November 27 registration
Aggregazione dei criteri
Assistente AI
Trascrizione
00:00:910Annalisa Cesaroni: I registered soon.
00:03:400Annalisa Cesaroni: Registration course, perfect. Medium se viro.
00:08:480Annalisa Cesaroni: Okay.
00:11:60Annalisa Cesaroni: Okay, so let's start.
00:14:640Annalisa Cesaroni: Si. Charto, charto. Si.
00:22:880Annalisa Cesaroni: Pericio infinito.
00:37:360Annalisa Cesaroni: Una sequenca prende una sequenca que aun limit.
00:41:820Annalisa Cesaroni: e a la anne.
00:58:270Annalisa Cesaroni: quindir.
01:12:780Annalisa Cesaroni: ci.
01:29:260Annalisa Cesaroni: Okay, so there are very limited affinity, but… C, C, see, C, per que…
01:33:710Annalisa Cesaroni: Oviamente, se una sequencer val infinito, ou dire que como unco… no? Seo una sequences, va a un limite finito, do elementi diversi de la secuenza maano che lindic sono molto grandi, anche con organ diversi, comun diverteramo semprevivecini a limite, no?
01:58:630Annalisa Cesaroni: We need a lowerropyovicini.
02:00:640Annalisa Cesaroni: quindi si in general,
02:10:370Annalisa Cesaroni: No, in general, debe a ver un limited finito, al trimenti, non sequenza di cosi. E han que se ha ordini diver si, oviamente qualifier ano ciao obviamente, se prendo una succacione que tende a un numero finito ele.
02:23:370Annalisa Cesaroni: echarto.
02:39:260Annalisa Cesaroni: Did I… Oh, shit.
02:43:610Annalisa Cesaroni: No, no, le… ostrito suche unifini as a limitina.
02:51:280Annalisa Cesaroni: Si, les do vacuum solestione di cosi.
03:07:370Annalisa Cesaroni: Hmm…
03:14:770Annalisa Cesaroni: But… because she, like I said, desocci.
03:28:680Annalisa Cesaroni: For no elimiter.
03:30:830Annalisa Cesaroni: considerande limited celavra, e punta questo. Pero probabe pou a ser que oste aguarda, nuna succesionetro, por… questes empeo.
03:52:720Annalisa Cesaroni: let's switch to English. For example, preniamo, so let's take the space of continuous function from 01, continuous function.
04:03:630Annalisa Cesaroni: So, function from 01 to R.
04:07:180Annalisa Cesaroni: F continues in 01.
04:10:830Annalisa Cesaroni: And, let's put this distance, the distance between F and G.
04:15:980Annalisa Cesaroni: Let's put the norm, this norm.
04:19:269Annalisa Cesaroni: F1, which is the integral between 0 and 1, of absolute value of F, the integral of the absolute value of F.
04:27:920Annalisa Cesaroni: this can be… can be proved that it is a norm. Why it is a norm? It is a norm because F1 equal to 0 if and only if F is equal to 0, because here I'm taking the absolute value of the integral, no? So, the integral
04:46:40Annalisa Cesaroni: between 0 and 1 of F of T, modulus of f of t in dt is equal to 0, if and only if f is equal to 0, because I'm taking the absolute wave. And moreover, lambda F is equal to modulus of lambda times norm of f, because I'm just taking the lambda outside the modulus.
05:05:370Annalisa Cesaroni: And moreover, F plus G1 less or equal than F1 plus G1.
05:13:250Annalisa Cesaroni: Because it's just the fact that modulus of F of T plus G of T is less or equal than modulus of F of T plus modulus of G of T, something like this, okay?
05:31:50Annalisa Cesaroni: And then, so, let's… so, this is… this is a norm, a norm on the space.
05:42:650Annalisa Cesaroni: on the space of a continuous function in 01, this is a linear space, because if I take two continuous functions, the sum is a continuous function, the multiplication by a constant is a continuous function, and that's all, no? It's a linear space. And this is a norm, so it induces a distance.
06:00:590Annalisa Cesaroni: between space… between, elements. F minus G,
06:04:610Annalisa Cesaroni: the distance between F and G, let's call it D1. D1FG is equal to the norm of F minus G.
06:15:910Annalisa Cesaroni: And, here we can… it is simple to construct a sequence which are Cauchy with respect to this distance, but which are not… which are not converging to any function in…
06:28:980Annalisa Cesaroni: the space. They are converging in a space which is a bit bigger, which is the space of the function which has bounded integral.
06:38:340Annalisa Cesaroni: So, at the end, for example, let's take a sequence FN of T, done with, like, like this. For example, between 0 and 1 half.
06:53:970Annalisa Cesaroni: It is, equal to 0, and then equal to 1, dunno.
07:02:990Annalisa Cesaroni: Okay, so Fn of t is equal to 0 for t less than 1F, then it is,
07:12:850Annalisa Cesaroni: 1 for T bigger than 1F plus 1 over n.
07:18:20Annalisa Cesaroni: And N is, what it is, N1T minus 1F, No, 1 over N?
07:30:970Annalisa Cesaroni: It is, what, T-1F, and then in,
07:39:920Annalisa Cesaroni: T equal to, 1F plus 1 over n, it has to be… so N was correct.
07:49:730Annalisa Cesaroni: T between 1F and 1 half plus 1 over n. It's continuous, this one, no?
07:56:00Annalisa Cesaroni: It's continuous, and actually, If I take FN, the distance between FN and FM, And… and then…
08:07:750Annalisa Cesaroni: Natural number, it is what?
08:11:380Annalisa Cesaroni: It is the integral between 0 and 1 of Fn of t minus FM .
08:17:740Annalisa Cesaroni: In the team.
08:19:330Annalisa Cesaroni: So, let's take, some… just to fix the idea, let's take M bigger than N, just to fix the idea. M bigger than N. So, 1 over N
08:30:440Annalisa Cesaroni: 1 over M is less than 1 over N, no?
08:34:620Annalisa Cesaroni: So… This is FN, and FM will be something like this.
08:41:130Annalisa Cesaroni: Hmm.
08:49:300Annalisa Cesaroni: Okay.
08:51:00Annalisa Cesaroni: It's becoming one, so maybe it's better to make the picture in the other.
08:57:360Annalisa Cesaroni: So it is, 0, 1… One of…
09:02:710Annalisa Cesaroni: 1F plus 1 over n, and then here I have 1F plus 1 over M, M bigger than n.
09:10:660Annalisa Cesaroni: Then I have that. The first one… FN is like this.
09:18:760Annalisa Cesaroni: And the second is like this.
09:24:590Annalisa Cesaroni: Okay.
09:31:700Annalisa Cesaroni: So, if I compute the distance between FN and FM,
09:37:760Annalisa Cesaroni: It is just the integral between 0 and 1 of Fn of t minus FM .
09:44:340Annalisa Cesaroni: And it is what?
09:46:510Annalisa Cesaroni: I have just to make the… the difference of the…
09:50:630Annalisa Cesaroni: So, here, the two areas are the same. So…
09:54:990Annalisa Cesaroni: from 1F plus 1 over n, they are exactly the same, they are 1, 1 minus 1. So it is just the integral between 1F and 1 half plus 1 over n, no? Of Fn of t minus FM of t.
10:10:400Annalisa Cesaroni: And then, one is just to… to compute everything, but, okay, it's something…
10:19:280Annalisa Cesaroni: going as what? Going as this, this quantity here. I have to make the difference between the two areas of the two-part, no? So, it will be something like, say.
10:32:440Annalisa Cesaroni: The… the area of the… all this red part will be what?
10:38:940Annalisa Cesaroni: We'll be… mmm… 1 over N minus 1 over M.
10:44:530Annalisa Cesaroni: times 1… Which is the area of this rectangular, plus, 1 half.
10:53:680Annalisa Cesaroni: The area of this triangular, which is 1 of 1 over M times 1, Minus… minus the area of…
11:06:610Annalisa Cesaroni: So, this is the, this sum is the area of the green part, no? Because it's just,
11:15:220Annalisa Cesaroni: The area of these… We…
11:18:430Annalisa Cesaroni: which is a rectangular of bases 1 over n minus 1 over M, and 8 one. And this one is the area of this triangle.
11:29:890Annalisa Cesaroni: which is 81 and basis 1 over M. So, 1 over M times 1 times 1F. And then, I have to eliminate the area
11:41:460Annalisa Cesaroni: Of this triangular.
11:43:890Annalisa Cesaroni: Which is what? Minus… it's a triangular of base 1 over n, so 1 half times 1 over n times 1.
11:53:750Annalisa Cesaroni: Something like this.
11:57:960Annalisa Cesaroni: Okay, so it is something like one-off…
12:02:00Annalisa Cesaroni: If I make all the computation, it is 1F1 over M minus 1 over M, something like this.
12:09:580Annalisa Cesaroni: The difference, because this minus this is one…
12:14:20Annalisa Cesaroni: Which is going to zero. So this sequence is a Cauchy sequence.
12:19:800Annalisa Cesaroni: Okay.
12:22:110Annalisa Cesaroni: It's just a silly computation, doesn't matter if… Okay, so FN is a Cauchy sequence.
12:29:760Annalisa Cesaroni: Mmm… but… It has not a limit in the…
12:35:140Annalisa Cesaroni: in the space of continuous function, because the limit of CN will be what? CN is done like this, 1F, and then like this. And as n goes to 0,
12:49:620Annalisa Cesaroni: then this quantity, this length goes to 0. So, at the end, FN is going where? It's going to the function which is 0 between 0 and 1 half, and 1 between 1 half and 1.
13:04:100Annalisa Cesaroni: Which is not a continuous function, okay?
13:08:550Annalisa Cesaroni: So, the limit of FN is something made like this. Zero between.
13:18:640Annalisa Cesaroni: And then Juan.
13:21:950Annalisa Cesaroni: Okay, this is… which is not continuous.
13:29:210Annalisa Cesaroni: But it is an element of… but this F is not continuous, but in any case, the integral between 0 and 1 of f of dt is finite.
13:42:250Annalisa Cesaroni: So… Okay, so actually, if I take this distance associated with this norm,
13:54:730Annalisa Cesaroni: C01 with this norm is not Bannock.
14:01:00Annalisa Cesaroni: It's not Barak, because, it's… it is not complete.
14:07:100Annalisa Cesaroni: But…
14:08:570Annalisa Cesaroni: It is not complete, because there are cachi sequences, which are converged into functions, which are not continuous, okay?
14:15:920Annalisa Cesaroni: But what is… the space which is continuous is what is called L101.
14:23:280Annalisa Cesaroni: L1, 01 are the function F from 01.
14:28:170Annalisa Cesaroni: to R, such that the integral is in 01 of f of t.
14:32:900Annalisa Cesaroni: Litti is finite.
14:36:20Annalisa Cesaroni: Which is bigger, which contains the space of continuous function.
14:41:710Annalisa Cesaroni: Okay, because a continuous function is finite integral, but there are also discontinuous functions, which are finite integral, okay?
14:50:150Annalisa Cesaroni: So, in principle, when I have a space where Cauchy sequences are not converging.
14:56:280Annalisa Cesaroni: there shouldn't exist some bigger space, which contains the space where I started.
15:02:460Annalisa Cesaroni: And which contains the limit, which is complete, and which contains the limit, okay?
15:08:490Annalisa Cesaroni: as for, what we consider Q with the Euclidean distance.
15:13:700Annalisa Cesaroni: what we… we realized that all… not all the Cauchy sequences in Q are converging, and so we enlarge the space, constructing R as the completion of Q. In the sense, we are adding two
15:27:360Annalisa Cesaroni: Q all the limit of the Cauchy sequences. Also here, we take the space of continuous function with this norm, which is an integral norm, and we
15:38:80Annalisa Cesaroni: add to this space all the limit of Cauchy sequence, and we end up with something which is like this. So, actually.
15:46:440Annalisa Cesaroni: the limit of every Cauchy sequence of a space of a Cauchy sequence of continuous function, a Cauchy sequence with respect to the L1 norm.
15:56:600Annalisa Cesaroni: Today, one arm is a function which satisfy this.
16:01:620Annalisa Cesaroni: So we, in some sense, once… once that we have a space and a norm on the space, or a metric, or a distance in the space, we can complete the space, adding to the space, adding all the limit of possible Cauchy sequence, okay?
16:22:200Annalisa Cesaroni: Okay.
16:24:880Annalisa Cesaroni: Now, we come back to our space of random variable with finite moment, because I want to…
16:31:640Annalisa Cesaroni: to say something about that, and to say that actually that spaces are, Banak, we can define. So, let's fix, once forever,
16:44:180Annalisa Cesaroni: No, bef… Okay. Now, before doing this, I want just to,
16:52:230Annalisa Cesaroni: to recall the Jensen inequality, that maybe you already… for sure you already know.
16:59:90Annalisa Cesaroni: Jensen inequality, that will be useful.
17:02:770Annalisa Cesaroni: The Jensen inequality in R, but it can be defined also in space bigger than R,
17:09:190Annalisa Cesaroni: So, let's take a function f from R to R convex.
17:16:99Annalisa Cesaroni: a convex… a convex function. What does it mean, convex? It means, in particular, that
17:23:200Annalisa Cesaroni: If I take two elements, F of… and I take the… Say, X1, X2, And they take,
17:35:510Annalisa Cesaroni: the, the…
17:37:810Annalisa Cesaroni: If I take two points, X1 and X2 in R, I take F of X1 and F of X2. I con- I consider the segment in R,
17:50:460Annalisa Cesaroni: Connecting X1, F of X1, and X2… F of X2, this segment is always before… above the graph of F. So, F convex means
18:07:540Annalisa Cesaroni: means that for every X1, X2 in R.
18:12:640Annalisa Cesaroni: the segment in R2.
18:17:420Annalisa Cesaroni: Connecting the two points.
18:20:710Annalisa Cesaroni: X1, F of X1.
18:24:140Annalisa Cesaroni: X2, F of X2.
18:26:850Annalisa Cesaroni: is above it.
18:29:160Annalisa Cesaroni: the graph.
18:33:890Annalisa Cesaroni: the graph of F. The curve graph of F, okay.
18:38:900Annalisa Cesaroni: This is the graph of F.
18:43:690Annalisa Cesaroni: This is the segment.
18:46:800Annalisa Cesaroni: Maybe the segment is exactly the graph of F. In this case, the function f is linear, okay? So it's convex in a degenerate way. Concave means the contrary, no? The segment. And, if f is,
19:03:210Annalisa Cesaroni: if F is, differentiable twice, twice differentiable, this means that actually, the…
19:14:150Annalisa Cesaroni: the second derivative of F is greater or equal than zero. Why? Because convexity means that… The, the…
19:30:220Annalisa Cesaroni: the slope of the tangents to the graph of F,
19:35:210Annalisa Cesaroni: Okay, which the slope of the tangent line to the graph of f in every point is just the derivative of the function at the point. If the slope of the tangents are increasing.
19:47:540Annalisa Cesaroni: then the function is convex, okay? So, what does it mean that the slope of the tangents are increasing, are going…
19:56:650Annalisa Cesaroni: Becoming more and more positive, for example, means that the first derivative is increasing.
20:04:100Annalisa Cesaroni: Which means that the second derivative is positive, okay?
20:07:500Annalisa Cesaroni: So, F sec on greater or equal than 0. And another characterization of convex function is the following.
20:14:950Annalisa Cesaroni: which will be, useful. For every X, one, Inar?
20:23:20Annalisa Cesaroni: If F' of X1 exists, It's also true NF prime.
20:31:590Annalisa Cesaroni: For, let's, for… let's write like this. For every exam, there exists P in R, P1 in R,
20:40:830Annalisa Cesaroni: P1 will be F' of X1 if it exists, otherwise it's called sub-differential, such that f of x is bigger or equal of F of X1 plus P1 x minus X1 for every X in R.
20:59:80Annalisa Cesaroni: Let's see what does it mean.
21:02:860Annalisa Cesaroni: It means the following.
21:05:720Annalisa Cesaroni: So, for every X1 fixed, Here, everything is fixed. Here, the X is this one.
21:13:180Annalisa Cesaroni: So I'm saying what? I'm saying that… let's… let's fix the point X one here.
21:21:460Annalisa Cesaroni: For example… I'm saying that there exists a P1, a number, okay?
21:31:200Annalisa Cesaroni: In RN, it will be a vector, but okay.
21:36:30Annalisa Cesaroni: a number such that if I take what it is.
21:40:970Annalisa Cesaroni: This is the… what is the graph of this function in X? The graph of this function in S is a line, straight line, passing through
21:53:200Annalisa Cesaroni: X1, F of X1.
21:58:90Annalisa Cesaroni: We slow puppy one.
22:00:00Annalisa Cesaroni: Okay.
22:01:590Annalisa Cesaroni: So, this is Y equal to F of X1 plus P1X minus X1.
22:09:750Annalisa Cesaroni: So, the tangent, the tangent straight line, if, if, F, F prime, F prime in, of X1 exists, okay?
22:19:800Annalisa Cesaroni: And I'm saying that… I'm saying that, actually, the graph of F is all above this… Straight line. Okay.
22:34:410Annalisa Cesaroni: So I'm saying that F of X is always above
22:39:560Annalisa Cesaroni: This… okay, geometrically. So, actually, when I have a tangent line to a convex function, then the tangent
22:48:820Annalisa Cesaroni: is such that the function is all on one side of the, two, hyperplane, two, two semi… semiplane in which the line divides the space, the, the, the plane. Okay.
23:08:840Annalisa Cesaroni: And, actually, this can be…
23:14:350Annalisa Cesaroni: can be written also for F, convex, but not regular. Okay, maybe P1 will be not unique, say, for example, let's consider this case.
23:25:710Annalisa Cesaroni: the case in which F of X is equal to modulus of x. Actually, there are a lot of…
23:32:770Annalisa Cesaroni: P of 0 can be written whatever, for example, 0. I have that, F of X,
23:39:680Annalisa Cesaroni: is bigger or equal than f of 0 plus 0 times X minus 0, for example, and also F of X is bigger than f of 0 plus 1 times X minus 0, and also f of x is bigger than f of 0 minus 1 x minus 0.
23:57:890Annalisa Cesaroni: Okay.
23:59:250Annalisa Cesaroni: Always… all these lines are… so, here I can take these lines, these lines, these lines, and all the lines below.
24:09:730Annalisa Cesaroni: All these lines make the same,
24:12:840Annalisa Cesaroni: Our own lines, which pass through 0, F of 0, and which are
24:17:890Annalisa Cesaroni: below the function everywhere, okay? So, for a convex function, it is always true that at every point of the graph, there exists at least one line.
24:30:400Annalisa Cesaroni: One straight line, such that the graph of the function is just on one part.
24:36:530Annalisa Cesaroni: It is not always the case for functions which are not convex, because if I have a function like this, and I take here the, for example, the…
24:48:150Annalisa Cesaroni: The tangent lines, sometimes the function will cross the tangent line, no, somewhere else.
24:57:530Annalisa Cesaroni: So, this is a peculiarity, a characterization, another characterization of convex function. Okay, and this is the characterization which is needed to prove the, which is used to prove the so-called Jensen inequality.
25:14:710Annalisa Cesaroni: Also, before going to Jensen inequality, also in dimension bigger than 1, it is all… it is true. If I have F from Rn to R convex.
25:25:240Annalisa Cesaroni: Then, for every X1 in Rn, there exists P1 in Rn, such that f of x is bigger or equal than F ,
25:35:330Annalisa Cesaroni: plus P1 scalar product X minus X1.
25:40:850Annalisa Cesaroni: This is the scalar product, scalar product in a… in a renn.
25:48:80Annalisa Cesaroni: Exactly the same. But in this case, in this case, what I have, I have that here I have RN,
25:57:60Annalisa Cesaroni: The graph of a function from Rn to R is an hypersurface in Rn plus 1,
26:04:290Annalisa Cesaroni: And this is an hyperplane in Rn plus 1, okay?
26:10:80Annalisa Cesaroni: So I'm saying that there is just an hyperplane.
26:13:820Annalisa Cesaroni: Staying below the graph of F at every point of the graph of F.
26:21:40Annalisa Cesaroni: Touching from below.
26:23:240Annalisa Cesaroni: And, this hyperplane, or the…
26:27:550Annalisa Cesaroni: the line as before, is called the support hyperplane. At every point, I have a support hyperplane, because it… it is like… like if…
26:37:310Annalisa Cesaroni: The, the function, the graph of the function is… Ehh…
26:42:890Annalisa Cesaroni: supported by the zipper plane. So, we have the upper plane, and above, I have the graph of the function.
26:49:290Annalisa Cesaroni: And this, these are all set every point, okay?
26:59:30Annalisa Cesaroni: Okay, so, let's,
27:04:870Annalisa Cesaroni: Actually, this property can be stated also for functions which are just continuous, not differentiable, okay? Because, in general, one says there exists P1, obviously, if F is differentiable, P1 will be the gradient of F at X1.
27:22:930Annalisa Cesaroni: So, the, the, the vector, which are, which has, as,
27:30:660Annalisa Cesaroni: elements as coordinates, the derivative in the i direction of the function f with respect… computed in X1. But if f is not differentiable.
27:42:290Annalisa Cesaroni: In general, the P1 is a vector, okay? If F is not differentiable, actually, one can expect that there are more than one P1 doing the same job, but okay.
27:54:250Annalisa Cesaroni: But at least there exists one, always one.
27:57:650Annalisa Cesaroni: Okay.
27:59:20Annalisa Cesaroni: And indeed, this P1 is called sub-differential, the sense that it is differential, defined everywhere, okay? Where the function is differentiable is the differential. Where the function is not differentiable, it's a sort of,
28:18:640Annalisa Cesaroni: Slope of every possible tangent line staying below the function, see?
28:25:410Annalisa Cesaroni: as here, as for the modulus, yeah, I have all possible.
28:31:280Annalisa Cesaroni: Okay, so Jensen inequality, Jensen inequality is the following.
28:37:490Annalisa Cesaroni: It says that, let F from R to R convex, Let's take mu.
28:44:430Annalisa Cesaroni: A probability, measure, so mu.
28:48:470Annalisa Cesaroni: A probability measure. What does it mean, a measure?
28:51:960Annalisa Cesaroni: Probability measure. So, it means that there is a boreal… it is a boreal measure with the measure of all the space equal to 1.
29:03:50Annalisa Cesaroni: So it is the, it is associated… it is the law of some random variable defining some…
29:11:290Annalisa Cesaroni: probability space, and I have the following. I have that,
29:17:690Annalisa Cesaroni: the integral over r of f of x D mu of X, huh?
29:24:630Annalisa Cesaroni: diminu, say, big is bigger or equal, always, of F computed at integral of r of X d mu.
29:38:120Annalisa Cesaroni: So, it is…
29:42:760Annalisa Cesaroni: Like this.
29:51:920Annalisa Cesaroni: And, mmm…
29:59:370Annalisa Cesaroni: If, let's… if X d mu of r is less than plus infinity.
30:05:580Annalisa Cesaroni: If, let's write like this, X d mu is equal to plus infinity, then integral over r of f of x d mu is plus infinity also.
30:16:480Annalisa Cesaroni: Let's right-click this. Okay.
30:19:300Annalisa Cesaroni: Obviously, here I'm computing F at some point, so it has… it has a meaning just if the point is a point in R, so it's an element, it's a number, okay?
30:36:190Annalisa Cesaroni: Okay, let's, let's start with… so we have A and B, two parts.
30:42:410Annalisa Cesaroni: If the integral of X d mu is less than plus infinity, then I have that the integral over r of f of x d mu is bigger or equal than F computed at
30:54:30Annalisa Cesaroni: this quantity.
30:56:90Annalisa Cesaroni: This is a number, right? Fixed number. I'm just taking X and integrating with respect to mu. I know as to how to integrate function with respect.
31:07:00Annalisa Cesaroni: And also, this is a number, okay.
31:10:370Annalisa Cesaroni: So, we…
31:12:680Annalisa Cesaroni: Let's prove B first. B is trivial, because… let's take, for example, in 0, that first, there exists P0 by convexity, such that f of x is bigger or equal than f of 0 plus speed 0 times X minus 0.
31:30:610Annalisa Cesaroni: Okay, there exists something like this, and then I integrate everything by mu, and I have integral.
31:38:20Annalisa Cesaroni: over r of f of x. D mu bigger or equal than integral over r of f of 0.
31:46:260Annalisa Cesaroni: d mu plus integral over r of P0X d mu.
31:55:410Annalisa Cesaroni: So, this is a number. F of 0 is just F of 0.
32:06:610Annalisa Cesaroni: This is a number, and then I have P0 integral over R of X d mu.
32:11:890Annalisa Cesaroni: So I have that. This is bigger or equal than this.
32:20:450Annalisa Cesaroni: Okay, because the F of 0 is a constant, a number.
32:27:800Annalisa Cesaroni: And, so, actually, what I have, I have that if this is class infinity, then also this is class infinity.
32:36:630Annalisa Cesaroni: This is just this, no? Because this is bigger or equal, okay? If P0 is equal to 0, maybe, and, I change the point, so, I put here, 0, but I can rewrite everything for…
32:56:350Annalisa Cesaroni: But I…
32:59:170Annalisa Cesaroni: Assuming that bit 0 is different from 0. If pit 0 is not different from 0, we take another point instead of 0. X1, such that the P1 is different from 0, there will be at least 1, otherwise F is just the linear function, okay?
33:16:190Annalisa Cesaroni: the constant function.
33:18:260Annalisa Cesaroni: Okay, so B is trivial, so if the integral over X of d mu is plus infinity, then also the integral of F in d mu is plus infinity. So…
33:32:10Annalisa Cesaroni: Okay, I'm seeing… That if P0 is equal to 0, we change point… See?
33:41:10Annalisa Cesaroni: Take X1, in R, such that P1 is different from 0.
33:48:990Annalisa Cesaroni: Such that f of x is bigger or equal than fx1 plus P1X minus X1, and they have the same, okay?
33:57:650Annalisa Cesaroni: So I have F of X d mu.
34:01:200Annalisa Cesaroni: greater or equal f of x1 d mu.
34:04:920Annalisa Cesaroni: plus P1 integral over R, P1X minus integral d mu.
34:11:260Annalisa Cesaroni: P1, X1, D mu.
34:14:330Annalisa Cesaroni: which is equal to F of X1
34:17:460Annalisa Cesaroni: Because the integral of d mu is equal to 1, plus P1 integral over r of X d mu minus P1X1, because this is a constant, and so I'm not integrating.
34:29:830Annalisa Cesaroni: Okay, so… at least one of the P, will be different from zero, so if it is…
34:37:449Annalisa Cesaroni: If it is going to plus… if it is plus infinity, then it is plus infini. Okay.
34:45:870Annalisa Cesaroni: Mmm…
34:55:190Annalisa Cesaroni: P1 bigger or equal than zero, otherwise.
34:58:760Annalisa Cesaroni: Okay, so let's,
35:01:160Annalisa Cesaroni: In any case, we can rule out this possibility. So, if the integral over X in the mu is plus infinity, also, the integral of F of X d mu is plus infinity. No, no problem with this. So, let's assume… assume that integral over…
35:18:450Annalisa Cesaroni: R of f of x, of X d mu is finite.
35:28:330Annalisa Cesaroni: It's finite. So, what we do? We do like this, we…
35:33:890Annalisa Cesaroni: We just… we… let's call it, I don't know, let's call it XCONZERO.
35:41:50Annalisa Cesaroni: Let's call this quantity X, the mu.
35:45:260Annalisa Cesaroni: X0 is a number.
35:47:350Annalisa Cesaroni: Okay.
35:48:830Annalisa Cesaroni: is a number.
35:51:850Annalisa Cesaroni: If you want, we can change the name of the…
35:57:410Annalisa Cesaroni: the variable of integration, which is mute, okay? It's the same, just not to make confusion with the X, which is a variable. It's a number. I'm integrating. Once that I integrate, it is a number. And so, I apply with this X0,
36:15:300Annalisa Cesaroni: the, convexity argument. So, F of X is bigger or equal than f of x0 plus, pit 0, some pit 0, x minus X0.
36:28:10Annalisa Cesaroni: P of X0, say, X minus at 0. This is true for every X.
36:36:100Annalisa Cesaroni: And then, I… I integrate with respect to mu, this inequality.
36:42:740Annalisa Cesaroni: So I integrate with respect to mu. I have f of x.
36:46:760Annalisa Cesaroni: bigger or equal than f of x0, plus P0 times P of X0, sorry, x minus X0, I integrate in r with respect to mu.
36:58:820Annalisa Cesaroni: mu integrating in X.
37:02:880Annalisa Cesaroni: Yes.
37:08:200Annalisa Cesaroni: Just the same, things that I rewrote here, no? This inequality, and then I iterate.
37:22:570Annalisa Cesaroni: And then it is what? This is just the integral over r of f of x diminu of X, bigger or equal than…
37:31:730Annalisa Cesaroni: What? This is a constant, f of x0.
37:36:580Annalisa Cesaroni: plus, this is what? Integral over R of PX0.
37:42:290Annalisa Cesaroni: X d mu of X.
37:45:410Annalisa Cesaroni: P of X0 can be put outside, because it's just, a number.
37:54:370Annalisa Cesaroni: minus X0.
37:59:360Annalisa Cesaroni: Okay.
38:04:270Annalisa Cesaroni: Now, I just remember what is X0. X0 is just…
38:09:130Annalisa Cesaroni: The integral of mu in the mu.
38:12:810Annalisa Cesaroni: of Y in the U of Y, of R.
38:17:250Annalisa Cesaroni: So it is integral over r of f of x.
38:20:700Annalisa Cesaroni: The mu of X bigger or equal than F, integral over R, Y diminu of y, or X as you want, the integral in the plus bit 0 of what? Integral over X of the mu of X over R,
38:37:110Annalisa Cesaroni: minus integral over R of Y diminu of y. This is actually the same thing.
38:44:110Annalisa Cesaroni: So it is zero.
38:45:870Annalisa Cesaroni: And so, we end up with the Jensen inequality.
38:50:980Annalisa Cesaroni: It's just the Jensen inequality, no?
38:55:290Annalisa Cesaroni: So it means that if random variable X has a defined expected value, every function convex that is convex or concave will have a defined expected value? Yes, sure, sure, yes.
39:12:190Annalisa Cesaroni: If X is in M1 random variable.
39:16:870Annalisa Cesaroni: F of X is the number 1, yeah.
39:21:340Annalisa Cesaroni: And, it's, but not, not,
39:26:890Annalisa Cesaroni: No, sorry, it is not in this way. If f of x is in M1, X is in M1. So, I'm saying the other way around, so I'm saying that
39:37:590Annalisa Cesaroni: If X is plus… X d mu is plus infinity, F of X d mu is plus infinity, and if this is finite, this is bigger than that, but maybe it's not,
39:49:430Annalisa Cesaroni: Yeah, it is defined, infinity, or… Yes, it's defined.
39:53:720Annalisa Cesaroni: At least define.
39:56:310Annalisa Cesaroni: Yes, indeed, we are going to use these in order to prove that, actually, if a random variable has bounded P moment, has bounded Q moment for every cube less than P.
40:10:390Annalisa Cesaroni: So…
40:11:570Annalisa Cesaroni: So now we define this… we fix a probability space. We define every… we fix a probability space, we take all the possible random variable on this probability space with values in R,
40:25:410Annalisa Cesaroni: So, we fix omega FP probability space.
40:35:800Annalisa Cesaroni: And, we defined M, MP,
40:39:640Annalisa Cesaroni: The space of random variable from omega to R, Random variable.
40:49:250Annalisa Cesaroni: Such that the moment is finite, the p moment is finite. So, the expected value of X to the power p.
40:58:940Annalisa Cesaroni: which is the integral over R of…
41:02:420Annalisa Cesaroni: X to the power p, the L of X, X.
41:06:580Annalisa Cesaroni: Is Vinet.
41:11:360Annalisa Cesaroni: And this can be defined for every P bigger or equal than 1, no?
41:39:350Annalisa Cesaroni: Then, I have the following. If…
41:43:370Annalisa Cesaroni: say, corollary of Jensen, corollary of Jensen inequality.
41:54:950Annalisa Cesaroni: If X is in MP, For some P bigger than 1, then…
42:02:570Annalisa Cesaroni: X is in MQ for every queue between 1 and P.
42:09:790Annalisa Cesaroni: So…
42:11:190Annalisa Cesaroni: If expected value of X to the power p is finite, then expected value of X to the power Q is finite for every Q.
42:22:820Annalisa Cesaroni: Between 1 and P.
42:25:400Annalisa Cesaroni: So, all the moment, okay.
42:38:630Annalisa Cesaroni: So, all these spaces are in…
42:44:20Annalisa Cesaroni: And it's catalated one on the other. I have M1, contain M2, contain M3, contain M4, etc.
42:52:840Annalisa Cesaroni: So, M1 contains M2, because
42:56:410Annalisa Cesaroni: I'm saying, if I take P equal to 2 here, I have that, if a random variable has a finite variance, then it has also finite mean.
43:08:320Annalisa Cesaroni: Okay.
43:09:460Annalisa Cesaroni: If a random variable is fine, so if a function is a random variable is in M2, then it is also in M1.
43:17:710Annalisa Cesaroni: Not whichever so. Maybe I have finite mean, but not finite variance, okay? But if I have finite variance, for sure I have finite mean. If I have finite third moment, I have finite second moment, finite first moment, and so on. So…
43:33:270Annalisa Cesaroni: they are drinking in some sense. M1 is the biggest. Biggest. The biggest is the space of all possible random variable, okay?
43:42:980Annalisa Cesaroni: Let's prove this theorem, corollary. Just a matter of writing… Down the, the…
43:50:420Annalisa Cesaroni: using the random… the Jensen inequality, where mu will be the low of X, okay?
43:59:50Annalisa Cesaroni: Low of X is in the probability space of R, no?
44:04:400Annalisa Cesaroni: We know this.
44:10:60Annalisa Cesaroni: the space of a probability of R.
44:13:500Annalisa Cesaroni: Or probability measure, say over. And, so let's prove it. Proof, take P bigger than 1.
44:21:590Annalisa Cesaroni: take every… let Q less than P and be greater or equal than 1.
44:28:650Annalisa Cesaroni: And,
44:36:780Annalisa Cesaroni: And take this function f of x, like this, x to the power p over Q.
44:47:850Annalisa Cesaroni: P overcome.
45:01:820Annalisa Cesaroni: Okay.
45:05:20Annalisa Cesaroni: And you see that P over Q is bigger than 1, okay? Because… and so this F is convex.
45:14:630Annalisa Cesaroni: F is convex. If P over Q is bigger than… bigger or equal than 2 is convex, smooth also. Otherwise, it's,
45:23:560Annalisa Cesaroni: can be… can have a… a problem in 0 of differential B… of second derivative, okay? Because F prime of X is what? It's P over Q, X, P minus… P over Q minus 1.
45:39:750Annalisa Cesaroni: X over… modulus of X, something like this, so it is, when I…
45:48:270Annalisa Cesaroni: When I have to compute,
45:51:150Annalisa Cesaroni: Okay, the second derivative… can I… I can have a problem in X equal to zero? Okay, but in any case, it's convex, okay?
46:01:190Annalisa Cesaroni: is convex, and then we can apply the, we can apply the Jensen inequality.
46:08:730Annalisa Cesaroni: what does it mean? What does it say, the Jensen inequality? It says that the integral over f of x d mu
46:15:740Annalisa Cesaroni: is bigger of this one, no?
46:21:370Annalisa Cesaroni: Let's,
46:42:730Annalisa Cesaroni: Let's, do… like this.
46:47:520Annalisa Cesaroni: Mmm…
47:02:280Annalisa Cesaroni: Let's make like this. It is a sort of Jensen inequality, not completely the same.
47:10:740Annalisa Cesaroni: I have… so, the expected value of X to the power p is what? Is the integral of X to the power p in the LX of X, no?
47:22:210Annalisa Cesaroni: Overall.
47:23:360Annalisa Cesaroni: Is this one.
47:25:350Annalisa Cesaroni: So, I could… I write this as integral over R of F,
47:33:860Annalisa Cesaroni: computed at X to the power Q.
47:37:520Annalisa Cesaroni: DLXX.
47:45:570Annalisa Cesaroni: It's correct, because F is taking a function, is taking an X, and computing the elevation to the power p over Q. So if I elevate P over Q, something elevated to the P, okay.
48:02:720Annalisa Cesaroni: This is just X to the power Q, elevated to the power P over Q.
48:08:410Annalisa Cesaroni: Which is X to the power P.
48:10:900Annalisa Cesaroni: Okay.
48:13:740Annalisa Cesaroni: Now, I have to use the same idea of the Jensen inequality, so…
48:18:800Annalisa Cesaroni: F is convex. F is convex.
48:23:620Annalisa Cesaroni: So, what I have, I have that F, X to the power Q, Is bigger or equal than…
48:32:790Annalisa Cesaroni: Then, then, for example, de integro.
48:51:470Annalisa Cesaroni: F of… let's call it X con 0.
48:59:180Annalisa Cesaroni: plus P0.
49:03:180Annalisa Cesaroni: X to the power cubed minus X con 0. And I have to choose what is the X con 0.
49:09:480Annalisa Cesaroni: where I want to… let's call it, maybe X from 0 is not, correct. Let's call it, Z, Z0.
49:17:780Annalisa Cesaroni: So…
49:19:30Annalisa Cesaroni: This is true for every X, and for z at 0 fixed, and pit 0 depending on pit 0 is pit 0, depends on Z at 0.
49:34:830Annalisa Cesaroni: So, let's take S for the, the…
49:38:670Annalisa Cesaroni: As for the Jensen inequality, let's take zet at 0 to be integral to X to the QLXX.
49:48:90Annalisa Cesaroni: If it is finite.
49:53:30Annalisa Cesaroni: Let's take this.
49:55:780Annalisa Cesaroni: Maybe with Y instead of X.
50:00:300Annalisa Cesaroni: If it is finite, with wine, just not to make confusion.
50:07:390Annalisa Cesaroni: So… Okay
50:13:370Annalisa Cesaroni: Okay.
50:17:230Annalisa Cesaroni: It is… if it is plus infinity, also this will be plus infinity, but it is not true that it is plus infinity, because
50:27:240Annalisa Cesaroni: I know that it is finest known. And so, actually, I have that F of XQ to the F of XQ, which is X to the power p, is bigger or equal than…
50:39:320Annalisa Cesaroni: F computed at… integral over r of X to the power cube, DLXX, sorry, Y, it was Y.
50:50:880Annalisa Cesaroni: Place… Pit 0,
50:53:940Annalisa Cesaroni: times what? Times x to the power p minus integral over r of y to the power Q.
51:01:10Annalisa Cesaroni: the LXY, something like this.
51:05:160Annalisa Cesaroni: X to the power Q, sorry, not P.
51:08:790Annalisa Cesaroni: Which is this one, no?
51:12:460Annalisa Cesaroni: And then I integrate everything with respect to the… I integrate with respect… integrate with respect to…
51:24:860Annalisa Cesaroni: the law.
51:27:680Annalisa Cesaroni: of X. And I have.
51:30:630Annalisa Cesaroni: Integral over r of x to the power p.
51:34:740Annalisa Cesaroni: the LX of X, so…
51:37:900Annalisa Cesaroni: Which is integral over R of F, X to the power Q, DLX of X.
51:45:360Annalisa Cesaroni: Bigger or equal of what?
51:48:560Annalisa Cesaroni: Of integral over.
51:51:700Annalisa Cesaroni: R of F of integral over R of…
51:59:590Annalisa Cesaroni: Bigger or equal, let's make it go to the other. Bigger or equal than this one, so F…
52:08:200Annalisa Cesaroni: Integral over R of Y to the power Q.
52:11:820Annalisa Cesaroni: the LXY.
52:17:50Annalisa Cesaroni: plus P0 integral over R of
52:21:790Annalisa Cesaroni: X to the power Q, D.
52:24:230Annalisa Cesaroni: DLX.
52:25:650Annalisa Cesaroni: of Y minus integral over r of y to the power Q, DLX of Y And then everything… So…
52:38:350Annalisa Cesaroni: Here I have… this is also integrated with respect to DLX, but it is a constant, okay?
52:52:60Annalisa Cesaroni: the LX of X. Also, this is integrating R with respect to the LX of X.
53:09:570Annalisa Cesaroni: But this is a constant with respect to X, so the integral is just 1. I take this constant outside this, the integral, it is 1. So it is what? Integral over R of Y to the power cube, DLXY to the power what? I have…
53:27:950Annalisa Cesaroni: to compute F, on these… these elements, so it is this element elevated to the power P over Q.
53:37:230Annalisa Cesaroni: Plus… This is exactly equal to this.
53:43:760Annalisa Cesaroni: Okay? And the integral over R of DLX of X is equal to 1, because here there is no X, you know? And so this quantity is 0.
53:54:960Annalisa Cesaroni: Because it… They are the same things.
53:58:660Annalisa Cesaroni: I end up with what? I end up having that the expected value of X to the power p.
54:06:10Annalisa Cesaroni: is bigger or equal than the expected value of X to the power Q, elevated to the power Q over P.
54:15:640Annalisa Cesaroni: So, the expected value of X to the power p is bigger or equal than this, which is the expected value of X to the power Q, and elevated outside to the power p over Q, which is F.
54:32:50Annalisa Cesaroni: Okay.
54:38:550Annalisa Cesaroni: So, in particular, this means that if it is finite, this is also finite.
54:47:120Annalisa Cesaroni: If this is fine, the one is fine, the other is fine.
54:51:40Annalisa Cesaroni: And actually, it says also something more. It says that
54:57:470Annalisa Cesaroni: the expected value of XP to the power 1 over P is bigger or equal for every P and Q.
55:06:860Annalisa Cesaroni: So… Moreover.
55:10:480Annalisa Cesaroni: Expected value of X to the power p, everything elevated to the power 1 over P is bigger or equal to expected value of X to the power Q.
55:19:800Annalisa Cesaroni: Elevating to the power 1 over Q, for every Q less than P.
55:41:70Annalisa Cesaroni: Okay.
55:44:640Annalisa Cesaroni: So…
55:52:940Annalisa Cesaroni: Okay, this is just computation, but okay, you don't need to know this computation just to know the end of this. It is sufficient to know these facts. Okay.
56:04:430Annalisa Cesaroni: It is 1, B, or Q in the second?
56:08:620Annalisa Cesaroni: This is P over Q. Here, I have 1 over Q, because I divided by… I elevate it to the power 1 over P, both, in both.
56:18:810Annalisa Cesaroni: You don't need to know all this computation, it is important just to know that the Jensen inequality also, and then you have that if a random variable has finite moment, P moment, it has also finite moment for every P less than… for every Q less than P.
56:34:180Annalisa Cesaroni: it is sufficient just to know this factor. And actually.
56:39:250Annalisa Cesaroni: This gives us also an idea of what can be a norm that we can put on the space MP. On the space MP, which is the space X random variable from omega to R, such that expected value of X to the P
56:56:660Annalisa Cesaroni: Is finite.
56:58:700Annalisa Cesaroni: I put this norm, I put the norm, XP, equal to expected value of X to the power p, everything to the power 1 over P.
57:09:480Annalisa Cesaroni: Why I put this norm? Because, actually, I have, okay, the space is characterized as the random variable for which this is finite, but then I have to elevate also to the power 1 over T in order to have the homogeneity of the norm, okay?
57:28:970Annalisa Cesaroni: Yeah, that, the norm… let's go… let's check that it is a norm.
57:40:780Annalisa Cesaroni: First, I have that XP, XP equal to 0, if and all if expected value of X to the power p is equal to 0, then
57:51:780Annalisa Cesaroni: It's also elevated to the peer root, but if the peer root is 0, also the element is 0, okay?
57:57:990Annalisa Cesaroni: And, this is, equal to 0 if and all if X is equal to 0 with probability 1.
58:09:750Annalisa Cesaroni: Almost surely.
58:13:860Annalisa Cesaroni: Which is the only notion of, equality that I have in a probability space, okay?
58:20:520Annalisa Cesaroni: Second, If I have the… if I take lambda in R, lambda times X, P, is just
58:29:590Annalisa Cesaroni: Expected value of…
58:33:530Annalisa Cesaroni: Lambda X to the power p, everything to the power 1 over P. But then it is expected value of lambda to the power p, x to the power p, everything to the power 1 over P. But lambda is a constant, so I can put outside the expected value.
58:51:50Annalisa Cesaroni: And so it is just the lambda to the power p, to the power 1 over P, expected value of X.
58:59:140Annalisa Cesaroni: Elevated to the P, to the power of 1 over P. So it is lambda.
59:05:90Annalisa Cesaroni: XP to the power 1 over P, so… The second,
59:11:90Annalisa Cesaroni: The second condition for being a norm is verified.
59:15:290Annalisa Cesaroni: So we need the third condition. The third condition is the triangular inequality, which is the more difficult one, okay?
59:41:970Annalisa Cesaroni: The third condition to be a norm will be, to have what? To have that the, if, third condition, if, if X and Y are in MP,
59:54:250Annalisa Cesaroni: Then, X plus YP has to be equal to… less or equal than XP plus YP.
00:03:300Annalisa Cesaroni: This is a bit tricky to prove, because it is just saying that expected value of X plus Y to the power p elevated to the power 1 over P is less or equal to expected value of X to the power p.
00:19:800Annalisa Cesaroni: Elevated to 1 over P plus expected value of Y to the power p, elevated to 1 over P.
00:26:670Annalisa Cesaroni: For example, for P equal to 2 means that the expected value of
00:32:520Annalisa Cesaroni: X plus Y to the square, everything 11…
00:37:420Annalisa Cesaroni: extracted root, yeah, less or equal than expected value of X squared.
00:43:950Annalisa Cesaroni: Square root of expected value of X squared plus square root of expected value of Y squared.
00:50:530Annalisa Cesaroni: Which is not completely obvious, no?
00:54:50Annalisa Cesaroni: Because, it is not completely obvious.
01:01:20Annalisa Cesaroni: That is also. In order to prove this, we need to introduce some inequality.
01:07:610Annalisa Cesaroni: Which is… which are also, important by their own.
01:12:740Annalisa Cesaroni: So… First of all, first of all, one can ask, also, if X… In order to prove this.
01:27:280Annalisa Cesaroni: We proved before some inequality, the, the, the so-called older inequality.
01:41:200Annalisa Cesaroni: older inequality. But before proving the older inequality, we proved the older inequality in the next,
01:48:770Annalisa Cesaroni: hour, but before, observation, observation, if X and Y are in MP, actually, X plus Y is an MP. Y, this is true.
02:06:370Annalisa Cesaroni: why this is true. In general, one… one could…
02:11:780Annalisa Cesaroni: Could also be that if X and Y have bounded the P moment.
02:18:60Annalisa Cesaroni: I don't know if the sum has bounded p moment, but the point is that X plus Y to the power P
02:26:180Annalisa Cesaroni: The elevation to the power P is, is a…
02:34:840Annalisa Cesaroni: Is a convex function.
02:36:760Annalisa Cesaroni: for P bigger than 1. P bigger than 1… So… FXX to the power P,
02:46:50Annalisa Cesaroni: is convex for P bigger or equal than 1, no? And so I have that F of X plus Y over 2
02:56:940Annalisa Cesaroni: If I, X plus, X plus Y,
03:04:20Annalisa Cesaroni: Y… X1 plus X2 over 2.
03:18:120Annalisa Cesaroni: What is F of X1, X2 over 2? Is F computed at the middle point between X1 and X2?
03:27:980Annalisa Cesaroni: Okay.
03:29:600Annalisa Cesaroni: X1 is there, here, and X2 is there. X1 plus X2 is the middle point between X1 and X2. And by convexity.
03:39:40Annalisa Cesaroni: This is surely… Less or equal than what?
03:44:80Annalisa Cesaroni: Then the point…
03:46:550Annalisa Cesaroni: which is on the segment joining X1 and X2, the middle point of the segment joining X1 and X2. So it is less or equal than FX1 plus FX2 over 2.
03:58:800Annalisa Cesaroni: by convexity.
04:00:300Annalisa Cesaroni: This is convexity.
04:01:890Annalisa Cesaroni: Convexity means that… let's make the picture a bit better.
04:10:680Annalisa Cesaroni: Convexity means that if I have X1 and X2, and I take the middle point, X1 plus the middle point in the middle, this one.
04:20:30Annalisa Cesaroni: This is the middle point between X1 and X2. The value of the function here is for sure less
04:28:200Annalisa Cesaroni: Below, than this point.
04:31:890Annalisa Cesaroni: which is the middle point between… in this segment. The middle point in this segment as Y variable.
04:41:90Annalisa Cesaroni: equal to the middle point of the segment FX1, FX2.
04:46:370Annalisa Cesaroni: Okay.
04:49:600Annalisa Cesaroni: let's call it, I don't know, A, A as coordinate X1 plus X2 over 2, and FX1 plus X2 over 2.
05:00:910Annalisa Cesaroni: Whereas B, as coordinate, X1 plus X2 over 2.
05:08:130Annalisa Cesaroni: Comma, FX1 plus FX2.
05:12:90Annalisa Cesaroni: over 2.
05:14:220Annalisa Cesaroni: Okay, so, actually, the convexity
05:18:120Annalisa Cesaroni: can be reduced to something like this. F computed at X1 plus X2 over 2 is less or equal than FX1 plus FX2 over 2.
05:26:950Annalisa Cesaroni: Okay.
05:28:560Annalisa Cesaroni: And so we apply this convexity here, taking,
05:36:190Annalisa Cesaroni: F of X equal to X to the power p. And so, what we have? We have that…
05:44:440Annalisa Cesaroni: So, we have that X1 plus X2 over 2 to the power p is less or equal than X1 to the power p plus x2 to the power p over 2.
05:58:00Annalisa Cesaroni: So, it is X1 plus X2 to the power p over 2 to the power p, less or equal, then, X1 to the power P plus X2 to the power p over 2. So, X1 plus X2 to the power p less or equal, then, 2P over 2 to P minus 1,
06:16:50Annalisa Cesaroni: X1 to the power P, X2 to the power p.
06:19:950Annalisa Cesaroni: Okay.
06:21:720Annalisa Cesaroni: So I multiply every…
06:23:680Annalisa Cesaroni: By 2 to the P, and I divide it by 2. Okay. And then, what does it mean? It means… and it is true for every X1 and X2 in R.
06:32:840Annalisa Cesaroni: So, in particular, if I take X and Y random variable.
06:39:310Annalisa Cesaroni: I have X plus Y to the power p is less or equal than 2P minus 1, X to the power P plus Y to the power p.
06:48:820Annalisa Cesaroni: And so the expected value of X plus Y to the power p is for sure less or equal than 2p minus 1.
06:58:320Annalisa Cesaroni: expected value of X to the power p, plus expected value of Y to the power p.
07:04:490Annalisa Cesaroni: Okay.
07:05:580Annalisa Cesaroni: It's just a linearity of the expected value. Okay, I take this…
07:11:640Annalisa Cesaroni: This is not yet the inequality that we want, because…
07:16:820Annalisa Cesaroni: We want the inequality that says that D is elevated to the power 1 over P is less or equal than D is elevated to the power 1 over P, plus D is elevated to the power 1 over P, okay? Something…
07:30:190Annalisa Cesaroni: A bit more, this is not the…
07:35:250Annalisa Cesaroni: the inequality that we want. It is a bit rough.
07:39:480Annalisa Cesaroni: still, but still says that if X and Y are in MP, then X plus Y is in MP, at least.
07:46:680Annalisa Cesaroni: Okay.
07:47:750Annalisa Cesaroni: This is true.
07:49:150Annalisa Cesaroni: If X and Y have bounded the P moment, then the sum has bounded P moment.
07:55:720Annalisa Cesaroni: At least, okay? But actually, this inequality is not sufficient.
08:01:730Annalisa Cesaroni: Because if you extract here the P root, you have to extract also here the P root.
08:08:460Annalisa Cesaroni: But the P root of a sum is not equal to the sum of the P root.
08:16:290Annalisa Cesaroni: So, actually, and then I have also this factor, too.
08:22:479Annalisa Cesaroni: to the power…
08:24:580Annalisa Cesaroni: Here I am an equal… I have also this factor, which is 2 to the power p minus 1, P root of this, which remains in some… it will… remains 2 to the power 1 minus 1 over P.
08:40:479Annalisa Cesaroni: So it remains something.
08:42:700Annalisa Cesaroni: Okay, so, actually, this convexity… the convexity to, modulus of X to the power p is sufficient to prove that, to have a rough estimate of the expected value of the…
08:56:979Annalisa Cesaroni: of the P moment of a sum of random variable, but just rough, just quite rough. If I have… want some more con… a more precise control, I have to use something more.
09:10:450Annalisa Cesaroni: Which is… which will be the older inequality, but maybe we can make a break now, 10 minutes break, and then continue.
09:36:760Annalisa Cesaroni: Indeed.
09:38:490Annalisa Cesaroni: Who are the methods of this movie?
10:04:330Annalisa Cesaroni: No blaming it for me.
11:00:20Annalisa Cesaroni: Okay, should be working. Registration… Okay, so, actually, the point is, of all these things, is…
11:11:590Annalisa Cesaroni: The characterization of convex function, which are
11:17:230Annalisa Cesaroni: Useful in some sense, because convex functions are one of the most used functions Analysis.
11:27:130Annalisa Cesaroni: And, the new use of this by… in order to attain some kind of inequality, integral inequality.
11:35:700Annalisa Cesaroni: Okay, so,
11:41:530Annalisa Cesaroni: Another inequality… another inequality is the, older. In order to get the older inequality, first of all, one proved the young, so-called young inequality.
11:54:110Annalisa Cesaroni: That is the following.
11:57:380Annalisa Cesaroni: Let's take P bigger than 1, and define a number, say, a real number bigger than 1, define Q, the conjugate of P,
12:13:410Annalisa Cesaroni: as the number Q bigger than 1, such that 1 over P plus 1 over Q is equal to 1.
12:21:290Annalisa Cesaroni: The conjugate is this one. For example, if P is equal to 2, Q is equal to 2.
12:27:720Annalisa Cesaroni: Because 1 half plus 1 half is equal to 1. If P is equal to 4, Q is equal to, 4 over 3.
12:37:720Annalisa Cesaroni: Okay, because 1 over 4 plus 3 over 4 is equal to 1. If P is equal to, I don't know, 3F,
12:45:580Annalisa Cesaroni: Q is equal to what? Q is equal to 3.
12:51:400Annalisa Cesaroni: 2 over 3 plus 1 over 3 is equal to 1. I have 2.
12:55:190Annalisa Cesaroni: The, the, the sum of the…
12:59:120Annalisa Cesaroni: Reciprocal has to be equal to 1, okay?
13:03:330Annalisa Cesaroni: And I have the 4, and the unique couple, which is such that P is equal to Q, is P equal to 2, and Q equal to 2, is the unique, no?
13:16:360Annalisa Cesaroni: And, I have the following for every,
13:22:80Annalisa Cesaroni: I don't know, call it for every, A and B in R,
13:27:420Annalisa Cesaroni: A and B positive, A and B positive, let's…
13:31:280Annalisa Cesaroni: positive in R, I have that the product of A times B is less or equal of what? Of A to the power P over P plus B to the power Q over Q.
13:46:340Annalisa Cesaroni: This is called the Young inequality.
13:52:300Annalisa Cesaroni: What does it mean?
13:54:00Annalisa Cesaroni: It is, for example, for P equal to Q equal to 2 is just A times B less or equal A squared over 2 plus B squared over 2, which is actually the classical,
14:07:700Annalisa Cesaroni: A plus B to the square, it is A squared plus B squared plus 2AB bigger or equal than 0. Okay?
14:16:940Annalisa Cesaroni: A minus, if you want, minus.
14:19:560Annalisa Cesaroni: Just to have everything, And so it is A squared plus B squared bigger or equal to 2AB.
14:28:990Annalisa Cesaroni: And that…
14:31:600Annalisa Cesaroni: AB, let's sort of equal A square, it's exactly this one, no? The fact that the square are positive.
14:39:820Annalisa Cesaroni: This is also, also for bigger, for other… for all the possible, couple of conjugate points, so, conjugate element. PQ is con… is,
14:53:500Annalisa Cesaroni: P and Q are related by this relation, okay?
14:56:950Annalisa Cesaroni: 1 over P plus 1 over 2. And, okay. We can skip the proof of these inequalities, just a matter of calculus 1, in the sense that
15:10:250Annalisa Cesaroni: For example, one can fix B positive.
15:14:460Annalisa Cesaroni: Obviously, if one of the two is, zero, this is trivial, no? Because it's zero, less or equal, then something positive. If one of the two… if our both zero, it's trivial, so it is not trivial just for every… for two, all the two,
15:31:860Annalisa Cesaroni: positive, and I define, for example, the function which… I fix… I fix B, then I define the function to every A associated, A to the P over P plus B to the Q over Q minus AB.
15:50:170Annalisa Cesaroni: QA and PAQ conjugate one of the other, no? We fix…
15:57:130Annalisa Cesaroni: PQ conjugate, we fix B, and we take this function, and we look for a minimum of this function.
16:05:830Annalisa Cesaroni: Why this function admits for A bigger or equal than 0? This function admits a minimum, because
16:13:330Annalisa Cesaroni: For A equal to 0, it is B to the power Q over Q. It starts from here. And then, for A going to plus infinity, this quantity is going to plus infinity, because
16:25:660Annalisa Cesaroni: This is going to plus infinity, this is going to minus infinity, but A to the P, we insert with respect to A, because P is bigger than 1, okay?
16:34:830Annalisa Cesaroni: So, this is starting from here, going to plus infinity, so somewhere… should have a minimum. Okay.
16:43:10Annalisa Cesaroni: Trivially.
16:44:660Annalisa Cesaroni: Okay, if you want, let's… let's write everything. So, F of A equal to AL to the P over P plus B over Q minus AB.
16:54:990Annalisa Cesaroni: F of 0 is equal to B.
16:58:750Annalisa Cesaroni: positive. And the limit as A goes to plus infinity of F of A is equal to plus infinity.
17:06:170Annalisa Cesaroni: Because A to the power P
17:09:160Annalisa Cesaroni: This term is winning with respect to this.
17:13:930Annalisa Cesaroni: Because… here I have a power P, which is bigger than 1, okay?
17:21:590Annalisa Cesaroni: P is bigger than 1.
17:23:430Annalisa Cesaroni: So, A to the power P, yeah, 1 over P plus…
17:28:350Annalisa Cesaroni: If you want, be to the…
17:31:700Annalisa Cesaroni: B8 to the power p minus 1, 0, this is going to 0, 0, this is going to plus infinity. Okay? Yeah.
17:40:290Annalisa Cesaroni: And then F… F is at minimum. Y is start positive, then goes to plus infinity, is continuous, should have a minimum.
17:50:100Annalisa Cesaroni: It's somewhere. Okay.
17:52:580Annalisa Cesaroni: F as a minimum.
17:55:160Annalisa Cesaroni: And how to… Find the minimum.
17:58:840Annalisa Cesaroni: I found… I just computed the derivative, and let's put the derivative equal to 0. If I have just one point where the derivative is equal to zero, then that's all. What is the derivative? A to the power p minus 1 minus B.
18:13:450Annalisa Cesaroni: I have to derive with respect to A, yeah? Here.
18:17:20Annalisa Cesaroni: And then, the minimum is A equal to B to the power 1 over P minus 1, something like this, okay?
18:25:890Annalisa Cesaroni: And F computed at this minimum, B to the power 1 is the minimum, the minimum, the point of minimum.
18:35:70Annalisa Cesaroni: Why I'm sure that it is the point of minimum? Because the function admits a minimum.
18:41:830Annalisa Cesaroni: The derivative is 0 just in one point, so this has to be the minimum, okay?
18:48:980Annalisa Cesaroni: And then I have that it is A1 over P, A to the power P over, B to the power P over P minus 1,
18:58:300Annalisa Cesaroni: plus B to the power Q over Q, plus… no, minus, sorry.
19:04:770Annalisa Cesaroni: minus B to the power 1 over P minus 1 times B to the 1.
19:10:220Annalisa Cesaroni: And the… One can check that all these,
19:15:960Annalisa Cesaroni: All these exponents are the same.
19:20:300Annalisa Cesaroni: So, Q is equal to P over P minus 1,
19:24:490Annalisa Cesaroni: Because 1 over Q is plus 1 over P is equal to 1, so if we take 1 over P on the other side, and we…
19:33:380Annalisa Cesaroni: Okay.
19:34:770Annalisa Cesaroni: So, actually, this is equal to this, and this is equal also to the sum of these two.
19:41:970Annalisa Cesaroni: Okay, so… I have F of BP1 over P minus 1.
19:51:570Annalisa Cesaroni: which is the minimum of F.
19:54:50Annalisa Cesaroni: is equal to B to the P over P minus 1 over P plus B to the Q over P minus B1 plus 1,
20:06:140Annalisa Cesaroni: P minus 1, which is B to the power Q, 1 over P plus 1 over Q,
20:13:460Annalisa Cesaroni: Sorry. Minus 1, which is 0.
20:19:310Annalisa Cesaroni: You're right, though.
20:21:370Annalisa Cesaroni: Oopsie… No, after the, mean as the minimum of F.
20:29:950Annalisa Cesaroni: is the minimum of F. The minimum value of F, no? This is the minimum point, point of minimum.
20:42:00Annalisa Cesaroni: So, what does it mean? It means that the minimum of F is 0, so what does it mean that F of A is bigger or equal than 0 for every A? So, A to the power P over P plus B to the power Q over Q minus AB is bigger or equal than 0, which is exactly the inequality that I wanted, okay?
21:02:170Annalisa Cesaroni: Now, from here, let's, write the older inequality. The older inequality is the following one.
21:11:220Annalisa Cesaroni: I take, X in MP.
21:15:580Annalisa Cesaroni: Why in MQ?
21:18:510Annalisa Cesaroni: Where 1 over P plus 1 over Q is equal to 1. So X a random variable, Y a random variable, which are… maybe if P and Q are both 2, they are in the same space, but otherwise they are in different space, okay?
21:34:530Annalisa Cesaroni: And then, I have the following. The expected value of X times Y the producto?
21:43:800Annalisa Cesaroni: is less or equal than the expected value of X to the pi over P.
21:50:410Annalisa Cesaroni: Elevated to the power 1 over P times the expected value of Y to the power Q, elevated to the power 1 over Q.
22:04:540Annalisa Cesaroni: So… It says something,
22:10:350Annalisa Cesaroni: It says something more about this X and Y, so I have that if X is a random variable with finite P moment, and also finite moment for every Q less than… for every value less than P. And Y is a random variable with finite Q moment, and P and Q are, one, the conjugate of the other.
22:33:750Annalisa Cesaroni: Then… a expected value of X times Y is well defined, and it is less or equal than this.
22:43:360Annalisa Cesaroni: If, if, for example, I take P equal to Q equal to 2, it just says that if X and Y are both
22:53:130Annalisa Cesaroni: in M2, the expected value of X, also without the modulus here, I can put without the modulus.
23:02:90Annalisa Cesaroni: It is the same. X times Y…
23:06:00Annalisa Cesaroni: Less or equal than the square root of…
23:09:880Annalisa Cesaroni: Expected value of X squared times expected value of Y squared.
23:16:440Annalisa Cesaroni: the product of the 2 square root can be written as a square root, okay? Square root times square root, if you want. It's the same.
23:30:320Annalisa Cesaroni: So, if I have, two random variable, both with finite,
23:36:730Annalisa Cesaroni: second moment, so in particular with finite mean and so with finite variance, because, okay, then the expected value of the product can be bounded in this way, can be bounded by the expected value of X squared.
23:52:440Annalisa Cesaroni: under the square root times the expected value of Y square under the square root. Okay.
24:01:570Annalisa Cesaroni: Let's prove this.
24:03:270Annalisa Cesaroni: And that's all. And, let's prove this. Let's prove this, prove… take A equal to For example.
24:14:30Annalisa Cesaroni: modulus of X over…
24:18:300Annalisa Cesaroni: Expected value of X to the power p, everything to the power 1 over P, B, modulus of Y over
24:25:910Annalisa Cesaroni: Expected value of Y to the power Q. Everything to the power 1 over Q.
24:32:210Annalisa Cesaroni: And apply young.
24:37:500Annalisa Cesaroni: So, I have that A times B, which is modulus of X,
24:42:100Annalisa Cesaroni: times modulus of Y over expected value of X to the power p 1 over P. Expected value of Y.
24:50:970Annalisa Cesaroni: to the power 1 over Q.
24:53:590Annalisa Cesaroni: Is la sur equal, then, A to the power P, so X.
25:00:170Annalisa Cesaroni: to the power P.
25:02:200Annalisa Cesaroni: over P.
25:03:970Annalisa Cesaroni: Expected value of X to the power p.
25:07:600Annalisa Cesaroni: Because here, if I elevate A to the power P, P, this is canceled, and then divided by P.
25:16:460Annalisa Cesaroni: So, yeah, this is A to the power P over P.
25:20:360Annalisa Cesaroni: And this is Y to the power Q over Q.
25:25:460Annalisa Cesaroni: expected value of Y to the power cube. That's…
25:30:260Annalisa Cesaroni: This is B to the power Q over Q.
25:36:160Annalisa Cesaroni: And then compute the respective value of everything.
25:47:210Annalisa Cesaroni: computed expected value, and we have what? Expected value of X times Y,
25:54:320Annalisa Cesaroni: Over… this is a constant, so it takes it from the expected value. Over, so, expected value of X to the power p, 1 over P. Expected value of Y to the power Q, 1 over Q.
26:09:50Annalisa Cesaroni: Less or equal to expected value of X to the power p.
26:14:590Annalisa Cesaroni: over P expected value of X to the power p, plus expected… Also, these are constants, so they exit from the expected value, okay?
26:25:560Annalisa Cesaroni: So I take the expected value.
26:28:200Annalisa Cesaroni: It's just the expected value of this, because these are constant.
26:38:410Annalisa Cesaroni: Q expected value of Y to the power Q.
26:46:970Annalisa Cesaroni: And so, you see that this is equal to this, and it cancels, this is equal to this, and it remains 1 over P plus 1 over Q, which is equal to 1, because P and Q are,
27:03:700Annalisa Cesaroni: are both, are conjugate one over the other. And so I have that. This…
27:09:230Annalisa Cesaroni: is less or equal than 1. So, expected value of X times Y is less or equal than
27:16:310Annalisa Cesaroni: So I multiply by the numerator, and I end up with the inequality that I wanted.
27:24:880Annalisa Cesaroni: Okay, I multiply by the numerator, and so I have…
27:29:810Annalisa Cesaroni: EX time Y is for sure less or equal than EX time Y, less or equal, here I multiply by the denominator, expected value of X to the power p, the power of 1 over P, expected value of Y to the power Q, 1 over Q.
27:50:370Annalisa Cesaroni: Because here, I multiply by this.
27:54:650Annalisa Cesaroni: The, the two, the two terms of the inequality, no?
28:01:270Annalisa Cesaroni: Here is canceled, and here is multiplied by this.
28:10:310Annalisa Cesaroni: And so this gives, for example, the fact that, okay, if I have X in MP and Y in MQ, X times Y are in M1.
28:22:20Annalisa Cesaroni: So, in particular, X in MP.
28:24:970Annalisa Cesaroni: And why? In MQ?
28:28:430Annalisa Cesaroni: X times Y is in M1. The product is in M1.
28:33:920Annalisa Cesaroni: for example, if I take X in M2, Y in M2, the product is not in M2 in general, but it is in M1. It is not true that the… if I have X with finite variance, Y of finite variance, the product cannot have… can have finite variance, or can have infinite variance, but the mean is finite, for sure.
28:57:650Annalisa Cesaroni: Could you give up a second?
28:59:470Annalisa Cesaroni: Oh, what?
29:06:840Annalisa Cesaroni: So it is just Yang applied to these two terms.
29:22:800Annalisa Cesaroni: Okay, and let's end up with, the fact that X and Y in MP then
29:30:950Annalisa Cesaroni: X plus YP, so the expected value of X plus Y
29:36:130Annalisa Cesaroni: to the P, to the power 1 over P is less or equal than…
29:42:90Annalisa Cesaroni: 2X to the P to the power 1 over P plus expected value of Y to the P, the power 1 over P.
29:51:440Annalisa Cesaroni: This can be obtained by using the, the older inequality. It's a bit,
30:01:210Annalisa Cesaroni: Nothing deep, but it's just a matter of, doing some computation, and so it is, so one right.
30:11:450Annalisa Cesaroni: One write the things like this. X plus Y to the power p, one right as X plus Y times X plus Y to the power P minus 1.
30:23:790Annalisa Cesaroni: Okay.
30:27:560Annalisa Cesaroni: So, in the end, Cora ol, Larry?
30:33:70Annalisa Cesaroni: Let's make like this. If X and Y are both in MP, Then…
30:40:570Annalisa Cesaroni: And so in particular, this is an arm.
30:44:990Annalisa Cesaroni: In particular, This is a norm.
30:51:870Annalisa Cesaroni: Since it satisfies the… triangular inequality.
30:57:390Annalisa Cesaroni: So, the LP norm… the MP norm of X plus Y is less or equal than the LP norm of X plus LP norm of Y.
31:08:350Annalisa Cesaroni: We define the norm of a random… the P norm of a random variable with finite p moment as the expected value of the P moment elevated to the power 1 over P, okay?
31:19:450Annalisa Cesaroni: How to prove this? One, just write X plus Y to the power p as X plus Y times X plus Y to the power p minus 1. And actually, and actually, this is in MP.
31:34:930Annalisa Cesaroni: And this is in M, P over P minus 1.
31:41:810Annalisa Cesaroni: Why this is true? Because if I take X plus Y to the power p minus 1, and I elevate it to the power p over p minus 1, I obtain X plus Y to the power p.
31:58:410Annalisa Cesaroni: X plus Y to the power p minus 1 is in M, P over P minus 1, because if I elevate this to the power p over p minus 1, P-1 is canceled, no? So X plus Y to the power p.
32:13:920Annalisa Cesaroni: Okay.
32:16:440Annalisa Cesaroni: expected value of X plus Y to the power p minus 1, elevated to the power p over p minus 1 is finite, because
32:25:80Annalisa Cesaroni: P minus 1 is canceled, it is X plus Y to the power p. That I know that is finite by what I've wrote before, no? I've wrote that the X plus Y to the power p is finite by this quantity.
32:41:720Annalisa Cesaroni: And then, actually, I have that these P and P over P minus 1 are conjugate aspirin to one of the other.
32:51:360Annalisa Cesaroni: Can you decay, because 1 over P, 1 over P plus… P minus 1 over P.
32:59:540Annalisa Cesaroni: is exactly 1.
33:01:910Annalisa Cesaroni: So they are to con… conjugate. Okay.
33:05:780Annalisa Cesaroni: And then I apply all their inequality, but okay. If you want, in the notes, it's written, it's just a matter of writing down the computation.
33:15:950Annalisa Cesaroni: Okay, so I applied the older inequality here, because this is in MP, this is in MP, in MQ, for Q, the conul-day despiron, okay?
33:30:720Annalisa Cesaroni: the first factor is in MP, so I apply the older inequality with this as
33:37:260Annalisa Cesaroni: what before I called X, and these with… what before I called Y? Okay, X times Y,
33:45:680Annalisa Cesaroni: Is less or equal.
33:47:550Annalisa Cesaroni: Okay.
33:53:220Annalisa Cesaroni: So, actually, it should be something like this. Expected value of X plus Y to the power p, so expected value of X times X plus Y.
34:04:370Annalisa Cesaroni: something like this. X plus Y to the power p minus 1, or P minus 1. And then I apply all the inequality to
34:13:00Annalisa Cesaroni: this product, okay? This is in MP, this is in MQ.
34:18:560Annalisa Cesaroni: What U is P over P minus 1.
34:21:810Annalisa Cesaroni: conjugate exponent.
34:23:910Annalisa Cesaroni: 1 over P plus 1 over Q is equal to 1. And so this is less or equal than expected value of this…
34:31:690Annalisa Cesaroni: If one, apply… and apply all their inequality, if one apply older inequality and make some computation.
34:40:780Annalisa Cesaroni: one end up with what we want to prove, actually. But it's just a matter of computation. So, at the end, with all their inequality, we end up with the fact that X plus Y p, so the expected value of X plus Y to the power P elevated to the power 1 over P, is less or equal to expected value.
35:00:20Annalisa Cesaroni: of X to the power Because, actually, here.
35:10:930Annalisa Cesaroni: Hmm.
35:14:100Annalisa Cesaroni: Yes, it should be something like this. Instead of taking this, I take this…
35:22:510Annalisa Cesaroni: Without the modulus, or something like this.
35:26:320Annalisa Cesaroni: X plus Y over, so the point is that, yes.
35:36:440Annalisa Cesaroni: Okay, there are some computation in the middle, and at the end, but the idea is that, is to use the fact that this product can be written… this… this is not a product, this quantity elevated to the power p, can be written as a product of two terms
35:54:600Annalisa Cesaroni: one in empty, the other in cube.
35:58:460Annalisa Cesaroni: And then we can apply Older. And actually, okay.
36:05:850Annalisa Cesaroni: Actually, we are not applying exactly to this, but to something
36:10:890Annalisa Cesaroni: Similar to this, okay? It's just a matter of computation, it is not the case to…
36:20:540Annalisa Cesaroni: It's just computation, just computation, one can… Trust that it's…
36:31:260Annalisa Cesaroni: And the result is obtained. Okay, so at the end, we have the following. Let's,
36:37:910Annalisa Cesaroni: Let's, put, let's recap.
36:42:290Annalisa Cesaroni: First, for every P bigger or equal than 1, I define MP, the space of a random variable.
36:50:920Annalisa Cesaroni: We'd find it… We'd find at the P moment.
36:57:260Annalisa Cesaroni: And I have M1 containing, M2 containing, etc. Second, on MP, I may define an arm.
37:11:130Annalisa Cesaroni: A norm, which is the…
37:14:160Annalisa Cesaroni: P normo, defined in the following way, defined in this way. Expected value of X to the power p, everything elevated to the power 1 over P.
37:23:580Annalisa Cesaroni: This is a norm.
37:26:860Annalisa Cesaroni: And I have a notion of… I have associated a distance. A distance will be what? The P between X and Y will be expected value of…
37:38:920Annalisa Cesaroni: X minus Y to the power P to the power 1 over P. This is the distance. And I have a notion of convergence.
37:47:90Annalisa Cesaroni: So, I say that Xn, a sequence of random variable, converts to F to X in MP, If…
37:57:160Annalisa Cesaroni: say, the expected value of XN minus X,
38:01:540Annalisa Cesaroni: to the power P is converging to 0.
38:07:840Annalisa Cesaroni: So, for example, Xn converts to X in M1 in mean if the expected value of modulus of Xn minus X is converging to 0. Xn converts to X in M2
38:21:440Annalisa Cesaroni: So, in mean square, if the expected value of Xi minus X converts to the square, converts to 0.
38:35:740Annalisa Cesaroni: So, on LP, I made a final norm. Actually, I proved that this is a norm. This is a norm because it is symmetric, homogeneous, and satisfies the triangular inequality.
38:49:670Annalisa Cesaroni: And the distance associated to this norm is just… I have that two random variable, which has fined P moment, as a distance
38:59:180Annalisa Cesaroni: Can't…
39:00:430Annalisa Cesaroni: A notion of distance between random variable can be given in this way. So, the distance is the expected value of the modulus of X minus y to the power p.
39:10:150Annalisa Cesaroni: elevated to the power 1 over P, obviously.
39:13:970Annalisa Cesaroni: So, which is actually the norm of X minus Y to the…
39:19:560Annalisa Cesaroni: The P norm of X minus Y, okay? The distance associated to a norm, the distance between two elements, is just the norm of the difference of the two elements.
39:29:50Annalisa Cesaroni: Okay, we have that the sum of random variable with finite moment is just a random variable with finite moment, etc.
39:39:700Annalisa Cesaroni: We have a notion of convergence of this space, and actually.
39:44:110Annalisa Cesaroni: So, and the notion of convergence is the following one.
39:47:720Annalisa Cesaroni: I have, for example, for P equal to 1, for P equal to 1, I say that Xn converts to X and X in mean. This means that the expected value of XN minus X
40:03:990Annalisa Cesaroni: is going to 0 as n goes to plus infinity.
40:08:290Annalisa Cesaroni: And for P equal to 2, I have that Xn converts to X in mean square, sometimes it's called mean square convergence. So, the expected value of Xn minus X to the square converts to 0.
40:24:20Annalisa Cesaroni: As n goes to plus infinity.
40:37:950Annalisa Cesaroni: Okay, and and actually, one can be proved is that MP with this norm is a Banach space.
40:47:90Annalisa Cesaroni: We are not going to prove, but it's a Banach space. So it is complete. Every C, every Cauchy sequence is converging into a random variable there.
40:58:20Annalisa Cesaroni: is a Bannock space. Every Cauchy sequence is converging.
41:12:860Annalisa Cesaroni: Okay, mmm… And then, so…
41:19:320Annalisa Cesaroni: Then, we have this random… this space of random variables, which are encapsulated… encapsulated one on the other. On MP, we define
41:30:710Annalisa Cesaroni: for every P bigger than 1, we define this norm, which is the expected value of X to the power p, elevated to the power 1 over P,
41:39:890Annalisa Cesaroni: and then the distance between two elements in the space as the expected value of X minus Y to the power p, elevated to the power 1 over P.
41:50:250Annalisa Cesaroni: And the notion of convergence. With this notion of convergence, we have that the space is Bannock. And then, third important thing is the older inequality. Older inequality says that
42:04:230Annalisa Cesaroni: Older inequality says that if X is in MP,
42:08:60Annalisa Cesaroni: In MP, you always put P above.
42:12:440Annalisa Cesaroni: And Y is in MQ, and 1 over P plus 1 over Q is equal to 1, Then.
42:19:940Annalisa Cesaroni: the expected value of X times Y
42:23:960Annalisa Cesaroni: which is always less or equal than the expected value of the modulus of X times Y.
42:31:500Annalisa Cesaroni: is less or equal than the expected value of X to the power p, to the power 1 over P, expected value of Y to the Q to the power 1 over Q. So, to the helping norm of X, X times the Q norm of Y, if you want.
43:00:840Annalisa Cesaroni: Okay, and these,
43:06:280Annalisa Cesaroni: This, for example, observation, observation, this says that if X and Y are both in M2,
43:15:210Annalisa Cesaroni: The expected value of X times Y is well defined, so…
43:19:200Annalisa Cesaroni: X times what X and Y.
43:22:780Annalisa Cesaroni: In M2,
43:24:260Annalisa Cesaroni: Since 1 over 2 plus 1 over 2 is equal to 1, so P equal to Q equal to 2, I have that expected value of X times Y is less or equal than the
43:35:330Annalisa Cesaroni: to norm of X, 2 norm of Y.
43:38:80Annalisa Cesaroni: So the square root of expected value of X to the square, square root of expected value of Y to the square.
43:46:370Annalisa Cesaroni: which is square root of expected value of X to the square times the square root of everything together, and so I have that expected value of X times Y
43:58:980Annalisa Cesaroni: To the square, if you want, is less or equal than
44:02:850Annalisa Cesaroni: Expected value of X to the square.
44:06:700Annalisa Cesaroni: times expected value of Y to the square, if you want.
44:12:10Annalisa Cesaroni: We elevate both the element of this inequality to the square. We eliminate the square root on the other side.
44:19:700Annalisa Cesaroni: Okay.
44:25:10Annalisa Cesaroni: So, for example, this means, for example, that if Xn is converging to X in mean square.
44:36:360Annalisa Cesaroni: What does it mean? It means that Xn minus X to the square is going to 0,
44:43:160Annalisa Cesaroni: mean square, so in M2, Then, Xn converts to F, X also in min.
44:51:880Annalisa Cesaroni: So, in M1, the expected value of XN minus X is going also to 0.
44:59:150Annalisa Cesaroni: Why this is true? Just applying this.
45:02:770Annalisa Cesaroni: take whatever Y if you want, and I have expected value of XN minus X times Y.
45:11:900Annalisa Cesaroni: Elevated to the power 2 less or equal expected value of XN minus X to the square.
45:20:590Annalisa Cesaroni: Product expected value of Y squared.
45:25:340Annalisa Cesaroni: Okay.
45:30:170Annalisa Cesaroni: for every Y, because I just apply these to XN minus X, and to Y whatever.
45:37:750Annalisa Cesaroni: I apply the previous inequality. So, for example, I can apply to Y, for example, I apply to… No.
45:48:450Annalisa Cesaroni: Y equal to who?
46:05:660Annalisa Cesaroni: Xn minus X over square root of Xn minus X, something like this.
46:15:550Annalisa Cesaroni: And I have what? I have the expected value of XN minus X.
46:22:440Annalisa Cesaroni: Because…
46:23:860Annalisa Cesaroni: Xn minus X times Y is just modulus of Xn minus X, less or equal the expected value of Xn minus X to the square.
46:35:290Annalisa Cesaroni: times 1, because it is, Y to the square is…
46:39:210Annalisa Cesaroni: Y is equal to 1, so Y to the square is equal to 1.
46:44:320Annalisa Cesaroni: And so, actually, I have that if it is, going to zero, this is also going to zero, no?
46:51:520Annalisa Cesaroni: So, actually, the older inequality says, in particular, that if I have a sequence which converts in M2, then the same sequence converts also in M1. The contrary is not true, because if I have a sequence which converts in M1, maybe the limit is not in M2, and so…
47:13:330Annalisa Cesaroni: Okay.
47:15:930Annalisa Cesaroni: So, we are giving… we are given some structure to this space of random variable.
47:21:540Annalisa Cesaroni: with the, so the idea is to give some structure to the space of random variable with finite moment. So the idea is to put on this space of random variable with finite moment a notion of distance, a notion of convergence, okay? What is the convergence of random variable associated to this? It's just the convergence of the… of the moment.
47:42:700Annalisa Cesaroni: Okay, the…
47:44:430Annalisa Cesaroni: Xn is converging to X if the moments of the random variable XN minus X are converging to 0. Okay? Something like this. So, we have convergence of the moment. So, we are putting some kind of…
48:01:150Annalisa Cesaroni: Functional structure on the space.
48:06:230Annalisa Cesaroni: And the… So…
48:09:890Annalisa Cesaroni: Actually, in this space, we… even if the elements are not points in… of the plane or point of the space, but the elements are a random variable, we can…
48:20:280Annalisa Cesaroni: do the calculus that we are used to do on finite dimension, just looking at the infinite dimension. And, in particular, the more interesting space in this case is M2,
48:36:170Annalisa Cesaroni: I'm too sorry.
48:38:130Annalisa Cesaroni: Which is the space of X random variable with finest second moment, because M2 has also this property, thanks to Walder inequality.
48:50:400Annalisa Cesaroni: Older inequality says that we can make a product of
48:56:890Annalisa Cesaroni: It… all the inequality says that it makes sense To compute the product.
49:06:580Annalisa Cesaroni: of X and Y, with X and Y in the space.
49:14:250Annalisa Cesaroni: Whereas, for different moments, it's not…
49:18:670Annalisa Cesaroni: This does not make sense, for example, to compute the product
49:23:320Annalisa Cesaroni: Okay, it makes sense, but it does not produce anything which we can control. If I take a function, a random variable which is in M1, and another random variable which is in M1, so with finite mean, and I compute the product of the two random variables, so the random variable product of the two…
49:43:250Annalisa Cesaroni: The product is the random variable we associate to every omega, X of omega, y of omega, no?
49:48:670Annalisa Cesaroni: the product, make.
49:51:240Annalisa Cesaroni: The product, in general, with the X and Y random… generic random variable.
49:57:470Annalisa Cesaroni: is something which is still a random variable, but it is a random variable about which we can say almost nothing also about the mean, the average, etc, the mean or the moments.
50:11:530Annalisa Cesaroni: Whereas for M2, we can define, we can define something which is,
50:18:300Annalisa Cesaroni: the… the… we have that XY in M2, we have that… the product SY is in M1, which is something…
50:26:940Annalisa Cesaroni: quite, important.
50:30:190Annalisa Cesaroni: And actually, also given by this older inequality, we can also define a sort of scalar product.
50:41:270Annalisa Cesaroni: A scholar producto?
50:43:430Annalisa Cesaroni: On this space, In this way, I have that the scalar product Hmm…
50:51:970Annalisa Cesaroni: So, a scalar product is what? It's a product which associates to every couple of elements in M2,
50:59:180Annalisa Cesaroni: A number, a scholar, so…
51:05:960Annalisa Cesaroni: It is not a product, it is a scalar product, in the sense that the outcome is a real number.
51:12:870Annalisa Cesaroni: So…
51:14:90Annalisa Cesaroni: The scalar product is what? If I take X here, Y here, the scalar product… I don't know, X…
51:25:260Annalisa Cesaroni: product Y, maybe it's not good to define like this. Let's define like this, XY,
51:34:140Annalisa Cesaroni: So I take the couple, and I go to the scalar product XY is equal to the expected value of X times Y, which is defined, well defined.
51:48:190Annalisa Cesaroni: Why it is important to have a Scalar product? Because the Scalar product is something which is similar to what we have in
51:57:430Annalisa Cesaroni: the space of infinitely dimensional space, in vectorial space. In RN, in R2, for example, no? In R2, in R3, in RN, we have the scalar product of two elements, no?
52:11:490Annalisa Cesaroni: And we can define what is the scalar product in R2. It's just…
52:17:940Annalisa Cesaroni: In R2, what is the scalar product of X1Y1, X2, Y2?
52:25:200Annalisa Cesaroni: Borinarin.
52:27:840Annalisa Cesaroni: In all the space RN, we have the scalar… we have scalar product, this… Scholar product.
52:36:900Annalisa Cesaroni: Which is just, if I have X1, Xn, Scholar product, Y1, YN.
52:46:20Annalisa Cesaroni: They…
52:46:890Annalisa Cesaroni: It is just what? If I have X1, Xn, an element over N. And Y1, YN, an element over N. What is the scalar product? It's just X1, Y1, plus X2, Y2 plus etc, etc, plus IXNYN.
53:05:490Annalisa Cesaroni: Okay.
53:09:10Annalisa Cesaroni: So, if I have, in R2, for example, the Scala product, Of these two.
53:30:730Annalisa Cesaroni: The scalar product is the cosine of the angle between the two vectors, no?
53:35:650Annalisa Cesaroni: Something like this.
53:39:480Annalisa Cesaroni: What does it mean that, for example, X1, X2, X1… sorry.
53:45:660Annalisa Cesaroni: I… Let's… Hmm.
53:55:710Annalisa Cesaroni: Using the scalar product, I can define the notion of orthogonality, for example. What does it mean that two vectors are orthogonal? Means that the scalar product is zero, okay? Two vectors.
54:11:640Annalisa Cesaroni: are orthogonal, So, to vector, let's call it, let's take the vector, X1,
54:20:340Annalisa Cesaroni: X2, XN in RN, and Y1.
54:26:240Annalisa Cesaroni: YN in RN are orthogonal, If the scalar product is zero.
54:42:130Annalisa Cesaroni: Okay.
54:44:750Annalisa Cesaroni: So, and when we… we have that, for example, if I have a point here and a point here.
54:54:140Annalisa Cesaroni: Okay, they are orthogonal if they are scalar product, if and only if the scalar product is zero. Okay, the scalar product is a way to…
55:01:770Annalisa Cesaroni: to measure the angle in R2, for example, between… and also in RN,
55:08:470Annalisa Cesaroni: to measure the angle between the vector. If the angle is, P over 2 is…
55:16:320Annalisa Cesaroni: then the scalar product is zero.
55:19:40Annalisa Cesaroni: Okay. It's a way to measure the angle, see?
55:23:640Annalisa Cesaroni: To measure in which way the two vectors are putting one with respect to the other.
55:33:460Annalisa Cesaroni: And actually, it's quite important to have this notion of orthogonality. And also, in this space, M2, we have the same notion of orthogonality. So, also in M2,
55:47:430Annalisa Cesaroni: We have a Scalar product.
55:49:460Annalisa Cesaroni: Which is actually… Related to the… Related to the norm, In the sense that,
55:59:270Annalisa Cesaroni: The scalar product of X with X is expected value of X squared.
56:05:570Annalisa Cesaroni: Because the scalar product of X and Y, which is… X2 to the square.
56:13:110Annalisa Cesaroni: And we may define… so, the scalar product of two vectors is just the expected value of X, of two… of two random variables is X times Y. And I can define the notion of orthogonality, so I say that X is orthogonal to Y,
56:29:640Annalisa Cesaroni: X and Y are orthogonal, a two orthogonal random variable.
56:35:730Annalisa Cesaroni: Perpendicular random variable.
56:42:730Annalisa Cesaroni: If… If the expected value of X times Y is equal to 0.
56:52:30Annalisa Cesaroni: Okay.
56:56:470Annalisa Cesaroni: And we will see next time, probably, that actually this notion of orthogonality is related to… can… can give some,
57:08:310Annalisa Cesaroni: is related about the fact that the two random variables, X and Y, are actually,
57:21:750Annalisa Cesaroni: are actually encoding different information, which are not… but we will see…
57:29:130Annalisa Cesaroni: Okay, so we define the… so we give this definition. Definition, X and Y are orthogonal, X is orthogonal to Y if, if and only if, X times Y has a, expected value, which is 0.
57:49:310Annalisa Cesaroni: I'm not putting the absolute value here, I'm taking expected value of X times Y,
57:54:960Annalisa Cesaroni: It is not said that the expected value of X time Y is equal to 0, because otherwise this will mean that X time y is equal to 0 almost everywhere, okay? Not this.
58:07:350Annalisa Cesaroni: So, for example, if X and Y are,
58:11:970Annalisa Cesaroni: If X and Y are independent, one or the other, and one of the two has mean zero.
58:18:360Annalisa Cesaroni: They are orthogonal.
58:20:670Annalisa Cesaroni: Okay.
58:22:130Annalisa Cesaroni: So, if X and Y are independent, example, independent.
58:30:840Annalisa Cesaroni: Then, X minus expected value of X, for example, is orthogonal to Y.
58:38:120Annalisa Cesaroni: or X is orthogonal to Y minus expected value of Y.
58:43:350Annalisa Cesaroni: Just in order to be sure that one of the two is a expected value, which is zero, no?
58:55:710Annalisa Cesaroni: So… More or less, it's a sort of independence, the orthogonality.
59:02:650Annalisa Cesaroni: Okay, some sense.
59:07:00Annalisa Cesaroni: So, if X and Y are independent, then actually up to assume that one of the two has mean zero, so up to eliminate… to subtract from one of the two the mean, then they are in…
59:26:620Annalisa Cesaroni: Oh, okay.
59:28:130Annalisa Cesaroni: Okay, so, another definition, definition, if I have S a subspace of M2, as subspace.
59:41:890Annalisa Cesaroni: I can define as orthogonal, of M2.
59:47:440Annalisa Cesaroni: As orthogonal, orthogonal subspace, Is what?
59:54:990Annalisa Cesaroni: is the space of X in M2, such that
00:01:340Annalisa Cesaroni: X is orthogonal to every element, Ines.
00:12:850Annalisa Cesaroni: So… Axonem to such that?
00:17:290Annalisa Cesaroni: For every Y… for every Y in S, I have that expected value of X times Y is equal to 0.
00:38:210Annalisa Cesaroni: In R2, what is the orthogonal subspace of,
00:43:390Annalisa Cesaroni: in R2, for example, just to maintain the geometric intuition. If I have some S, which is, say, a line.
00:52:470Annalisa Cesaroni: in R2.
00:54:960Annalisa Cesaroni: a line passing through the origin, because I have to… I need to have a subspace, so 0 as…
01:02:260Annalisa Cesaroni: be an element of S. What is S orthogonal? S orthogonal will be the line passing through 0 and orthogonal to S, okay? Just this.
01:18:110Annalisa Cesaroni: Just this one.
01:19:480Annalisa Cesaroni: So, example. Let's compute an orthogonal space associated to some, to some space in M2.
01:31:630Annalisa Cesaroni: Let's compute, for example.
01:36:620Annalisa Cesaroni: the easiest possible one. Let's take S to be the subspace as example, So, let's compute one of…
01:48:790Annalisa Cesaroni: So, in R2, S is a line passing through 0,
01:58:870Annalisa Cesaroni: And as orthogonal is the line, perpendicular line.
02:06:610Annalisa Cesaroni: to S, passing through 0.
02:10:620Annalisa Cesaroni: Why they have to pass through zero? Because they have to be subspaces.
02:16:240Annalisa Cesaroni: So, our two is… total as a vectorial space, okay? So, S as a vectorial space, as to
02:25:960Annalisa Cesaroni: be a linear space, so it has to, contain zero, okay?
02:34:650Annalisa Cesaroni: Let's make this example, S the random variable X in M2 with, mean equal to 0.
02:44:550Annalisa Cesaroni: Not with the absolute value, but mean equal to zero.
02:52:330Annalisa Cesaroni: And, so, in particular, S is a subspace of,
02:59:210Annalisa Cesaroni: of M2, since if I have two random variables with mean 0, the sum as mean 0, okay?
03:06:70Annalisa Cesaroni: X, Y in S, expected value of X plus Y is just expected value of X plus expected value of Y.
03:14:690Annalisa Cesaroni: 0 plus 0, 0.
03:17:270Annalisa Cesaroni: And, so it is a random variable. Obviously, the random value, which is 0 everywhere, as means 0, so that zero is there, 0 is in S, and also the expected value of lambda times X is lambda, expected value of X, so if it is 0,
03:33:690Annalisa Cesaroni: Also, lambda times 0 is 0. Okay? So it is a subspace.
03:41:380Annalisa Cesaroni: Subspace means it is close with respect to sum and with respect to multiplication by constant.
03:47:440Annalisa Cesaroni: We want to compute as orthogonal. S- orthogonal are all the Y in M2.
03:54:50Annalisa Cesaroni: Such that Y is orthogonal to every element, is orthogonal to every X in S.
04:03:770Annalisa Cesaroni: Okay, how to compute this? Okay, S orthogonal, R.
04:08:690Annalisa Cesaroni: the random variable, which are orthogonal to every random variable which means zero, okay?
04:17:900Annalisa Cesaroni: to everything.
04:21:670Annalisa Cesaroni: to every random variable would mean zero. So…
04:25:310Annalisa Cesaroni: Y is in S orthogonal if and all if for every, for every X in S. So, for every X such that expected value of X is equal to 0, it also that expected value of X times Y is equal to 0.
04:41:640Annalisa Cesaroni: Okay.
04:44:270Annalisa Cesaroni: This is the effect.
04:56:110Annalisa Cesaroni: Okay, so it has to be true for every…
05:01:360Annalisa Cesaroni: First of all, first of all, we remember that, actually, if I multiply X in S by a constant, then it is, again, a random variable, and it is. So, if I take Y is equal to a constant, Y constant, random variable.
05:20:890Annalisa Cesaroni: If… If Y?
05:23:870Annalisa Cesaroni: is constant.
05:26:300Annalisa Cesaroni: So, Y of omega is equal to C for every omega, for some C, whatever. So, expected value of Y times X is expected value of C times X, which is C expected value of X, which is equal to 0 for every X.
05:45:390Annalisa Cesaroni: for every X with the expected value of X equal to 0.
05:49:470Annalisa Cesaroni: So the constant random variable are orthogonal to the space. So… Why, constant?
05:58:790Annalisa Cesaroni: random variables.
06:04:530Annalisa Cesaroni: are in S orthogonal, because actually, they are orthogonal. Okay.
06:11:30Annalisa Cesaroni: to every, the cost and random variable are orthogonal to every element in S.
06:16:810Annalisa Cesaroni: So, I… I can ask. Okay, but there are other random variables which are not constant and which are orthogonal to all this.
06:27:920Annalisa Cesaroni: So, we have just the constant random variable. So, assume that there exists Y,
06:34:160Annalisa Cesaroni: in S orthogonal, which is maybe not constant. Okay.
06:42:760Annalisa Cesaroni: So, this means that for every X with expected value of X equal to 0, the expected value of Y times X is equal to 0. This means this, no?
06:54:230Annalisa Cesaroni: That Y is in as orthogonal if and only if for every X with expected value of X equal to 0, it holds that.
07:03:50Annalisa Cesaroni: the two, the two, these two things are equivalent, no? By definition. Y is in a sortogonal if and only if for every X with mean equal to 0,
07:14:220Annalisa Cesaroni: So… Let's take Y minus expected value of Y.
07:20:910Annalisa Cesaroni: This is a random variable, which is Y minus a constant, okay?
07:27:270Annalisa Cesaroni: This is in S. Why? Because the expected value of Y minus expected value of Y is what? Expected value of Y minus expected value of expected value of Y. The expected value of a constant is the constant.
07:42:460Annalisa Cesaroni: So, it is equal to expected value of Y minus expected value of y equal to 0.
07:49:190Annalisa Cesaroni: Okay.
07:50:160Annalisa Cesaroni: So, this is an element in S. So, this is one of these X.
07:56:240Annalisa Cesaroni: Here, for which this condition has to hold.
08:00:650Annalisa Cesaroni: Okay, so in place of X here, I can put this Y minus expected value of Y.
08:07:420Annalisa Cesaroni: in place of this X, I can put this, because also this is in S.
08:13:390Annalisa Cesaroni: Okay.
08:14:730Annalisa Cesaroni: For every X… for every X in S, you know?
08:18:600Annalisa Cesaroni: That's to be true.
08:20:810Annalisa Cesaroni: Which means expected… So, I have that…
08:25:350Annalisa Cesaroni: Expected value of Y times Y minus expected value of Y has to be equal to 0.
08:31:790Annalisa Cesaroni: if Y is in a orthogonal, because
08:35:220Annalisa Cesaroni: This is in S. This is an element in S.
08:39:320Annalisa Cesaroni: And so, Y is orthogonal to every element in S. But what does it mean, this? It means that 0
08:46:780Annalisa Cesaroni: is equal to expected value of Y times Y minus expected value of Y.
08:53:650Annalisa Cesaroni: Which is what? Which is expected value of… let's put… let's make the product Y squared minus Y expected value of Y.
09:05:350Annalisa Cesaroni: Okay.
09:07:420Annalisa Cesaroni: So, what it is, it is.
09:10:200Annalisa Cesaroni: expected value of Y squared, minus expected value of Y, expected value of Y.
09:18:160Annalisa Cesaroni: then expected value of Y is a constant, can be put outside, so it is expected value of Y squared.
09:24:730Annalisa Cesaroni: minus expected value of Y, times suspected value of Y.
09:31:880Annalisa Cesaroni: Because I put these outside.
09:34:400Annalisa Cesaroni: Okay
09:35:740Annalisa Cesaroni: So, what does it mean? That Y has to have this quantity equal to 0, so expected value of Y squared.
09:46:10Annalisa Cesaroni: minus expected value of Y, To the square has to be equal to 0.
09:53:430Annalisa Cesaroni: So… Which are the random variable which has this property, just the constant, okay?
10:00:460Annalisa Cesaroni: Just a constant one.
10:04:910Annalisa Cesaroni: Okay.
10:07:60Annalisa Cesaroni: So, Y has to be… has to be constant.
10:15:120Annalisa Cesaroni: Because, when…
10:18:830Annalisa Cesaroni: is the same computation when we compute the variance, so let's take expected value of Y minus expected value of Y to the square.
10:29:440Annalisa Cesaroni: one make this computation, it is expected value of Y squared.
10:34:550Annalisa Cesaroni: minus 2 expected value of Y squared plus expected value of Y-square.
10:42:490Annalisa Cesaroni: One compute everything, and so…
10:45:550Annalisa Cesaroni: these minuses, and so I have that it is equal to zero, so expected these…
10:55:620Annalisa Cesaroni: has to be… Okay, it's just, let's make the computation.
11:02:380Annalisa Cesaroni: Let's make the computation. So it is…
11:06:520Annalisa Cesaroni: We know this, so we compute expected value of Y minus expected value of Y to the square.
11:13:550Annalisa Cesaroni: It is expected value of Y squared minus 2Y, expected value of Y, plus expected value of Y squared, and so use the linearity, no? So, expected value of Y squared.
11:27:420Annalisa Cesaroni: minus 2 expected value of Y times expected value of Y, so expected value of Y to the square, plus expected value of Y to the square.
11:38:700Annalisa Cesaroni: And so, here I sum this… these two.
11:42:340Annalisa Cesaroni: And I have… expected value of Y squared minus expected value of Y
11:50:860Annalisa Cesaroni: everything to the square. And it has to be equal to zero by this. Equal to zero, which means that…
11:59:300Annalisa Cesaroni: this quantity.
12:01:340Annalisa Cesaroni: Has to be almost everywhere constant equal to zero.
12:05:680Annalisa Cesaroni: So, in particular, Y minus expected value of Y has to be equal to 0 with probability 1.
12:15:640Annalisa Cesaroni: Which means that… Y is equal to expected value of Y, so Y is a constant.
12:24:120Annalisa Cesaroni: Okay.
12:25:820Annalisa Cesaroni: So, actually, if S are the function… are the random variable with the expected value of X equal to 0, as orthogonal are just the constant random variable.
12:38:470Annalisa Cesaroni: Because we show that, we show that, actually, if I have, mmm…
12:49:570Annalisa Cesaroni: a random variable which is orthogonal to all these, then actually, it has to be constant. It has to be equal to this is mean. Okay.
13:02:330Annalisa Cesaroni: Okay, next time we will do the, the…
13:08:130Annalisa Cesaroni: Some other result about this kind of, this space, with this, Scalar product, we… we…
13:20:790Annalisa Cesaroni: We will see that, actually, these are Spanish Albano space, which are called the Hilbert spaces.
13:27:110Annalisa Cesaroni: So… a Banak space.
13:31:660Annalisa Cesaroni: with a Scalar product is… with a Scalar product is called… Ilbert space.
13:47:400Annalisa Cesaroni: So, M2 is an Iilbert space. M2 is an Iilbert space, not just Banach.
13:55:740Annalisa Cesaroni: And we will see next time the, the…
14:00:710Annalisa Cesaroni: some result about these inverse spaces, which has some relation, for example, in the space of random variable with the notion of,
14:23:570Annalisa Cesaroni: Veher.
14:38:750Annalisa Cesaroni: With the notion of conditional expectation, sorry. Okay, we will see that actually, in Hilbert spaces, we can define a notion of projection that in M2 is exactly the notion of conditional expectation for a random variable, okay?
14:58:660Annalisa Cesaroni: conditional expectation of a random variable is to compute the… Say, an orthogonal projection, say.
15:07:740Annalisa Cesaroni: Let me see this.
15:09:20Annalisa Cesaroni: Next time.
15:10:910Annalisa Cesaroni: Okay.
15:14:290Annalisa Cesaroni: Okay, we'll land here.