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Transcript
00:04:250Paolo Guiotto: Not by.
00:07:750Paolo Guiotto: We have not yet started, but… You have a problem?
00:17:590Paolo Guiotto: Okay, so yesterday we introduced the definition of Fourier transform of a Borel measure, in fact. In particular, we are interested to the case when the Borel measure is a law of a random variable.
00:35:290Paolo Guiotto: And, the definition of Foulet transform is,
00:39:750Paolo Guiotto: inspired by the case when the variable is absolutely continuous. In that case.
00:48:760Paolo Guiotto: The Fourier transform of the density can be seen as the integral of the character E minus height CX with respect to the
01:01:340Paolo Guiotto: law of x. Now, this integral can be defined, whatever is the law of x, so not necessarily
01:10:260Paolo Guiotto: having a density, and this is called the Fourier transform of the measure mu, where mu is the law of X in particular, but the concept is independent of a probabilistic,
01:25:180Paolo Guiotto: structure. We may say that this is how it's defined in analysis, the Fourier transform of a Boreal measure.
01:37:460Paolo Guiotto: We have checked that it is always well defined, because the function, the character, is integral with respect to the measure mu, since the measure mu is a probability measure, so there is no… even if the function is not going to zero at infinite, there is no problem, because
01:56:730Paolo Guiotto: We are integrating with respect to a probability measure.
02:00:470Paolo Guiotto: If the measure, is the law of a random variable, with a density, then the Fourier transform of that measure, the law of X, coincides with the Fourier transform of X. And this was the, origin of this definition.
02:19:10Paolo Guiotto: It is interesting to notice that,
02:21:650Paolo Guiotto: The starting object, mu is a measure, the arrival object, mu hat, is a function, so…
02:29:480Paolo Guiotto: it's, in principle, something that should be easier to handle with respect to the measure. We computed the
02:37:750Paolo Guiotto: The… mmm… the characteristic function, the Fourier transform of,
02:45:490Paolo Guiotto: I'm using this term, characteristic function because I'm going to introduce in a moment, differentiate transform. Let's keep the Fourier transform, because there is a slight difference between the two definitions.
02:56:710Paolo Guiotto: for the Gaussian, distributions, we have done… the last calculation was for the Fourier transform of the low for a Gaussian multivariate, and that's the formula.
03:11:00Paolo Guiotto: So, we start from this point, so let's still go back a second to introduce, basically a new definition, which is still the same thing, basically. So, if, mu
03:27:900Paolo Guiotto: is a probability measure… on R.
03:37:350Paolo Guiotto: with the Borel sigma algebra, or if we have the multidimensional case, we have Rn with the Borel sigma algebra of Rn.
03:50:230Paolo Guiotto: D, Fourier transformer, mu cat.
03:55:600Paolo Guiotto: C… is, by definition, the integral on R, E minus iCX.
04:03:680Paolo Guiotto: integrated with respect to the measure mu individable X.
04:08:640Paolo Guiotto: This is the Fourier transformer of This measure mu.
04:15:280Paolo Guiotto: In particular.
04:19:589Paolo Guiotto: If mu is the law of a random variable, the Fourier transform of this is well-defined. So, the hat of this mu X is well
04:33:50Paolo Guiotto: defined.
04:36:500Paolo Guiotto: And it is equal to… well, mu hat of mu X hat C is the integral on R of E minus IC.
04:49:730Paolo Guiotto: X d mu X, but we know that integrating a function, we have this general formula, which says that if you integrate a function P of variable X,
05:10:270Paolo Guiotto: This is always a surprise.
05:13:270Paolo Guiotto: we have seen that this is the expected value of phi of X.
05:24:580Paolo Guiotto: And, so in particular, this means that this is the expectation of
05:31:340Paolo Guiotto: E minus ICX. This is for the, scar case. For the vector case, it is E minus IC dot X.
05:45:630Paolo Guiotto: if X is, multivariate.
05:55:110Paolo Guiotto: Now, this function here deserves a special name, not actually this one, but the function with the change design.
06:04:20Paolo Guiotto: Exponents, so we have this definition.
06:09:110Paolo Guiotto: given… a random variable, X, That can be… scalar, or multivariate, We define
06:28:250Paolo Guiotto: the… characteristic function.
06:39:650Paolo Guiotto: function.
06:42:100Paolo Guiotto: of XM.
06:44:580Paolo Guiotto: as this function.
06:47:320Paolo Guiotto: It is usually denoted with phi sub X.
06:51:10Paolo Guiotto: of C is, by definition, the expected value of exponential not minus iCX, but plus ICX, so ICX. This is for the scarcase.
07:06:360Paolo Guiotto: If, X is multivariate.
07:14:960Paolo Guiotto: The characteristic function is a function defined as the expected value of exponential xi.
07:25:30Paolo Guiotto: dot X. So in the first case, C is a real variable. In the second case, C is a vector of Rn.
07:37:640Paolo Guiotto: So, as you can see,
07:41:700Paolo Guiotto: And this is how you find this concept in probability textbooks. So they…
07:50:790Paolo Guiotto: They do not even mention, normally, the Fooley transformer as an operation on measures, but they define this, the characteristic function. But, of course, there is a strict connection, which is the characteristic function
08:09:940Paolo Guiotto: of the random variable X is, with the notation of the Fourier transform, is the Fourier transform of the low of X evaluated at minus C.
08:22:30Paolo Guiotto: There is nothing to do to eliminate this.
08:26:830Paolo Guiotto: So, in analysis, the Fourier transform is denoted with the minus sign, because this comes from this color product, etc. That minus is, because this is… this would be the conjugate of this.
08:41:730Paolo Guiotto: So, this is a product between… so when you have the function F, F's color something. Okay, so this is the origin of the minus from analysis. In probability, they prefer to use the sine plus
08:55:680Paolo Guiotto: So there is just a very little care to take. So, for example.
09:02:210Paolo Guiotto: If we have X, which is a multivariate portion, so let's do the general formula, where M is a vector of Rn, and C is positive, symmetric.
09:20:370Paolo Guiotto: The characteristic function of the Goshen.
09:25:200Paolo Guiotto: Well, you reminded that you computed just yesterday.
09:29:190Paolo Guiotto: this, the Fourier transform for, the law of, multivariate Gaussian. It is that formula. When you change Xi by minus C, to get the characteristic function.
09:44:960Paolo Guiotto: The, second term, 6C.C, does not change, because you have the minus 2 times, so it becomes a plus. And the first one changed signs, so you get this.
09:57:620Paolo Guiotto: You're getting.
09:59:160Paolo Guiotto: E to IC dot M minus 1 half CC dot C. That's the characteristic function for the multivariate quotient.
10:16:370Paolo Guiotto: We can compute, for example,
10:22:780Paolo Guiotto: Let's do some examples. So, if X is a uniform.
10:27:760Paolo Guiotto: random variable with the distribution on the interval AB.
10:34:00Paolo Guiotto: I remind you that this means that this is an absolutely continuous random variable with density, so X is absolutely
10:47:780Paolo Guiotto: Continos.
10:49:810Paolo Guiotto: We don't… density.
10:56:220Paolo Guiotto: Which is, basically the indicator of the interval AB.
11:01:20Paolo Guiotto: So, FX of little x is the indicator of interval AB,
11:09:780Paolo Guiotto: scaled to make this an integral 1 function by the length of the integral. So, divided by 1 over… divided by B minus A.
11:20:630Paolo Guiotto: So in this case, if you want to compute the characteristic function ph.
11:28:810Paolo Guiotto: So, of course, I should say it is the expected value of exponential ICX, but since I have a density.
11:39:230Paolo Guiotto: I go straight with this formula that whenever you do the expected value of P , it's the integral.
11:47:540Paolo Guiotto: of the function phi with respect to the distribu- the law of x, which is F of X times the x for this case. So I have the integral on R,
11:59:780Paolo Guiotto: of E to ICX, huh?
12:03:220Paolo Guiotto: times the density, FX, X DXN.
12:08:990Paolo Guiotto: So, integral on R, e to ic, 1 over B minus A, indicator.
12:19:440Paolo Guiotto: of the interval ABXDX. Basically, what we are doing is the Fourier transform of a rectangle, now a function which is 1 on a interval of 0 elsewhere.
12:32:200Paolo Guiotto: However, this… the rectangle is, the rectangle is the function which is symmetric with respect to the audience. However, the calculation is…
12:41:420Paolo Guiotto: Very similar, I have 1 over B minus A, the integral becomes the integral from A to B of E to IC.
12:48:610Paolo Guiotto: X.
12:49:950Paolo Guiotto: DXM.
12:52:520Paolo Guiotto: Now, integrating this exponential means I have to distinguish the case C equals 0 from C different from 0. So for C equals 0, I have 1 over B minus A integral from A to B of E to 0.
13:10:970Paolo Guiotto: So, E20 is 1.
13:13:770Paolo Guiotto: the X, and I get B minus A in the integral of a B minus A. Wow.
13:20:750Paolo Guiotto: Foxy different from zero.
13:24:20Paolo Guiotto: I can see this function, the exponential, as the derivative with respect to X of the same exponential into x divided by the factor xi.
13:39:440Paolo Guiotto: Because when you do the derivative, you get exactly P2ICX.
13:45:320Paolo Guiotto: So, since this is a constant with respect to the integration variable, I can write outside, I see.
13:52:470Paolo Guiotto: B minus A.
13:54:740Paolo Guiotto: And then we have the evaluation of EICX.
14:00:720Paolo Guiotto: from X equals A to X equals B.
14:05:890Paolo Guiotto: So we get 1 over ICB minus A.
14:13:400Paolo Guiotto: E2ICB minus E2ICA.
14:19:880Paolo Guiotto: And that's the characteristic function for this random variable.
14:28:770Paolo Guiotto: Well, the idea is that to each random variable we say that we always associate the… let's say that we are talking of a numerical random variable. We have similar concepts for the
14:43:190Paolo Guiotto: multivariate random variable. We always associate a low, which is a probability measure, a CDF, which is a concrete function.
14:53:800Paolo Guiotto: Sometimes we can have the density, if it is absolutely continuous, but not always.
14:59:560Paolo Guiotto: And we can also associate a characteristic function.
15:03:760Paolo Guiotto: Normally, the idea is that it is better to deal with functions than with measures, so knowing the CDF, or if available, the density, or the characteristic function should be better to work with the random variable.
15:21:960Paolo Guiotto: Rather than working with the law, which is not better specified probability measure, or going back to the space, working with the abstract probability bit.
15:35:550Paolo Guiotto: Which might be on a space where we do not have any metric structure and things like that.
15:42:50Paolo Guiotto: But which one is better depends from the cases.
15:51:490Paolo Guiotto: I will illustrate an example in a moment. For example, with Gaussian, things, the characteristic function is very good, because
16:03:470Paolo Guiotto: as we will see, the Fourier transform, the characteristic function is unique, so if two random variables, they have the same characteristic function.
16:17:370Paolo Guiotto: They also have the same distribution, so they are the same under the probabilistic point of view.
16:24:700Paolo Guiotto: So if you are able to compute the characteristic function and see that it is the characteristic function of a random value, you know, you recognize, you can say it is a random value with that distribution because of the uniqueness. And this works nicely with Gaussian, for example.
16:42:710Paolo Guiotto: But with other variables, it's almost impossible to compute the Fourier transform, okay? So, for example, we can do… because these calculations have been done previously when we have seen the Fourier transform. If X is an exponential.
17:02:220Paolo Guiotto: With parameter, lambda, let's say.
17:06:150Paolo Guiotto: remind that this means that X is absolutely continuous, with density.
17:19:670Paolo Guiotto: FXX is E to minus lambda X divided by lambda, indicator 0 plus infinity.
17:30:510Paolo Guiotto: X.
17:36:70Paolo Guiotto: So, in this case, we have that the characteristic function PXC of this is, again, the expected value of EICX. Since we have the density, this is the integral.
17:53:130Paolo Guiotto: on R of E, I, C, X,
17:58:610Paolo Guiotto: the density of X, so E minus lambda X over lambda indicator, 0 plus infinity, X.
18:08:430Paolo Guiotto: DX.
18:10:100Paolo Guiotto: So this means that we have to integrate on 0 plus infinity.
18:14:530Paolo Guiotto: I can write de facto 1 over lambda outside P2IX minus lambda times X, the X.
18:25:830Paolo Guiotto: In this case, since lambda is a positive number, yeah.
18:34:320Paolo Guiotto: We see that this number is never equal 0, because it's a complex number with the real part minus lambda, imaginary part C.
18:45:850Paolo Guiotto: To be zero, these two values should be 0. Now, C can be zero, but lambda is never zero, so that number is different from zero.
18:54:750Paolo Guiotto: So we can look at this as the derivative with respect to C of the same function, e to ic minus lambda X, divided by that coefficient.
19:06:280Paolo Guiotto: IC minus lambda. So we can integrate, saying that it is 1 over lambda.
19:12:790Paolo Guiotto: We have the evaluation of this thing, EIXC minus lambda times X divided IXC minus lambda. This from X equals 0 to X equals
19:26:410Paolo Guiotto: plus infinity. Now, this is a factor I can just write
19:31:550Paolo Guiotto: outside of this evaluation, 1 over lambda IC minus lambda.
19:42:380Paolo Guiotto: I'm… I'm…
19:51:830Paolo Guiotto: I suspect that this is not divided by lambda, but times lambda.
19:58:550Paolo Guiotto: It should be this one, because when we do the integral, this would be the integral of
20:04:360Paolo Guiotto: So the evaluation E minus lambda X between… yeah, it is not divided by lambda, but multiplied by lambda, sorry.
20:12:850Paolo Guiotto: So… Oops.
20:17:80Paolo Guiotto: This is de facto.
20:21:300Paolo Guiotto: So here, the lambda is a numerator.
20:26:30Paolo Guiotto: Then we have to evaluate this exponential, EIC minus lambda at, say, let's write, plus infinity minus the value at 0, and it is E to 0.
20:39:610Paolo Guiotto: at plus infinity, you see that you have a part which is E to IXC, let's say X,
20:48:690Paolo Guiotto: and the part which is E minus lambda X. This is a modulus 1 number, and this is going to zero. So when you do the limit of this thing, you go to 0, so that value will be 0. And therefore, the final result is lambda…
21:07:870Paolo Guiotto: This is 1 minus 1, so this parenthesis is minus 1. So if you want, we can write lambda minus iC
21:17:540Paolo Guiotto: And this is the value of the characteristic function.
21:22:650Paolo Guiotto: But apart from very few cases, it's impossible to compute
21:31:410Paolo Guiotto: The, the characteristic function of, any random variables.
21:37:750Paolo Guiotto: Well… some remarks… On these, some properties of these, remarks.
21:48:150Paolo Guiotto: on that.
21:49:490Paolo Guiotto: CX.
21:51:570Paolo Guiotto: A first remark is that we can easily see that whatever is X, the value of the characteristic function at 0 must be always equal to 1.
22:04:510Paolo Guiotto: And this is because if you compute phi X at 0, this is the expected value of e to i0x.
22:16:300Paolo Guiotto: Now, this is 0, so e to 0 is 1. Expectation of 1 is 1, because we are on a probability space.
22:26:920Paolo Guiotto: guide.
22:27:910Paolo Guiotto: So this is, 1.
22:31:880Paolo Guiotto: This is a first property. A second general property is that, and this is basically part of the Riemann-Lebag theorem, same kind of argument, this function, the characteristic function, is a continuous function in real line.
22:52:120Paolo Guiotto: This follows… I do not repeat the proof, because it's the same proof we have seen for the L1 Fourier transform, okay? So this…
23:04:950Paolo Guiotto: follows… from… De… dominated the… convergence.
23:18:290Paolo Guiotto: that you apply to the, to the expected value, which is an integral at the end. Now, you remind that the expectation, no, phi X of C, is the integral on omega of e to icp.
23:37:960Paolo Guiotto: So, let's do the proof. When… so, to check continuity, you take a sequence CN that goes to some point C star, and you want to prove, this is the goal.
23:53:220Paolo Guiotto: that the value phi X at point CN goes to phi at X at point C star.
24:02:870Paolo Guiotto: So this is the problem.
24:05:660Paolo Guiotto: But if you write what is phi X of CN, you see that this is an integral of omega of EICNX.
24:19:180Paolo Guiotto: in the probability P. So, as you can see, there is a sequence into the integral, okay? So, let's say that this is NF
24:29:470Paolo Guiotto: and of the variable, which is the variable that you have here. So here, it would be the omega that it is here. So these are a function of variable omega.
24:42:540Paolo Guiotto: Now, what happens here?
24:45:430Paolo Guiotto: Yeah, we noticed that, huh?
24:52:640Paolo Guiotto: So, to apply the dominated convergence, I mean, number one.
24:58:100Paolo Guiotto: to show that the point-wise limit exists at least for almost every omega under probability P. But what is the limit? Fn of omega
25:10:860Paolo Guiotto: being this E to IXNX of omega, you do the limit here, index CN. Since XN goes to that value C star.
25:22:620Paolo Guiotto: You can say that whatever is omega, provided this quantity is defined, so since this is a random variable, it will be defined for almost every omega with respect to the probability P. This will go to E2IC star X omega.
25:41:680Paolo Guiotto: let's say, for every omega, such that there exists a capital X of omega, this means for… well, let's say P,
25:55:400Paolo Guiotto: almost every omega are in probabilistic jargon, P, almost surely.
26:05:960Paolo Guiotto: Okay. Number two, we need an integral dominant for the sequence FN, for the absolute values.
26:16:880Paolo Guiotto: But, when you take an omega, you do the absolute value, you are doing the absolute value of EI CN capital X of omega.
26:26:760Paolo Guiotto: Now, our capital X is real-valued, so this is a real number, C is a real number, CN is a real number, so that's an exponential of an imaginary number. So it's a unitary number in C, so the modulus is 1, constantly.
26:46:660Paolo Guiotto: And if you take this function 1, the function constantly equal to 1, which is, of course, independent of the n, which is the index for which I do the limit.
26:57:620Paolo Guiotto: This is an L1 function for the space omega, because we have a probability space, so constants are integrable automatically.
27:07:400Paolo Guiotto: No? So, dominated convergence, applies…
27:14:160Paolo Guiotto: And it says that we can pass the limit into the integral. So we can say that when we do the limit in n of the values phi at point CN, which is the limit in n.
27:26:620Paolo Guiotto: of the integral on omega of iCln.
27:32:840Paolo Guiotto: X.
27:38:280Paolo Guiotto: We can travel.
27:39:790Paolo Guiotto: the limiting side.
27:41:770Paolo Guiotto: And we get that this is the limit, the integral of the limit of the same thing, but we just say that 0.1, that when you do the limit in n of the quantities inside, you get that value, E2IC star X.
27:58:210Paolo Guiotto: So, at the end, we get integral of omega of e to iC star.
28:03:490Paolo Guiotto: capital X, DP, and that's the value of the characteristic function at point C star.
28:10:770Paolo Guiotto: So we show that whatever is the sequence CN that goes to a certain point, which is obituary here, C star.
28:20:900Paolo Guiotto: The characteristic function on point CN will go always to the value of the characteristic function at point C star, and this exactly means that the limit when the variable C goes to C star of CXC
28:40:260Paolo Guiotto: Is the value of the characteristic function at point C star.
28:48:150Paolo Guiotto: Basically, we made a proof of this spot.
29:01:400Paolo Guiotto: Okay, so another important property of the characteristic function is that PRX is bounded?
29:13:870Paolo Guiotto: And this is easy, because if you take the absolute value of phi X,
29:20:700Paolo Guiotto: It is the modulus of the integral of… let's keep the expectations. So, the modulus of the expectation of EICX,
29:31:200Paolo Guiotto: Since the expectation is an integral, we can carry inside the model. This is less little than the expected value.
29:39:700Paolo Guiotto: of the modulus.
29:43:580Paolo Guiotto: Now, that's again equal by T1.
29:46:70Paolo Guiotto: We'll be glad.
29:51:830Paolo Guiotto: So we can be more precise.
29:55:130Paolo Guiotto: This is… that's the return one forever to see.
29:59:570Paolo Guiotto: Now, you reminded that for the L1 Fourier transform, the Fourier transform also goes to 0 to infinity. We have never proved this factor. At least, we have proved that if we assume some regularity for the function.
30:14:520Paolo Guiotto: for which we do the Fourier transform, but that's a general factor, okay? The Fourier transform goes to zero at infinity. Well, this is different here, because this is not necessarily true.
30:27:800Paolo Guiotto: So, in general.
30:37:800Paolo Guiotto: CF…
30:44:160Paolo Guiotto: when soon goes to the background, you see finished.
30:47:120Paolo Guiotto: So, to show this, we have an example, a simple, very simple example. Let's take as X a random variable constantly equal to some value X0.
30:59:730Paolo Guiotto: So in this case, what is the characteristic function? Here, it's triggeredly equal to…
31:06:790Paolo Guiotto: expected value of e to ic, but since capital X is constant equal to X0, this is the expectation of e to ic, this little x0. Now, this is a constant.
31:25:690Paolo Guiotto: Constant in what?
31:27:850Paolo Guiotto: in the integration variable, not the expected value is an integral in the variable of the space omega. So you are integrating a constant, you get a constant. So this will be E2iCX0.
31:42:990Paolo Guiotto: So, as you can see, this is not going to zero. It's not even having a limit when C goes to plus infinity, so it does not have, it does not go anywhere when C goes to plus minus infinity.
31:59:450Paolo Guiotto: So, for example, in particular, if X is constantly equal to 0, you get that de-characteristic function of this, you take X0 equals 0, is constantly equal to 1. So, it's flat. It doesn't go to 0.
32:18:480Paolo Guiotto: So this is an important difference.
32:23:740Paolo Guiotto: Another important feature that comes from the property that we have seen for the Fourier transform, I'm talking about the derivative of the Fourier transform.
32:37:650Paolo Guiotto: you remind that the derivative of the Fourier transform is a Fourier transform of something like variable times F, this formula. Now, something similar happens here, and we have this. So, if the variable X is L1,
32:54:150Paolo Guiotto: omega, so this means that the expected value of modulus of X is finite.
33:02:530Paolo Guiotto: Then, the characteristic function is differentiable.
33:06:980Paolo Guiotto: And… so there exists the derivative of the characteristic function, and this is,
33:14:540Paolo Guiotto: Of course, this, this result will come out from differentiating under integral sine. It is the expected value of IXEIC capital X.
33:29:840Paolo Guiotto: And, in particular, We see that if we take X equals 0,
33:38:150Paolo Guiotto: The derivative of the characteristic function at zero this kills the exponential.
33:45:970Paolo Guiotto: you get e to 0, so constant equal to 1, so it is equal to I, the expected value of X.
33:54:960Paolo Guiotto: We have an extension of this, more in general.
34:02:850Paolo Guiotto: which is due to the iteration of this property. If the expected value of modulus X to power n is finite.
34:17:610Paolo Guiotto: This means that, technically, we would write X is an LN omega variable.
34:26:489Paolo Guiotto: Here, N is natural.
34:29:570Paolo Guiotto: 1, 2, 3, etc. Great or equal than 1.
34:34:840Paolo Guiotto: So, if we have that the nth power of X has finite expectation, then there exists the nth derivative of the characteristic function.
34:51:969Paolo Guiotto: with respect to the variable X,
34:55:550Paolo Guiotto: And now, as you may imagine, it's just an iteration of this formula, so if you do another derivative, you get another IX. If you do another derivative, you get another IX. So, once you have done n derivatives, you will have IX to power n into the expectation, so you get this.
35:13:860Paolo Guiotto: Expected value of IX to power n e to ic.
35:22:990Paolo Guiotto: And, in particular.
35:26:650Paolo Guiotto: for C equals 0, we get this formula, that the antiderivative of C at point… of the antiderivative of the characteristic function at 0, when you put X equals 0 into the formula, it is equal to i to n, the expected value
35:45:520Paolo Guiotto: of X to N.
35:48:760Paolo Guiotto: Now, these quantities, we expected values of ballots of random variable, are called the moments of the random variable.
35:59:550Paolo Guiotto: So the first of all the moment, N equals 1, is just the mean value. The second moment is not the bias, but it's, strictly related to the bias, because it is e to the X squared. So to get the bias, you have to subtract the square of the mean value, and so on.
36:18:410Paolo Guiotto: So in some sense, from the characteristic function, by computing derivatives, we can get these quantities through this formula. Now, just to, understand the
36:32:710Paolo Guiotto: Basically, it's the same kind of argument we have already seen for the L1 Fullerton form. Let's just quickly justify the first derivative.
36:45:270Paolo Guiotto: So let's… justify…
36:52:110Paolo Guiotto: The first derivative.
36:54:400Paolo Guiotto: Now, of course, to show that.
37:00:870Paolo Guiotto: there exists the derivative with respect to C of the characteristic function.
37:06:650Paolo Guiotto: we want to differentiate the expected value of EICX, right?
37:14:710Paolo Guiotto: So, the idea is that since the expected value is an integral over omega, what we want to do, the tool we have to do this derivative, is to carry inside the derivative.
37:28:930Paolo Guiotto: If this can be done, so, let's put, for the question here, we have that this will be the expected value of the derivative, respect to C, of this EICX.
37:41:610Paolo Guiotto: And of course, if you do the derivative, you get that derivative with respect to C, this is I capital X, that multiplies the factor of C times the exponential, or if you want, the derivative of the exponential is the exponential times the derivative of the factor.
38:00:130Paolo Guiotto: And that would be the conclusion, the formula we have here, right? Now, to check that this can be, this is possible.
38:11:930Paolo Guiotto: is possible.
38:15:660Paolo Guiotto: If… reminder, there are two conditions. Number one, we need that the derivative can be computed, and in fact, this is possible, we have done. So there exists the derivative with respect to C of the function with respect to the parameter.
38:33:440Paolo Guiotto: And this, in fact, is I, capital X, E-I-CX. This is for every serial.
38:42:20Paolo Guiotto: There is no condition, and for the omegas, for which capital X of omega is well-defined, and therefore, omega, almost surely.
38:54:720Paolo Guiotto: Number 2.
38:56:640Paolo Guiotto: we need a bound for this quantity, a bound which is independent on the parameter, of the parameter C. So we need to take the derivative, respect to C, of that EI CX,
39:10:80Paolo Guiotto: which is, by the way, I, capital X, E, I, C, X. And we need to bound with an L1 function independently of the parameter.
39:20:250Paolo Guiotto: But if you take the models, you get modulus of I, models of X, and modulus E, I, C, X.
39:28:200Paolo Guiotto: modulus of i is 1, modulus of X is itself, and this one is again a modulus 1 number. So at the end, this is equal to modulus of X, which is supposed to be L1 by assumption. You remind that in assumptions, we made this.
39:48:190Paolo Guiotto: Requirement. X is in L1, no? Expectation of modules of X, fine.
39:53:680Paolo Guiotto: So, this is a dominant for the derivative independent of the parameter, so for every value C real.
40:01:910Paolo Guiotto: And, again, almost surely in omega. Omega, almost surely.
40:08:150Paolo Guiotto: Therefore, we have the conditions to apply the theorem, and the conclusion follows.
40:26:330Paolo Guiotto: Okay,
40:29:670Paolo Guiotto: remarkable, property of this characteristic function that, is there are… for example, it is possible to show that this is different, again, from the L1 Fourier transform.
40:44:670Paolo Guiotto: You remind that for the L1 Fourier transform, we have not seen the L2.
40:50:240Paolo Guiotto: But for the L1 Fooley transform, we do not have a characterization of the image of the Fourier transform, so we don't know… we do not have a test, there is not a test, truth or false, that says
41:05:780Paolo Guiotto: This function is the Fourier transform of some function, okay?
41:10:620Paolo Guiotto: So we know that to be the Fourier transform of an L1 function, you must be continuous, you must go to 0 at infinity.
41:20:720Paolo Guiotto: You must be bounded, but we can have functions that… it's possible to prove that we can have functions that are continuous, bounded, going to zero at infinity, which are not fully transformed of anything.
41:36:980Paolo Guiotto: Okay?
41:38:170Paolo Guiotto: And here, for example, in the exam, some of you… there was a question.
41:45:80Paolo Guiotto: Concerning the… you never mind the last exercise, second point, the question was, suppose that you have a function who has a fully original.
41:55:10Paolo Guiotto: What can you say about the square of that function?
41:59:680Paolo Guiotto: Does the… this… this H square has a fully original or not?
42:07:200Paolo Guiotto: And some of you probably confused what we proved with the inversion formula. Let's say that if a function, if you start from an L1 function.
42:22:180Paolo Guiotto: which has also a Fourier transform in L1, then that function has a Fourier original.
42:27:700Paolo Guiotto: The Fourier transform is invertible. Now, that's in sufficient condition.
42:31:700Paolo Guiotto: In that case, you could have, and some of you have done this, taking H squared, you discover that if H is the hat of F, H squared is F hat times fet, so it's the convolution
42:46:00Paolo Guiotto: It's the transform of the convolution, so the convolution is a free original. It's an L1 because it's a convolution of two L1 functions, is L1, so that's fine. You don't need to justify to check that the function, or to add the condition on the func… that the function must be in L1, okay? That's not required.
43:05:970Paolo Guiotto: Okay, let's close the parentheses. But in any case, we do not have a characterization, a test that says you are a Fourier transform of something if and only if this happens.
43:17:920Paolo Guiotto: For the characteristic functions, there is a test of this type. So it is possible to say that a characteristic function, a function is a characteristic function of sample, if and only if a certain condition happens. We do not see… this is called the Bochner theorem, we do not see this here. But I want to discuss, in any case.
43:42:90Paolo Guiotto: uniqueness. So the fact that if two functions… two random variables have the same
43:49:650Paolo Guiotto: characteristic function, then they must be the same random variable. They must have the same distribution, the same law. So this is the uniqueness
44:03:690Paolo Guiotto: TRM.
44:07:630Paolo Guiotto: So, we know that if, X and Y are… So… Random variables, such that…
44:24:750Paolo Guiotto: They have the same law, so mu X is equal to mu Y,
44:30:570Paolo Guiotto: Then, they have the same, the same characteristic functions, then phi X is equal to phi Y.
44:41:630Paolo Guiotto: And this because, you know, this is trivial, tx of c, in terms of the mu X, is the integral on R, P, I, C, X, D mu X.
44:58:520Paolo Guiotto: No?
44:59:530Paolo Guiotto: But since this measure is equal to the measure muy.
45:05:410Paolo Guiotto: And so, changing letters, this is the integral on R of E to ICY d mu Why?
45:16:730Paolo Guiotto: in the variable Y, this is the characteristic function of
45:21:880Paolo Guiotto: Y at point C. So they two are the same, so this is a trivial fact.
45:27:280Paolo Guiotto: The key point is that also the vice versa holds.
45:33:70Paolo Guiotto: So, this is an important TRM.
45:38:690Paolo Guiotto: If phi X is equal to phi Y,
45:43:110Paolo Guiotto: then the two distributions are the same. Nu X is equal to mu Y.
45:49:210Paolo Guiotto: So this is important, because whenever you know that, for example, the characteristic function of a random variable is the same of the characteristic function of a Gaussian, you can deduce that the random variable is Gaussian. Let me just, first, before we see this factor, show an example of this.
46:09:650Paolo Guiotto: So, suppose that X is a multivariate Goshen.
46:17:440Paolo Guiotto: So, where M is a vector of RN, C positive and symmetric.
46:28:70Paolo Guiotto: Then we take a vector A, fixed, over N, and we define this variable y equal A dot X. So, A1X1 plus A2X2 plus
46:45:930Paolo Guiotto: etc. ANXN.
46:52:120Paolo Guiotto: a linear combination of these, components of this Gaussian array.
46:59:180Paolo Guiotto: Now, it turns out that, Y is motion, and we determine the distribution. So, let's say, let's put as a problem, show that…
47:12:580Paolo Guiotto: Why?
47:13:630Paolo Guiotto: is Goshen, And… determine its… Density.
47:27:290Paolo Guiotto: Well, it means a balance, basically.
47:29:910Paolo Guiotto: Now, I could determine mean and variance just doing the calculation of the mean and the variance of this. It's not particularly complicated, but this wouldn't say that the variable is quotient.
47:42:590Paolo Guiotto: they would say, what is the mean and what is the variance. So, if I want to show that it is Gaussian, I need to show that the density of this Y is the density of a Gaussian random variable. So, in principle, I could solve this problem directly, so let's say a direct solution.
48:05:540Paolo Guiotto: it seems feasible, because I needed to compute this… the density of Y, okay? Now, the problem is that Y is a function of X, but in this case, we are not on a change
48:25:480Paolo Guiotto: of distribution, because, you see, we start from X, which is an array, and we end to Y, which is a number. So we do not have a map of X into Y, Rn to Rn, but rather Rn to R. So it's an unusual situation.
48:44:260Paolo Guiotto: So, there could be two ways to attack this. One way could be, okay, let's complete this map by adding some other components, which is a little bit tricky.
48:55:990Paolo Guiotto: Or we could say, okay, let's, instead of starting from density, let's start from the CDF. I could say that the CDF, which is always defined of Y, is the probability that Y is less or equal than little y.
49:14:700Paolo Guiotto: So I should compute the probability that A dot X is less or equal than Y.
49:21:140Paolo Guiotto: And this could be done by using the… because this involves the array X, this could be done by using the density, so I would have to compute the integral on the domain A dot little x less or equal than Y,
49:39:160Paolo Guiotto: of the Gaussian density, so 1 over root of 2 pi to the n determinant of matrix C,
49:49:70Paolo Guiotto: E2 minus 1 half C minus 1, say, oh, sorry.
49:54:870Paolo Guiotto: X minus M, Scala X minus M, in DS.
50:02:230Paolo Guiotto: Now, I should compute this multiple integral.
50:09:220Paolo Guiotto: The problem is that this seems to be tricky. It's… this is enough plane, no, you see? A dot. Well, let's assume that, of course, A is non-trivial.
50:20:390Paolo Guiotto: So, different from zero. Otherwise, if A is 0, A.x is constantly equal to 0, so the situation is different.
50:30:610Paolo Guiotto: If A is non-zero, that's enough plane. No? Because, you see, it's A1X1 plus A2X2 plus A3XE, etc, less or equal than something. It's an F plane. But how do we compute this integral?
50:48:420Paolo Guiotto: That's the point.
50:50:150Paolo Guiotto: Let's use the characteristic function, and let's see how the calculation becomes really, really easy.
50:58:270Paolo Guiotto: That's why… an example of what I said above,
51:04:910Paolo Guiotto: Y for Gaussians, it is a good tool, the characteristic function. Because if I want to compute the characteristic function of Y,
51:14:840Paolo Guiotto: Well, this is the expected value of EICY.
51:21:640Paolo Guiotto: Right?
51:23:150Paolo Guiotto: But why is A dot X?
51:28:00Paolo Guiotto: So we can say that this is the expected value of EICA dot X.
51:36:620Paolo Guiotto: That reminds of the characteristic function of X.
51:41:630Paolo Guiotto: And in fact, I may think this coefficient, this C, is scalar, because Y is a numerical variable, so the variable of the characteristic function is number.
51:55:360Paolo Guiotto: I can give this C to the vector A. Now, CA is what? CA is the vector with components CA1, CA2, CA3, CAN.
52:11:360Paolo Guiotto: Okay? So now, you see that here we have the characteristic function of the array X evaluated at this factor, CA, right?
52:26:460Paolo Guiotto: Now, we know what is the characteristic function of a multivariate Gaussian.
52:31:130Paolo Guiotto: We, we're reminded, here the formula.
52:35:990Paolo Guiotto: is EIG… well, let's say, at the vector of C, our vector is AHC, CA, sorry. So, let's say that…
52:49:560Paolo Guiotto: Since X is normal with mean M and covariance matrix C,
52:56:850Paolo Guiotto: D ph, let's use a different letter, eta is EI eta dot M minus 1 part C eta dot eta. No? This is the CDF of the multivariate X. So.
53:16:920Paolo Guiotto: CX, evaluated at CAE, where this is a scholar.
53:23:650Paolo Guiotto: And this is a vector.
53:27:350Paolo Guiotto: And that vector is, as I said above, the product, scala times vector, will be EI, eta is CA, this dot M minus 1 half C, eta is again C, A dot CA.
53:47:150Paolo Guiotto: If you want to let's put some pains, but it's useless.
53:51:40Paolo Guiotto: Now, from this, we get that this is E2I
53:56:140Paolo Guiotto: you know that this color product is linear, so I can take back out this coefficient and see this as C that multiplies this colored product between A and M, right?
54:09:780Paolo Guiotto: minus 1 half. Again, also here, this color can be written outside of the scalar product, and reminded C is a matrix, so it's a linear…
54:22:420Paolo Guiotto: operation. So you can take disk C, take out of the product here, and go out of C. So I get XC from this factor, another XC from the other one, so this becomes C squared CA dot A.
54:38:440Paolo Guiotto: Now, if you look at this, this is phi Y of X, right? How it depends on C.
54:47:380Paolo Guiotto: What you read is this. It is E2 It is E to I.
54:55:410Paolo Guiotto: C.
54:57:250Paolo Guiotto: times something, a number. Let's, let's give a name, alpha for a second.
55:04:440Paolo Guiotto: minus 1 half a number, this one, CA dot A, which is a positive number, because C is strictly positive, no? A is non-zero, so that quantity is positive. I could call it sigma square, so this is sigma square.
55:22:870Paolo Guiotto: Xi Square.
55:25:990Paolo Guiotto: But that's the characteristic function of a Gaussian
55:31:210Paolo Guiotto: Okay, so this is the, the characteristic function.
55:36:670Paolo Guiotto: characteristic.
55:39:740Paolo Guiotto: function.
55:41:790Paolo Guiotto: of a scalar normal variable with mean alpha and the variance sigma squared.
55:52:140Paolo Guiotto: So, since I have the uniqueness, my variable has this distribution. So, Y…
55:58:310Paolo Guiotto: is a normal with mean alpha and variance sigma squared, so that means, the mean is A dot M, and the variance is CA dot A.
56:13:450Paolo Guiotto: So if you want the density, we can now write FY
56:18:550Paolo Guiotto: Y is 1 over root of 2 pi sigma squared is CA dot A.
56:28:270Paolo Guiotto: Times the exponential, minus 1 half.
56:32:720Paolo Guiotto: We have X minus the mean, which is A dot M,
56:38:290Paolo Guiotto: divided by the sigma square, which is CA dot A.
56:43:660Paolo Guiotto: I'm sorry, the Y is the lesser in here the square. And that's the density.
56:51:190Paolo Guiotto: I… I'm sure that if you…
56:53:650Paolo Guiotto: try to compute this integral, this could be a little bit more complicated to, to do this.
57:03:910Paolo Guiotto: So this is an example, and especially with Gaussians, it works nicely because of certain reasons. And from this fact, also, we can draw certain conclusions.
57:18:950Paolo Guiotto: Okay, so let's, let's,
57:23:100Paolo Guiotto: see this, this important theorem, the uniqueness theorem, so that if you have the same characteristic function, you must have the same distribution, okay?
57:37:680Paolo Guiotto: Now, that…
57:39:480Paolo Guiotto: I want to do this also, this proof, because it is a little bit based on what we have seen for… it is actually based on the Fourier transform we have seen for the L1 case.
57:50:550Paolo Guiotto: Okay, so to prove the theorem, we needed to start with the LM.
57:58:970Paolo Guiotto: Whoa.
58:00:30Paolo Guiotto: I will open a quick parenthesis, because this is something we have not seen for the Fourier transformer. That is a particular case of dilemma we want to see. This is called the duality Lemma.
58:18:730Paolo Guiotto: it is, let's say, a fundamental step in building the L2 Fourier transform. It says that if you have two functions, F and G in L1,
58:30:50Paolo Guiotto: you can… you have this formula, that if you take the integral of F times the written form of G,
58:37:930Paolo Guiotto: So here, since we have… we have to put some letters for the variable, so let's say that we use the letter X, and we will use the letter X also for the Fourier transform of G, okay? Well, we can call the variable of the Fourier transform as we are.
58:55:530Paolo Guiotto: Now, the nice thing is that you can move the head from one factor to the other. So this is equal to the head of f at point XG of X.
59:08:420Paolo Guiotto: Yes.
59:11:770Paolo Guiotto: It's just straightforward. You write down the left-hand side, and you can see that you can move the D-hat on the other side. Indeed, so let's see the quick proof.
59:27:430Paolo Guiotto: If you start with this, f of x g hat of x.
59:34:90Paolo Guiotto: Now, I, of course, replace here the definition of Fourier transformer. That's the traditional Fourier transformer, L1 Fourier transformer.
59:43:900Paolo Guiotto: By the way, notice that there is no problem with the definition of the integer, because in this particular theorem, you should say that it's integral, right?
59:54:290Paolo Guiotto: Now, F is for me and one, and B6 is the latent form of sheep.
00:00:360Paolo Guiotto: Which is a bounded function, particular.
00:04:670Paolo Guiotto: So this problem is bounded by F, and therefore it is integral, okay? So there is no problem with the integral. Now, I will replace to G hat this… its Fourier transform, so we have to be careful, because we are used to write G hat of C, let's say. So this is the integral of G,
00:27:500Paolo Guiotto: we cannot use the variable X, because this is already used for… so let's use the letter Y for the variable. E minus i
00:37:340Paolo Guiotto: I should put here the XC, but Xi is the letter, that I put in the argument of different forms, so this is actually X.
00:45:580Paolo Guiotto: Y, DY. So, what we have here is an integral on R, f of x.
00:51:510Paolo Guiotto: integral on R, G, Y, E minus IXY, this is DY, and this is DX.
01:03:150Paolo Guiotto: Okay.
01:04:250Paolo Guiotto: Now, what we do is we flip the order of the two integrations. This can be done because of the Fabini theorem. I don't want to
01:13:300Paolo Guiotto: spend the time on technical details. So, if you flip the order, so this means that,
01:22:120Paolo Guiotto: you do first the integration in X, and then dot that one in Y, so it means that you have F of X,
01:30:350Paolo Guiotto: This remains inside because it depends on both the variables, on X and Y, so I have to… still to write there.
01:38:100Paolo Guiotto: So this is now integration in X, and then outside, we have,
01:42:650Paolo Guiotto: times G of Y in DY.
01:46:480Paolo Guiotto: But what do you see here?
01:48:720Paolo Guiotto: This is the Fourier transform of F, no? Evaluated at this point, this is the variable, you see? Because X is the integration variable, now it's Y the variable of this thing. So this is the F at point Y. So at the end, we have integral on R, F at Y,
02:08:280Paolo Guiotto: G of YDY.
02:11:90Paolo Guiotto: And that's the end of this book.
02:13:280Paolo Guiotto: Okay? So the idea is that you can move the head from one factor to the other of a product.
02:21:90Paolo Guiotto: Now, this, now extends to this lemma, which is, Extension.
02:33:830Paolo Guiotto: Now, imagine that we have mu, Borel, Probability.
02:42:300Paolo Guiotto: measure.
02:44:930Paolo Guiotto: and, say, F and L1 function.
02:51:430Paolo Guiotto: FNL1… And L1, be careful, because here, it is L1R,
03:00:500Paolo Guiotto: So, with respect to the Lebec measure, LeBague measure.
03:09:140Paolo Guiotto: Then, the duality lemma states that if you do the integral on R of the Fourier transform of F in the measure mu.
03:19:930Paolo Guiotto: And notice that, so this is the…
03:22:850Paolo Guiotto: and wants to get those FOMO better, okay?
03:26:290Paolo Guiotto: So the traditional L1 Fourier transform. L1 Fourier transform level.
03:33:680Paolo Guiotto: So, in particular, since mu is a probability measure, and this guy is continuous and bounded, this integral is well-defined, because bounded functions are always integral with respect to probability measures. There is no problem with the definition of this. So…
03:52:270Paolo Guiotto: I should check that the effect is in that one perspective, the measure view.
03:57:600Paolo Guiotto: But since it is bounded, and it is, it is measurable by the continuity, and by boundedness is integral. Now, it happens that you can flip the hat on the other factor, well, you may say, but there is one as the factor. No, imagine that the other factor is the measure mu.
04:17:440Paolo Guiotto: And this becomes the integral still on R of F,
04:21:340Paolo Guiotto: mu hat in the Liebagum measure. So, let's say that DX, we call this.
04:29:550Paolo Guiotto: Notice that also this second lead, the government.
04:33:630Paolo Guiotto: Because here, we assume that that is 1, the back measure. D factor of mu is the mu.
04:40:390Paolo Guiotto: latest form of the measurement that we know is continuous boundary. So, this factor here is continuous amount of times vector, this is controlled by act, which is the integral part. So, everything makes sense.
04:55:200Paolo Guiotto: Now, this is a calculation similar to the previous one, so just a question of applying the definition. So, proof…
05:06:700Paolo Guiotto: we start again from the integral of F at, Vimeo…
05:12:980Paolo Guiotto: We write this as integral of f at. So, let's say that this is F at of, X,
05:22:270Paolo Guiotto: D mu X, okay? So we are integrating, and the letter we use for the variable is X. This is the integral on R of F of Y
05:33:380Paolo Guiotto: E to minus iXY in DY, exactly the same way we did above.
05:42:700Paolo Guiotto: So now we have this is a double integral, so there is an integration. The innermost integration is that one, F of YE minus iXYDY. Then we have to integrate all this with respect to the measure mu in the variable X.
06:01:390Paolo Guiotto: Now, here, as you…
06:05:210Paolo Guiotto: Imagine we do the same operation, so we flip the two integrations. So there is an extension of the Fubini tonality theorem that allows this. So imagine that we flip the two integrations.
06:19:850Paolo Guiotto: So we do first… for the moment, I will write the same function.
06:24:630Paolo Guiotto: But now, let's integrate first in EXA.
06:28:350Paolo Guiotto: and then in Y.
06:31:20Paolo Guiotto: Since F of Y here is independent of X, which is the integration variable for that integral, this is a content, so it comes out. So we have integral on R of FY, integral on R, E minus i.
06:48:380Paolo Guiotto: X, Y, D mu, all this in DY. And the quantity that you see here is, by definition, what we call the Fourier transform of, of mu. So this is mu hat evaluated at Y.
07:07:870Paolo Guiotto: So at the end, we have integral on R, F, Y mu hat Y in DY, which is exactly what
07:19:430Paolo Guiotto: Was in the statement, okay?
07:26:320Paolo Guiotto: You see?
07:28:40Paolo Guiotto: Okay, so it's a similar proof, and basically it's a straightforward proof. We put inside the definition of the Fourier transform, and we flip the order of the integration to see what happens. Now, we can see the proof of the main theorem.
07:48:740Paolo Guiotto: Uniqueness.
07:53:00Paolo Guiotto: Dior.
07:56:790Paolo Guiotto: So, the hypothesis is that we have two random variables for which their respective characteristic functions are the same.
08:10:80Paolo Guiotto: And the thesis is to convince you that the distribution they lose, must be the same.
08:23:399Paolo Guiotto: Now, remind that, there is, this, connection between… because this formula, the duality formula, is written for a measure, Borel measure mu, that, in particular, we want to apply to the
08:42:350Paolo Guiotto: low of X and Y. So let's write this formula here for mu X and mu Y, and let's see what happens.
08:53:10Paolo Guiotto: So, from… duality, We have, huh?
09:04:00Paolo Guiotto: this. So we start with mu equals mu X. So I write integral of f hat d mu X,
09:14:50Paolo Guiotto: integral on r of f at d mu X,
09:19:740Paolo Guiotto: is equal to… by duality, I can move the head on the other factor, which is the measure, so this is F of X
09:31:560Paolo Guiotto: mu.
09:33:270Paolo Guiotto: X hat of X, DX, right? This is duality.
09:43:470Paolo Guiotto: Now, new X hat is more or less CX. You remind that there is a relation… unfortunately, the Fourier transform of probabilists is different from the Fourier transform of
10:01:650Paolo Guiotto: analysis, because of the sign in the argument, but basically they are the same.
10:08:670Paolo Guiotto: You see, the relation is that one is the other with the argument change design. So I can say that if you have the mu X head at point C, this will be the phi X at point minus C, okay? So let's use this.
10:31:270Paolo Guiotto: So here, this is the fiance.
10:35:660Paolo Guiotto: at point minus X.
10:39:50Paolo Guiotto: Now, here is where the assumption comes. Phi X is equal to phi Y. So, this is the hypothesis.
10:49:340Paolo Guiotto: And so this is equal to phi Y of minus X.
10:54:940Paolo Guiotto: And going back, phi Y of minus X will be the hat of muy evaluated at X.
11:03:800Paolo Guiotto: So, at the end of all this, I can say that this integral is equal to the integral on r of f of x, the hat of muy
11:14:180Paolo Guiotto: X in the X.
11:16:710Paolo Guiotto: And because of the same reason, duality.
11:20:760Paolo Guiotto: I can remove back the hat on the first factor, So, duality.
11:27:370Paolo Guiotto: And say that this is the integral of F at respect to the measure muy.
11:34:710Paolo Guiotto: So…
11:35:910Paolo Guiotto: the combination of all this has produced this identity, that… so if phi X is equal to phi Y, we have that the integral of F at
11:49:150Paolo Guiotto: in D mUX,
11:51:210Paolo Guiotto: is equal to the integral of F at in the muy, and this is for every F in L1 realign with respect to the LeBague measure. So, say, let's call it DX.
12:07:730Paolo Guiotto: Now, this looks like the conclusion.
12:10:970Paolo Guiotto: Because it seems like if I'm saying all integrals respect mu and X are the same are all integrals respect muy, the particular I need to stay together.
12:23:00Paolo Guiotto: Instead of sector at the indicator, you would have that integral of the indicator is the measure of the sector equal to the measure of the set.
12:33:00Paolo Guiotto: Okay? So, this should be more or less very close to be the conclusion.
12:39:340Paolo Guiotto: The unique problem is a technical problem. I cannot take an indicator yet. Imagine that you put an indicator of an electron. So…
12:49:740Paolo Guiotto: The idea is, for the next step.
12:52:840Paolo Guiotto: is to replace this by indicator of a generic interval, AD.
13:00:880Paolo Guiotto: But can I do that, really?
13:03:690Paolo Guiotto: Can I say that, if, I take the indicator, the indicator is the head of something.
13:13:620Paolo Guiotto: The answer is…
13:24:410Paolo Guiotto: Can you… can you say that this is the hat of some effort for some effort?
13:30:740Paolo Guiotto: F, of course, in L1, R with respect to the Lebec measure.
13:40:170Paolo Guiotto: You know, what are the properties of the Fourier transform of an L1 function? We say that the function must be…
13:47:660Paolo Guiotto: Continuous.
13:49:690Paolo Guiotto: And that function is not continuous, so it cannot be the Fullerton form, okay?
13:55:510Paolo Guiotto: Since, F.
14:01:640Paolo Guiotto: Hat must be continuous.
14:04:640Paolo Guiotto: an indicator cannot be
14:10:820Paolo Guiotto: the Fourier transform of an L1 function. So, I cannot find that function. So, this is the technical problem. I now will tell you how to solve this. It's a technical point, it's not a…
14:27:520Paolo Guiotto: conceptual point. Well, the first remark is that,
14:34:830Paolo Guiotto: From this formula, let's give a name to this.
14:39:570Paolo Guiotto: Ballet.
14:41:340Paolo Guiotto: So, from the first point is that from this, formula.
14:49:400Paolo Guiotto: It follows that the integral on R, I will use the letter phi, phi denu X,
14:57:440Paolo Guiotto: is equal to the integral on R of phi denu y. This, for every phi, which is good enough
15:06:770Paolo Guiotto: So, here, I will take this English force class. Remind that class of extremely regular functions that go, to zero very fast at infinity together with their derivatives.
15:23:360Paolo Guiotto: This because, this… Because,
15:33:520Paolo Guiotto: If you take a Schwartz function.
15:37:170Paolo Guiotto: Now, you remind that,
15:42:490Paolo Guiotto: Well, I can always say that ph is definitely the Fourier transform. There is an L1 function, such that ph is the Fourier transform of that function. Why?
15:55:590Paolo Guiotto: Here, this is because of the inversion formula. That is, in the…
16:07:380Paolo Guiotto: And here, I will return on the question we discussed before.
16:14:120Paolo Guiotto: If I want to prove that the existence of a Fourier original, I have a sufficient condition, which is, if you know that the function phi and its transform, phi hat, are both in L1, and this is true, because phi is a Schwartz function, and so
16:33:650Paolo Guiotto: Also, the transform of phi is a Schwarz function. We proved this fact. The Foulette transform maps the Schwarz space into itself, and Schwarz functions
16:45:910Paolo Guiotto: are in L1, because they are… they came fast at infinity, and moreover, they are regular. So, since these two are in L1, we know that inversion formula applies
16:58:00Paolo Guiotto: Inversion formula.
17:01:150Paolo Guiotto: applies.
17:05:250Paolo Guiotto: And the inversion formula says that your function p, at point x is 1 over 2 pi, the double hat of P evaluated at minus X.
17:18:320Paolo Guiotto: But the right-hand side is a hat, is the hat of 1 over 2 pi, the hat of phi.
17:26:870Paolo Guiotto: with the argument changed of sine evaluated at X.
17:32:900Paolo Guiotto: So you see that phi is the Fourier transform of this guy, F, which is an L1 function, because, of course, the factor 1 over 2 pi does not change the feature of this fiat. Fiat is in L1, okay?
17:49:640Paolo Guiotto: So, since, the function, the Schwarz function, is the Fourier transform of something, and for functions which are Fourier transforms, the identity falls, this means that also for Schwarz functions, the identity falls.
18:06:340Paolo Guiotto: So we now know a little bit more than this. This said, the integral of, the integral with respect to UX and UI of the same
18:18:430Paolo Guiotto: function, which is the transform of F, is 1. Now I know that the two measures have the same
18:31:590Paolo Guiotto: But…
18:33:90Paolo Guiotto: So, now, the point is that, can I, replace here, do one more step, Schwartz function in that indicator? That seems, impossible, because the indicator is discontinuous, and the Schwarz function is right.
18:49:400Paolo Guiotto: But the idea here can be the following. Now, Take an indicator.
18:56:470Paolo Guiotto: I will… I will, I will describe qualitatively. If you look in the notes, you can see the analytical construction of this. Now, take an indicator of an interval, generic interval AB.
19:10:400Paolo Guiotto: Now, the indicator is a function made like that. It is 0 here, it is 1 here, and 0 here.
19:17:630Paolo Guiotto: The idea is to regularize this indicator by using functions of this nature. You say that they are zero.
19:26:270Paolo Guiotto: A bit at left of point A, so let's say A minus 1 over N,
19:35:790Paolo Guiotto: Then they move regularly up to 1, so very smoothly. Now, the point is, is it possible and how to do that?
19:45:500Paolo Guiotto: Trust me, it is possible to do in a very regular way, see infinity way, so with all derivatives, and then doing the same here at right. So, moving down regularly in such a way that you are 0 after B plus 1 over n.
20:04:140Paolo Guiotto: call this function fianna, and suppose that we can do, with the function which is in the Schwartz space, okay? Now, the idea is that when it goes to infinity, these blue functions should insert to the red function.
20:20:180Paolo Guiotto: This would be quantized automatically.
20:27:450Paolo Guiotto: So, such that diffian are Schwarz functions, they are pointwise converging to the indicator of the interval AB, X by X. So, for every X, not almost every, even. So, for every, you can prove this.
20:47:650Paolo Guiotto: Then, if you apply The, the… let's call it the two ballots.
20:57:120Paolo Guiotto: step.
20:58:680Paolo Guiotto: you have that the integral on r of phi n in d mu X
21:05:940Paolo Guiotto: by the previous argument, the two ballots, is the integral on R of phi n in d muy.
21:16:450Paolo Guiotto: for everyone.
21:19:660Paolo Guiotto: And to Phoenicia…
21:25:450Paolo Guiotto: To finish, we pass to the living.
21:28:50Paolo Guiotto: And we use the dominated converters.
21:30:920Paolo Guiotto: You don't need any more, because by construction, as you can see, these functions are all positive and bounded by 1.
21:41:140Paolo Guiotto: Okay, and therefore, I would say that they are all controlled by… I drew a controller for all these functions, let's say…
21:53:560Paolo Guiotto: By doing green, huh?
21:56:310Paolo Guiotto: But actually, I could say that, since in the idea, this A minus N to…
22:06:470Paolo Guiotto: So, let's say that all the Fiena.
22:17:180Paolo Guiotto: Also, the fee N are smaller than the fee one.
22:21:750Paolo Guiotto: So I have a bound independent of N, which is a Schwartz function, so it is an L1 function.
22:28:820Paolo Guiotto: So… By passing to the limit, this will go to the integral on R of phi d mu X,
22:37:330Paolo Guiotto: I dominated.
22:39:120Paolo Guiotto: convergence, and for the same reason, this will go to the integral on R,
22:45:150Paolo Guiotto: Actually, the fee is the indicator.
22:48:570Paolo Guiotto: Indicator AB, And here I have the same indicator, AB, indemu Y.
22:58:20Paolo Guiotto: But this will be the measure, mu X, of the interval ABE,
23:03:770Paolo Guiotto: And this one will be the measure Y of the interval AB.
23:09:990Paolo Guiotto: And since the limit is unique, the conclusion will be that this identity holds, and this is the third
23:18:10Paolo Guiotto: step, so we prove that mu X on AB
23:24:800Paolo Guiotto: is equal to muy on AB, for every interval AB.
23:34:530Paolo Guiotto: Now, as you may imagine, once two Borel measure.
23:38:810Paolo Guiotto: So mesh, which are defined on the sigma generated by the universe.
23:46:480Paolo Guiotto: When the two coincide, I'm originally supposed to be made, like, that we coincide on every set, okay? So, we accept this,
23:57:750Paolo Guiotto: So, from this, it follows that MUX
24:01:660Paolo Guiotto: of E is equal to muy
24:05:70Paolo Guiotto: of E for every e Borel set.
24:09:540Paolo Guiotto: of R, and this finishes the proof, okay?
24:17:590Paolo Guiotto: Okay, let's say that, this is, this is,
24:25:720Paolo Guiotto: All we need… we will return on the characteristic function later, when we will do convergence, because characteristic function is related to, way, sequences of random variable converge.
24:41:290Paolo Guiotto: At the end of the chapter, there are a few exercises.
24:46:990Paolo Guiotto: I uploaded today, I recognize that there are errors in these exercises, so I corrected, and I also… there were a couple of exercises that
24:58:720Paolo Guiotto: where the assignment was not complete at all, was just the beginning of the exercise. So I corrected and republished the notes, and also I added the next two or three
25:11:460Paolo Guiotto: But, so please check, and do some of the exercises, maybe we have a…
25:19:860Paolo Guiotto: Some minutes to do some of them.
25:23:570Paolo Guiotto: So, do exercises.
25:27:660Paolo Guiotto: From 5… 3, 1, 2…
25:34:730Paolo Guiotto: well, I may say there are some of them which are theoretical. We have not yet, 2-9.
25:44:780Paolo Guiotto: We still need to pass through independence before we can really start working with probabilities. So, yeah, we are still a bit limited. But in any case, you can do some of these exercises. So, for example.
26:01:790Paolo Guiotto: Some of them are… the first three are calculation of the characteristic functions, but this is because… this is basically an older exercise on computing Fourier transforms.
26:13:780Paolo Guiotto: So… Let's take the 5… Did he… 5…
26:23:880Paolo Guiotto: This is an exercise, on,
26:28:620Paolo Guiotto: densities, so it's, let's say, a theoretical exercise. Let's, take X, Y, B, absolutely continuous.
26:40:880Paolo Guiotto: random variables.
26:44:250Paolo Guiotto: with density.
26:51:620Paolo Guiotto: F, X, F, Y.
26:57:120Paolo Guiotto: Check that this identity holds.
27:04:580Paolo Guiotto: It is integral on R of phi XC, so the characteristic function of X, times the density of Y. Since we are here computing an integral in the variable C, I will call
27:22:650Paolo Guiotto: the argument of FY also C, because that's the integration variable. E minus IC… Why?
27:32:800Paolo Guiotto: Dixie… It's just… I think it's another version of duality.
27:38:840Paolo Guiotto: Equality integral on R of phy, at Y minus X, FX.
27:49:320Paolo Guiotto: Yes.
27:50:660Paolo Guiotto: There is a mix of things here. You see, there is a convolution on the right-hand side.
27:56:770Paolo Guiotto: And there is a transformer at the left-hand side. Well, let's see what is it.
28:02:520Paolo Guiotto: So, solution, let's start from the… the integral ph F… Y… C… E minus ICY.
28:22:290Paolo Guiotto: the, C.
28:26:320Paolo Guiotto: This looks like the kind of calculation we have done a minute ago with the duality. Now, the unique ingredient we need to probably to write is this phi X.
28:38:440Paolo Guiotto: PXCE is the characteristic function of, capital X, so EMA E to ICX.
28:49:550Paolo Guiotto: Which is, since X is absolutely continuous, and we have a density, this is the integral.
28:55:810Paolo Guiotto: of, that function, EICX, times the density FXEDx.
29:05:380Paolo Guiotto: Bright.
29:06:910Paolo Guiotto: So let's plug this into this integral, we get a double integration. So we have an integration of this integral.
29:17:850Paolo Guiotto: E-I-C… F… X… the AXA.
29:25:250Paolo Guiotto: times FYC… E minus IXCY.
29:33:540Paolo Guiotto: in Dixie.
29:38:970Paolo Guiotto: Well, probably we have to flip the order of the two integrations. If we look at the final
29:48:420Paolo Guiotto: integral that contains FX. So, probably, let's flip the two integrations.
29:55:550Paolo Guiotto: So we have an integral.
30:00:810Paolo Guiotto: So if we do first the integration in C, and last the integration in X, so, this…
30:10:860Paolo Guiotto: Excellent.
30:12:120Paolo Guiotto: We… here we have, F, Y, XC…
30:18:460Paolo Guiotto: So definitely we have to keep inside the E minus iCY.
30:23:520Paolo Guiotto: But also, I have to take this inside, because it contains seeds.
30:30:710Paolo Guiotto: So, outside, I have only FXX.
30:35:90Paolo Guiotto: Okay.
30:36:470Paolo Guiotto: Now, coupling these two exponentials, we have EIC, X minus Y.
30:48:310Paolo Guiotto: So this is the integral on R of FY.
30:52:450Paolo Guiotto: C… Dixie.
31:01:340Paolo Guiotto: And this looks like, the same of the characteristic function of Y, no? Because you have the density. The unique difference is that,
31:12:700Paolo Guiotto: So the phi Y, let's say that phi Y, I do not use the letter X C, because it's already, used there. Let's say of eta, this is the integral on R of FY etta
31:30:710Paolo Guiotto: And no, sorry, FY… let's use the letter C, as in that integral.
31:36:720Paolo Guiotto: E, I, EI, let's say eta C dxi. This would be the characteristic function for Y. As you can see, it's the same if we take eta equal this value.
31:55:760Paolo Guiotto: So this is the FYI.
31:58:720Paolo Guiotto: of X minus Y.
32:02:710Paolo Guiotto: So, since all this… is this. We get, finally, that this is the integral on r of phi Y
32:15:30Paolo Guiotto: X minus Y.
32:17:690Paolo Guiotto: F… X minus Y or Y minus X, so…
32:24:950Paolo Guiotto: Where did that minus comes from? That minus comes from the origin.
32:31:90Paolo Guiotto: So, this is FX.
32:35:720Paolo Guiotto: QVX.
32:40:20Paolo Guiotto: So there is a mistake in the…
32:43:780Paolo Guiotto: That is X minus Y here.
32:52:300Paolo Guiotto: If I'm not wrong, right? You see?
32:55:470Paolo Guiotto: Lovely.
32:56:280Paolo Guiotto: I guessed it.
33:00:40Paolo Guiotto: Okay, so it's 12.02, we stop here, do the exercises, and… And see you on Friday.
33:12:80Paolo Guiotto: Who knows?
33:14:910Paolo Guiotto: Oh.
33:16:780Paolo Guiotto: And I don't see… is not from… I'm sorry.
33:25:900Paolo Guiotto: I see now. I'm sorry.
33:29:510Paolo Guiotto: It's too late for that.