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Transcript
00:05:200Paolo Guiotto: Okay, so today we start, And you…
00:10:780Paolo Guiotto: Topic that is convergence of random variables.
00:16:850Paolo Guiotto: So, we will discuss, convergence,
00:23:370Paolo Guiotto: Or a sequence of random variables, so let… Omega F… B… be a probability space.
00:38:730Paolo Guiotto: And we assume to have a sequence, say, capital XN,
00:43:310Paolo Guiotto: Of, random variables, for the moment, just measurable functions.
00:48:230Paolo Guiotto: sequence… of random variables.
00:56:190Paolo Guiotto: But we want to discuss,
00:58:970Paolo Guiotto: To introduce the most important, concept of convergence for these sequences.
01:07:600Paolo Guiotto: Now, we can consider this as measurable functions, so we have at least two types of convergence, which are naturally defined here. So, let's say natural…
01:23:80Paolo Guiotto: convergence.
01:24:890Paolo Guiotto: are, for example, I can consider the sequence XN as convergent to some X point-wise, while in the language of measurable function, means almost everywhere, or almost surely here.
01:41:320Paolo Guiotto: So this means that the probability that XM goes to X
01:47:500Paolo Guiotto: is 1. In fact, in probability, this is almost… also called probability 1 convergence, or convergence with probability 1. So, convergence…
02:03:470Paolo Guiotto: we… probability… 1.
02:10:60Paolo Guiotto: So we already explored this kind of convergence with measurable functions, so we know something about this, and we will use it.
02:21:920Paolo Guiotto: There is something, let's say, essentially new that comes from the fact that we are dealing with the probability space, so I find it measures space, and…
02:33:710Paolo Guiotto: with the fact that that could be also independence playing a role here. So we will see how this reflects on this convergence. And the other one is the LP convergence. So this is usually… so this, in this case, means that, let's say, with the notations of probability, so we know that this means the integral of the
02:58:460Paolo Guiotto: So, with the language of norms, this means,
03:03:410Paolo Guiotto: Norm of XN minus X in P norm goes to 0, and this would be the integral of omega of modus xn minus X to power P in DP.
03:15:400Paolo Guiotto: goes to zero. In the language of probability, you know that we do not write integrals, so this means that the expected value of modulus Xn minus X to power P
03:28:520Paolo Guiotto: goes to zero. So these are called convergences in mean, because there is the mean value involved, okay? So this is the second, type of convergence we can already,
03:47:180Paolo Guiotto: We have already discussed previously. There are other two types of convergence. One is the convergence in probability.
03:57:730Paolo Guiotto: convergence.
04:02:230Paolo Guiotto: in probability.
04:06:110Paolo Guiotto: Which is,
04:08:400Paolo Guiotto: could be formulated also in the general language of measures. It's not specific of probability, but in probability, it makes sense. So if you look at the point-wise convergence, the point-wise convergence means that
04:24:160Paolo Guiotto: If we are a little bit more formal, pointwise, almost surely, pointwise convergence means that
04:35:780Paolo Guiotto: Means that, the set of, omega, in capital omega, such that,
04:46:480Paolo Guiotto: this thing, a limit, when n goes to infinity of Xn of omega equal X of omega.
04:58:480Paolo Guiotto: This holds for almost, every omega, almost surely, omega in capital Omega.
05:05:870Paolo Guiotto: That is, we say that the probability of the set of omegas for V in capital omega for which the limit in n of the Xn omega is X of omega.
05:20:480Paolo Guiotto: is equal to 1.
05:23:670Paolo Guiotto: Because this means that
05:25:420Paolo Guiotto: the set of omegas for which this is false is a measure 0 set, a probability 0 of n. And this is what we wrote above quickly. It is probability where the limit in n of the Xn is equal to X,
05:41:710Paolo Guiotto: is 1, or the same thing is probability that XN goes to X is equal to 1, okay? This is the convergence, the point-wise convergence, with probability 1.
05:59:660Paolo Guiotto: Well, convergence in probability says something different, but similar to this, apparently. We take the probability that XN is, say, far from X, or a certain quantity epsilon that will be positive.
06:16:320Paolo Guiotto: And the idea is that we say that Xn converges in probability if this probability goes to zero. So the probability that Xn is far from X by a bound epsilon.
06:31:940Paolo Guiotto: vanish when n goes to infinity.
06:34:600Paolo Guiotto: Well, as you may imagine, there should be some relation with this and this one that seems to be stronger, and in fact, as we will see, this is stronger and implies this one, okay?
06:48:500Paolo Guiotto: So it's a weaker definition of convergence, and it is actually weaker, as we will see also of the LP convergence. Also, this one implies the convergence in probability.
07:01:660Paolo Guiotto: And there is a final, definition, which is interesting, for probability, which is convergence, in distribution.
07:16:890Paolo Guiotto: So, this one, has a number of equivalent formulations.
07:26:210Paolo Guiotto: Well, one of them is this one, that you look at the characteristic function of XN, and this pointwise converges to the characteristic function of X.
07:44:700Paolo Guiotto: So, it's, you know that the characteristic function is the Fourier transform, so it's, indifferent, where the Xena are defined.
07:54:320Paolo Guiotto: You can have that two random variables on two different probability spaces have the same…
08:01:610Paolo Guiotto: characteristic function. So this means that in this final definition, the variables X and could be even defined on different probability spaces, so it's a very weak sense of convergence.
08:17:230Paolo Guiotto: Equivalently, as we will see, this is equivalent to the following, that when we do the expected value of a function, say, phi of Xn, if the function phi is nice enough, this goes to the expected value of phi of X.
08:38:240Paolo Guiotto: So this is… these are the four definitions.
08:41:580Paolo Guiotto: And this one is weaker than the previous one.
08:46:190Paolo Guiotto: So basically, we are going to do a process from strong definition to weaker definition, the last two, and there are a few others, but let's say that these four are the most important.
08:59:980Paolo Guiotto: So let's start from the LP and the point-wise convergence. So the first is… so let's start with the LP convergence.
09:13:480Paolo Guiotto: We already know what is it, so the definition is that XN converges in LP.
09:24:670Paolo Guiotto: 2X if and only if, if you want to be writing probabilistic notations, expectation of modules XN minus X to power P, this goes to 0.
09:39:690Paolo Guiotto: Now, here, P is any exponent from 1 to plus infinity, so we may also include the L-infinity convergence, which is a particular… let's say that it is a case, a case different from the LP, because it is not involved in an integral.
09:59:150Paolo Guiotto: And we know that for P equal plus infinity, the LP convergence implies the pointwise convergence is stronger. It's a uniform convergence, sort of uniform convergence. While for P… for all the other P, from 1 to plus infinity, but excluded plus infinity.
10:17:920Paolo Guiotto: we… it is non-true that DLP convergence implies the point-wise conversion. We have seen examples. The same example we have seen with those, I will refresh in a moment, with those indicators works also here, because we can look at that as a probability space.
10:33:570Paolo Guiotto: So, we… we do a couple of remarks. We noticed that here, since we are on a probability space, since…
10:47:60Paolo Guiotto: we… On a probability.
10:53:840Paolo Guiotto: space,
10:55:870Paolo Guiotto: These convergences are ordered, in the sense that if you have that XN converges in LP to X,
11:06:600Paolo Guiotto: and you take a Q, which is less than P,
11:11:610Paolo Guiotto: So, greater or equal than 1, then Xn will converge also in LQ norm to X.
11:20:200Paolo Guiotto: So, L2 convergence implies L1 convergence, L4 implies L2, and so on, no? So, for example,
11:30:930Paolo Guiotto: which is a very important case, the L2 convergence implies the L1 convergence.
11:41:450Paolo Guiotto: Well, this can be proved as a consequence of the generalization of the Cauchy-Swartz inequality, which is the holder inequality.
11:49:340Paolo Guiotto: Or we can get from the Janssen inequality that we have seen for the conditional expectation, but we have a particular case of that inequality. This…
12:04:320Paolo Guiotto: Scan.
12:05:760Paolo Guiotto: be obtained at the…
12:11:580Paolo Guiotto: I was… a consequence.
12:19:140Paolo Guiotto: of Janssenins.
12:24:860Paolo Guiotto: inequality.
12:26:990Paolo Guiotto: So you reminded that we proved this, that, if, phi is a convex function.
12:37:660Paolo Guiotto: Then,
12:40:260Paolo Guiotto: Actually, here we… we need the… yeah, we have that, fee of the conditional expectation.
12:50:50Paolo Guiotto: of, of, X, given, a sub sigma algebra G.
12:57:170Paolo Guiotto: is less or equal than expect the conditional expectation of phi of X even G.
13:06:580Paolo Guiotto: Now, if you take the generated sigma algebra, the trivial sigma algebra, made of empty and the full space.
13:15:920Paolo Guiotto: So…
13:16:870Paolo Guiotto: This is if you want a particular case of a sigma algebra generated by a partition. The partition is made by one single set, the omega set itself. In this case, you can easily see that the conditional expectation of X given G is just the expectation of X.
13:38:750Paolo Guiotto: Okay.
13:40:300Paolo Guiotto: So, you have that the inequality boils down to this phi of the expected value of X,
13:47:470Paolo Guiotto: is less or equal than expected value of phi of X.
13:54:440Paolo Guiotto: And this is the, let's say, the classical form of the Janssen inequality.
14:00:150Paolo Guiotto: So now, what has this to do with the LP convergence? So this yields to an inequality between the LP norms. From this, it follows this proposition.
14:15:580Paolo Guiotto: the, if,
14:17:730Paolo Guiotto: say Q is greater or equal than 1, and it is less than P, less than plus infinity, we have that, D, Q norm
14:29:710Paolo Guiotto: of X is controlled by the P norm All next, so…
14:37:220Paolo Guiotto: So, the higher is… so the norms are really ordered, no? When you increase the exponent, you increase the norm, huh?
14:45:830Paolo Guiotto: And this is because…
14:49:650Paolo Guiotto: If we take the Q norm, this is to power Q for a second, this is the expected value of modulus of X to power Q.
15:13:450Paolo Guiotto: So,
15:19:390Paolo Guiotto: Yeah, and the Pinotment, the no… the Pinotment…
15:23:280Paolo Guiotto: to power P is the expected value of modulus of X to power P.
15:32:300Paolo Guiotto: Do we need the concave version of this?
15:37:840Paolo Guiotto: Okay, let's see what happens.
15:40:250Paolo Guiotto: So,
15:46:940Paolo Guiotto: So, the Q norm…
15:52:910Paolo Guiotto: Yeah, I now want to do this. I want to transform this exponent into this one.
15:59:400Paolo Guiotto: So if I do this, I raise all this to the exponent P over Q, since p is greater than Q, so take the function phi of X equal to X
16:11:920Paolo Guiotto: to exponent P over Q. Since P over Q is greater than 1, this function is convex, so for X positive.
16:21:110Paolo Guiotto: It's a function made like a power with an exponent greater than 1, so this is the plot of phi. So phi is convex.
16:31:190Paolo Guiotto: If we apply a fee, to the Q norma, so I have that fee of the Q norma
16:42:560Paolo Guiotto: to power Q, since this is raising this quantity, the Q norm to power Q to exponent P over Q, this is the Q norm to power P.
16:57:690Paolo Guiotto: On the other side, this is a phi of the expectation of modulus X to power Q.
17:06:560Paolo Guiotto: And because of Janssen inequality, this… we can switch, let's say, expectation with phi, so this becomes less or equal than the expected value of phi of modulus of X to the Q.
17:22:940Paolo Guiotto: But, this is a modulus of X to Q raised to P over Q,
17:30:780Paolo Guiotto: And therefore, this is expected value of modulus of X to P, which is the P norm of X.
17:41:140Paolo Guiotto: to power P. So, at the end, I get that this, the Q norm of X to power P is less or equal than this, which is the
17:55:130Paolo Guiotto: P norm of X to power P, and taking the P root, I get, finally, that the Q norm of X is less or equal than the P norm of X.
18:09:300Paolo Guiotto: Okay, so this says, basically, that the strongest is that one with the highest exponent, and the weakest is that one with the lower. And so, in particular, the 1L1 convergence is the weakest among the
18:26:810Paolo Guiotto: LP convergences, so… And, of course, this is more trivial, also.
18:35:620Paolo Guiotto: Any, let's say, Q normal is controlled by the infinity
18:41:500Paolo Guiotto: Norma. This is, trivial, because the Q norma is,
18:49:540Paolo Guiotto: is the expected value of modulus of X to power Q, all this to 1 over Q.
18:59:180Paolo Guiotto: And since here this is controlled by the infinity normal, modulus of X is always controlled by its infinity normal.
19:10:110Paolo Guiotto: this to power Q.
19:12:190Paolo Guiotto: So if I replace that quantity with the norm, infinite norm to power Q, I increase the function, I increase the integral, the expectation, so this will be less or equal than the expected value of infinity norm of X to power Q.
19:29:400Paolo Guiotto: all this raised to 1 over Q. But this is a constant, so the expectation of a constant is the value of the constant, so you have infinity norm
19:39:140Paolo Guiotto: of X to power Q, all this to 1 over Q, and this is the Q norm, the infinite norm of
19:46:320Paolo Guiotto: X.
19:47:510Paolo Guiotto: So, this yields this bound. So, the infinity non-convergence is, of course, stronger than all LP convergences.
19:59:220Paolo Guiotto: Now, we remind that, so in particular, from this, it follows that if you have infinity convergence, so if XN
20:11:720Paolo Guiotto: converges in an infinity, norm to X, then we have that Xn converges also, almost surely, to X.
20:26:330Paolo Guiotto: Because this is stronger.
20:28:740Paolo Guiotto: But…
20:30:740Paolo Guiotto: If Xn converges in some LP norm to X, here P is less than plus infinity, greater or equal than 1, in general, we do not have the point-wise convergence.
20:51:830Paolo Guiotto: Okay, so, And the example is, is the same. We have, done,
21:00:630Paolo Guiotto: In the first part of this course, so we take as omega, the probability space, the interval 0, 1. F is the Lebag sigma algebra, and the probability is the Lebag measure on that interval.
21:15:820Paolo Guiotto: And, the functions are the sequence XN,
21:21:60Paolo Guiotto: is the sequence built in this way? You remember that you take the interval 0, 1, you divide in two parts, and you build the first two functions of the sequence.
21:36:740Paolo Guiotto: So this is 1 here and 0 here.
21:41:540Paolo Guiotto: This is 0 here and 1 here.
21:46:690Paolo Guiotto: So this is, let's say… now we should… Let's call this X1.
21:54:720Paolo Guiotto: No, you see X0, maybe X1 this. Then we have X2, we divide the interval in four parts, 4 equal parts.
22:05:970Paolo Guiotto: So this is the first of this block.
22:10:410Paolo Guiotto: And then we have the second, which is, one on the second,
22:15:790Paolo Guiotto: Quarter of the interval, and so on.
22:20:360Paolo Guiotto: Okay, so this is the sequence. We can write analytically, but…
22:25:670Paolo Guiotto: It's just a waste of time doing exactly here, because we understand that, first of all, this sequence, XN goes in L1 norm.
22:39:880Paolo Guiotto: to zero.
22:41:250Paolo Guiotto: And that's because if we need just to compute the expected value of modulus Xn minus 0, which is the modulus of Xn, Xn is positive, which is the expectation of XN,
22:55:650Paolo Guiotto: But, as you can see, the expectation of XN is just the integral of that variable, so it's the area between XN and the axis, so it reduces to this thing. So, for the first two, the expected value.
23:14:40Paolo Guiotto: yeah, of, this, X0, 1 is equal to, 1 half for the next four, X, two.
23:25:620Paolo Guiotto: 3, 4, 5. This is equal to 1 fourth, and so on. As you can see, it decays rapidly to 0.
23:37:210Paolo Guiotto: But if you look at the sequence XN, it is never point-wise convergence to anything.
23:45:770Paolo Guiotto: So, for every omega in 01, so in the sample space, which is the interval 0, 1. So, this is the example that shows you can have LP convergence, but not the pointwise convergence.
24:02:110Paolo Guiotto: However, we have this remarkable factor that we have,
24:06:540Paolo Guiotto: mentioned for measure theory, and talks also here, of course, that, if, XN converges in LP norm.
24:19:830Paolo Guiotto: To some X, there is always a subsequence, X and K, drawn from, DXN.
24:29:30Paolo Guiotto: Such that for this sequence, X and K, we have convergence with the probability 1, or almost true convergence to X.
24:38:820Paolo Guiotto: In the case of this sequence, you just take the first element or feature of the blocks, so this one, then you take this one, then you will take the X6, and so on, okay?
24:51:990Paolo Guiotto: And that one will go pointwise at every omega to zeros, except omega equals zero, where it goes to 1.
25:00:690Paolo Guiotto: So, in conclusion, we can say that, in general, the LP convergence, this should be kept in mind, that if you have LP convergence, in general, does not imply…
25:17:650Paolo Guiotto: The point-wise convergence with probability 1.
25:22:440Paolo Guiotto: But it implies, if we accept to
25:26:30Paolo Guiotto: to extract a subsequence, but it implies that to exist a subsequence X and K.
25:34:660Paolo Guiotto: in the XN, such that X and K goes almost surely to X.
25:44:80Paolo Guiotto: Okay, now, about the almost pure convergence, we also know that it does not imply DLP convergence.
26:01:520Paolo Guiotto: Sweet.
26:02:820Paolo Guiotto: Also… no… That's it.
26:09:880Paolo Guiotto: the, XN, that… XN.
26:16:560Paolo Guiotto: Almost surely.
26:18:320Paolo Guiotto: converging to some X does not imply any LP convergence.
26:27:730Paolo Guiotto: I… also here, we have, still the simple example in the same… even…
26:33:910Paolo Guiotto: Simpler than the previous one. So, as above, take 0, 1,
26:39:330Paolo Guiotto: with the sigma above Lebec class, and the probability measure, the Lebec measure.
26:47:110Paolo Guiotto: And the idea, we take a sequence made like this. 0, 1,
26:52:480Paolo Guiotto: we divide 01 into a big part where the variable is 0, and the small part where the variable is big, big enough to have that the integral is too big. So, for example, put this N square.
27:10:760Paolo Guiotto: So, in this case, you have that this XN would be
27:16:10Paolo Guiotto: the indicator of the interval 0, 1 over n squared, that is 1 over n, which is multiplied by n squared out here.
27:28:510Paolo Guiotto: Now, in this case, you can see that Xn of omega goes to zero.
27:36:370Paolo Guiotto: for every omega of 0, 1, except for omega equals 0, where it goes to plus infinity, but since it goes for every omega, and the measure is the LeBague measure, this means that it goes almost surely. So this means that XN goes almost surely.
27:55:40Paolo Guiotto: to zero.
27:56:670Paolo Guiotto: But if you do the weakest LP convergence is the L1, so if you don't have L1, you cannot have L2, because we say the
28:05:910Paolo Guiotto: the LP norms are ordered.
28:09:820Paolo Guiotto: So if you have a convergence sum.
28:12:210Paolo Guiotto: LP norm with P greater than 1, you have also L1. So if L1 does not hold, you cannot have any LP with P greater than 1.
28:21:550Paolo Guiotto: So here, we see that in the convergence, if there is a limit, since there is a point-wise limit, it must be zero.
28:31:830Paolo Guiotto: Because if you have an LP convergence, so if Xn would converge, say, in L1 to X, then necessarily there is… there would be a subsequence X and K, such that X and K would go almost surely to that X. But we know that XN goes almost surely to 0, so also the subsequence would go to 0.
28:55:610Paolo Guiotto: So the unique limit, possible limit, is 0.
28:58:580Paolo Guiotto: So this says that XN should go in L1 cents to 0. But if you compute the expected value of modulus Xn minus 0, which is the
29:13:690Paolo Guiotto: the test for convergence in L1, this is expected value of Xn, Xn is positive, so we get this, and this is a standard integral, so 1 over n times n squared, so it is equal to n that goes to plus infinity.
29:33:770Paolo Guiotto: So this implies that XN cannot converge in Elvis sense to 0. And since 0 is the unique possible candidate, this means that this sequence cannot be convergent.
29:47:950Paolo Guiotto: So, this shows that,
29:52:440Paolo Guiotto: So we have seen that. LP convergence does not imply almost sure convergence, and almost sure convergence does not imply LP convergence, though they are not equivalent.
30:01:590Paolo Guiotto: There is this, this sort of weak relation. LP convergence implies almost true convergence for a subset.
30:13:100Paolo Guiotto: Now, it is very important to explore a bit more the point-wise convergence, so… Let's… See… better…
30:29:890Paolo Guiotto: on the… Which… Conditions… we… have… Almost sure are probability ones.
30:50:300Paolo Guiotto: convergence.
30:52:10Paolo Guiotto: Now, the point is that usually, when you give a random variable, we will never give the analytical formula, for example.
30:58:410Paolo Guiotto: So normally, it's not possible to compute the limit just because you see omega by omega, what happens, you see?
31:05:750Paolo Guiotto: So, you'd rather tell what is the distribution and things like that, okay? So, it's important to understand how could we verify this, this, convergence. So, we may say that, so returning to the definition, so XN
31:24:590Paolo Guiotto: goes almost surely to, say, X, if and or if, as we said, the probability of limit n of the Xn omega, well, let's… let's do not write the omega, XN is equal to X,
31:43:110Paolo Guiotto: Is equal to 1, this probability.
31:47:80Paolo Guiotto: Now, or,
31:55:100Paolo Guiotto: Okay, let's see if we can… we can write this thing here in positive, no?
32:00:230Paolo Guiotto: Now, one limit, that's a point-wise limit, right? So it means that you take an omega, and you look at the limit of Xn of omega equal to X of omega. When this happens, well.
32:15:110Paolo Guiotto: This is a limit of a numerical sequence of values, no? So, you say that the limit in N of the sequence AN is equal to L,
32:25:970Paolo Guiotto: Find it.
32:28:500Paolo Guiotto: If and only if the following property holds. This is the first year cargo, so, you know, for every epsilon.
32:35:850Paolo Guiotto: That means how much close to the limit you want to go. There exists an initial index n, which is, in general, do not forget that, in general, this depends on epsilon.
32:49:700Paolo Guiotto: And…
32:50:490Paolo Guiotto: possibly on the limit, but since we're here, we are talking of a unique limit, it seems that this is superfluous. Remind that we are going soon to talk about the limit that depends on omega, so…
33:00:860Paolo Guiotto: actually, this will be something which is valuable, depends on the limit, and depends on omega, on the omega. However, let's keep for the moment just this notation, natural, such that, from that natural index.
33:16:540Paolo Guiotto: on, we have that this distance is less or equal than epsilon for every n larger than this capital N.
33:24:780Paolo Guiotto: So, if we want to rewrite this, we may say that this property, star, the limit of Xn omega is X omega, if for every epsilon.
33:36:440Paolo Guiotto: There exists a natural N.
33:39:930Paolo Guiotto: Such that the distance between xn of omega and X of omega is less than epsilon for every n greater or equal than capital N.
33:52:600Paolo Guiotto: Now, this is the property, no?
33:55:900Paolo Guiotto: But since we have now to put this into this condition, probability of the set of omegas where this thing happens is 1, so I want to try to write this as a set, in such a way that I can say the set of omegas that verify this condition has probability equal 1.
34:14:989Paolo Guiotto: So, this means, literally, there's probability of the set of omegas in capital omega, such that
34:23:290Paolo Guiotto: All this is true, so for every epsilon positive, there exists a natural n, such that the distance between XN omega and X omega is less or equal than epsilon for every n greater or equal than capital N. All this
34:41:449Paolo Guiotto: is 1. The probability of this size is 1.
34:45:719Paolo Guiotto: Now, let's try… to decompose this writing, because if you have an omega that belongs to this set.
34:55:969Paolo Guiotto: Okay? You see that you have this first for every epsilon.
35:02:10Paolo Guiotto: For every epsilon, there exists an N such that this condition is verified another for every.
35:09:230Paolo Guiotto: So, what should be reasonable to do? Let's start from the set of omegas in omega that verify the inequality. The distance between X and omega and the limit X omega is less or equal than epsilon.
35:25:590Paolo Guiotto: Okay, if you are in this set.
35:28:860Paolo Guiotto: If you are in this set, it means just that if you are an omega, which is in this set, it means that Xn of omega
35:37:450Paolo Guiotto: has a distance to the limit no larger than epsilon. So, if you are in this set for a single n, for a single epsilon, it doesn't mean that the limit is… exists and it is equal to X of omega, you see?
35:50:330Paolo Guiotto: This is just a set of omegas that verify this inequality. Now, if I wanted this inequality to be verified for every n larger than this capital N,
36:01:810Paolo Guiotto: It means the mi omega must be in all these sets, for every n larger than capital N, so it means that it must be in an intersection
36:12:80Paolo Guiotto: of these sets for n greater or equal than capital N. Now, if the omega is here, it means that for that omega, it belongs in all this set, so these qualities are verified for every n larger than this set.
36:29:560Paolo Guiotto: So, I am now more close to this self, but I'm not yet. I'm just refined this condition.
36:36:150Paolo Guiotto: Okay, but that capital N exists. So, for which N? It means for one N, okay? So, this means that there is an N for which I am in this set. So, this means I am in the union of these sets, union with respect to the capital N.
36:55:730Paolo Guiotto: Now, if I am in this sector with the union, it means… imagine I am owner of this set. In the section of this, it means that I am in at least one of them, so there exists an N.
37:08:780Paolo Guiotto: For which I am in missing the second, for which I am in all of them for every end lasts the death end.
37:15:540Paolo Guiotto: Okay, so now this means that if you are here, the omega verifies the existing n for which the inequality, Xn minus X is less than epsilon for every n little greater than
37:31:100Paolo Guiotto: capital N. Now, there is another constraint here. For every epsilon, this must be. So my omega must be in this set for every epsilon. Now, there is an epsilon here, this means that there is an intersection of the positive epsilon.
37:49:400Paolo Guiotto: And my omega must be in all this if I want that, and it's here. You see? Because if you are here, for every epsilon.
38:00:00Paolo Guiotto: You are in this set, so it means that you are in the intersection of all these sets.
38:06:960Paolo Guiotto: Now, if you think about there is no other condition, so that intersection, union intersection is exactly the
38:14:210Paolo Guiotto: event we are describing there, the event for which the sequence XN has limit X. So, the event the sequence XN has limit N can be written in this way. So, we have to prove that.
38:29:730Paolo Guiotto: So, probability, that… Intersection of an epsilon positive.
38:35:640Paolo Guiotto: of the union over n of the intersection when little n is larger than capital n, of, we can shortly say modulus Xn minus X is less or equal than epsilon. This probability is 1.
38:53:780Paolo Guiotto: Okay, so this is equivalent, an equivalent formulation to the statement XN converges almost surely to S. So this is equivalent to…
39:07:440Paolo Guiotto: Xn converges almost surely, or with probability 1 to X.
39:14:890Paolo Guiotto: Okay.
39:16:980Paolo Guiotto: Now, we need a little refinement, yeah, because
39:21:710Paolo Guiotto: These two operations, so this intersection is a countable set operation, because it involves the n that here is a discrete index, it's a natural 1, 2, 3, 4.
39:34:820Paolo Guiotto: So this is a count of our operation, which is good with the sigma algebra and with measure structure. Also, this one, because this capital N is an initial index, is a natural.
39:47:630Paolo Guiotto: This one is a real number, so it wouldn't be good, because this is an uncountable operation.
39:53:90Paolo Guiotto: But the idea you understand is that it is sufficient that here we take N epsilon, that B is small enough, so we don't need to take all the epsilon. We can't just do a selection of epsilon that, with the root condition goes to zero. For example, we could replace that epsilon
40:11:830Paolo Guiotto: with a discrete epsilon, for example, 1 over K here, and take here all K, let's say positive, but K natural.
40:23:00Paolo Guiotto: In such a way that we discretize. You can understand what we have on the same side.
40:28:620Paolo Guiotto: It is clear that if you are here, intersection for every action of union, or for intersection, etc, you are in this right sector, because if you are for every action, you will be for this particular excellence, no?
40:43:890Paolo Guiotto: But vice versa, if you verify that quantity for a… just for numbers of type 1 over K, so 1, 1 half, 1 third, 1 fourth, and this can be as small as you like.
40:56:140Paolo Guiotto: whatever is absolute you take, there will always be a 1 over K, which is less than your epsilon. So, if you are in the dissection with K, you definitely are in the dissection for the epsilon. So, basically, this means that we can also
41:10:780Paolo Guiotto: Recast this in this way.
41:13:840Paolo Guiotto: probability of the intersection of a K positive, so to 1, this is a discrete k, so let's make as k from 1 to infinity.
41:24:820Paolo Guiotto: of union for n from 1 to infinity of the intersection for little n from capital n to infinity of these events, so distance between Xn and X less or equal than epsilon. All this probability must be equal to 1.
41:45:940Paolo Guiotto: Now, this is equivalent that if the probability of this event is 1,
41:53:860Paolo Guiotto: The probability of its complementary must be zero, because we are in a probability space, so this is equivalent of saying that the probability of the complementary
42:04:440Paolo Guiotto: So, complementary.
42:12:430Paolo Guiotto: is equal to 0. Now, let's write what is this complementary. Well, we have to do the complementary of all this operation. So, what does it mean? So we have the probability of… let's start from the last… so, from the first, which is the last action operation.
42:28:730Paolo Guiotto: Now, what is the fundamental of an intersection? It's a union, so this becomes probability of union.
42:36:600Paolo Guiotto: for k going from 1 to infinity of. What is the probability of the union? It's an intersection, so intersection…
42:45:60Paolo Guiotto: for capital N going from 1 to infinity. And then, what is the probability of, again, an intersection is a union. So, union…
42:55:80Paolo Guiotto: for little n greater or equal than capital n. And now we have to do the complementary of the sets with the brackets. Yeah, no? So, that means complementary of all these is the union of intersection of unions of complementaries of this.
43:13:310Paolo Guiotto: Okay?
43:14:610Paolo Guiotto: So I write complementary. So we have complementary of modules, exam.
43:21:790Paolo Guiotto: minus X is less or equal than epsilon, complementary.
43:26:950Paolo Guiotto: But the set where, the event where the distance is not less than epsilon is the event where the distance is greater than epsilon. So the complementary of this is the event where the distance between X and X is larger than epsilon.
43:42:790Paolo Guiotto: So finally, The conclusion is, this. So…
43:48:70Paolo Guiotto: The sequence XN converges, almost surely.
43:52:200Paolo Guiotto: to X, if and only if.
43:54:780Paolo Guiotto: The probability of this big event, union over K of the intersection of a capital N of the unions over n greater or equal than capital N, I do not rewrite K from 1 to… well, let's rewrite k from 1 to infinity, n from 1 to infinity.
44:13:670Paolo Guiotto: and little n greater than capital n of distance between XN and X larger than epsilon, this probability must be equal to zero.
44:28:400Paolo Guiotto: Now, what is the point here?
44:31:190Paolo Guiotto: So you see that we have a union here, probability of a union, you would say it is the sum of the probabilities, but that's not a disjoint union, not necessarily.
44:42:870Paolo Guiotto: However, if you look at the yellow set, what I circled in yellow, now, to have that probability of the union of yellow sets is zero, you must necessarily have that
44:58:540Paolo Guiotto: And it is actually an if and a leaf, that the yellow set have Probability equal to zero.
45:05:850Paolo Guiotto: Because if they have probability equals zero, their union, even if it is not disjoint, I can use sub-additivity, no? Probability of their union is less than sum of the probabilities. If they have measure 0, probability zero, the total is 0.
45:24:50Paolo Guiotto: But vice versa, if this is 0, and it is the union of this, they cannot have positive probability. Now you understand, because the union contains each of them, no? So what I'm saying is that here.
45:37:160Paolo Guiotto: We see that, here.
45:42:30Paolo Guiotto: We have that probability of union in K, whatever is the set, let's call EK,
45:48:910Paolo Guiotto: is equal to zero, and this is if and only if the probability of the Ks are all equal zero for every K.
45:58:00Paolo Guiotto: This because if… the probability of the union of the K is 0,
46:06:440Paolo Guiotto: Since this contains each of the sets, each of the K, so this contains, say, EJ for every J,
46:15:940Paolo Guiotto: then I will have that the monotonicity, the probability of EJ will be less or equal than the probability of the union of all the EK
46:28:740Paolo Guiotto: which is supposed to be 0, but since the probability is greater or equal than zero, I would get that, finally, the probability of the EJ is 0 for every J, right?
46:42:940Paolo Guiotto: Okay, so this implies that… this means that if the probability of the union is zero, then each of the components has probability 0. And vice versa. If the probability of the EJ are equal to zero for every J,
46:58:290Paolo Guiotto: Then, the probability of the union of a K, of the K, by sub-additivity.
47:06:740Paolo Guiotto: which always holds for a measure. This is less or equal than sum of the P of EK.
47:14:50Paolo Guiotto: which are supposed to be zero, so we are summing zeros, we get zero. But since the probability cannot be less than zero, the probability must be equal to zero. So probability of…
47:26:130Paolo Guiotto: the union… of the UK, Is equal to zero.
47:32:250Paolo Guiotto: So, this, just to explain why, here, we have this situation. We have the probability of that union to have convergence, this is equivalent to pointwise convergence. So, you must have that all these circles set have probability equals zero. So, I can say, once again, that
47:51:870Paolo Guiotto: XN converges, almost surely, to XN,
47:57:200Paolo Guiotto: If, and only if, the probability of each of these circled sets, and here I cannot decompose
48:05:850Paolo Guiotto: anymore this, so I can say intersection. When capital N goes from 1 to infinity of union, when n is larger or equal than capital N of these events, so modulus Xn minus X,
48:22:410Paolo Guiotto: Greater or equal than epsilon, this thing has probability zero, or it is an impossible event.
48:30:640Paolo Guiotto: Now, this type of Saturn It's particularly important, because let's see the structure of this operation.
48:41:150Paolo Guiotto: So let's call this EN again. So this is the set, is the intersection 1 to infinity of the union when n is larger or equal than n of sets EN.
48:55:850Paolo Guiotto: What is this set of omega? What is this event?
48:59:490Paolo Guiotto: So, if you take an omega, which is in this intersection.
49:03:710Paolo Guiotto: of the unions, it means that that omega will be in all the elements of the intersection, so that omega will belong to the unions for N larger than capital n of the EN. And this will hold for every n.
49:21:340Paolo Guiotto: So, we will say that this means that for every tabular N, there will be at least one of the EN, E little n.
49:30:380Paolo Guiotto: where omega is. So, for every capital N to exists an N larger than capital N, such that omega belongs to that set EN.
49:42:200Paolo Guiotto: So what is saying this?
49:44:490Paolo Guiotto: This is saying that this is the set of the omegas.
49:49:740Paolo Guiotto: That belongs… let's verify this condition, but verifying this condition means that literally what is written is, whatever is this capital N, there is a little n larger than human, for which omega is in here. So, fixed capital N equals 1, okay?
50:09:80Paolo Guiotto: So if I fix n equal 1, so…
50:13:910Paolo Guiotto: or n equal 1, I say there exists a little n, we will call it N1. There exists an N1 larger than 1.
50:25:970Paolo Guiotto: Such that, Omega belongs to… En1.
50:32:700Paolo Guiotto: Now, fixer n equal to… There exists an N2,
50:42:10Paolo Guiotto: Well, instead of fixing n equal to, take n equal n1.
50:50:400Paolo Guiotto: Or at least N1 plus 1. You can always do, because this says per every n, okay?
50:56:730Paolo Guiotto: Now, take this capital N, which means take a capital N which is larger than the initial N, no? So what I'm saying is, in some sense, this figure, this is the list of naturals, no?
51:13:530Paolo Guiotto: So this says that I started with capital N equals 1, I have, there exists an N1, but I don't know where is it, N1, and omega belongs to the corresponding set.
51:24:610Paolo Guiotto: EN1.
51:26:300Paolo Guiotto: Now I take, as capital N the next number after N1, which is N1 plus 1.
51:33:560Paolo Guiotto: It says that there is just a natural n, which is larger than this capital N, so it will be greater than this one, maybe this one, just this one, but in general, it will be at the right, so there will be another natural that I call N2, where I can say that my omega belongs into this EN2.
51:53:930Paolo Guiotto: And now you can repeat the game, so such that omega belongs to EN2. Take the new N as N2,
52:03:100Paolo Guiotto: plus 1.
52:04:690Paolo Guiotto: Notice that DN2, by construction, is greater or equal than capital N, which is greater than N1. So, as the figure suggests, N2 is bigger than N1.
52:18:540Paolo Guiotto: So the same happens here. Now I take the next integer, n2 plus 1, as the initial N, and there is a little n larger than this capital N, so there exists an N3, which will be greater than capital N, or equal.
52:35:460Paolo Guiotto: But capital N is already greater than N2, so it will be greater than N2, so it will be about here, N3, for which omega belongs to this set of omega N3.
52:48:640Paolo Guiotto: And so on. You can repeat this repeat many times. So what does it mean?
52:53:840Paolo Guiotto: Now, this set, and this is characterizing, basically, the elements of the set, this is the set of omega that belongs to infinitely many of the sets End.
53:08:220Paolo Guiotto: So not the same for each omega. You can be in different infinitely many sets. Maybe you are in E2, E4, E6, E8, and so on, or maybe you are in E1, E5, E10, E20, or you are with a random sequence of natural, but the point is that this set is the set of omegas that belongs to infinitely many
53:32:810Paolo Guiotto: of the sets EN, okay? So this particular set.
53:39:190Paolo Guiotto: Intersection of N, union, when n is greater or equal than n of y n.
53:45:520Paolo Guiotto: is D.
53:48:170Paolo Guiotto: set of… Omegas, in capital Omegas, touch that.
53:56:700Paolo Guiotto: They… belong.
54:02:320Paolo Guiotto: 2.
54:03:390Paolo Guiotto: infinitely.
54:07:90Paolo Guiotto: Mt. Feel.
54:10:940Paolo Guiotto: Delete.
54:12:880Paolo Guiotto: Manny.
54:15:710Paolo Guiotto: Ian.
54:22:760Paolo Guiotto: So this is this set, and it deserves a special name. This set is, called the Lim Soup…
54:34:960Paolo Guiotto: of the EN.
54:37:340Paolo Guiotto: This definition comes from limits. I don't know if you've never seen the limb soup as the limit operation of sequences. It doesn't matter. We can take just the definition here.
54:49:930Paolo Guiotto: Because the analogy is just with the definition of a limit, superior limit.
54:56:860Paolo Guiotto: But if you have never seen, it doesn't matter. So the point is that this is the set made by the omegas that belongs to infinity many, so not necessarily to all the yen, you see?
55:10:310Paolo Guiotto: So, I'm not saying that the omegas are in all the N. That would be the intersection of the N.
55:16:900Paolo Guiotto: Okay, but here you are not the intersection, you can see now. You are in the intersection of these units, so it means that for every annually, there's at least one of them.
55:27:880Paolo Guiotto: The point is that you may expect, maybe I am always in E1. No, because the fact that this union is made by a year n greater than N, and then is growing from 1 to infinity.
55:40:980Paolo Guiotto: forces that if you are in this set, you must be infinity many times in DCA. But this infinity many times changes omega by omega, okay?
55:51:860Paolo Guiotto: So for an omega, you have a sequence of N1, N2, N3, and so on, where you are. For another omega, you have another sequence, and so on. So it's not always the same.
56:01:810Paolo Guiotto: So I'm not saying you are always in these infinitely many, this infinitely many chains with Omega.
56:11:10Paolo Guiotto: Okay, now, the point is, how can we prove that the probability is equal to zero? Here, there is a nice first result, which is called the first Borrel-Cantellile lemma.
56:24:300Paolo Guiotto: So, first… Bora.
56:31:720Paolo Guiotto: It's gonna leave.
56:36:00Paolo Guiotto: Dilemma.
56:40:860Paolo Guiotto: Well, this says that, if, there's some…
56:48:730Paolo Guiotto: of the probabilities of these sets yen, is find it.
56:54:900Paolo Guiotto: Then, the probability of the limb subset of the yen.
57:02:580Paolo Guiotto: Is equal to zero.
57:04:500Paolo Guiotto: It's actually a simple,
57:06:990Paolo Guiotto: this… this first, Borel-Cantellama is a simple remark, but it's remarkable, because it says that, okay, so the focus is now, remind, I want to prove… if I want to prove almost sure convergence, I should be able to check that that probability is zero, no?
57:24:510Paolo Guiotto: Normally, computing exactly a probability is very hard, if not impossible.
57:29:870Paolo Guiotto: But this says that this moves the focus from this set to assessing the probabilities of these events.
57:39:730Paolo Guiotto: And if you look at this event, how this event is made, the probability of models that differs greater than absolute should remind that, for example, we have something like the Chebyshev inequality that assesses probabilities when the random variable is bigger, no?
57:54:970Paolo Guiotto: It's bigger than some threshold epsilon.
57:58:70Paolo Guiotto: No? This is what is normally used here, and the variations of combative equality.
58:04:280Paolo Guiotto: So we can assess the probability of the N. If they are small enough, because you understand that we need… they are small to have this condition verified, they are positive, so they form a medical series.
58:18:860Paolo Guiotto: We want that this series be converted, so they must go to zero first, this is not sufficient, they must go to zero fast enough, okay, to make the infinite sum find.
58:31:350Paolo Guiotto: However, this is, let's say, a numerical series of numbers. If we are able to assess these probabilities, we should be able to discuss whether this sound is finite or not. If it is finite, we have the desired conclusion. Now, this is, as I said, quite easy, because
58:52:360Paolo Guiotto: We may notice that, by assumption, we have that, so, if we take… So I say, proof.
59:05:890Paolo Guiotto: We noticed that…
59:11:390Paolo Guiotto: If we take the probability of this union.
59:16:590Paolo Guiotto: the union of little n greater or equal than capital n of the EN.
59:23:190Paolo Guiotto: By sub-additivity.
59:27:890Paolo Guiotto: This is less or equal than the sum for N larger than capital n to… so this would be the sum from capital N to plus infinity of these probabilities, the probabilities of the EN, right?
59:46:420Paolo Guiotto: But now, since the sum of the probabilities is supposed to be finite, what happens to this sum, which is still an infinite sum, but you see.
59:58:440Paolo Guiotto: This is the sum that starts from the index sub capital N here.
00:05:950Paolo Guiotto: So, if I have that the total sum
00:10:600Paolo Guiotto: from 0 to plus infinity of the probability of the yen. This is finite.
00:17:910Paolo Guiotto: So let's call it S, this sum. Well, this means that S is the limit
00:25:630Paolo Guiotto: for, say, let me use just capital N,
00:30:420Paolo Guiotto: going to plus infinity of the sums from 0 to, let me just use capital n minus 1, of the probabilities of the N. This is the definition of convergence. Now, how do you define the sum of a series? You define sums, and then you send the number of terms to infinity, no? This is the definition of sum.
00:53:950Paolo Guiotto: But if this sum goes to the total sums of the probabilities, the difference will go to zero, so S minus sum
01:06:140Paolo Guiotto: n from 0 to capital n minus 1 of the probabilities of EN, this difference will go to 0, which is equivalent.
01:15:940Paolo Guiotto: But S is the total sum, sum from 0 to infinity of the probabilities of the EN.
01:23:900Paolo Guiotto: Minus the sum of the first n minus 1 probabilities, the difference between this and this, yields exactly the sum for capital N to plus infinity of the probabilities of the n.
01:37:330Paolo Guiotto: And now we know that this goes to zero.
01:40:90Paolo Guiotto: Okay? In other words, this is the tail of the series. If the series is convergent, the tail must go to zero. Cannot go anywhere else. Okay? So, in particular.
01:52:930Paolo Guiotto: Now, I know that this guy here is going to zero when n is going to plus infinity.
02:00:520Paolo Guiotto: And therefore, this probability is going to zero.
02:04:230Paolo Guiotto: So, the probability.
02:07:970Paolo Guiotto: the probability of the union, N larger than capital N of the EN
02:14:340Paolo Guiotto: This goes to 0 when capital n goes to plus infinity.
02:19:480Paolo Guiotto: Now, not only it goes to zero, but if you look at this sequence, How is made, this sequence?
02:29:640Paolo Guiotto: This is a union, not a union.
02:33:430Paolo Guiotto: Of the set, when this little n is larger than capital N. I imagine that you pass from capital N to capital N plus 1.
02:43:960Paolo Guiotto: All it's say. Let's pass from 1 to…
02:46:730Paolo Guiotto: Well, the union for n equal greater or equal than, from 0 to 1. N greater or equal than 0, and the union for n greater or equal than 1. I eliminated one set.
02:57:930Paolo Guiotto: So, what, what happens to the union?
03:00:990Paolo Guiotto: the union has reduced, because I took out one set from this union. So, in general, when you increase this capital N, this per minute decreases. So this is a decreasing sequence of sets. So we can also look at this as a decreasing sequence of sets.
03:19:980Paolo Guiotto: And therefore, this smells of the continuity from above, no? So, and in fact.
03:27:450Paolo Guiotto: I have that the probability of the limit set… what is the limit set? The limit set for decreasing sequence is the intersection, right? So it is the intersection of N, of the union of N greater than
03:43:880Paolo Guiotto: capital N of the yen, so this is the limit set, the probability of the limit, by continuity.
03:52:260Paolo Guiotto: from.
03:54:380Paolo Guiotto: above,
03:55:870Paolo Guiotto: which is not always verified, but for probability measures, it's always verified. For finite measures, it's always verified. This will be the limit
04:05:940Paolo Guiotto: in capital N of the probability of the elements of the sequence, which is this one, it's not the sequence of the yen. It's the sequence of the unions of the yen, so union.
04:16:320Paolo Guiotto: little n larger than capital n of the yen.
04:19:940Paolo Guiotto: But we just come to see that this limit is zero.
04:24:220Paolo Guiotto: So we get that this is zero, and this is the conclusion.
04:30:220Paolo Guiotto: So this is important because,
04:33:480Paolo Guiotto: It, yields a condition, a sufficient condition, under which we can say that this,
04:41:460Paolo Guiotto: limit probability is zero. Now, you may wonder what happens if the sum is plus infinity? Of course, we cannot use this, but what is interesting is that if we add the little assumption, that's the second Borencantelli lemma.
05:07:590Paolo Guiotto: So, we… the little assumption, which is important in application, is that we assume that D events are independent.
05:13:930Paolo Guiotto: If the EM are independent.
05:21:800Paolo Guiotto: Now, the sum of the probabilities of the yen equal to plus infinity
05:28:840Paolo Guiotto: implies that the probability of the limb subset
05:35:00Paolo Guiotto: Now, it turns out that this means 1.
05:38:790Paolo Guiotto: So this is, by the way, we have a sort of remark here.
05:44:440Paolo Guiotto: remark.
05:46:830Paolo Guiotto: So, assuming that the yen are independent, if the yen independent.
05:56:990Paolo Guiotto: Now, since the sum of the probabilities, since, the sum…
06:04:370Paolo Guiotto: Of the probabilities of the yen.
06:08:200Paolo Guiotto: can be only either finite or plus infinite, because it's a sum of positive numbers. There is always the sum of this, no? It's a series of positive numbers, so it is always convergent or divergent. There is no other possibility. Since is…
06:26:420Paolo Guiotto: Always.
06:30:140Paolo Guiotto: either.
06:35:100Paolo Guiotto: convergent.
06:37:80Paolo Guiotto: or divergent.
06:39:730Paolo Guiotto: So this means that either this sum of the probabilities is finite, or infinite.
06:51:540Paolo Guiotto: So one of these two cases is necessarily true.
06:56:180Paolo Guiotto: Well, in the first case, we would be, in particular, it is not required for the first lemma that the yen are independent, but if they are, it's still true, no? We are in the case, when the sum is final, we would have the probability of the limb soup would be equal to zero. So this would say that
07:16:320Paolo Guiotto: Probability of limb soup.
07:20:90Paolo Guiotto: So this is not an if and all-if, but it's an explication.
07:24:520Paolo Guiotto: Probability of limb soup of the yen would be zero.
07:29:600Paolo Guiotto: While here, because of the second Boracantali lemma, we would have that the probability is 1.
07:45:100Paolo Guiotto: So, in this case, we would have a sort of dichotomy, no? Either this is an impossible event, or this is a certain event. There cannot be a halfway possibility.
07:58:690Paolo Guiotto: So, in other words, if… if we think about today yen as the set where
08:05:690Paolo Guiotto: we will apply later, returning back to our discussion of pointwise number events, the N at these events, no? If the events are independent, so this basically is related to independence of the variables XN at the end, it means that
08:25:649Paolo Guiotto: Remind that that set means when probability of limb soup is zero means convergence.
08:31:779Paolo Guiotto: But that's the main probability of limp soup is 1.
08:35:69Paolo Guiotto: Well, it's the opposite, no, because it's the… if you say that the probability is sufficient 1 means that this event is a probability 1 event, but this was the event where the sequence is not… is not converted, right?
08:49:10Paolo Guiotto: So, if I'm saying that the probability
08:51:910Paolo Guiotto: of the event where the sequence is not convergent. This one, it means that the sequence is never convergent, you see?
08:59:340Paolo Guiotto: So, it happens this interesting phenomenon that if you have a sequence of independent primary variables, either the limit, pointwise limit, exists with probability 1, or it exists with probability 0.
09:12:160Paolo Guiotto: So you cannot have that for some omega is convergent, and for some other, it's not convergent. No, it's drastic. So, it's either it is always convergent, or it is never convergent.
09:22:800Paolo Guiotto: This is, one of the 01 Kolmogorov law that tells something like this for this type of events. However, I would… I would not factor in this, but it's an example.
09:34:870Paolo Guiotto: Let's see the proof of this second… Boray Pantelli.
09:41:260Paolo Guiotto: Le m.
09:46:529Paolo Guiotto: Now, to prove that this, probability of limb soup is 1, now, normally, it is,
09:56:580Paolo Guiotto: It is, it is always, reasonable.
10:01:850Paolo Guiotto: to try to prove that it is zero, since I have to prove this one.
10:06:290Paolo Guiotto: what should be zero is the complementary, okay? Because the idea is that, if I have to prove that something, like a probability, is zero, I need to assess that probability and show that it must be small.
10:19:860Paolo Guiotto: Okay, which seems to be easier than proving that the probability is close to 1, which means the probability
10:27:800Paolo Guiotto: you should do a lower bound assessment, saying that that probability must be greater than something slightly less than 1. So, it's better, normally, as a strategy, try to prove that the probability is 0. So, since here we have to prove that probability 1, what we do is this, probability of the limb soup.
10:50:190Paolo Guiotto: of the yen.
10:51:980Paolo Guiotto: Is 1 if and only if the probability of the complementary of this set Lim Sup of EN.
11:01:530Paolo Guiotto: Complementary.
11:03:490Paolo Guiotto: is zero.
11:04:970Paolo Guiotto: And this is what? Well, remind that the linso is the intersection of a capital N, the unions of N greater or equal than capital N of the EN, right?
11:15:340Paolo Guiotto: So, the complementary means the probability of
11:19:210Paolo Guiotto: Union in capital N, intersection when n is greater or equal than capital N of the EN complementaries.
11:28:890Paolo Guiotto: Okay?
11:31:100Paolo Guiotto: So, since I want to check that this is 0,
11:35:370Paolo Guiotto: I can use the argument I used above, no? This is a union that should… union of this, that should be zero. And this happens if and only if
11:45:380Paolo Guiotto: Each of the sets I circled here has probability 0. This happens if and if the probability of the intersection for n larger than capital n of the n
11:59:760Paolo Guiotto: Complementary is equal to zero.
12:02:500Paolo Guiotto: So let's prove this.
12:04:290Paolo Guiotto: Now, this is the probability of an interception of events, which are the complementaries of the yen.
12:13:650Paolo Guiotto: Now, we know that D and E are independent, so if we have a finite intersection, we can split this into the product of the probability, so that's what we are going to do.
12:23:930Paolo Guiotto: Now, since here we have an infinite intersection, we have a first to truncate to a finite intersection. So what do we do? Well, since this thing is the intersection when n is larger than capital n of the EN complementaries.
12:40:240Paolo Guiotto: I want to take this intersection. Intersection for
12:44:240Paolo Guiotto: N, from capital N to, let's stop at, let's say, capital M, of these sets, EN complementary.
12:54:850Paolo Guiotto: So this is, let's write this in the same form. So this is the intersection from n to plus infinity of this. And this is an intersection where we stop at index M. Now, what is the relation between these two? You see that
13:10:780Paolo Guiotto: when I increase M, so I could do… let's take the limit. When I increase M, what happens?
13:18:760Paolo Guiotto: Here, I intersect with more elements, so what happens to the intersection?
13:27:740Paolo Guiotto: If you do an intersection with a new set, you normally will decrease now, because intersections
13:33:310Paolo Guiotto: diminished elements. They do not add, no, they require more, no, more conditions. So, in general, if you add one more set into this intersection, whatever they are, this means that you have less elements. So, this is a decreasing sequence with them.
13:51:200Paolo Guiotto: So they are going down to the full intersection, so it's decreasing sequence, and therefore it works because we have a probability, so we can say that the probability of the full intersection from n to plus infinity of the N complementaries, because of the continuity from
14:11:90Paolo Guiotto: Above… This is the limit when M
14:17:340Paolo Guiotto: goes to plus infinity of the probability of the intersection for N
14:24:110Paolo Guiotto: Going from capital N to capital M of the EN complementary.
14:29:80Paolo Guiotto: Okay.
14:31:70Paolo Guiotto: Now, since this… so let's keep limit with M going to plus infinity. By independence.
14:40:20Paolo Guiotto: Here is where we use this assumption.
14:43:990Paolo Guiotto: Here we have probability of… it is clear that if the events are independent, also the complementaries are independent, so we have that this transforms into a product for N going to capital N to capital M of probability of EN
15:02:110Paolo Guiotto: Complemented.
15:04:40Paolo Guiotto: Now, this is the complementary, so it is 1 minus.
15:07:440Paolo Guiotto: The probability of the end.
15:11:180Paolo Guiotto: Now, we have to convince that this limit… remind that the goal is to prove that this limit is zero, because we want to prove that the probability of that intersection is zero.
15:22:140Paolo Guiotto: Okay, so how can we do that? So we have limit…
15:25:850Paolo Guiotto: M going to plus infinity of the product for little m going from capital N to capital M of 1 minus the probability of EN.
15:41:720Paolo Guiotto: Now, the trick here…
15:44:790Paolo Guiotto: is, to transform that product into a sum by using, basically logarithms. So now, can I introduce… how can I introduce this? Well, I say that X is equal to E log of X. That's what they say.
16:01:420Paolo Guiotto: So this becomes a limit for M going to plus infinity.
16:06:630Paolo Guiotto: of exponential log of the product, little n from capital N to capital M, 1 minus probability of EN.
16:19:850Paolo Guiotto: And so this now transforms into the sum, so limit m going to plus infinity.
16:27:100Paolo Guiotto: E to sum for N going from capital N to capital M of log 1 minus probability of PN.
16:40:240Paolo Guiotto: Now we are almost at the end, because now we restored the limit here, we carry in the exponent, so this becomes the exponential of the limit, so it is some
16:51:130Paolo Guiotto: for n going from capital N to plus infinity of log of 1 minus the probability of the set y n.
17:00:820Paolo Guiotto: I want to say that this is,
17:04:800Paolo Guiotto: what is the idea, no? The idea is that,
17:10:510Paolo Guiotto: I say roughly because I have to be a bit more refined. But we know that this is a probability, so it is a number between 0 and 1, right? And there, I have something log of 1 plus X, when x is close to 0, in some sense. Well, actually, it is not necessarily close to 0 because
17:27:940Paolo Guiotto: the assumption is that the sum is infinite, so they could be even equal to 1. But you think, if they are equal to 1, or if they are close to 1, you see that this value would be close to 0, so log would be very bad.
17:42:270Paolo Guiotto: Because it would be, too,
17:47:850Paolo Guiotto: Yeah, in that case, we would have minus infinity.
17:58:40Paolo Guiotto: So, because I want to say that this is approximately X. If this is correct, let's see, this would be more or less like the sum of minus the probabilities of,
18:12:80Paolo Guiotto: Yeah, yeah, I was saying, yeah, I have to prove that this is 0, right? So we have to prove that this is zero. This is the limit which we want. So, the idea is that if these numbers are close to 1, they are going to be zero one, because they are probabilinx.
18:31:800Paolo Guiotto: If they are close to 1, so they are bigger probabilities, this argument would be close to 0, so log would be close to negative infinity.
18:43:20Paolo Guiotto: And that would be exponential of minus infinity zero, no? But, on the other side, this number could be even smaller.
18:52:140Paolo Guiotto: And the sum of the properties is divergent, no? So, something like the harmonic sum. The sum of 1 over n is divergent, but they are going to 0. If they are going to 0, we know that the acidity of log and n log of 1 plus X is more or less like X, no?
19:10:540Paolo Guiotto: So, in that case, I would have that these are, like, a log of 1 minus P means that they are minus P, so the sum of these logs would be the sum of these minus the probabilities, or minus the sum of the probabilities.
19:24:320Paolo Guiotto: But by assumption, that sum is divergent, I would have a minus plus infinity, so again, E2 minus infinity, so again, limit would be zero. Now, to make this a precise argument.
19:39:140Paolo Guiotto: we have to make this a precise inequality, so we need a clear argument. Now, the point is that we have this remarkable inequality.
19:56:660Paolo Guiotto: That says that the log of 1 plus X,
20:01:20Paolo Guiotto: So I don't use the asymptotics, I use just a very, say.
20:06:710Paolo Guiotto: in some sense, bad inequalities. That says, log of 1 plus X is less than X. I'm saying that log of 1 plus X is this, so this is minus 1. Here, the log goes to minus infinity.
20:23:540Paolo Guiotto: Index is this.
20:26:830Paolo Guiotto: Now, this inequality comes just from the concavity of the function log 1 plus X.
20:33:510Paolo Guiotto: Because this function, phi of x equals log of 1 plus X,
20:40:150Paolo Guiotto: is concave.
20:44:20Paolo Guiotto: you see from log is concave, that's a log, no? It's just that the argument is translated of one unit, or if you want, you compute the second derivative, and you see that the second derivative
20:57:850Paolo Guiotto: The first one is 1 over 1 plus X, the second is minus 1 over 1 plus X squared, so you see that it is negative.
21:05:770Paolo Guiotto: So, the function is concave.
21:08:90Paolo Guiotto: A property of concavity is that, as convexity, you have that the function is above each of the tangents.
21:17:210Paolo Guiotto: For concavity, we have the opposite. So, they are below each of the tangents. So, the condition is that P of X will be less or equal
21:30:150Paolo Guiotto: each of the tangents, I choose the tangent at X equals 0. So this is phi of 0 plus phi prime of 0 times X.
21:40:900Paolo Guiotto: If you now plug the ingredients, you put X equals 0, phi of 0 is log of 1, so 0.
21:47:280Paolo Guiotto: And the phi prime is… phi prime is 1 over 1 plus X at 0 is 1.
21:54:440Paolo Guiotto: And so you get here the conclusion. Log of 1 plus X is less or equal than X. So this is the remarkable inequality.
22:04:440Paolo Guiotto: Now, if we use the remarkable inequality here, and this holds whatever is X between larger than minus 1, provided the logarithm is defined, so we use here, so with X equal minus the probability.
22:21:410Paolo Guiotto: So, in particular, we will have that log of 1 minus the probability of yen.
22:28:600Paolo Guiotto: This is less or equal than minus the probability of EN.
22:33:480Paolo Guiotto: And therefore, Since at the exponent, you have the sum of these logs, you have that, the sum.
22:41:630Paolo Guiotto: For, what is it?
22:44:230Paolo Guiotto: n going from capital N to class infinity of the log of 1 minus the probability of EN,
22:53:990Paolo Guiotto: This will be less or equal than sum of minus the probabilities, or minus sum for n going from capital N plus infinity of the probabilities of the N.
23:05:370Paolo Guiotto: But now, we use the assumption. We know that this sum is infinite.
23:11:370Paolo Guiotto: Yeah, we know that the total sum is infinite, but since the… no, you know that the total sum from zero to infinity of the probabilities, this one is supposed to be infinite, right?
23:24:720Paolo Guiotto: But you can always split into the sum from 0 to capital N minus 1 of the probabilities.
23:32:600Paolo Guiotto: And then you have the tail, sum from capital N to plus infinity of the probabilities.
23:39:820Paolo Guiotto: This one, it's definitely a finite quantity, because it's a finite sum of probabilities, so it's a real number.
23:47:340Paolo Guiotto: It's a positive number.
23:48:960Paolo Guiotto: And therefore, if you want that this identity makes plus infinity, necessarily this must be equal to plus infinity.
23:56:920Paolo Guiotto: So, this explains why that day is my… is passing 50, so minus infinity, yeah.
24:03:120Paolo Guiotto: Therefore, that sum is minus infinity, and the exponential, we have that this exponential, e to minus infinity.
24:12:580Paolo Guiotto: So finally, we return here, so we get that probability of the intersection for n greater or equal than n to plus infinity of the yen complementaries, which is, after calculation, exponential of the sum
24:29:10Paolo Guiotto: of, for X, N going from capital N to plus infinity of the logs.
24:34:560Paolo Guiotto: of 1 minus the probability of EN.
24:38:770Paolo Guiotto: We just say that this is…
24:41:510Paolo Guiotto: because of the inequality, less or equal E to sum for n going from capital N to infinity of minus the probabilities of the n.
24:51:180Paolo Guiotto: Which is e to minus infinity equals 0, and that's what we wanted, okay?
24:58:10Paolo Guiotto: So this finishes the proof of this theorem.
25:08:820Paolo Guiotto: Well, a little remark.
25:13:750Paolo Guiotto: The conclusion of the second Burejk Canteli lemma is false if the events are not independent.
25:21:50Paolo Guiotto: if, the EN not… Independent.
25:30:70Paolo Guiotto: We… even… even with… sum of the probabilities of the yen.
25:40:380Paolo Guiotto: Equal plus infinity.
25:43:140Paolo Guiotto: We may have that the probability of the limb soup is not… is not equal to 1.
25:50:750Paolo Guiotto: We… May… Well, I have…
25:57:200Paolo Guiotto: That the probability of the limb soup
26:02:500Paolo Guiotto: of Diana is less than once.
26:06:620Paolo Guiotto: And the example is somehow trivial.
26:11:60Paolo Guiotto: take all the N equal to the same set E. Take a set E, for which you know that
26:19:980Paolo Guiotto: with… Probability of E, which is positive, but less than once.
26:30:650Paolo Guiotto: So, clearly, the N are not independent, you know? Unless we can say that D… Yen.
26:41:820Paolo Guiotto: independent.
26:45:940Paolo Guiotto: If and all if, let's check for 2, but then, if you want, you can check for any finite number, but just to do not, to not complicate things, or if you want. The probability that EN1, intersection EN2, intersection, etc, ENK,
27:05:810Paolo Guiotto: Is equal to the product, the probabilities of,
27:09:910Paolo Guiotto: E and J for J going from 1 to K.
27:15:380Paolo Guiotto: Well, to have this, since the intersection… the set is always the same, so the intersection of all these yarn is the set E, so this is key.
27:27:70Paolo Guiotto: So you must have that.
27:44:520Paolo Guiotto: Depends if the probability of E is 0,
27:47:780Paolo Guiotto: or if the probability of V is 1.
27:50:540Paolo Guiotto: And there is no other possibility, because if the probability… this happens either, if the probability of E is equal to 0,
28:00:740Paolo Guiotto: Or…
28:01:890Paolo Guiotto: if probability is different from 0, you would divide, and you would get that probability of E to exponent k minus 1 should be equal to 1, so this is possible if and only if probability of E is equal to 1, okay?
28:19:50Paolo Guiotto: So, the unique possibility to have that these sets, no, made by the same set, repeated infinity many times.
28:27:200Paolo Guiotto: The unique possibility that they are independent is when the probability is 0 or 1, okay? So they are not independent in general. So if the probability of that E is between 0 and 1, they cannot be independent, okay? So…
28:46:10Paolo Guiotto: If the probability of E It's between 0 and 1.
28:52:50Paolo Guiotto: So it can be any sum, so I don't know, the probability space is 0, 1, you take an interval with the measure 1 half, no? That's at E.
29:00:340Paolo Guiotto: In general, the sets EN not.
29:06:190Paolo Guiotto: independent. So we have a setup which is with not independent events. In this case, what happens to the probability of the limb soup? In this case.
29:22:650Paolo Guiotto: If you look at the limb superset.
29:26:870Paolo Guiotto: of the yen? Well, this by definition, is the intersection on capital N of the union when n is larger than capital n of the EN.
29:36:940Paolo Guiotto: But now, here, all these sets are the same set E, so when you do unions of E repeated infinity many times, you get E, no? So this is E, and here you are doing intersections of E, infinity many times, so you get, again, E.
29:52:310Paolo Guiotto: So this is the set, and therefore the probability of the limb soup set
29:59:890Paolo Guiotto: of DZN is the probability of E, which is not 0 and neither 1.
30:08:40Paolo Guiotto: In particular, the conclusion of the break and tally is no longer true when the events are not independent.
30:19:70Paolo Guiotto: Well, just let's do a simple example. It takes one minute, because time is basically over.
30:26:670Paolo Guiotto: example, so suppose that, XN says is a sequence of, Bernoulli…
30:39:370Paolo Guiotto: random variables.
30:42:130Paolo Guiotto: with these probabilities. The probability that XN is 1, so normally with Bernoulli, we mean variables that takes values 0 or 1.
30:52:600Paolo Guiotto: with certain probabilities, so let's say value… probability that XN is 1 is PN, and the probability that XN is 0 is 1 minus PN.
31:08:800Paolo Guiotto: It says that, show that.
31:15:500Paolo Guiotto: If the sum of this BN is finite.
31:20:330Paolo Guiotto: then Xn goes to 0 almost surely, so with probability 1.
31:31:890Paolo Guiotto: Okay, so what we do is, we apply the machinery we have seen here.
31:39:80Paolo Guiotto: So, we want to prove, we want to prove, returning back to this… discussion…
31:48:450Paolo Guiotto: with the limit. So, we prove that XN goes almost surely to X, so in particular here will be 0, if and only if the probability of that set, the limb soup, of modulus XN,
32:04:430Paolo Guiotto: minus X, X is 0. Larger than epsilon is, is equal…
32:13:740Paolo Guiotto: Well, here, the text should be 1 over K, sorry.
32:20:210Paolo Guiotto: 1 over K is greater than epsilon, but however, we can keep the epsilon, it's the same. So,
32:29:970Paolo Guiotto: XN.
32:34:100Paolo Guiotto: goes almost surely to zero, if and only if the probability of the limb soup
32:44:310Paolo Guiotto: of this event, modulus of Xn minus the limit, which is 0, greater or equal than epsilon.
32:52:140Paolo Guiotto: this is equal to zero for every epsilon positive. But this means the probability that, Lim Supa.
33:04:400Paolo Guiotto: Well, this is XN. Xn takes value 0, 1, so basically Xn greater than epsilon.
33:12:220Paolo Guiotto: So this is the… probability we have to show that it is equal to zero. Now, by…
33:21:20Paolo Guiotto: the… First… But I can tell, dilemma… this… Happens.
33:36:60Paolo Guiotto: If we are able to prove that these are the sets yen to which we apply the Borel-Cantelle lemma, if the series of the probabilities of the N
33:49:70Paolo Guiotto: is fine, is convergence. That is, if the series of the probability that XM is greater or equal than epsilon is finite, and this for every epsilon poly.
34:05:400Paolo Guiotto: But what is that probability?
34:08:180Paolo Guiotto: This probability, in this case, Since epsilon here is positive. Xn takes value 0 or 1.
34:18:480Paolo Guiotto: Okay? It's a Bernoulli. So, depends on epsilon, because if epsilon is 1 million, XN cannot be greater than 1 million, so we can have that. If epsilon
34:30:120Paolo Guiotto: is, say, greater than 1, the probability will be 0, because Xn can take its only values 0, 1, so it cannot be greater than X1, so it would be just 0.
34:41:230Paolo Guiotto: If epsilon is less or equal 1, but positive.
34:45:340Paolo Guiotto: Xn is greater than epsilon only if XN is 1, because there are just two values, not that XN takes either 0 or 1.
34:55:20Paolo Guiotto: So, if you are strictly greater than epsilon, it means that you cannot be 0, and so the only possibility is that XN be equal to 1, but that's the probability PN. So, we have that, so…
35:09:100Paolo Guiotto: for epsilon greater than 1, the series of the probabilities XN larger than epsilon is just a series of zeros, and so it gives zero, and it is converged.
35:23:930Paolo Guiotto: For epsilon, less or equal than 1, but positive, then the series of the probabilities Xn greater than epsilon is equal to the series of the numbers PN, which are supposed to be
35:36:40Paolo Guiotto: Fine by assumption.
35:41:250Paolo Guiotto: So, in any case, we get that whatever is epsilon, the sum of that probabilities is finite, so the sum of the probabilities where Xn is greater or equal than epsilon is finite, whatever is epsilon positive.
35:59:370Paolo Guiotto: And therefore, by the Borek-Cantell dilemma, so first…
36:04:10Paolo Guiotto: For I can tell you, the probability of the lean soup
36:10:740Paolo Guiotto: of the XN greater or equal, then epsilon is 0.
36:15:330Paolo Guiotto: But as we said above, this is the same of saying modulus Xn minus the limit 0 is greater or equal than epsilon. So this means that, Xn goes to
36:29:500Paolo Guiotto: Zero, almost surely.
36:35:500Paolo Guiotto: Okay.
36:37:590Paolo Guiotto: Let me check if, we can… I can leave you something to verify here.
36:52:230Paolo Guiotto: Well, basically, all the exercises involves also the probability… the convergence in probability and distribution, we have not yet seen.
37:01:450Paolo Guiotto: But there is something you could do, D.
37:05:460Paolo Guiotto: do exercise… Due.
37:09:820Paolo Guiotto: 8, 5, 4…
37:19:470Paolo Guiotto: Now, these are with the convergence in distribution.
37:24:620Paolo Guiotto: But do this one, and, maybe, 853…
37:31:450Paolo Guiotto: Apart for convergence in probability and distribution. It is NLP, almost surely, probability.
37:38:760Paolo Guiotto: So, not probable.
37:42:360Paolo Guiotto: convergence.
37:44:370Paolo Guiotto: and distribution.
37:47:100Paolo Guiotto: We will see our webinar team.
37:49:810Paolo Guiotto: Tomorrow, we have class, right? So we'll be seeing tomorrow, Jess.
37:53:840Paolo Guiotto: Okay.
37:55:330Paolo Guiotto: Have a nice day.
37:59:00Paolo Guiotto: And by the.