set will be less equal than this probability.
01:24:710Paolo Guiotto: Okay? And now I take the limit. So the limit of this will be less or equal than the limit of that. This is the next step.
01:32:410Paolo Guiotto: I realized that there is a probability of something there.
01:37:580Paolo Guiotto: And these sets, if we look carefully, when we increase N, they decrease as sets, because they are made of unions, of tail unions. So, unions or sets, they give all indexes from N to plus 50.
01:51:360Paolo Guiotto: So when I pass from n to n plus 1, I lose one of these sets, so the union becomes smaller.
01:58:490Paolo Guiotto: So that's a decreasing sequence of family. I have a limit of probabilities for a decreasing sequence. I can look up the continuity from above. This says that it is the probability of the limit itself.
02:10:550Paolo Guiotto: lead set for decreasing sequence means the intersection of this. So, the intersection of these sets, but this is exactly the set for which we know the probability 0. And this dance means that
02:23:430Paolo Guiotto: as we wanted, the limit of the probabilities that Xn is away from X for this quantity epsilon is, is going to zero. Now.
02:35:230Paolo Guiotto: Of course, huh?
02:36:860Paolo Guiotto: So, warning…
02:40:400Paolo Guiotto: This… these two concepts of convergence are not equivalent. So, the convergence in probability is really weaker.
02:49:890Paolo Guiotto: Does not imply the convergence The almost sure convergence.
02:56:630Paolo Guiotto: Now, the example of this is the same example we have for the L1 convergence that does not imply the LP convergence. So, the example is you take omega, the interval 0, 1,
03:11:670Paolo Guiotto: The, F is, the sigma algebra generated by… The vows, of 01.
03:20:720Paolo Guiotto: And then you take as probability deliberate measure.
03:23:980Paolo Guiotto: Then you construct the sequence? You want a sequence that converges in probability, but it is not convergent, almost surely. So, as we said, we take that sequence. I do not write analytically, so it is…
03:38:480Paolo Guiotto: 1, 0, one, here 0 here, so this is the interval 0, 1, then we have,
03:48:610Paolo Guiotto: The second function, 0, 1, This is 1 half.
03:55:60Paolo Guiotto: And then… and so on. You know very well what these functions are.
04:00:840Paolo Guiotto: So the next one will be 1 here and 0 elsewhere, etc.
04:06:790Paolo Guiotto: Now, why this sequence goes to… of course, the limit will be zero in probability, because if you do the check, this XN
04:18:130Paolo Guiotto: goes to zero in probability, because if you do, what is the probability that modulus of Xn minus the limit 0 be greater or equal than epsil? No?
04:30:410Paolo Guiotto: So you see that this is, first of all, XN minus 0 is XN, and since XN is positive.
04:39:870Paolo Guiotto: So these are the XN, this is the X0, X1, X2, and so on.
04:45:450Paolo Guiotto: This becomes the probability that Xn is greater or equal than epsilon.
04:51:850Paolo Guiotto: Now, if epsilon is too big, like this, there is no omega for which XL is greater than epsilon, so we can say that this will be 0 when epsilon is greater than 1. But when epsilon is between 0 and 1,
05:08:950Paolo Guiotto: Well, no, zero cannot be taken, because the value of epsilon is always strictly positive.
05:16:530Paolo Guiotto: So imagine we are here with epsilon.
05:21:200Paolo Guiotto: So it's the same epsilon, and this is the same 1, no? So basically, what you get is that the probability when XN is greater than epsilon coincides with the probability where XN is equal to 1.
05:36:610Paolo Guiotto: But that probability is the size of the interval on which you have, the function is equal to 1. So this will be, so, let's say 0 for epsilon
05:49:240Paolo Guiotto: greater than 1. Otherwise, you have… I just write the list of values, so you start, you have one half, no. What is the measure of the set? The probability is the Lebec measure, so the set of omegas for which the variable is above epsilon is this one, is this interval. So it's the interval 0, 1.5, and therefore, its probability is the Lebec measure.
06:12:920Paolo Guiotto: Will be 1 half.
06:14:270Paolo Guiotto: For the second one will be one half.
06:17:390Paolo Guiotto: For the third one, this one, when this X2 is greater than epsilon, here, no? On the interval 0, 1 fourth. So it is the probability of this interval, which is the length, because the measure is the, is the Lebec measure. And then we will have four values, 1 fourth.
06:36:520Paolo Guiotto: For the next, the three functions. Then, when we switch to the case of the function where we divide the innate parts, the interval 0, 1, so this is 1 over 8, our function will be 1 only here and 0 here. So when I ask when it is greater than epsilon with epsilon less than 1, when I am on that
06:59:890Paolo Guiotto: the interval 0, 1 over 8, so the probability would be 1 over 8, and so on. This will be 16 times, no, sorry, 8 times, then there will be 1 over 16, 16 times, then there will be 1 over 32, 32 times, and so on. This is the sequence of values. Now, what you see is that for epsilon greater than 1,
07:22:950Paolo Guiotto: the probability that the XN is away from 0 for at least this epsilon is constantly equal to zero, so it goes to 0. It's a constant, it goes to zero.
07:35:600Paolo Guiotto: If you take an epsilon less than 1, no, that probability is non-zero, but it's this sequence, 1 half, 1 half, 1 fourth, one-fourth, 1 fourth, one-fourth, and so on, so clearly it goes to zero.
07:50:190Paolo Guiotto: So, whatever is the choice of epsilon, that probability goes to zero, so this means that the probability of distance between Xn and 0 greater than epsilon, this goes to zero, this for every epsilon positive.
08:07:580Paolo Guiotto: And this exactly means that this sequence, Xn, goes to 0 in probability.
08:14:970Paolo Guiotto: But as already we know, this sequence is never convergent.
08:19:689Paolo Guiotto: But… Oz.
08:23:370Paolo Guiotto: Wait. No.
08:26:979Paolo Guiotto: This XN, is never convergent to anything, actually.
08:37:439Paolo Guiotto: Let's say this, even for every omega in the sample space, there is no way to have convergence for this sequence.
08:46:580Paolo Guiotto: Exactly as it happens for the LP convergence. We know that LP convergence does not imply the almost sure convergence. It's the same example, the same problem.
09:01:700Paolo Guiotto: But what is true is that there exists a subsequence which is point-wise convergence. The same happens with this type of convergence. It's similar proof, we have not seen the other proof, we will not see this one.
09:15:229Paolo Guiotto: there is this factor, which might be useful to keep in mind, that if we have a sequence XN of random variables on omega, so you see here, you don't need to talk about integrability because there is no expectation here, such that this sequence XN
09:35:810Paolo Guiotto: goes in probability to some X.
09:39:310Paolo Guiotto: Then, there exists a sub-sequence X and K, into this XN.
09:47:69Paolo Guiotto: Such that the X and K, this one, goes almost surely to the desired limit X.
09:55:170Paolo Guiotto: And if you look at the example we have done, it is exactly the same subsequence we have shown for the L1 convergence. So you take just the first element of each group, no? Which is the indicator of this interval 0, 1 over 2 to some exponent.
10:15:240Paolo Guiotto: Okay.
10:16:650Paolo Guiotto: The other convergence we have seen, so now we have… L1, LP convergence.
10:23:560Paolo Guiotto: almost sure, and convergence in probabilities. What is the relation between L1 convergence, LP, more in general, convergence, and convergence in probabilities? Since the LP convergences are all stronger than the L1,
10:42:870Paolo Guiotto: we will just limit to the L1 convergence, because if you have LP, you have L1. If L1 implies probability, you are done. And in fact, we have this proposition that says this is easy.
10:55:940Paolo Guiotto: that if you have that XN goes in L1 to X, also for every LP, the same conclusion, then Xn goes also in probability to X.
11:10:530Paolo Guiotto: This is just a consequence, of course. Here, maybe we have to add the sum requirement, that is, now the sequence XN cannot just be a sequence of measurable functions, because L1 demands the expectation finites, so they must be L1 variables, but that's just a detail.
11:31:560Paolo Guiotto: Now, for the proof, that's a consequence of the, of the Chebyshev inequality, because here we have a probability that modulus Xn minus X is greater or equal than epsilon.
11:47:210Paolo Guiotto: Remind that to have convergence in probability, I have to show that this quantity goes to zero, right? Now, I assess this in terms of what? By Chebyshev.
11:59:00Paolo Guiotto: Inequality, this is 1 over epsilon, the expected value of this quantity here, so the modulus of Xn minus X.
12:11:710Paolo Guiotto: And this is exactly the L1 norma of Xn minus X.
12:17:520Paolo Guiotto: So if this goes to 0, Then…
12:21:180Paolo Guiotto: Also, this will go to zero.
12:24:620Paolo Guiotto: And that's it for the proof.
12:28:00Paolo Guiotto: Okay, so this is very, very direct, easy. So, L1 implies convergence in probability, but of course, they won't be equivalent. So, warning.
12:43:910Paolo Guiotto: The convergence in probability
12:49:490Paolo Guiotto: Does not imply, even if the variables are… of course, the convergence in probability applies to just random variables without any
12:59:720Paolo Guiotto: condition on the expectation, assuming even that the random variables are, have a finite, I have mean values, so I'm in L1,
13:10:960Paolo Guiotto: This does not imply that XN converges in L1 to X.
13:23:380Paolo Guiotto: Yeah. I think that we can use the same example we have used to show that point-wise convergence does not imply L1 convergence. So again, take the sample space, the interval 0, 1.
13:40:60Paolo Guiotto: usual setup, the usual… So it's up…
13:47:270Paolo Guiotto: Now, we take a sequence of… let's see if we can see this as the indicators on the interval 0, 1.
13:56:330Paolo Guiotto: On, small intervals.
13:59:430Paolo Guiotto: So they are 0 everywhere, except a small interval where they are bigger. I want… they are bigger in such a way that the expected value of the XN are finite, but they do not converge in L1 to anything. So I will do this, take 1 over N here, and probably n squared here.
14:19:190Paolo Guiotto: So this is the X AND.
14:21:800Paolo Guiotto: Now, if you look at this XN, clearly this XN belongs to L1 omega, no? And the expected value of the modulus of XN
14:34:640Paolo Guiotto: Which is, equal to the expected value of XN, because Xn is positive.
14:44:270Paolo Guiotto: This is what? Well, you see, the variable is 0 everywhere, except on the interval 0, 1 over n, while it is constant equal to n squared. So it would be n squared times the probability of that interval, which is 1 over n, and this makes N. So this goes to plus infinity.
15:03:260Paolo Guiotto: So this says at once also that they are, yes, in L1, but since the norm of DXN
15:09:810Paolo Guiotto: The L1 norm goes to plus infinity. This sequence is unbounded in L1, so it cannot be convergent.
15:16:800Paolo Guiotto: So the sequence XN is unbounded.
15:23:280Paolo Guiotto: in L1 omega.
15:26:970Paolo Guiotto: So, it cannot be… You cannot.
15:32:930Paolo Guiotto: Beat.
15:35:510Paolo Guiotto: convergent.
15:37:300Paolo Guiotto: to… And it… X in L1.
15:44:460Paolo Guiotto: Nonetheless, we showed that it converges in probability to zero.
15:50:10Paolo Guiotto: So… Mom.
15:53:690Paolo Guiotto: Les?
15:57:380Paolo Guiotto: Xn goes in probability to 0.
16:04:550Paolo Guiotto: Why? Because it's the simple calculation. Again, we compute the probability that XN is away from 0 for this value epsilon.
16:16:460Paolo Guiotto: Of course, this means that
16:18:840Paolo Guiotto: This is the probability that absolute value of X is greater than epsilon. Xn is positive, so at the end, I can say this is the probability that Xn is greater or equal than epsilon.
16:32:590Paolo Guiotto: So now you fix an epsilon. Where is it here in this figure? Well, I can say, since I have to do the limit, I can imagine that n can be taken big as I want, so the general situation will be that my epsilon
16:46:160Paolo Guiotto: is below n squared, because when I take N going to infinity, sooner or later, n squared will be greater than epsilon. Epsilon here is fixed, no?
16:56:200Paolo Guiotto: So, fix an epsilon positive, speaks… Epsilon positive.
17:02:800Paolo Guiotto: And, if you want to be formally correct, and let N…
17:07:950Paolo Guiotto: be such that n squared is greater or equal than epsilon, okay? So this will be definitely true, because n is going to plus infinite, so sooner or later, this will be true.
17:22:590Paolo Guiotto: But if you are in this case, so you… you look at DXN when DXN is greater than this epsilon, is above on this interval. So the probability of that interval is… the probability is the LeBague measure, it's still the length of the interval.
17:39:500Paolo Guiotto: So this is equal to 1 over n.
17:43:950Paolo Guiotto: And therefore, when you send N to plus infinity, this quantity goes to zero.
17:50:550Paolo Guiotto: And this explains that the sequence is convergent in probability.
17:56:00Paolo Guiotto: Okay, so let's make a sort of diagram that says that if we have that the sequence XN converges in L1 to X,
18:11:240Paolo Guiotto: Then, now we know that it converges in probability.
18:17:10Paolo Guiotto: 2X. And the same happens if we have the almost sure convergence.
18:24:100Paolo Guiotto: Or the probability 1 convergence.
18:27:140Paolo Guiotto: But there is no relation between these two, so…
18:31:270Paolo Guiotto: There is no way to say that one implies the other.
18:36:540Paolo Guiotto: There is… no one of the two is weaker than the other, so they are not, related. The most can be said is that here, there exists, maybe, we could say.
18:52:380Paolo Guiotto: Maybe, no.
18:57:230Paolo Guiotto: Okay, let's say that here, this implies that, There exists a subsequence X and the key…
19:07:350Paolo Guiotto: Such that, X and K
19:11:370Paolo Guiotto: goes almost surely to X, so this can be side. Also, here we have that the vice versa of these implications are not
19:24:360Paolo Guiotto: But that's not true, so this is not true, and this is non-true as well.
19:29:430Paolo Guiotto: But we can say that, there is sort of similar implication here. From convergence in probability, we have that, again, there exists a subsequence, X and K, such that X and K converges almost surely to X.
19:49:700Paolo Guiotto: Okay, so this is more or less the picture of this.
19:57:520Paolo Guiotto: Well, let's see some… problem.
20:05:220Paolo Guiotto: Here, I… let's consider this one. This is, taken… Wrong.
20:12:240Paolo Guiotto: some external source. We have a sequence YN, of, independent, random variables.
20:26:160Paolo Guiotto: They are distributed exponentially, so YN is exponential.
20:32:880Paolo Guiotto: With a certain parameter lambda N.
20:36:730Paolo Guiotto: Lambda N, for the moment, is not better specified positive number.
20:41:630Paolo Guiotto: No, no, it says something, that lambda n goes to plus infinity. So, well, lambda n goes to plus infinity.
20:52:250Paolo Guiotto: Okay, so the first question is, check that…
20:57:990Paolo Guiotto: The sequence YN goes to zero in probability.
21:04:20Paolo Guiotto: Question 2… It asks, when?
21:11:780Paolo Guiotto: Do we have that YN goes to 0?
21:15:230Paolo Guiotto: Almost surely.
21:17:40Paolo Guiotto: Or with probability 1. And it asks to examine these cases.
21:23:690Paolo Guiotto: do… Cases.
21:27:740Paolo Guiotto: When lambda N is N,
21:31:170Paolo Guiotto: When lambda n is logarithm of n, and when lambda n is logarithm of n squared.
21:43:160Paolo Guiotto: Okay.
21:47:280Paolo Guiotto: So, here, we are going to use the definition for the first question. So, to check that YN goes to 0 in probability, we have to show that
21:59:490Paolo Guiotto: According to the definition.
22:01:880Paolo Guiotto: The probability that
22:04:550Paolo Guiotto: modulus of YN minus the limit 0 greater than epsilon, this thing goes to zero when n goes to plus infinity.
22:16:610Paolo Guiotto: Now, let's compute that probability. Probability where modulus of YN minus 0 greater or equal than epsilon. Well, first of all, this is modulus of YN.
22:29:750Paolo Guiotto: Great or equal than…
22:31:690Paolo Guiotto: epsilon. Now, remind that YN is… here we have an information about the distribution. It says it is exponential. So, when I say that YN is exponential, with the parameter lambda n.
22:49:80Paolo Guiotto: This, equivalently means that we have,
22:53:30Paolo Guiotto: Of course, a CDF, but we have also a density, so there exists a density FYN of Y.
23:00:850Paolo Guiotto: And this density is E minus D lambda, so in this case, lambda NY, indicator 0 plus infinity for Y.
23:15:70Paolo Guiotto: So, in particular, since there is this indicator that tells that the density is zero and y is negative, in particular, the probability that YN
23:26:300Paolo Guiotto: Is less than zero is equal to zero.
23:29:970Paolo Guiotto: And so this means that YN is positive, so that probability boils down to the probability where YN is greater or equal than absence.
23:42:750Paolo Guiotto: Now, since we have the density, this is the integral, on the range epsilon to plus infinity.
23:52:310Paolo Guiotto: Dewey.
23:58:820Paolo Guiotto: One second, let's see if we can do…
24:04:910Paolo Guiotto: Let's see if it can do.
24:08:790Paolo Guiotto: I don't want that itself.
24:13:180Paolo Guiotto: That is before time.
24:18:730Paolo Guiotto: Okay, let me try to…
27:28:280Paolo Guiotto: Oh, I wasn't mute before.
27:31:710Paolo Guiotto: Who knows? Okay.
27:34:270Paolo Guiotto: We are recording, so let's return here.
27:39:910Paolo Guiotto: Good. So…
27:41:810Paolo Guiotto: I was saying that if you know the CDF, this is 1 minus the probability that Y is… YN is less than epsilon. Now, since the variable is absolutely continuous, the CDF is, in particular, continuous. Even more, it is almost everywhere differentiable. This is to say that this is the same of the probability that Y
28:06:750Paolo Guiotto: again.
28:07:410Paolo Guiotto: is less or equal than epsilon, which is the CDF of YN.
28:12:70Paolo Guiotto: at epsilon. So if you remind of the CDF, you don't need to compute the integral, you can go directly. Otherwise, you compute the integral, which is not such a big problem. So we have, what do we have? We have minus 1 over lambda n
28:30:230Paolo Guiotto: E minus lambda NY between epsilon and plus infinity.
28:38:430Paolo Guiotto: Is it correct?
28:43:630Paolo Guiotto: There's something wrong, probably.
28:51:90Paolo Guiotto: Because the integral must be 1 as if I integrate from 0 and plus infinity, and so this means that there is an error with the density.
29:02:30Paolo Guiotto: So what I wrote here is the correct… no, there should be a lambda here.
29:09:950Paolo Guiotto: Because otherwise, I do not get one when I do the total integral. So there is a lambda here, lambda N,
29:17:380Paolo Guiotto: And so this is just the minus this.
29:21:170Paolo Guiotto: Otherwise, if you take epsilon equals 0, you would get something different from 1, so now it is correct. So at plus infinity, I get 0, lambda n remains that are positive numbers. At epsilon, I get minus, so with the minus plus E minus lambda n epsilon.
29:41:380Paolo Guiotto: Now, this is what the…
29:44:50Paolo Guiotto: This is the probability that Yn minus 0 in absolute value is greater or equal than epsilon. So, let's write a probability that distance between yn and 0, greater or equal than epsilon, is equal, exactly equal to E minus lambda n times.
30:03:450Paolo Guiotto: Absolutely. Now, what happens if we send N to infinity?
30:07:790Paolo Guiotto: You see that n is here, into these lambda Ns. What do we know about lambda N? Since…
30:16:520Paolo Guiotto: We know that, the lambda in.
30:20:290Paolo Guiotto: Are going to plus infinity.
30:26:40Paolo Guiotto: This means that that exponent, E minus lambda n lambda N epsilon, since this is positive, the exponent is going to minus infinity, and the exponential is going to zero. And this is the conclusion, so YN goes to zero in probability.
30:47:420Paolo Guiotto: We can now discuss the second question.
30:51:320Paolo Guiotto: Which is, the almost sure convergence of the sequence YN, no?
31:00:100Paolo Guiotto: Now, we may say that, we may notice that even if we don't know what is the limit.
31:06:310Paolo Guiotto: Since we know that almost sure convergence always implies probability convergence with the same limit, the unit possibility to have an almost sure limit is zero, okay? So we noticed that.
31:20:640Paolo Guiotto: YN, if YN converges almost surely to some Y, then necessarily that YN would converge also in probability to the same Y.
31:34:180Paolo Guiotto: Well, actually, I should convince you that the limiting probability is unique to tell this. Well, okay, exercise…
31:44:960Paolo Guiotto: check.
31:47:780Paolo Guiotto: that… Limit… in probability, is unique.
31:59:740Paolo Guiotto: There is not the proof in notes, but try to do, you just have to work with the definition. Maybe we will see this solution, on Friday.
32:11:150Paolo Guiotto: And we already know that the sequence YN converges in probability to 0, so Y must be equal to 0.
32:19:610Paolo Guiotto: So, we investigate if Y converges almost purely to 0. Now, to have this, we have already, many times, checked this. This means that the probability
32:33:310Paolo Guiotto: of the limb subset intersection in capital n union for n greater or equal than capital n of distance between YN and the limit, so modulus y n greater than epsilon. This probability must be equal to zero.
32:53:250Paolo Guiotto: Okay, now, since this is the limb soup set.
32:58:220Paolo Guiotto: of these events, modulus y n greater or equal than epsilon.
33:04:410Paolo Guiotto: So, the idea is, as usual, to apply the Borean telilemma to assess this probability.
33:10:600Paolo Guiotto: So we know that the probability that modulus y n is greater or equal than epsilon, is the quantity we computed a minute ago, is just that exponential, E minus lambda N epsilon, so E minus lambda N.
33:26:260Paolo Guiotto: Epsilon.
33:28:50Paolo Guiotto: And moreover, the events modulus YN, greater equal than epsilon.
33:36:340Paolo Guiotto: These events depend on the YN, and since the YN are independent.
33:42:720Paolo Guiotto: Yes, independent random variables, so… independent.
33:50:130Paolo Guiotto: We can say that That probability of the limb soup So by…
33:57:920Paolo Guiotto: what I can tell you, the probability of the limb soup of these EN
34:04:90Paolo Guiotto: is 0. In this case, if you go back to the statement.
34:09:50Paolo Guiotto: Of the bread can tell you? Where is it?
34:13:230Paolo Guiotto: It's a… Rolls up…
34:17:300Paolo Guiotto: Okay, you see that. In this case, we are in the case when the events are independent, so we have basically an alternative.
34:27:630Paolo Guiotto: If the series of the P probabilities is finite, then that probability is zero. If the series is infinite, we can use also the second statement. The first one does not demand
34:39:150Paolo Guiotto: to have EN independent, but the second requires this. And in that case, we have that the probability is 1, so the conclusion would be never.
34:48:490Paolo Guiotto: So here we see that, in other words, the probability of the limb subset is 0 if and only if the sum of the probabilities of DN is finite. Because if the sum is infinite, we can apply the second part that says the probability of the limb soup would be 1, not 0.
35:08:60Paolo Guiotto: So the unique possibility to have probability 0 is, in other words, this becomes an if and only if in that case, you see?
35:16:460Paolo Guiotto: So… We can say that. What is it?
35:23:950Paolo Guiotto: So, this if and only if the sum of the probabilities of the n
35:31:840Paolo Guiotto: is finite. So, if and only if the sum of these quantities, E minus lambda and epsilon, is finite.
35:42:420Paolo Guiotto: So, this is the condition that must be verified in order to have the pointwise
36:06:820Paolo Guiotto: converges almost surely to zero, if and if the series of E minus lambda N is n epsilon is convergent.
36:17:270Paolo Guiotto: But this means what? This means the series, I can rewrite this as E minus epsilon to the N. This is a geometric series.
36:27:30Paolo Guiotto: geometric.
36:29:540Paolo Guiotto: Serious.
36:31:250Paolo Guiotto: Where you see that this quantity here is positive, and…
36:36:150Paolo Guiotto: Less than 1, because epsilon here is positive.
36:41:20Paolo Guiotto: And this is the condition that ensures that CDS is convergent.
36:46:60Paolo Guiotto: So this is verified in this case, okay? So we can say that for this lambda N, this sequence goes to zero almost surely. If you take lambda n, the second case, for example, lambda n
37:00:820Paolo Guiotto: equal log of n. Let's see what happens. So you see that now we are getting a slower lambda N,
37:09:980Paolo Guiotto: Respect to the previous case, we have that YN goes to 0, almost surely.
37:15:720Paolo Guiotto: If and only if we plug this lambda N into that formula, we get E minus log of n
37:23:170Paolo Guiotto: Times, let's put the epsilon here.
37:26:420Paolo Guiotto: Now, what is this? I can rewrite this, giving the minus epsilon at the exponent of N, in such a way this becomes E to log of n to minus epsilon.
37:38:900Paolo Guiotto: or in other words, E and log, they cancel. This is sum of n to minus epsilon.
37:45:510Paolo Guiotto: Which means, sum of 1 over n to epsilon.
37:51:470Paolo Guiotto: Okay? Now, when this series is convergent, because this is an harmonic series, this is finite if and only if that number epsilon is a little bit big, so epsilon must be greater than 1.
38:07:500Paolo Guiotto: So what is the conclusion? Is it true that YN is almost surely convergent to 0?
38:14:710Paolo Guiotto: No, because that limb superset is a probability zero event only when epsilon is greater than 1. For the other epsilon, that probability is 1.
38:27:190Paolo Guiotto: So it means that it is never convergent. So in this case, we could say that YN is not convergent to 0 with
38:40:420Paolo Guiotto: Probability 1.
38:42:790Paolo Guiotto: You may ask, but for some epsilon, it is true. No, because to have convergence, almost sure convergence, you must have that for every epsilon, that set, the limb soup set.
38:56:940Paolo Guiotto: where is it? Must have probability 0. If, for one single epsilon, you do not have probability equals zero, this means that there is no convergence.
39:07:770Paolo Guiotto: Okay? So, this property, probability of this set equals zero, must be verified for every epsilon positive, not just for some epsilon.
39:18:170Paolo Guiotto: or if there is some epsilon that does not verify this, it's not a problem. It's a problem. So this is saying, it is true when epsilon is greater than 1, it is false for all the other epsilon. Do the remaining case, so lambda n equal log N,
39:36:540Paolo Guiotto: square. This is a bit faster than this one, and you check whether anything changed or not.
39:44:70Paolo Guiotto: Okay.
39:45:490Paolo Guiotto: Let's see on Friday.
39:48:350Paolo Guiotto: I will, modify something in notes, adding some exercises. Let me live with… Some exercise to do, so…
40:06:750Paolo Guiotto: Sort.
40:08:300Paolo Guiotto: Well, this was basically an extension at 852, so I will not give you any more.
40:13:810Paolo Guiotto: do… 855… The next one…
40:23:380Paolo Guiotto: asks in distribution, this one asks for with the convergence in distribution. I will add the, plus exercises I will write here below, okay?
40:35:60Paolo Guiotto: Okay, let's stop here.