AI Assistant
Transcript
00:02:110Paolo Guiotto: So we start with this.
00:05:630Paolo Guiotto: the exercises.
00:09:460Paolo Guiotto: So, Exercise 19, huh?
00:15:970Paolo Guiotto: Says we have a probability space,
00:23:960Paolo Guiotto: So, with the, so Omega… S… Or what would be… space, and G… contain enough…
00:42:130Paolo Guiotto: sub… Sigma Algebra.
00:50:100Paolo Guiotto: Well, there is a first question, which is a… theoretical, check.
00:55:860Paolo Guiotto: It's just a question concerning about definitions and theory. It says X is in L2,
01:02:560Paolo Guiotto: So what is the conditional expectation of X given G?
01:09:560Paolo Guiotto: How it's characterized through that.
01:15:10Paolo Guiotto: The norm… the L2 norm of the conditional expectation of X given G.
01:22:550Paolo Guiotto: is less than or equal than the L2 norm.
01:26:920Paolo Guiotto: of X. Well, this was the first definition of conditional expectation. The conditional expectation of X given G.
01:39:970Paolo Guiotto: is, the… Orthogonal.
01:48:00Paolo Guiotto: Rejection.
01:51:670Paolo Guiotto: on… The… Subspace.
02:01:220Paolo Guiotto: on the subspace, say, U, defined by the L2 random variables, which are G measurable.
02:13:500Paolo Guiotto: off… Gilbert Space…
02:23:500Paolo Guiotto: H equal L2 omega of all the F measurable.
02:29:520Paolo Guiotto: L2 and non-variables.
02:33:450Paolo Guiotto: source. It is characterized
02:43:300Paolo Guiotto: Bye.
02:45:190Paolo Guiotto: the… Or tonality cognition.
02:56:340Paolo Guiotto: Which is, if you take X and you subtract your orthogonal projection, so your orthogonal projection of X given G, this, is perpendicular to Y,
03:10:160Paolo Guiotto: So this product is equal to 0 for every Y in L2.
03:16:680Paolo Guiotto: G, so for every Y, L2 and G measure 1.
03:22:990Paolo Guiotto: And, but this is the characterization.
03:29:180Paolo Guiotto: It is characterized by this. Now, about the inequality.
03:36:50Paolo Guiotto: Well, since, we can write X as X minus the orthogonal projection.
03:51:130Paolo Guiotto: Plus your total projection.
03:56:940Paolo Guiotto: And these two.
03:59:190Paolo Guiotto: are perpendicular, because this one is an element of the projection space, so it belongs to L2.
04:07:710Paolo Guiotto: G, so it is… the two elements are perpendicular, because by the orthogonality condition, this is orthogonal to any Y, which is L through G, according to the Pythagorean
04:27:450Paolo Guiotto: CRM.
04:29:690Paolo Guiotto: the, L2 Norma.
04:32:950Paolo Guiotto: square.
04:34:320Paolo Guiotto: is the… since the two are orthogonal, the sum of these two norms, E minus,
04:40:310Paolo Guiotto: It's conditional expectation, or orthogonal projection squared, plus the norm of the conditional expectation of X given G.
04:51:910Paolo Guiotto: And from this, we get the conclusion, because this, of course, is greater or equal than zero, and so this is greater or equal than the norm of the conditional expectation of X given G.
05:08:220Paolo Guiotto: in L2 squared.
05:11:920Paolo Guiotto: So this is about the first question.
05:15:710Paolo Guiotto: The second question is, it says, we have now two random variables, XY in L2.
05:25:00Paolo Guiotto: Such that, you know that the conditional expectation of X given Y
05:32:150Paolo Guiotto: when we say X given Y, we… yeah.
05:36:670Paolo Guiotto: Yeah.
05:38:530Paolo Guiotto: byte, E of, XZ equals E of, YZL, and then, Z equals, Y.
05:48:380Paolo Guiotto: Could, the inequality.
05:50:780Paolo Guiotto: To take, you say? Yeah, we write, E of XL equals, E of Y gel.
06:00:320Paolo Guiotto: And we take, Z equals Y. Where is Z? I do not understand. What do you mean?
06:07:360Paolo Guiotto: Like, why being the… Y is the Y you see there in the orthogonality condition. Taking Y as being the conditional expectation of, X knowing G. Yeah, I see, I see. Yeah, you say, we take as Y the conditional expectation of X given G.
06:24:740Paolo Guiotto: So that condition becomes scalar Prototex.
06:28:350Paolo Guiotto: times the conditional expectation minus the norm of the conditional expectation square equals zero. So, the conditional expectation square is equal to X color of the conditional expectation used to choose words, yes. It is correct.
06:47:730Paolo Guiotto: Okay, so now, let's assume that these are two random variables such that the conditional expectation of X given Y is Y. X given Y, this means the sigma algebra generated by this random variable Y, and at the same time, the conditional expectation of Y given X is equal to X.
07:07:200Paolo Guiotto: Then it says that the two random variables, X and Y, must be the same.
07:12:430Paolo Guiotto: Almost surely.
07:15:700Paolo Guiotto: Okay.
07:22:670Paolo Guiotto: Yeah.
07:23:970Paolo Guiotto: Yes, we use, this identity here. We could use this identity here.
07:32:450Paolo Guiotto: that says, by, Pythagorean, TRM.
07:42:00Paolo Guiotto: we have that the norm of X in L2 squared, it is equal to X minus the conditional expectation of X given NG, and we take as G the sigma algebra generated by Y.
07:56:860Paolo Guiotto: in L2, plus the norm of the conditional expectation of X given Y.
08:03:960Paolo Guiotto: in L2 squared. Because by the assumption this is Y, and so we get that this is a norm of X minus Y in L2 squared plus norm of Y in L2 squared.
08:19:810Paolo Guiotto: So, the goal is to prove that X is equal to Y, so there is a little hint that says check that norm of X minus Y in L2 is equal to 0. So, from this identity, we get that the norm of X
08:35:520Paolo Guiotto: minus Y in L2 square, it is equal to the norm of X in L2 square.
08:42:120Paolo Guiotto: Minus the norm of Y in L2 square.
08:46:420Paolo Guiotto: Now, here we used the first property, that the conditional expectation of X given Y is what? But if we flip the order of the two, we can use the second one.
09:01:80Paolo Guiotto: So, by… the fact that the conditional expectation of Y given X is X,
09:11:260Paolo Guiotto: we get, basically, we flip X and Y, and we get that norm of Y minus X in L2 square. It is equal to norm of Y in L2 square minus norm of X,
09:24:390Paolo Guiotto: in L2 square.
09:26:860Paolo Guiotto: But these two quantities are the same, huh?
09:30:370Paolo Guiotto: So when we sum the two, we do the first plus the second, we get that left 2 times the norm of X minus Y in L2 square, equal, and as you can see, when we sum these two, we get 0.
09:44:400Paolo Guiotto: So from this, we get norm of X minus Y.
09:47:910Paolo Guiotto: equals 0, and L2 equals 0, and so the conclusion, X equals Y almost surely, because norm zero, the vanishing of the alton means that that difference is zero almost surely.
10:02:870Paolo Guiotto: And therefore, they are equally. This is a final question that says, now we have three variables, X, Y,
10:09:980Paolo Guiotto: and Z in L2 omega.
10:14:150Paolo Guiotto: And we know that the conditional expectation of X given Y
10:20:280Paolo Guiotto: is equal to Y. The conditional expectation of Y given Z is equal to Z,
10:28:340Paolo Guiotto: And the conditional expectation of Z given X, it is equal to X.
10:35:110Paolo Guiotto: Now, the conclusion is to prove that X is equal to Y, which is equal to Z, almost surely.
10:44:360Paolo Guiotto: So we, we basically used the same idea.
10:50:100Paolo Guiotto: Because, as, in, 2,
10:55:770Paolo Guiotto: So, if we use the first property, the conditional expectation of X given Y is equal to Y, we did use that. The L2 norm of X minus Y squared, it is equal to norm of X squared.
11:10:670Paolo Guiotto: minus norm of Y squared. This is…
11:15:920Paolo Guiotto: what we proved, no? This was the consequence of this, was exactly here, no? So we proved that the norm of X in L2 square is equal to norm of X minus Y squared plus norm of Y, and from this, we get this
11:33:810Paolo Guiotto: identity, norm of X minus Y squared, it is equal to norm of X squared minus norm of y squared.
11:40:970Paolo Guiotto: We apply the same for the second, and for the second, so from, let's say, 1…
11:47:950Paolo Guiotto: So this is, I toy… and 3i…
11:54:320Paolo Guiotto: So, from I, we get this. From 2I,
11:58:770Paolo Guiotto: we got, that norm of Y minus Z2 square.
12:03:910Paolo Guiotto: is equal to norm of Y2 squared minus norm of Z2 squared.
12:11:70Paolo Guiotto: And similarly, from the third one.
12:15:540Paolo Guiotto: which is the conditional expectation, Z given X is equal to X, we get that, Z minus X L2 squared is equal to normal Z in L2.
12:29:550Paolo Guiotto: Squared plus norm and minus, sorry.
12:33:470Paolo Guiotto: minus norm of X in L2 squared.
12:37:770Paolo Guiotto: And then, as above, we sum these three identities, and we get that the sum of the norms, X minus Y2
12:48:670Paolo Guiotto: squared plus norm of Y minus Z2 squared plus norm of X z minus X, which is the same of X minus Z.
13:00:510Paolo Guiotto: L2 square… well, as you can see, this cancels with this, this cancels with this, and this with this, so at the end, you get 0.
13:12:170Paolo Guiotto: This is the sum of three positive quantities equal to 0. All of these quantities must be equal to zero, so norm of X minus Y
13:21:650Paolo Guiotto: 2.
13:22:910Paolo Guiotto: square, equal norm of XY minus Z.
13:27:530Paolo Guiotto: to square the equal norm of Z minus X, 2.
13:32:270Paolo Guiotto: squared equals 0. And therefore, from this, we get that X minus Y is equal to 0 almost everywhere. So, formally, we should say it is equal to 0 on omega minus a measured zero set, let's call it N1.
13:50:670Paolo Guiotto: with probability of N1 is equal to 0.
13:54:890Paolo Guiotto: The second one tells that Y minus Z is equal to 0 on omega minus possibly another set, and 2, where, however, the probability is equal to 0.
14:07:380Paolo Guiotto: And the third one says that Z minus X is equal, again, to 0 on omega, except for a set N3, where again, the probability of N3 is 0.
14:19:450Paolo Guiotto: So, all these three are valid.
14:22:890Paolo Guiotto: So, say A, B, and C.
14:27:370Paolo Guiotto: A, B, and C.
14:30:850Paolo Guiotto: hold… Simultaneously.
14:41:60Paolo Guiotto: on… Omega minus all the exceptions together, so we say N1, Union N2, Union N3.
14:52:530Paolo Guiotto: And since the probability of this set, N1, union, N2, union, N3,
15:00:900Paolo Guiotto: by sub-additivity is less than the sum of the probabilities, and 1 plus probability, and 2 plus probability, and 3. And these 3 are 0, so this is a 0.
15:14:930Paolo Guiotto: We got the data.
15:17:730Paolo Guiotto: That, probability equals zero set, so the three identities holds for all omega except a measure of zero set, and this means exactly the conclusion, because we have, so,
15:32:840Paolo Guiotto: X equals Y, which is equal to Z on omega.
15:38:730Paolo Guiotto: minus N, where n is this union, and the probability of N is equal to 0.
15:45:920Paolo Guiotto: So this means, all this means, Almost surely.
15:51:390Paolo Guiotto: and descends D.
15:53:440Paolo Guiotto: Exercise.
15:57:450Paolo Guiotto: Do you have any questions?
16:02:370Paolo Guiotto: So this was not a particularly probabilistic exercise, because basically it was based on the orthogonal projection on L2.
16:12:550Paolo Guiotto: The next one is exercise 20.
16:19:420Paolo Guiotto: We have this function capital F of X equal E,
16:24:590Paolo Guiotto: 2 minus E2 minus X. All this is the exponent, right? X is real.
16:34:750Paolo Guiotto: So, number one is, check…
16:39:250Paolo Guiotto: that this capital F is a CDF.
16:45:220Paolo Guiotto: Cumulative distribution function.
16:49:180Paolo Guiotto: Well, to be… Cdf… F. Mas.
16:59:290Paolo Guiotto: fulfill… the… Following…
17:09:380Paolo Guiotto: Properties…
17:16:210Paolo Guiotto: So, first, when I say the limit at minus infinity must be 0 at plus infinity. It must be equal to 1. Second, the function f must be
17:29:240Paolo Guiotto: increasing, not necessarily strictly, and third, F must be continuous, right continuous.
17:46:500Paolo Guiotto: Well, this can be easily checked, because f at minus infinity means the limit when x goes to minus infinity of f of x.
17:58:250Paolo Guiotto: And this is the limit when x goes to minus infinity.
18:04:760Paolo Guiotto: Well, notice that, in particular, from these two combined, it follows that F is always between 0 and 1. So, this should be saved, because the CDF is…
18:17:610Paolo Guiotto: A function defined on the real line with values between 0, 1, such that blah blah blah, but these two together imply this one.
18:26:410Paolo Guiotto: So, when X goes to minus infinity, what happens? This exponent goes to plus infinity, so the exponential goes to E to plus infinity plus infinity with the minus, minus infinity, so the exponential at the end goes to 0.
18:45:140Paolo Guiotto: And similarly, f at plus infinity is the limit for X going
18:51:650Paolo Guiotto: to plus infinity of E minus E minus X. Now again, when x goes to plus infinity, this exponent goes to minus infinity, so the exponential goes to zero. With the minus, it will go to 0. The argument of the final exponential is going to 0, so the exponential goes to E to 0, which is equal to 1.
19:17:970Paolo Guiotto: S is increasing, that's evident because, you know, when X increases, minus X decreases, so the exponential e to minus X exponential is an increasing function, so when the argument decreases.
19:35:970Paolo Guiotto: the exponential decreases as well, so this is decreasing. With the minus, minus E to minus X is increasing, and therefore.
19:45:20Paolo Guiotto: E of minus E to minus X will be increasing, because the exponential keeps… is in increasing function. So, the conclusion is that F is increasing.
19:59:10Paolo Guiotto: Finally, I've write continuous. Well, actually, this is a continuous function, because it is a composition of continuous functions. So, clearly.
20:11:400Paolo Guiotto: This capital F is continuous.
20:16:790Paolo Guiotto: being good.
20:19:40Paolo Guiotto: composition.
20:25:490Paolo Guiotto: of continuous… functions So, it is more than right continuous. It is also differentiable, if you want, it is…
20:37:60Paolo Guiotto: It is, it is absolutely continuous. Okay.
20:42:140Paolo Guiotto: Now, let now, the problem says, let now, XN
20:51:50Paolo Guiotto: be a sequence of independent, identically distributed random variables.
21:01:790Paolo Guiotto: with this CDF, with the… F, Xana…
21:11:710Paolo Guiotto: of X equal 1 minus E minus X indicator of… 0 plus infinity X.
21:23:590Paolo Guiotto: Which is, if I'm not wrong, an exponential distribution.
21:28:230Paolo Guiotto: Okay, number… question number two is, determine… the… CDF… of Wayne.
21:40:400Paolo Guiotto: Which is defined as the maximum of X1XN.
21:52:120Paolo Guiotto: So, what we have to compute, it is F, YN, capital FYN, let's say, of Y, which is, by definition, if you write the definition in terms of the YN, it is the probability that YN be less or equal than Y.
22:12:60Paolo Guiotto: Now, this YN is the maximum of these quantities.
22:16:190Paolo Guiotto: We computed sometimes this.
22:18:580Paolo Guiotto: So, the maximum of X1, XN, this maximum will be… must be less or equal than Y.
22:27:750Paolo Guiotto: And this is possible if and only if each of the capital X is less than Y. So this is X1 must be less than Y, and X2 must be less than little y, and so on.
22:41:570Paolo Guiotto: until XN less or equal than little y. And we have the probability of this set.
22:47:280Paolo Guiotto: Now, these commas means that we are doing an intersection.
22:51:680Paolo Guiotto: Right? So this is a probability that intersection of sets like XJ less or equal than Y for J going from 1 to N.
23:05:970Paolo Guiotto: And now, by independence, of the, of the XJ.
23:14:810Paolo Guiotto: We know that if they are independent, the probability of the intersection splits into the product of the probabilities, so we write product J from 1 to N, probability of XJ less or equal than Y,
23:28:850Paolo Guiotto: This one is the CDF of XJ at point Y, so we have this product. They are all the same, so we have a product for J
23:41:510Paolo Guiotto: equal 1 to n of, according to the definition, which is, this one, now we use.
23:49:760Paolo Guiotto: So, with X equals Y, so we have 1 minus E to minus Y.
23:55:930Paolo Guiotto: times indicator… For, of 0 plus infinity Y.
24:05:950Paolo Guiotto: Well, these, are the same factors, repeated n times, so we get the nth power, so we have 1 minus E to minus y to the n.
24:17:240Paolo Guiotto: And then the indicator is just a 01. When you rise to the N, you'll get still 0, 1, so it's the same thing. Indicator of 0 plus infinity Y. So this is the,
24:31:930Paolo Guiotto: the final… the conclusion, so this is the F, Y, N, why is this one.
24:40:110Paolo Guiotto: the CDF pop.
24:44:190Paolo Guiotto: of the Yard.
24:45:760Paolo Guiotto: Question 3.
24:48:810Paolo Guiotto: Use 2 to prove that.
24:51:540Paolo Guiotto: prove… That, now, it says you take YN, you subtract the log of n.
25:00:510Paolo Guiotto: And this is a sequence of random variables that converges in distribution.
25:11:690Paolo Guiotto: Determining also the, limit distribution.
25:21:450Paolo Guiotto: Okay, so now, you know that when we study convergence, especially for
25:27:340Paolo Guiotto: convergence and distribution, we have a…
25:30:00Paolo Guiotto: Different possibilities to check convergence, no?
25:33:790Paolo Guiotto: So, one is that, for example, the point-wise convergence of the characteristic function.
25:40:650Paolo Guiotto: Here, we do not have characteristic functions, but we have the CDF. So we have, maybe, to look at the condition for CDF. So let's remind what is it, that condition.
25:53:160Paolo Guiotto: Let's remind… that, As ZN converges in distribution.
26:05:220Paolo Guiotto: to some Z, if… well, there is a condition about the CDF, which is a pointwise convergence, so FZN, well, let's say Z, converges to FZ.
26:22:200Paolo Guiotto: Zed.
26:24:370Paolo Guiotto: Well, this is not necessarily true for every Z, differently for the characteristic function.
26:31:310Paolo Guiotto: But it must be, and it characterizes, if it holds for… at every Z for which the final, the limit CDF is continuous. So, for every Z such that FZ
26:46:940Paolo Guiotto: is continuous.
26:50:00Paolo Guiotto: at point Z.
26:52:620Paolo Guiotto: Now, the point is that we don't know what is, at this moment, for us, this is the ZN, so we don't know what is Z we have to determine. So what we do is, let's see what happens when we do the pointwise limit, and then, depending on what is the limit, we will see if it is a pointwise limit for every Z.
27:12:900Paolo Guiotto: Which is a continuity point for the, for the limit CDF. So… Let's compute…
27:25:230Paolo Guiotto: the limit… in N of F, the variable is YN minus log N.
27:34:620Paolo Guiotto: of Y.
27:37:180Paolo Guiotto: Well, actually, this is not even the FYN, because the variable is YN minus log n, but that's not a big problem, because FYN minus log of n, this is the random variable Y, this would be the probability that YN minus log of n
27:56:430Paolo Guiotto: is less or equal than Y.
27:58:910Paolo Guiotto: So, clearly, we carry this log of n on the other side, and we get the probability that YN is less than Y plus log of n.
28:09:450Paolo Guiotto: So, we know this. This is the FYN.
28:13:960Paolo Guiotto: Evaluated at point Y plus log of n.
28:18:510Paolo Guiotto: And for this, we have a formula. It is that one. So we have E to minus E minus Y, where Y is this value, so minus Y plus log of n.
28:34:450Paolo Guiotto: all this to the n indicator interval 0 plus infinity of Y plus log of n. This is literally what we have when we replace this expression there.
28:48:950Paolo Guiotto: Now, let's rearrange a bit this. For example, you see that there is exponential that goes together with log, so we may write this as E2 minus E2 minus y.
29:01:980Paolo Guiotto: And then maybe we give the minus inside the argument, so log of n to minus 1, all this to the power n, then we have the indicator, 0 plus infinity of Y plus log n.
29:17:790Paolo Guiotto: Well, we can do something also on the indicator in a second. So, first of all, this is a 1 over N, so we have 1 minus E2 minus Y divided N.
29:31:150Paolo Guiotto: 2DN,
29:33:90Paolo Guiotto: This, indicator… well, the indicator, on the interval 0 plus infinity of Y plus log of n.
29:43:440Paolo Guiotto: This is 1 if and only if Y plus log of n is greater or equal than 0.
29:49:280Paolo Guiotto: In terms of Y, so if I look at this as an indicator in Y,
29:55:150Paolo Guiotto: Y plus log of n greater or equal than zero means Y greater than minus log of n, so this is the indicator from minus log of n to plus infinity of y.
30:07:820Paolo Guiotto: So I put here indicator minus log N to infinity, of Y.
30:16:110Paolo Guiotto: Okay, so this is the CDF of YN minus log N. Now we have to do the limit in n of this. Y is fixed here.
30:25:510Paolo Guiotto: So it is clear that there are two variables, so it's a little bit messy, but we have to look at this as a limit in N. So…
30:33:910Paolo Guiotto: the limit in N of F Yn minus log of n.
30:41:240Paolo Guiotto: Why?
30:42:510Paolo Guiotto: is the limit in n when n goes to plus infinity, of course. This 1 minus E to minus y over n to the n
30:56:490Paolo Guiotto: And then we have this indicator, minus log N to plus infinity, Evaluated at what?
31:04:330Paolo Guiotto: Now, what happens when n goes to plus infinity? As you can see, this is a remarkable structure, something like 1 plus alpha over n to the n.
31:15:180Paolo Guiotto: And we know that this goes to E to alpha, whatever is alpha real.
31:22:510Paolo Guiotto: So this expression goes to E, and the alpha is this one, of minus E to minus y.
31:31:690Paolo Guiotto: And what about this indicator? Now, Y is fixed, okay? So, wherever is Y,
31:38:940Paolo Guiotto: positive, negative, it doesn't matter. When n goes to plus infinity, this log of n goes to… the minus log of n goes to minus infinity, so I can say that…
31:50:600Paolo Guiotto: For n large, minus log N, it will be at left of Y, so that quantity is constantly equal to 1 for n large. So the quantity, indicator of minus log n
32:05:850Paolo Guiotto: plus infinity of Y,
32:09:30Paolo Guiotto: is constantly equal to 1 in n for every n larger than some initial N and 0. It depends on Y, okay, but Y is fixed here. So I can say, for every Y fixed.
32:24:490Paolo Guiotto: sooner or later, this quantity becomes equal to 1. So this allows to say that the limit in N
32:33:480Paolo Guiotto: of the FYN minus log of n.
32:38:350Paolo Guiotto: Log of n. At Y, at the end, is E to minus E to minus Y, and this for every Y real.
32:50:190Paolo Guiotto: Now, this one, as we discussed above, is a CDF. So, since it is a CDF, there is also a random variable, Z, well, let's say Z, for which the,
33:05:990Paolo Guiotto: what you could call, but for… about these notations, maybe Y.
33:11:460Paolo Guiotto: But no, let's call it Z. There exists a random variable Z for which that's the CDF. And since we know that this is a continuous function.
33:21:460Paolo Guiotto: Now, in mind that to check, the continuity, the convergence, the convergence in distribution using this condition.
33:31:160Paolo Guiotto: We have to check that the point-wise limit exists for every point where the limit CDF is continuous, but this limit CDF is continuous everywhere, and the limit holds for every Y.
33:44:10Paolo Guiotto: So, it means that the sequence YN minus log of n converges in distribution to that random variable Z.
33:59:420Paolo Guiotto: Okay, and that's all for the exercise number 20.
34:04:390Paolo Guiotto: Do you have any questions?
34:10:929Paolo Guiotto: Okay, so let's now move to Exercise 21.
34:19:690Paolo Guiotto: So this is an exercise,
34:23:520Paolo Guiotto: Yeah, here we have also to remind some property of the Fourier transformer.
34:30:739Paolo Guiotto: It says, first there is a gen- general, property, so that says that, show that,
34:41:719Paolo Guiotto: if X, Y, R, absolutely.
34:47:310Paolo Guiotto: Continuos.
34:48:960Paolo Guiotto: So this means that they have a density, okay? So there exists, they have densities, F, X, F, Y, that are, I remind you, L1 functions in the real liner, positive, F, X, FY.
35:04:720Paolo Guiotto: Are both positive, and their integrals are equal to 1.
35:13:470Paolo Guiotto: These are the characteristics. Of course, there are the densities, and it means that the probability that X belongs to some set E is the integral over E of the density FXDX, and the same for Y.
35:33:00Paolo Guiotto: Now, it says, prove that also X plus Y Jeez.
35:39:220Paolo Guiotto: absolutely continuous.
35:43:340Paolo Guiotto: And… The density of the sum
35:50:40Paolo Guiotto: Is the convolution between the two densities.
35:57:10Paolo Guiotto: Okay.
36:01:330Paolo Guiotto: I don't remind if we have proved in general this, but however, let's see why.
36:06:960Paolo Guiotto: Well,
36:09:900Paolo Guiotto: we could proceed, say, in two ways. One is the, say, trivial way. The exercise says check that this is the density, so it is enough to check that the quantity at the right-hand side does what
36:25:880Paolo Guiotto: the quantity at the left-hand side should do, because you already know the answer. Or if you don't know the answer, you should start saying, okay, let's compute the probability that X plus Y belongs to something. Let's see what happens. Well, let's do this one. So, probability that X plus Y belongs to E,
36:47:90Paolo Guiotto: Now, the goal is to prove
36:51:870Paolo Guiotto: that, there exists a function, the density of X plus Y, such that positive L1
37:01:960Paolo Guiotto: Integral equal 1.
37:07:300Paolo Guiotto: Such that this integral is the integral on e of that function, f of x plus y. Let's say… the letter, we use letter Z, DZ. Okay? The goal is this one. So what can be said? Now, X plus Y belongs to E,
37:26:750Paolo Guiotto: I could think that this is, something like, well, I could see this as,
37:36:880Paolo Guiotto: a function of the pair XY, no? So it's like phi of XY belongs to E.
37:46:880Paolo Guiotto: Now, I write this because I have, basically, the joint distribution of XY. I have because it is supposed that…
37:58:460Paolo Guiotto: I forgot to write are absolutely independent, otherwise the conclusion is not true. So I know that the XY are absolutely continuous and independent. This is important because it says
38:11:880Paolo Guiotto: also the pair, XY, has a density, and the density is the product of the two densities. So we could say that it is the integral on XY, the density of XY,
38:29:50Paolo Guiotto: The joint density, which is, by the way, by independence.
38:33:790Paolo Guiotto: F, X, X, F, Y, Y.
38:38:450Paolo Guiotto: on the set of pairs, XY, such that X plus Y belongs to E.
38:47:960Paolo Guiotto: This is literally what is it. So this is the integral on pairs XY, such that X plus Y belongs to E. Let me just put a bit of order here. FXXFY.
39:03:290Paolo Guiotto: Y DX, DY.
39:07:860Paolo Guiotto: Now, I have to transform this into a one variable integral, because the variable X plus Y is a scalar.
39:15:180Paolo Guiotto: variable, and I need to read the density, so I have to transform… to make something to… to transform this integration into an integration of a unique… the Unix set E. Now, as you can see, there is a variable that belongs to E. It is X plus Y.
39:32:80Paolo Guiotto: So this suggests to say, okay, let's do a change of variables, so we call, I don't know, U the X plus Y,
39:41:540Paolo Guiotto: In such a way that, well, actually, keeping the analogy with that letter, let's call it Z instead, that this will belong to E.
39:51:00Paolo Guiotto: And let's keep one of the other variables the same. So, for example, we… the second variable, W, we use X. So this is the change of variable in F2, no? Now, when we do this change of variable, we have that. This becomes, this integration, Z belongs to E,
40:11:00Paolo Guiotto: And about W, W is X, which is arbitrary in, in R, so, X in R.
40:20:630Paolo Guiotto: Sorry, this is E.
40:24:410Paolo Guiotto: Then, here we have FX, X remains, well, let's say we called, we called XW, sorry. So this is Z, this is W.
40:37:330Paolo Guiotto: So this will be W. FY of Y… well, Y disappear? Because Y becomes Z minus X, X is W, so Z minus W.
40:49:00Paolo Guiotto: So, Z minus W. Now, there is the modulus of the determinant of the change variable, but this is the transformation, no?
40:57:830Paolo Guiotto: So this is the map that gives XZW in function of XY.
41:04:550Paolo Guiotto: well, we should inverse this, which is the inversion is X equal W, and Y equal Z minus X, so Z minus W. Now, this is the map of which I should compute the determinant
41:22:840Paolo Guiotto: Of, this is the, say, the phi-1, phi minus 1,
41:27:270Paolo Guiotto: prime, the Jacobian matrix of the changeover variable to have this transformed into an integral in DZDW.
41:35:240Paolo Guiotto: Well, the determinant of this is easy, because W, phi minus 1 prime.
41:41:490Paolo Guiotto: is the Jacobian matrix of this transformation, so you have to compute the derivatives with respect to Z and W of this map. For the first component, derivative with respect to Z is 0, with respect to W is 1. For the second component, it's 1 minus 1.
41:56:640Paolo Guiotto: So when you compute the determinant of this determinant of this matrix, you get 0 times minus 1, 0, minus 1 times 1, so it is minus 1, with the absolute value, you get 1. So there are nothing changed.
42:10:950Paolo Guiotto: And here, we now split the integration into two integrations. So, we can say that this is the integration. First, for example, on W, in R, of FXW, F,
42:27:130Paolo Guiotto: YZ minus W in D, W.
42:32:500Paolo Guiotto: And then we integrate this thing in Z, where Z belongs to the range E.
42:39:400Paolo Guiotto: Now you read here, inside, there is a function, a function of the variable Z, which is exactly the convolution. This is FX star, FY at point Z.
42:54:320Paolo Guiotto: That, does this when you integrate on E. This function, you get the probability that X plus Y belongs to E. So that's automatically, by definition, the density of,
43:08:290Paolo Guiotto: So this turns out to be the density of X plus Y.
43:11:820Paolo Guiotto: And notice that we don't need to verify anything here, because since this is valid automatically, that f is positive, the integral will be 1, etc. However, you could also check directly that this is true.
43:26:460Paolo Guiotto: Okay.
43:28:470Paolo Guiotto: This is, about the question one.
43:32:410Paolo Guiotto: Now, the problem says, select now… Xena.
43:41:250Paolo Guiotto: And natural, greater or equal than 1.
43:48:240Paolo Guiotto: Independent, identically distributed random variables.
43:53:580Paolo Guiotto: with, common densities, which is a… this is,
44:02:420Paolo Guiotto: Basically, it's equivalent saying that each Xn has an exponential distribution. It says this is the density of XN, it's this lambda E2 minus lambda X indicator, 0 plus infinity of X.
44:19:40Paolo Guiotto: So, basically, we are saying each XN is an exponential of parameter lambda.
44:32:410Paolo Guiotto: It says, number one… determine… determines… the distribution.
44:48:330Paolo Guiotto: of X1.
44:51:580Paolo Guiotto: plus XN.
44:55:600Paolo Guiotto: It says, use… use Fourier transform. Let's see how.
45:03:960Paolo Guiotto: The distribution, so basically this is a sum of n independent random variables, so this was the formula for two independent random variables.
45:19:330Paolo Guiotto: It will be the same for n independent random variables, so it will be the consecutive convolution of things. So, let's see if we,
45:32:330Paolo Guiotto: I think we computed the density because of, of,
45:40:200Paolo Guiotto: So we can say that the density of X1, if we want a formal
45:46:380Paolo Guiotto: argument could be the following, no? So you look at this, the density between the sum X1 plus XN minus 1, and the last one XL. These two are independent, because the first one
46:02:460Paolo Guiotto: The first block depends on X1, Xn minus 1. The second one is Xn. Xn is independent of the other, so it is independent of the sum, so we can say that this is the convolution between FX1 plus A2, etc, Xn minus 1,
46:20:990Paolo Guiotto: N, F, X, N.
46:24:910Paolo Guiotto: And repeating this operation after n steps, we have that this is FX1 convolution with FX2, convolution with FX3, and so on, convolution with FXN. Now, all this, at the end, are the same function.
46:42:770Paolo Guiotto: So now, to understand what is this,
46:45:840Paolo Guiotto: it is convenient to compute the Fourier transform. So I noticed that the Fourier transform of the density of the sum, X1 plus XN, is the Fourier transform of this convolution product, FX1 star.
47:04:460Paolo Guiotto: FXN.
47:07:800Paolo Guiotto: Now, this is the traditional L1 Fooley transform. We know that Fooley transforms convolutions into algebraic products, so this will be FX1 heter times FX2 hat, etc, FXN hat.
47:27:350Paolo Guiotto: And what about the FXN at… well, they are all the same, because, you see, they are identically distributed, and the density does not depend on N. So let's, let's say that this is a unique F at to the N.
47:47:370Paolo Guiotto: And let's see what is this F at. F at, say, at point C is equal to…
47:54:570Paolo Guiotto: Well, this is not exactly one of the functions for which we have a different form in the table, at least. We have not written this. It looks like an exponential, but it is,
48:08:400Paolo Guiotto: Only on the positive side, so let's do the calculation directly. So we have the integral on r of the function, which is lambda e to minus lambda X indicator, 0 plus infinity X.
48:23:380Paolo Guiotto: times e to minus i, here we are talking about the traditional Fourier transform, so I use this notation, i minus iCX dx. So this is the integral from 0 to plus infinity.
48:39:480Paolo Guiotto: of, well, let's say lambda can be written outside E2 minus lambda plus IXC times X DX.
48:50:840Paolo Guiotto: Since this number cannot be 0,
48:56:140Paolo Guiotto: Because here, I guess, I forgot to say that lambda will be a positive value.
49:04:520Paolo Guiotto: So we can integrate this, saying that this comes from the derivative of e to minus lambda plus xc.
49:15:170Paolo Guiotto: X divided by this coefficient, minus lambda plus IC.
49:21:550Paolo Guiotto: This is evaluated at plus infinity minus the value at zero.
49:27:40Paolo Guiotto: at plus infinity module C minus lambda plus ik X.
49:34:600Paolo Guiotto: What is it? Well, this part here, E2 minus xx, is the usual unitary complex number, so the absolute value is minus lambda X, and when you send the X to plus infinity, this quantity goes to 0 because lambda
49:51:520Paolo Guiotto: is positive here. So we get that the evaluation reduces to the evaluation at 0, the exponential becomes 1, the sine is minus, so this becomes 1 over lambda plus xi. So at the end, this is lambda over lambda plus xi. This is the Fourier transform.
50:13:350Paolo Guiotto: Now, let's go back to the Fourier transform of the convolution.
50:18:560Paolo Guiotto: So the, so the Fourier transform, sorry, the Fourier transform of the sum X1 plus Xn, which is our goal, the same variable x, it is equal to this
50:33:380Paolo Guiotto: function here.
50:35:250Paolo Guiotto: raised to power n, so lambda over lambda plus IC to the end.
50:43:300Paolo Guiotto: We have to go back to…
50:45:750Paolo Guiotto: We want this guy here, no? The function inside. So, we have somehow to invert this formula.
50:55:940Paolo Guiotto: Well, here we could notice that this is lambda to the N,
51:00:910Paolo Guiotto: 1 over lambda plus IC to the n.
51:06:850Paolo Guiotto: Now…
51:07:910Paolo Guiotto: I know that the hat of the function I'm looking for, it is this one, so now I want to extract FX1 plus
51:19:980Paolo Guiotto: Xena.
51:21:680Paolo Guiotto: So I could apply another time the Fourier transformer, using version formula, and computed the Fourier transform of the right-hand side. Or.
51:38:350Paolo Guiotto: Or…
51:44:250Paolo Guiotto: Yeah. I could notice that this thing is derivative.
51:51:760Paolo Guiotto: More or less. So you know that when you do the derivative with respect to C of 1 over lambda plus IC,
52:00:690Paolo Guiotto: So this is the derivative with respect to Xi of lambda plus xi to exponent minus 1. This is minus…
52:09:410Paolo Guiotto: 1 over lambda plus iC, the exponent now becomes 2 here, and then there is derivative with respect to X of the argument, which is just I, so I have this
52:28:800Paolo Guiotto: So if I do the second derivative with respect to X, so if I derive 2 times this.
52:36:680Paolo Guiotto: It means I am now deriving this quantity.
52:41:480Paolo Guiotto: So the minus remains minus, minus i. Then I have the derivative respect to C of lambda plus IXC2 minus 2. So this becomes… you notice the minus 2 comes down with the minus makes plus.
52:58:740Paolo Guiotto: I, 2, then we have lambda plus IC2 minus 3, and then there is the derivative of the argument, which is another I, so let's say I square.
53:10:380Paolo Guiotto: Let's right this way, because we will understand.
53:14:740Paolo Guiotto: what happens? So, the next time, if I do the third derivative with respect to X of lambda plus, like, C to exponent minus 1, the next time we come down a minus, so I have a minus, there is a 3 that with the 2 yields a 3 factorial.
53:34:470Paolo Guiotto: Then there will be another, a new i, so I have i cubed lambda plus iC to exponent minus 4.
53:43:240Paolo Guiotto: And then the derivative, the fourth derivative of this lambda plus xi to minus 1, it is now plus…
53:52:250Paolo Guiotto: The next time you have a minus 4, no, so it becomes plus, but 4 times 3 factorial, it is 4 factorial. I to the power 4, lambda plus IXC2 minus 5. So, if I want the exponent minus n that I have here, so this is lambda to be n lambda plus IXC to
54:15:560Paolo Guiotto: minus N. To arrive to minus n, I have to differentiate n minus 1 times this thing,
54:27:70Paolo Guiotto: When I differentiate n minus one time, notice the indexes here. You see that you have 4 here, 4 here, 3 here, 3 here. So there is a plus-minus, and you see that it is plus-minus, it is plus when the order of the derivative is even.
54:44:570Paolo Guiotto: And, minus 1, it is odd, so it is just minus 1 to the order of the derivative, which is minus…
54:52:50Paolo Guiotto: N minus 1 here, then there is n minus 1 factorial, then I2DN minus 1, and then finally lambda plus IC2 minus N.
55:07:00Paolo Guiotto: So this is to say that our lambda plus XC2 minus n
55:16:140Paolo Guiotto: is what? I have to divide by this.
55:20:510Paolo Guiotto: So when I divide by this, it is plus minus 1, it doesn't change, so let's write minus 1 to n minus 1. Then we divide by the factorial of n minus 1.
55:31:770Paolo Guiotto: And then we divide, by, i to the n minus 1,
55:37:740Paolo Guiotto: which is like multiplying by 1 over i. 1 over i is minus i.
55:44:910Paolo Guiotto: You see, there would be a 1 over i to n minus 1, which is a minus i to n minus 1. This is interesting because it kills that sine, so lambda plus iC to the minus n.
56:00:500Paolo Guiotto: So this, minus…
56:03:840Paolo Guiotto: goes together with this one, so they kill it, and so it remains i to n minus 1.
56:11:10Paolo Guiotto: divided n minus 1 factorial lambda plus xi to minus N.
56:22:140Paolo Guiotto: No, no, sorry, at WMS, there is not this one, there is the derivative here, because I'm extracting this term.
56:29:610Paolo Guiotto: Sorry, I am extracting this one, so this is, what I have to write here is the derivative and minus 1 times of
56:38:380Paolo Guiotto: lambda plus sidec to minus 1.
56:41:770Paolo Guiotto: So this is not this.
56:44:30Paolo Guiotto: But… the derivative N minus 1 time lambda plus xi 2 minus 1.
56:52:280Paolo Guiotto: All this mess, to do what?
56:55:770Paolo Guiotto: Because I have that disc.
57:01:740Paolo Guiotto: multiplied by lambda to the n is the Fourier's form of what I'm looking for. So let's finally multiply by lambda to the n, lambda to the n, and this is the hat of FX1 plus
57:17:350Paolo Guiotto: Xen.
57:19:290Paolo Guiotto: So I now can extract F by inversion formula.
57:36:550Paolo Guiotto: Or if you want, you don't need necessarily to use inversion formula, because
57:42:420Paolo Guiotto: We computed the Fourier transform of F at, right? This was the Fourier transform of F at Oxy.
57:51:770Paolo Guiotto: And you see that, it is this one.
57:55:980Paolo Guiotto: So that, the clearance form of F at is equal to lambda
58:02:820Paolo Guiotto: times lambda plus xi to minus 1.
58:06:830Paolo Guiotto: So I can say that this…
58:10:70Paolo Guiotto: is I don't need to use the inversion formula. I can just do one step and say, I…
58:19:730Paolo Guiotto: to n minus 1, lambda to the n divided n minus 1 factorial.
58:26:20Paolo Guiotto: Tender is the nth minus 1 derivative of this. Well, I need a lambda, right?
58:36:140Paolo Guiotto: a lambda here, so I borrow a lambda from this, so I put minus 1 here. Lambda.
58:43:800Paolo Guiotto: times lambda plus xi minus 1. Now this is the Fourier transform of that function, the exponential.
58:54:810Paolo Guiotto: evaluated at Xi.
58:57:530Paolo Guiotto: Why this is convenient? Because we have the derivative of the Fourier transform, and the derivative of the Fourier transform
59:07:320Paolo Guiotto: It is…
59:09:330Paolo Guiotto: So, i to n minus 1… well, let's put together that i and lambda first. I lambda to exponent n minus 1 divided n minus 1 factorial. This is the derivative, the derivative of the Fourier transform
59:26:610Paolo Guiotto: that's not the Fourier transform of the derivative, it's the derivative of the Fourier transform, it's the Fourier transform
59:35:160Paolo Guiotto: of, is minus plus… do you remember the formula?
59:42:60Paolo Guiotto: Is it minus i, the variable F?
59:48:590Paolo Guiotto: Isn't it?
59:50:360Paolo Guiotto: It's minus or plus.
59:55:50Paolo Guiotto: Yeah, because when you do the derivative, you differentiate an integral sine and the exponential as a minus. Yes. So it is minus. So, now, if you repeat n minus 1 times, what happens? You have to repeat minus n minus 1 times this thing, so you get this factor to the exponent n minus 1 times F.
00:15:260Paolo Guiotto: So, returning back to this, we have that d hat of the sum of the density of the sum
00:27:180Paolo Guiotto: evaluated at point x is equal to… we have a coefficient, which is that i lambda to n minus 1 divided by n minus 1 factorial, and then we have…
00:42:480Paolo Guiotto: that derivative, which is the head of minus i, the variable, 2N minus 1, times F,
00:52:420Paolo Guiotto: What is going on?
00:55:810Paolo Guiotto: at point C.
00:59:110Paolo Guiotto: Now, this is a coefficient, so we can carry inside, we can put under the Fourier transform sign, all this, and this says that the density, FX1 plus FXN,
01:13:880Paolo Guiotto: at x is this thing, i lambda to n minus 1 divided n minus 1 factorial, then we have minus i.
01:27:440Paolo Guiotto: The variable is X to the n minus 1,
01:31:890Paolo Guiotto: F of X, where f of x is the exponential lambda E minus lambda X indicator, 0 plus infinity of X.
01:45:370Paolo Guiotto: We can simplify the eye, because you see that we have…
01:50:270Paolo Guiotto: that minus i here together with this i, they are both raised to exponent n minus 1, so I can put together, and I have minus i squared, which is 1. So at the end, we get lambda to n minus 1,
02:07:810Paolo Guiotto: There is another lambda here, so it is lambda to the N.
02:12:440Paolo Guiotto: divided by n minus 1 factorial.
02:18:760Paolo Guiotto: X to n minus 1.
02:24:30Paolo Guiotto: E to minus lambda X indicator 0 plus infinity.
02:30:110Paolo Guiotto: of X. And this is the…
02:34:20Paolo Guiotto: It's been tough, this path, yeah.
02:36:820Paolo Guiotto: This is the density of the… SUM F… X1 plus… Excellent.
02:50:180Paolo Guiotto: Okay, so let's quickly review. So what we did was, we had to compute the density of the sum.
03:00:630Paolo Guiotto: The initial part of the exercise suggests that this will be the convolution, and in fact, it is the convolution n times of the single variable densities, which are the same function, because by assumption, all the variables have the same distribution.
03:16:550Paolo Guiotto: So we had to compute the, convolution of these functions.
03:23:40Paolo Guiotto: The exercise suggests to use the Fourier transform, so we take the Fourier transform of this F here, because when we compute the Fourier transform of a convolution, we have a remarkable property that discounts, the algebraic product of the Fourier transforms.
03:39:820Paolo Guiotto: In this case, since they are the same function, which is the unit Fourier transform of this thing, we have that function raised to the N.
03:50:330Paolo Guiotto: we computed the Fourier transform of that F, and we obtained this one. So, putting this into the formula, we get that this… the density of the sum has Fourier transformed this function here.
04:07:660Paolo Guiotto: Now we wanted to determine DF.
04:10:820Paolo Guiotto: So, the possibilities are we apply the inversion formula, so we do an at both sides, but then we have to compute the Fourier transform of that thing.
04:19:200Paolo Guiotto: Okay?
04:20:420Paolo Guiotto: Or, we try to see the right-hand side has a Fourier transform of something.
04:25:670Paolo Guiotto: And the trick here is to notice that this denominator, so once we write the fraction like that, this denominator can be seen as coming from a derivative of this one that is, at the end, the Fourier transform of the function f.
04:40:340Paolo Guiotto: In fact, we have seen that after a few steps, that if we differentiate n minus 1 times this function, we get, apart for the coefficient, that fraction, 1 over lambda plus x c to the n.
04:54:440Paolo Guiotto: So this means that extracting this, lambda plus XC2 minus n is, apart for a constant, the derivative of this, and this is a Fourier transform.
05:06:630Paolo Guiotto: So,
05:09:410Paolo Guiotto: returning back to the initial F, the F of the Fourier transform of the density of the sum, it turns out to be apart for a coefficient, the derivative of a Fourier transform. We now use the remarkable property of the derivative of Fourier transform, which is the Fourier transform of multiplication by a factor minus iD variable.
05:33:600Paolo Guiotto: We repeated this N minus 1 times, so we have this raised to power n minus 1. And therefore, we see that the hat of the density of the sum is a unique big hat of something which is explicit.
05:48:280Paolo Guiotto: Because we have all the ingredients, that F is the initial density, we used, so it is the exponential density here. And so, at the end, this is this formula.
05:58:350Paolo Guiotto: There is a final question.
06:01:80Paolo Guiotto: Really, these exercises are much more complicated than the previous ones.
06:10:240Paolo Guiotto: Now, it says, let N, independent, of the Xena.
06:19:20Paolo Guiotto: And it says that the distribution is probability that this n is equal to a natural n.
06:26:850Paolo Guiotto: is equal to 1 minus P to N minus 1 times P, where P is a number between 0 and 1.
06:36:10Paolo Guiotto: And did this size determine the distribution?
06:39:950Paolo Guiotto: distribution Of…
06:45:240Paolo Guiotto: X1 plus X2, etc, plus XN. But be careful, because N is a random variable, so that quantity is a random variable, the number of terms depends on the… on the sum.
06:58:960Paolo Guiotto: So, here, to compute this, we, so, we wanted to find the density of this variable, but to find the density here.
07:12:50Paolo Guiotto: it… we start from the CDF, so we compute the probability that X1 plus XN
07:20:960Paolo Guiotto: be less or equal than some value X. Now, the point is that, it seems to be complicated.
07:28:910Paolo Guiotto: But the point is that you could say, since n is a random quantity that takes integer values, you could split this into the probability of the union, that is a disjoint union.
07:44:210Paolo Guiotto: Where X1 plus XN is less or equal than little x, and capital N is little n.
07:54:10Paolo Guiotto: No? Because that little n… if capital N is 1, that is X1 less or equal than X. If capital N is equal to 2, that's X1 plus X2 less or equal than X, etc.
08:05:770Paolo Guiotto: So, since that capital N can take the values 1, 2, 3, 4, etc, we split this into the union, which is a disjoint union, because the capital N cannot be two different values.
08:18:20Paolo Guiotto: Now, this split becomes a sum of probabilities, and since n is independent of this, I can transform this into the product. So I have X1 plus
08:30:529Paolo Guiotto: Xn, less or equal than X, times the probability that capital N is equal to N.
08:36:899Paolo Guiotto: This one is known, it is 1 minus P to N minus 1 times P.
08:44:430Paolo Guiotto: But also, the other one is known because, well, here we have the density, and it is the probability that this variable is less than X, so it is the integral from, if you want, minus infinity to X, of the density that we have
09:00:450Paolo Guiotto: written here. So, we have to write lambda to the n of… N minus 1 factorial.
09:13:80Paolo Guiotto: Then we have, well, X, let's say Y.
09:17:850Paolo Guiotto: 2N minus 1, because X is used. E2 minus lambda y.
09:24:859Paolo Guiotto: indicators 0 plus infinity of Y in DY.
09:31:399Paolo Guiotto: This is the term I have to multiply by this.
09:35:180Paolo Guiotto: Now, let's arrange and see what is it. So, it is sum for n going from 0 to infinity. The integration is from 0 to plus infinity, so this means that this integral is from 0 to X.
09:49:630Paolo Guiotto: then we can say lambda to the n over n minus 1 factorial, Y to n minus 1, E minus lambda y, there is no more the indicator.
10:01:920Paolo Guiotto: All this is multiplied by this number, 1 minus P to N minus 1 times P.
10:10:470Paolo Guiotto: So this is, what could we do? We could put this inside, and also carry the sum inside, because these quantities are positive.
10:22:220Paolo Guiotto: And the monotone convergence for series applies, you can carry inside the sum.
10:29:820Paolo Guiotto: And this becomes this, sums for n going from 0 to infinity.
10:34:960Paolo Guiotto: Well, the sum is, for these terms, so let's, put together…
10:53:670Paolo Guiotto: So, we put together, maybe…
10:59:650Paolo Guiotto: That is it.
11:06:620Paolo Guiotto: I should give this 1 minus P inside.
11:13:770Paolo Guiotto: Wait a second…
11:18:700Paolo Guiotto: Yeah, idiot. The sum must start from 1, because there is at least one term, so that index is starting from 1.
11:29:50Paolo Guiotto: Because, cannot be 0, you know? Otherwise, X1 plus X0. So, because I was wondering what, what is,
11:39:110Paolo Guiotto: what happens with N equals 0, but there is no 0. N, from 1, so let's put together the… this way. So, we have lambda E minus lambda y.
11:53:710Paolo Guiotto: then, so I'm splitting this in this way, in such a way I can use the exponent n-1. So you have lambda Y1 minus P, all this number is raised to n minus 1, divided n minus 1 factorial.
12:10:870Paolo Guiotto: If I'm not wrong, DY all this times P at the end.
12:15:300Paolo Guiotto: Now, this is the sum…
12:18:130Paolo Guiotto: If you rescale the indexer, n goes from 1 to infinity, n minus 1 goes from 0 to infinity. Of lambda Y lambda
12:28:670Paolo Guiotto: Y… 1 minus P to the n of n factorial.
12:37:220Paolo Guiotto: And this is the sum of X to the n of, say, T to the N.
12:44:80Paolo Guiotto: divided then factorial, it is e to t. So at the end, this is E to lambda y1 minus P.
12:54:790Paolo Guiotto: So this is a P, integral 0 to X, e to lambda y 1 minus p, there is that lambda we can put outside, it's just a constant, and there is this second exponential minus lambda y dy.
13:13:960Paolo Guiotto: Okay, so this is the probability that this is less or equal than X. So, if I call SN the variable X1 plus X capital N, this is probability that S capital N is less than X.
13:31:910Paolo Guiotto: So it is the CDF of SN evaluated at point x, and it is equal to P lambda integral 0 to X of that function, that exponential.
13:45:430Paolo Guiotto: You see that the lambda y and lambda Y, these two can be simplified, so it remains, actually, E to minus…
13:57:430Paolo Guiotto: lambda PYDY. We could compute, but since we want the density, density, if this is the function, the density is the derivative. So the density of SNX
14:12:550Paolo Guiotto: is the derivative with respect to X of the CDF.
14:18:230Paolo Guiotto: And this is P lambda.
14:20:750Paolo Guiotto: When we differentiate this thing, we get E minus lambda PX.
14:27:260Paolo Guiotto: this for X positive, for X negative is 0.
14:33:350Paolo Guiotto: Because the CDF is 0 for X equals 0, and so for x less than 0, it must be 0.
14:40:670Paolo Guiotto: And so this is the conclusion. What is the CDF of this function?
14:51:630Paolo Guiotto: Okay, so this is clearly… Much more complex than the previous ones.
15:00:400Paolo Guiotto: Okay, so…
15:08:260Paolo Guiotto: There is a second, I… we have time, until,
15:14:630Paolo Guiotto: If you are okay, we can do until 10-20.
15:18:610Paolo Guiotto: at least to give a look to the other problems. I do not remind exactly the numbers.
15:27:290Paolo Guiotto: Okay, so I won't be able to do all of them, so would you like to see any one of them, specifically?
15:40:240Paolo Guiotto: The designer?
15:42:20Paolo Guiotto: The 9 is…
15:48:110Paolo Guiotto: The 9 is on the Brownian motion, and…
15:51:730Paolo Guiotto: The other one was? 25. 25 is 9, and… 17. 17 is… Yes, that's,
16:05:40Paolo Guiotto: Okay, let's see, quickly the 9, huh?
16:11:670Paolo Guiotto: Exercise 9,
16:13:900Paolo Guiotto: Here, it says WT… The equator 0 is a Brownian motion, and we define this, B of T,
16:25:440Paolo Guiotto: is by definition TW1 over T for T positive.
16:31:520Paolo Guiotto: And 0 for t equals 0.
16:35:60Paolo Guiotto: So, question one, the, B of T… R.
16:41:160Paolo Guiotto: Goshen.
16:45:620Paolo Guiotto: The determining their distribution.
16:48:320Paolo Guiotto: and the increments of B of D So, B… It says BT1.
16:55:910Paolo Guiotto: Bt2 minus BT1.
17:00:770Paolo Guiotto: etc.
17:02:300Paolo Guiotto: Well, it means for consecutive times, so T1 is less than T2, less than T3, etc.
17:10:460Paolo Guiotto: Independent.
17:15:550Paolo Guiotto: And finally, there is another question inside the trajectories are continuous, And zero excluded, plus infinity.
17:28:760Paolo Guiotto: With probability 1.
17:33:860Paolo Guiotto: So these are basically three questions that are almost the check of the properties that also be of P in the Brownian motion. What is missing is the continuity at zero, which is true, but it's more delicate than what
17:47:490Paolo Guiotto: what is required here. So, B of T are Gaussian. Well, we know that, in general, WS is normal, mean 0, variance S. So, W1 over T will be normal, mean 0, and variance 1 over T.
18:04:640Paolo Guiotto: And when you multiply a variable by a scalar, so, for example, by t here, well, the mean is still zero, and the variance is multiplied by the square of the equation.
18:22:10Paolo Guiotto: So this is a T squared times 1 over T,
18:25:630Paolo Guiotto: So you get normal mean 0 and variance T. If you want, you can check also by the characteristics function, because if you take the phi of W1 over T, XC, this is, sorry…
18:40:610Paolo Guiotto: yeah, T1 over T. It is expectation of EIC
18:47:590Paolo Guiotto: TW of 1 over T. Then you give the T to the scarc, and this is the characteristic function of the variable W1 over T, evaluated at Txi.
19:00:990Paolo Guiotto: W1 over T is normal, means 0, and the variance 1 over T, this, because we know that is a Brownian motion. So this will be E to minus 1 half. We have the variance, which is 1 over T, times the square of the argument Txi.
19:20:630Paolo Guiotto: So we get, at the end, E minus 1 half T squared over T, T, C squared. And that's, exactly, says that TW1 over T is normal, mean 0, and variance T.
19:39:430Paolo Guiotto: About the increments, if, we have times T1, less than T2, less than T3, etc.
19:49:440Paolo Guiotto: Well, you see that, the increments of the Brownian motion, the BT1,
19:56:410Paolo Guiotto: Of the brown mash, of this B. Bt1, BD, BT2 minus BT1, etc.
20:06:240Paolo Guiotto: Well, right in the definition, this is T1W1 over T1. This is T2W, T2,
20:16:220Paolo Guiotto: W1 over T2 minus T1W1 over T1, so they are not exactly increments of the Brownian motion.
20:29:240Paolo Guiotto: But, we may, We may, so, it's… it is,
20:44:580Paolo Guiotto: So, moreover, you see that 1 over T1 actually would be greater than 1 over T2. So,
20:55:800Paolo Guiotto: We should find a way to check that these are independent, no? So how can we do?
21:01:970Paolo Guiotto: Well, let's show just for the two, okay? For simplicity. So this is, to show that they are independent.
21:13:540Paolo Guiotto: we could, well, since these are Gaussian variables, it is convenient to use the characteristic functions. So, we compute the characteristic function of BT1 of the pair, Bt1, BT2,
21:29:110Paolo Guiotto: minus BT1.
21:32:680Paolo Guiotto: So this is a B variant, characteristic function, so it's a function of two variables, say, c eta, and we wonder if this is a B, phi BT1 of C times phi.
21:45:190Paolo Guiotto: the, the increment.
21:47:950Paolo Guiotto: B.
21:49:820Paolo Guiotto: T2 minus BT1 of eta. Let's see if this is true. If this is true, we have that they are independent.
21:59:660Paolo Guiotto: Now, if we write this, this is… this is just a calculation. This is E2I. We have the variable xita times these two vectors, BT1, BT2 minus Bt1. We already write as T1W1 over T1,
22:19:390Paolo Guiotto: and T2W1 over T2 minus T1W1 over T1.
22:29:760Paolo Guiotto: Now, of course, we have to try to recreate the increments of the Brownian motion, because it's on them that we have some information.
22:40:630Paolo Guiotto: Okay, so if we write this argument in the exponent, this is CT1W1 over T1 plus eta.
22:52:860Paolo Guiotto: T2W1 over T2 minus T1W1 over T1.
23:03:510Paolo Guiotto: Now, reminder, times are ordered in that way.
23:08:970Paolo Guiotto: Yeah. So, we should… so you see that these times… this time, 1 over 2 comes first, then this. So, we can say that, if we split things in this way…
23:28:70Paolo Guiotto: Or we could have recreated this also here.
23:31:470Paolo Guiotto: Aware.
23:35:40Paolo Guiotto: So…
23:38:150Paolo Guiotto: I want to recreate the increment. The increment will be W1 over T1 minus W1 over T2, so I have to put the same coefficient here. So I need to add and subtract T1 here.
23:55:250Paolo Guiotto: and split this. So this comes… E to I…
24:09:670Paolo Guiotto: No, no, sorry, no, because, no. Well, let's write, let's write, let's write, analytically, everything that multiplies WT1 over T1 and W1 over T2. So this is C, T1,
24:28:180Paolo Guiotto: So, forget the second of this.
24:32:160Paolo Guiotto: minus eta.
24:35:620Paolo Guiotto: T1W1 over T1, but not wrong.
24:42:200Paolo Guiotto: plus eta T2, W1 over T2.
24:48:50Paolo Guiotto: Yes. Now, I… it's here that I take this, and I subtract W1 over T2, and I add the W1 over T2.
24:59:800Paolo Guiotto: Because I'm trying to recreate the increment in the future in such a way that they are independent. So I have expectation of EI, then I have this number, C, E1 minus eta.
25:15:440Paolo Guiotto: So, C minus eta T1.
25:21:250Paolo Guiotto: T1, and then I have the increment W1 over T1 minus W1 over T2.
25:31:520Paolo Guiotto: And then, what remains,
25:36:840Paolo Guiotto: about this, so this W, 1 over T2…
25:41:950Paolo Guiotto: is now multiplied by, C minus eta T1, apart from i.
25:48:860Paolo Guiotto: C minus eta T1.
25:53:300Paolo Guiotto: And then I have plus eta T2, if I'm not wrong.
26:00:570Paolo Guiotto: Now, these two are independent, so…
26:04:160Paolo Guiotto: I can split the expectation here.
26:08:60Paolo Guiotto: And, they are Gaussian also.
26:12:50Paolo Guiotto: So this is a standard Gaussian with mean zero variance, the difference between the two times. So this is E minus 1 half the variance, which is the difference between the two times.
26:28:540Paolo Guiotto: The times are ordered, 1 over T1 is greater, yes, so 1 over T1 minus 1 over T2.
26:37:900Paolo Guiotto: And then there is the square of this number here, so C minus eta squared T1 squared. This is the first, and the second here is similar.
26:49:760Paolo Guiotto: We have E minus 1 half, the variance is 1 over T2, and then we have the square of that, so C minus eta.
26:59:330Paolo Guiotto: T1 plus eta T2 squared.
27:05:530Paolo Guiotto: So now, if the calculation is correct, we should get this.
27:10:500Paolo Guiotto: Okay.
27:13:530Paolo Guiotto: So, let's see. E minus 1 alpha, this is T1, T2.
27:20:990Paolo Guiotto: T2 minus T1, C minus eta squared, T1 squared, then we put together the two exponents. There is a common factor of 1 half, so let's…
27:33:570Paolo Guiotto: right here, so plus 1 over T2.
27:37:600Paolo Guiotto: Then we do that square, we have…
27:40:930Paolo Guiotto: C minus eta squared T1 squared plus eta squared T2 squared, and then we have the double product plus
27:51:860Paolo Guiotto: 2… eta c minus eta T1, T2.
28:00:620Paolo Guiotto: Now, what is this mess?
28:11:870Paolo Guiotto: Do you see anything? So, we have this, cancel this.
28:25:20Paolo Guiotto: So, the only thing that comes, we have to put together these two. So, we have divided these two. We have a factor of C minus eta squared.
28:37:20Paolo Guiotto: And numerator is… you see T1, T2…
28:42:440Paolo Guiotto: that comes from this times this, then minus T1 square, but there we have plus T1 square, right? This cancels.
28:52:00Paolo Guiotto: The T2 simplifies now, so everything here comes at T1C minus eta square, this part.
29:01:650Paolo Guiotto: Then we have… plus T2 square over T2 square, 1 eta square.
29:11:880Paolo Guiotto: And this is so… no, no, no, it's, T2, it a square, because there is the square, yeah.
29:19:120Paolo Guiotto: And then we have plus 2 eta c minus eta. T1, T2 divided T2, this is T1.
29:29:490Paolo Guiotto: So, that comes E minus 1 half this expression.
29:36:660Paolo Guiotto: T1C minus eta squared plus T2 eta squared plus 2, eta, C minus eta.
29:49:260Paolo Guiotto: T1.
29:53:900Paolo Guiotto: And now…
30:02:880Paolo Guiotto: Remind that we have to split into a function of clear function of theta, so this is C squared…
30:10:300Paolo Guiotto: plus eta squared minus 2C eta.
30:15:570Paolo Guiotto: So, the mixed term is…
30:18:190Paolo Guiotto: minus 2 theta times T1, right? And here we have plus 2 theta times T1 dicil.
30:27:890Paolo Guiotto: So, there is no mixed term. I have e to minus 1 half. Let's isolate the terms with the C squared. There is only this one, and we get T1C square.
30:41:610Paolo Guiotto: And then let's isolate the term with eta squared. So, for eta squared, the factor is T1 here.
30:51:640Paolo Guiotto: plus T2… Here. But then we have…
30:56:970Paolo Guiotto: 2 eta minus theta T1. So, minus… 2… T1.
31:05:80Paolo Guiotto: Right?
31:06:900Paolo Guiotto: And what's that? That's T2 minus T1.
31:11:610Paolo Guiotto: So at the end, we got…
31:13:370Paolo Guiotto: E minus 1 half T1C squared, E minus 1 half T2 minus T1 eta squared. That's what we had to obtain, because this says that
31:27:990Paolo Guiotto: Bt1, it is Gaussian means zero, variance T1, so this is the phi of BT1.
31:37:830Paolo Guiotto: C.
31:39:20Paolo Guiotto: And the increment BT2 minus BT1 is Gaussian, mean zero variance T2 minus T1. So this is P of BT2 minus BT1.
31:50:220Paolo Guiotto: of IPTA.
31:51:630Paolo Guiotto: We can repeat… no, it's not yet time.
31:57:170Paolo Guiotto: Yes, but it's not yet done, it's standard. Okay. So this, shows that the increments are independent,
32:08:40Paolo Guiotto: Finally, about the continuity, well, this is evident, because B of T omega is just TW1 over T omega. We know that the Brownian motion trajectories are continuous from 0 plus infinity
32:27:980Paolo Guiotto: with the probability 1, okay?
32:31:840Paolo Guiotto: So…
32:33:420Paolo Guiotto: this is a continuous function for every T positive, not for t equals 0, where it's not defined, and I multiply by this, which is, again, a continuous function, so I get continuity.
32:46:550Paolo Guiotto: The second question of this… is check that B of T
32:52:950Paolo Guiotto: goes in probability to zero when t goes to 0 plus. So it's a sort of weak form of the continuity. In fact, it would be possible to prove that it goes to zero most surely, so we have also the continuity at zero. But that's much more complicated.
33:12:650Paolo Guiotto: To prove this, we have to prove that the probability that modulus B of T is greater or equal than epsilon, this thing goes to 0 when t goes to zero positive.
33:26:830Paolo Guiotto: But this can be, estimated, because B of T, we remind that we know that this is normal, means zero variance T.
33:35:320Paolo Guiotto: So, we can use the distribution. This is the probability, that modulus B of T is greater or equal than epsilon.
33:46:510Paolo Guiotto: Well, it is the integral, since the density of this variable is e to minus X squared over 2t divided the root of 2 pi…
34:03:710Paolo Guiotto: It's not yet done.
34:06:380Paolo Guiotto: Thanks.
34:07:650Paolo Guiotto: I'm supposed to sound perfect.
34:09:00Paolo Guiotto: Yeah, but it's not yet, but this is another class.
34:12:510Paolo Guiotto: Okay.
34:17:500Paolo Guiotto: So, so we should integrate on the variable modulus of X greater than epsilon.
34:27:320Paolo Guiotto: So, this is the probability. So, by symmetry, this is 2 times the integral from epsilon to plus infinity E minus X squared over 2t divided the root of 2 pi t dx.
34:42:270Paolo Guiotto: And now we have to check that when T goes to 0, this quantity goes to, 0.
34:49:40Paolo Guiotto: Well, doing the limiting directly, it would be a problem, because T going to 0, this denominator goes to 0. Well, clearly, it is zero positive, so this would make the exponential, having the exponent going to minus infinity, so very likely it goes to zero, but…
35:06:210Paolo Guiotto: We could transform this by a simple change of variables, so to eliminate T in the integration variable. So let's set U equal U over the X over the root of t.
35:17:250Paolo Guiotto: So this means that this becomes 2, the integral. In the integration, this becomes 2 u squared over 2, and the scaling factor is just 2 pi, because this is absorbed by the change of variables, so this is the u.
35:31:30Paolo Guiotto: And about the range, since X goes from epsilon to plus infinity, this we got from epsilon over root of T,
35:39:910Paolo Guiotto: To class infinity.
35:43:860Paolo Guiotto: But then, now it is clear what happens, because when T goes to 0, positive.
35:48:660Paolo Guiotto: This quantity here goes to plus infinity, and clearly the integral will go to zero.
35:55:630Paolo Guiotto: understands this exercise.
36:01:120Paolo Guiotto: Okay, I would stop here. In any case, I will publish the solution of all the other exercises, even included those that we have not seen.
36:10:740Paolo Guiotto: And that's it, okay?
36:16:70Paolo Guiotto: So let's stop the recording.