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Transcript
00:08:800Paolo Guiotto: Good morning.
00:13:220Paolo Guiotto: Okay, I start with the… An exercise left from yesterday on mapping random vectors.
00:27:700Paolo Guiotto: This is the 436, so we have, XYB variant.
00:35:330Paolo Guiotto: random variable.
00:37:530Paolo Guiotto: we'd, density.
00:44:100Paolo Guiotto: F, X, Y… equal to E minus X minus 2 absolute value of Y, Indicator, zero, passing frequency.
00:58:490Paolo Guiotto: Excellent.
01:02:240Paolo Guiotto: Number 1… Check it.
01:07:520Paolo Guiotto: that… FXY is a probability density.
01:18:170Paolo Guiotto: Number 2… define Z… as X squared plus Y.
01:28:590Paolo Guiotto: and W… as 3X squared minus Y.
01:37:330Paolo Guiotto: So, show… That's… prepare the variable. ZW is… Absolutely. Continuous.
01:50:780Paolo Guiotto: And, but… determine… its density.
02:00:990Paolo Guiotto: FZW.
02:05:720Paolo Guiotto: Number 3, compute the probability that Z plus W.
02:11:170Paolo Guiotto: did positive.
02:15:40Paolo Guiotto: And number four, compute.
02:18:130Paolo Guiotto: Also, the densities… of… Z in W.
02:29:410Paolo Guiotto: It's quite, say, standard exercise.
02:36:600Paolo Guiotto: So, first we have to check that this is a probability density. We notice that to be a probability density to be…
02:47:470Paolo Guiotto: It… probability… density… this F, X, Y, Moss.
02:59:910Paolo Guiotto: B.
03:01:640Paolo Guiotto: So that… two conditions. Number one, must be positive.
03:06:880Paolo Guiotto: Greater or equal than zero, not necessarily street 3. And this is evident because it is exponential times the indicator.
03:14:740Paolo Guiotto: So we can just say… True.
03:19:320Paolo Guiotto: And the second, the integral on F2 of the function fxy.
03:26:100Paolo Guiotto: Must be equal to 1.
03:30:110Paolo Guiotto: Now, if we do the integral on F2 of this F,
03:34:440Paolo Guiotto: This is E minus X minus 2 absolute value of Y indicator, 0 plus infinity, 4X.
03:46:700Paolo Guiotto: the XDY.
03:50:800Paolo Guiotto: Well, there is nothing particular to say, that we apply the reduction formula, integrating first, it's indifferent. Well, we can integrate first in X, for example.
04:03:310Paolo Guiotto: And then in Y…
04:05:230Paolo Guiotto: the function, which is E minus X, then that would be E minus 2 absolute value of Y, which is a constant in the X integration, so we can write outside them.
04:16:269Paolo Guiotto: For the range X, this condition means x between 0 plus infinity, and since there is no condition for y, minus infinity to plus infinity.
04:27:630Paolo Guiotto: Now, this integral, this is the derivative with respect to X of minus e to minus X.
04:36:310Paolo Guiotto: So the integral will be the evaluation of minus C minus X, From 0 to plus infinity.
04:45:310Paolo Guiotto: At zero… at plus infinity, we get 0.
04:49:470Paolo Guiotto: minus… at 0, we get minus E201, so minus 1.
04:55:590Paolo Guiotto: So we have 1 for the integral.
04:59:190Paolo Guiotto: This is now the integral for minus infinity plus infinity of E minus 2 modulus y times that integral
05:09:220Paolo Guiotto: as… Is equal to 1, so we have to integrate that.
05:16:250Paolo Guiotto: Well, since there is the modulus, this function is even, so we could say that it is 2 times the integral from 0 to plus infinity, E minus 2Y.
05:28:00Paolo Guiotto: the wife… And that's the easy, easy but to one, okay? Sure.
05:34:440Paolo Guiotto: This has been checked.
05:38:290Paolo Guiotto: Number two, we have these, two variables.
05:43:700Paolo Guiotto: Z and Y defined by these formulas. So clearly, this is a function of XY. We have that ZW, the pair ZW, is a function of the pair XY.
05:59:650Paolo Guiotto: Where… This function, phi.
06:04:750Paolo Guiotto: is mapping R2 into R2, which is defined as, just taking lowercase letters, X squared plus Y, the first component.
06:15:790Paolo Guiotto: 3, X squared minus Y, the second component.
06:20:980Paolo Guiotto: Now, as you can see, there is this X square that says that this is not injective, no, because if you
06:28:40Paolo Guiotto: I take X equals 1 and X equals minus 1, they go into the same point.
06:32:240Paolo Guiotto: So, in principle, this is not invertible.
06:36:500Paolo Guiotto: However…
06:39:740Paolo Guiotto: So, yesterday, I gave you, let's say, the version of the formula saying that this map fee… where is it?
06:53:510Paolo Guiotto: is a map from RN to RN, and it is invaluable, etc.
06:59:340Paolo Guiotto: Actually, we don't need that to feed via dejection on our end, because suppose that the variable X, for example, X is an array, okay?
07:10:960Paolo Guiotto: So, if X is an array, and for example, the components of X are positive, it means that that capital X
07:19:880Paolo Guiotto: if we are in two dimensions, so let's say that X is a parent of two random variables. If the two components are positive, it means that that vector X belongs to the first quarter.
07:31:860Paolo Guiotto: So, it's important that phi be a ejection from where the values of X are taken to the… to the… to the… to the codomain. So, here we could notice that our X
07:49:360Paolo Guiotto: If you look at the density.
07:51:770Paolo Guiotto: This, in particular, says that the density is zero for X negative, so this will mean that the probability that X is negative is 0. Let's see this. We notice…
08:08:990Paolo Guiotto: That, huh?
08:11:340Paolo Guiotto: The probability that X is negative
08:15:810Paolo Guiotto: is equal to water. So, if I have to write this probability in terms of the joint density, I could say this is the probability that the
08:24:760Paolo Guiotto: pair XY belongs to minus infinity to 0 times R,
08:32:49Paolo Guiotto: in such a way that I can use the law of x, or the density, to compute this probability. This will be the integral on this set, minus infinity to 0 times R of the density FXY
08:46:520Paolo Guiotto: X, Y, BX.
08:49:640Paolo Guiotto: the Y.
08:51:290Paolo Guiotto: But, you remind that the definition of the density for this problem is that we have that exponentially multiples of Y indicator on 0 plus infinity for X.
09:09:700Paolo Guiotto: DXDY.
09:12:310Paolo Guiotto: But in this domain, the indicator is 0, because for this domain, X would be negative, so the indicator of positive number would be zero, so that would be constantly equal to zero, so this number is 0.
09:28:130Paolo Guiotto: So, it means that DX is positive. So, in other words, the vector XY belongs to this set, which is 0 plus infinity for X, and for Y, there is no restriction R.
09:49:10Paolo Guiotto: Now, if we call, this, D,
09:55:430Paolo Guiotto: On this set, this map could be considered… it becomes injective because the X is positive. So I want to show that on this set D, this fee becomes a rejection. So let's consider…
10:10:80Paolo Guiotto: Sincer.
10:12:810Paolo Guiotto: XY belongs to D.
10:16:300Paolo Guiotto: weed, probability 1.
10:22:720Paolo Guiotto: We can.
10:24:690Paolo Guiotto: consider…
10:28:650Paolo Guiotto: our map fee as a map from D to whatever is its image we will see later. So, in this case, it is, by definition, subjective, but it is also injective, and actually, we will see in that table, we will compute the inverse.
10:46:740Paolo Guiotto: In this way.
10:52:740Paolo Guiotto: B is.
10:54:900Paolo Guiotto: Clearly.
10:58:200Paolo Guiotto: Subjective.
11:05:340Paolo Guiotto: And the… also… injective.
11:14:900Paolo Guiotto: So, it is in vertical.
11:20:490Paolo Guiotto: Well, if you want, we prove injective, but in fact, we need, to… to apply the change,
11:28:90Paolo Guiotto: the… the change of density formula, we need the inverse map phi minus 1, so we do more… we need more than just checking injectivity. But you can see that, suppose that,
11:45:880Paolo Guiotto: Well, let's prove directly… That's… overlooked.
11:52:340Paolo Guiotto: that, ease.
11:56:590Paolo Guiotto: investors.
12:00:420Paolo Guiotto: So we… we do this. So, if we call Z equal, ZY… let's say ZW equal phi of XY, this means that, Z is, the two components are X squared plus Y,
12:21:800Paolo Guiotto: And W is 3X squared minus Y.
12:26:860Paolo Guiotto: So now, if I want to invert this to express XY functions of ZW,
12:35:270Paolo Guiotto: I could notice that if you do the sum of these two equations, you eliminate Y, so you get that Z plus W is equal to 34X squared.
12:48:470Paolo Guiotto: To… to get the… we could, square, sorry.
13:01:510Paolo Guiotto: I don't…
13:02:800Paolo Guiotto: No, we need also to have the equation for Y. So, I can do, for example, 3 times the first equation, and it becomes 3Z equal 3X squared plus 3Y.
13:19:620Paolo Guiotto: Then do the first minus the second. This minus this one.
13:26:190Paolo Guiotto: So we get 3Z minus W, the 3X squared disappear, we get 4Y. So this yields this.
13:36:950Paolo Guiotto: Y is equal 3Z minus W divided by 4, and X square is equal to Z plus W divided by 4.
13:49:310Paolo Guiotto: Now, since our point XY has X positive, because we are considering this map fee as if it starts, this is the plane XY,
14:03:100Paolo Guiotto: in this set, the set where X is positive and Y is any real, so this is the domain. And this map here, is, is going into a plane ZW.
14:23:680Paolo Guiotto: Where, we have…
14:27:380Paolo Guiotto: So that X is the root… well, actually, you may wonder, what is the image? Let's take one second here. Since this must be verified, you see that the second one, in the condition, this quantity must be positive.
14:46:630Paolo Guiotto: Okay, because it's a square. So this, in particular means that Z plus W must be positive, so W greater than or equal than minus Z. So it means that if ZW is phi of XY,
15:00:640Paolo Guiotto: necessarily, W must be greater or equal than minus Z. So in this plane, the plane ZW, this is W equal minus Z. W equal minus Z. W above means here.
15:17:70Paolo Guiotto: So the point, if XY is here, phi of XY is there.
15:31:670Paolo Guiotto: And so this is the image of the draw.
15:37:310Paolo Guiotto: this map phi. So, in this case, we can say that Y is 3Z minus W divided 4, and X is the root
15:50:650Paolo Guiotto: I say the positive root because X is positive. Now, otherwise X squared equals something should yield two values for X, but since X must be positive, we take on it the positive root, so let's remind here that X must be positive.
16:06:570Paolo Guiotto: So Z, root of Z plus W divided 4, or root of Z plus W divided 2.
16:13:570Paolo Guiotto: Out of the…
16:22:200Paolo Guiotto: Okay.
16:24:90Paolo Guiotto: So the map you see here is, we said that ZY is phi of XY if and only if XY is this thing. So XY is the vector made at components root of Z plus W, divided to
16:42:370Paolo Guiotto: 3Z minus W divided 4, and this is the phi minus 1 map of ZW.
16:54:450Paolo Guiotto: Okay, so now we have that this fee is invertible, and of course, it will be differentiable.
17:04:599Paolo Guiotto: well, actually, that could be a little technical problem on Zed…
17:10:869Paolo Guiotto: on the W's for which the argument of the root is zero, so on that line, on the line as W equal minus Z, this… this fee, this map, would be not differentiable, because the root is not differentiable when the argument is zero. But that's a set of points which is negligible for…
17:30:330Paolo Guiotto: the Lebec measure, which is the measure under which we consider the densities, so we can say that the density of ZW
17:42:300Paolo Guiotto: at point ZW is the density of XY at the point XY that corresponds to XW, which is that point, root of Z plus W divided by 2, and second component, 3Z minus W divided by 4,
18:01:530Paolo Guiotto: times the modulus of the determinant of phi minus 1 prime ZW.
18:11:310Paolo Guiotto: If we want, we compute the Jacobian matrix of phi minus 1,
18:15:740Paolo Guiotto: Or in alternative, we can do this calculation directly from phi. Let… let me… because maybe, in this case, it's a bit, maybe not too much. So, we can…
18:32:150Paolo Guiotto: compute.
18:35:990Paolo Guiotto: the determinant of the Jacobian matrix of V minus 1 directly
18:46:420Paolo Guiotto: So we take… this is a phi minus 1 that you see here.
18:52:230Paolo Guiotto: It's a function of ZW. We have a 2x2 matrix, where there are the gradients of the two components. So let me just set this calculation, but I won't do the calculation this way.
19:05:790Paolo Guiotto: So, phi minus 1 prime, ZW,
19:09:820Paolo Guiotto: will be the matrix made of the gradient of root of Z plus W.
19:15:640Paolo Guiotto: divided by 2, and the gradient of 3Z minus W divided by 4.
19:22:410Paolo Guiotto: It's not a particularly complicated calculation.
19:26:120Paolo Guiotto: The reason why I don't do this is just to show you an alternative way, okay? So you can do this.
19:33:180Paolo Guiotto: Or…
19:34:510Paolo Guiotto: noticing that there is a relation between the Jacobian matrix of phi minus 1 and the Jacobian matrix of phi.
19:43:160Paolo Guiotto: Since, the Jacobian metrics.
19:50:210Paolo Guiotto: So it's the derivative of the inverse, no? In one variable calculus, this is the reciprocal of the derivative of the direct function.
20:00:40Paolo Guiotto: In several variable calculus, the relation is the following. You do the Jacobian matrix of phi, and you take the inverse matrix of that.
20:10:610Paolo Guiotto: Okay?
20:11:970Paolo Guiotto: Now, we don't need to compute this matrix, that would be a little bit of a mess, not particularly for a 2x2 matrix, but since we need to do the determinant of this matrix, we notice that the determinant of this is exactly the reciprocal of the determinant of the direct map.
20:29:770Paolo Guiotto: Fee, the problematics of fee.
20:34:610Paolo Guiotto: That's why it's maybe a bit easier. Well, let's see. Phi prime, the unique thing you have to be careful is that phi is a function of XY.
20:45:790Paolo Guiotto: And we want, at the end, a result in ZW. So let's start with phi prime. Phi prime of XY is the Jacobian matrix of phi. Phi is the direct map which is written here. If you want, this is…
21:01:840Paolo Guiotto: the direct map for you.
21:04:10Paolo Guiotto: So the two components are X squared plus Y, and 3X squared minus Y. So we have to put down there the two gradients of this. So we have the gradient of X squared plus Y, and the gradient of 3X squared minus Y.
21:23:900Paolo Guiotto: Now, these gradients are easy, because they are polynomials, so derivative with respect to X2X, derivative with respect to Y1, derivative with respect to X6x, derivative with respect to Y minus 1. So that's the Jacobian matrix of phi prime.
21:40:20Paolo Guiotto: The determinant of this matrix, phi prime, is equal to, so minus 2X minus 6X at minus 8X.
21:54:50Paolo Guiotto: So… the determinant
21:58:210Paolo Guiotto: of the inverse… of the Jacobian matrix of the inverse is 1 over the determinant of the Jacobian matrix of the direct map, so this will be 1 over minus 8X. As you can see, this is something that depends on X.
22:16:10Paolo Guiotto: But this function is a function of ZW, so how do you write ZW here? Well, you remind that, since we inverted, X is this quantity here.
22:28:350Paolo Guiotto: Okay? So, this is minus 1 over 8 root of Z plus W divided by 2. So, at the end, what is it? Minus 1 over 4 root of Z plus W.
22:47:710Paolo Guiotto: Okay?
22:49:680Paolo Guiotto: Now, of course, if you do directly the calculation, you get the same, it's different, this. And so, to return to the density, so the conclusion is that the density of the pair ZW, point ZW, is
23:09:530Paolo Guiotto: the density of XY evaluated in that pair, so, yeah, exponential.
23:17:500Paolo Guiotto: minus E minus X, so root of Z plus W divided by 4, if I'm not stronger.
23:26:30Paolo Guiotto: Now, divide by 2.
23:31:910Paolo Guiotto: minus 2 modulus of Y, right?
23:36:350Paolo Guiotto: Modulus of Y is 3Z minus W over 4, so modulus 3Z minus W over 4.
23:44:790Paolo Guiotto: Then we have the indicators, 0 plus infinity of X, which is, which is constantly 1, because that is the positive root of this number, so this is identically 1.
24:00:60Paolo Guiotto: Times the modulus of the determinant of the inverse
24:05:410Paolo Guiotto: Well, let's write down here, otherwise you have to write 2 times times modules of determinant of the Jacobian matrix of the inverse, but we have seen that this is…
24:16:710Paolo Guiotto: the determinant of the Jacobian matrix of P minus 1 is minus 1 over 4, the root, so with the modulus, is 1 over 4, the root of Z plus W.
24:31:80Paolo Guiotto: But that's the density of this thing.
24:38:430Paolo Guiotto: going to be a mess, please. Number three.
24:42:440Paolo Guiotto: Probability that Z plus W is positive.
24:52:50Paolo Guiotto: Well…
24:53:810Paolo Guiotto: We could say that this is the integral of the density of ZWZW, on the domain where Z plus W is positive, which is a half plane.
25:09:640Paolo Guiotto: But this probably would be a mess, okay?
25:14:170Paolo Guiotto: Because the density seems to be…
25:17:650Paolo Guiotto: However, I get that since we have to compute the density of Z and W, we have to, to take the problem of,
25:26:850Paolo Guiotto: computing the integrals, or however, we will see later how to do. In alternative, I could remind that since Z is, what is, X squared plus Y,
25:40:840Paolo Guiotto: And W is 3X squared minus Y.
25:46:660Paolo Guiotto: So Z plus W is, in fact, 4X squared.
25:51:640Paolo Guiotto: So, I'm computing the probability that Z plus W is positive, means the probability that 4X square is positive, which is clearly equal to 1.
26:04:390Paolo Guiotto: Without any calculation, okay?
26:07:640Paolo Guiotto: So, number 4 now… Compute the marginal density of ZFW.
26:16:900Paolo Guiotto: Here, so, first of all, since, ZW is absolutely continuous.
26:30:20Paolo Guiotto: the two marginals, Z and W's, are absolutely continuous as well. So, we know that there is a density, and in principle, the density of Z
26:42:690Paolo Guiotto: Should be determined by doing the integral on the other variable of the joint density.
26:56:390Paolo Guiotto: Now, is this…
27:06:230Paolo Guiotto: Is this reasonable, or not?
27:22:500Paolo Guiotto: I don't know, it's not a good shape, this thing, okay?
27:29:710Paolo Guiotto: So… Let's see if we can do differently.
27:37:510Paolo Guiotto: So that is the variable… X squared plus… Wives…
27:50:320Paolo Guiotto: I would say, maybe, let's try to determine the CDF of this by using the density of X.
27:58:120Paolo Guiotto: And why?
28:00:760Paolo Guiotto: And then doing the derivative, we get the density of Z, no? So, I have that the CBF of Z,
28:09:120Paolo Guiotto: Is the probability
28:11:650Paolo Guiotto: that this capital Z is less or equal than this little z, so the probability that X squared plus Y
28:20:510Paolo Guiotto: is less or equal than that.
28:26:230Paolo Guiotto: Now, this, can be expressed
28:30:30Paolo Guiotto: as the integral on the set of points XY, where X squared plus Y is less or equal than Z of the density with respect to XY, the XDY.
28:48:460Paolo Guiotto: So the density, this is easier, because it is E2 minus X, E minus 2 modulus y, and then there is the indicator 0 plus infinity of X.
29:04:780Paolo Guiotto: Okay, so…
29:09:490Paolo Guiotto: since this, this condition Z here is… you have to look at Z as a constant, okay? And we are just integrating in the two variables XY, that's a double integral, so we set this double integration.
29:25:240Paolo Guiotto: Now, it is better if we start integration in X or in Z.
29:36:80Paolo Guiotto: I would say, since this condition tells Y is less or equal than Z minus X squared, and I don't have any restriction on Y, because Y…
29:48:220Paolo Guiotto: is just any real. I would say, let's first integrate in Y, and therefore we will have the integration in Y is from minus infinity to Z minus X squared, the key density, E minus X, E minus 2 modulus y
30:04:950Paolo Guiotto: the indicators, 0 plus infinity X.
30:09:970Paolo Guiotto: this is in Y, then we will integrate in X.
30:14:260Paolo Guiotto: Now, we carry outside everything, well, 4X,
30:23:50Paolo Guiotto: Yeah, there is no restriction here in this constraint, and so the unique restriction is that X must be positive, so we can just say it's from 0 to plus infinity, and this is…
30:37:210Paolo Guiotto: already 1. So we have integral 0 plus infinity of E minus X, integral minus infinity to Z minus X squared, E minus 2 modulus of Y, DY, then DX.
30:57:290Paolo Guiotto: Okay, now let's focus on the integral here.
31:01:50Paolo Guiotto: Z minus X squared, E minus 2 modules of Y.
31:08:280Paolo Guiotto: Because the point is, is this positive or negative, because of the modulus? So, let's say that, if,
31:18:530Paolo Guiotto: That is negative, so Z minus X squared negative means X squared greater than Z, so the other case will be X squared less than Z.
31:30:610Paolo Guiotto: So this is the case when Z minus X squared is negative, right?
31:37:120Paolo Guiotto: So let's say less than or equal, it equals to 0. That's negative, so we have the integral for minus infinity to that value, Z minus X squared, of the modulus of y is minus y, always, so we have e to 2Y dy.
31:56:890Paolo Guiotto: In the second case.
32:00:60Paolo Guiotto: Since Z minus X squared would be positive, I would split the integration from minus infinity to that value, z minus X squared, into the integration from minus infinity to 0.
32:15:110Paolo Guiotto: So minus infinity to 0 over E, this becomes 2Y, EY.
32:21:930Paolo Guiotto: Then we have the piece from 0 to Z minus X squared of E minus 2YDY.
32:31:70Paolo Guiotto: So we have to do these integrals.
32:34:620Paolo Guiotto: They are not, complicated.
32:38:620Paolo Guiotto: So, because the E2Y is the derivative of E2Y over 2, so we have the evaluation of E2Y over 2.
32:50:180Paolo Guiotto: E toy over 2, from minus infinity to Z minus X squared.
32:56:520Paolo Guiotto: This is for X squared greater than Z.
33:08:390Paolo Guiotto: Okay…
33:10:140Paolo Guiotto: So, at minus infinity, we get 0. So, the unique value is the value at, so 1 half E22Z minus 2X squared.
33:25:440Paolo Guiotto: while 4X square less than Z,
33:31:00Paolo Guiotto: That integral from minus infinity to 0 is, again, that's the derivative of e to Z over 2.
33:38:680Paolo Guiotto: So we have the evaluation E2Y over 2 from minus infinity to 0. Then here we have the evaluation of E minus 2Y over minus 2 from 0 to Z minus X squared.
33:56:320Paolo Guiotto: This evaluation yields at zero, we get 1 half at the minus immediately 0, so 1 half minus 1 half times…
34:08:429Paolo Guiotto: E2 minus 2Z and minus X squared.
34:13:900Paolo Guiotto: Minus at 0, we get 1.
34:18:860Paolo Guiotto: So let's clean up this a bit.
34:23:230Paolo Guiotto: So this is the integral in the inside integral, so we get that integral from minus infinity
34:33:199Paolo Guiotto: to Z minus… X squared… It looks easier, but it may be…
34:40:310Paolo Guiotto: Do not show you his name.
34:42:600Paolo Guiotto: This is equal to… One path.
34:50:320Paolo Guiotto: E to 2Z minus 2X squared is for X squared greater or equal to Z.
34:58:950Paolo Guiotto: Adjusting this is one… minus… one half… E2… minus 2Z plus…
35:11:230Paolo Guiotto: to X squared, if X square is less than itself.
35:28:540Paolo Guiotto: You have to be… I'm thinking now, because,
35:32:870Paolo Guiotto: So this expression should be written inside, but, so…
35:38:430Paolo Guiotto: Since the inside integration is in Excel.
35:47:490Paolo Guiotto: You see that, what they want to… I'm thinking about, how can I say about this in terms of X?
35:57:400Paolo Guiotto: Because, remind that our X is positive. Here, Z is, for the moment, any number from minus infinity to plus infinity.
36:08:210Paolo Guiotto: So, if ZED, for example, is negative.
36:12:830Paolo Guiotto: The first condition is… is always true, the second is never true, you see? If Z is positive, then the two is two conditions, so I have to split the case Z positive, Z negative here.
36:29:640Paolo Guiotto: to continue.
36:31:240Paolo Guiotto: So, if Z is negative, F…
36:37:610Paolo Guiotto: Z of little z is the integral, this is the value of FZ, right?
36:46:360Paolo Guiotto: Oh.
36:48:490Paolo Guiotto: This is FZ of Z.
36:52:590Paolo Guiotto: No, sorry, yes, it is this one, but I want to… What is going on?
37:00:870Paolo Guiotto: book.
37:01:910Paolo Guiotto: I want to write down here, FZ of Z. Because I have to do this integral from 0 to plus infinity of E minus X times that integral. So,
37:15:560Paolo Guiotto: Zed.
37:17:320Paolo Guiotto: So I have, this is the integral from 0 to plus infinity, E minus X.
37:26:230Paolo Guiotto: Then there is that value. Now, since Z here is negative.
37:33:500Paolo Guiotto: This… this cannot be the condition.
37:37:660Paolo Guiotto: And this is always verified, so the value of that integral, the innermost integral, would be this quantity here. So I have to put here 1 half E22z minus 2x squared, this in DX.
38:01:780Paolo Guiotto: Yes, sir?
38:03:740Paolo Guiotto: And now, how do we compute this? Because this is 1F e to 2Z integral 0 to plus infinity E minus X minus 2X squared. This is
38:18:780Paolo Guiotto: a Gaussian integral, but… Okay, okay, this can be seen as…
38:34:500Paolo Guiotto: let's… this… this sounds like a Gaussian integral because of the X squared. So, to put in the standard form, I will… I will say that this is… I factorize the minus 2, I get X square…
38:52:530Paolo Guiotto: Xpr plus 1 half X.
38:58:660Paolo Guiotto: I want to complete the square, to transform this into a square.
39:02:860Paolo Guiotto: I have to look at this as a double product, so I put a 1 fourth here, it's the same.
39:09:780Paolo Guiotto: And therefore, I need the…
39:14:610Paolo Guiotto: So this is, X squared 2AX, I need A squared, so…
39:20:950Paolo Guiotto: 1 over 16 minus 1 over 16, in such a way that this B equals E minus 2,
39:30:80Paolo Guiotto: X, plus, 1 fourth square.
39:37:90Paolo Guiotto: This, yields, this term here.
39:43:450Paolo Guiotto: And then we have minus 1 over 16 out here.
39:47:970Paolo Guiotto: It's an audible exercise where I found this.
39:51:550Paolo Guiotto: So we have E minus 2X plus 1 fourth square, and the e to 2 times e to 1 over 8 can be written outside.
40:05:510Paolo Guiotto: It's just a constant. So we have, e to 1 over 8 half e to 2x, integral from 0 to plus infinity, e to minus
40:18:460Paolo Guiotto: Let's put in, well…
40:25:520Paolo Guiotto: Okay, let's… let's play with this form.
40:32:10Paolo Guiotto: DX, so…
40:35:950Paolo Guiotto: Now, if I put the 2 inside, this becomes a root of 2 that multiplies this, and I change variable. I call y equal the root of 2, X plus 1
40:52:210Paolo Guiotto: Forte.
40:57:420Paolo Guiotto: I want to have a… I want to have a 2 to the denominator, sorry. Let's put a 2 denominator, this means that we have 4 here, right? So this goes inside as a 2.
41:09:210Paolo Guiotto: So, the change of variable is this one, 2x plus 1 fourth. This becomes E to 1 over 8 half e to 2x.
41:20:150Paolo Guiotto: The integral, let's transform the integral. E to minus y square over 2. The X DY, this is 1 half DY.
41:31:600Paolo Guiotto: And the range for y, when x goes from 0 to plus infinity, for X equals 0, you get to 1 half to plus infinity, this.
41:43:970Paolo Guiotto: Now, this, one half we put D here, Now, this is a constant.
41:54:730Paolo Guiotto: This would be if I add here a root of 2 pi, root of 2 pi.
42:03:870Paolo Guiotto: This is the CDF of the standard Gaussian, not actually… because it's 1 minus…
42:10:300Paolo Guiotto: 1 minus phi of 1 half, where this phi of t is the integral from minus infinity to t of e minus y squared over 2DY divided root of 2 pi.
42:29:460Paolo Guiotto: So we got this thing.
42:32:560Paolo Guiotto: So we got… E to 1 over 8 divided for this value, 1 minus…
42:42:530Paolo Guiotto: the CVF of the standard of Goshen, times e to 2x, And this is the… There is something…
42:59:940Paolo Guiotto: No, this was a Z. There was something wrong. There is a Z here. Sorry, I brought X, but it's a Z. This comes from this, you see, this is the exponential that I… I carried outside.
43:16:70Paolo Guiotto: It was a little bit strange that we had this
43:21:150Paolo Guiotto: So this is the value of FZ Z for Z negative.
43:32:980Paolo Guiotto: For Z positive, It's strange, this exercise. I don't know where I found it.
43:40:600Paolo Guiotto: Or, sometimes, you know, that…
43:45:950Paolo Guiotto: If you don't write the solution, this happens.
43:50:40Paolo Guiotto: Okay. However, let's… do the second case when Z is positive.
43:59:670Paolo Guiotto: When Z is positive, Fair enough.
44:09:520Paolo Guiotto: This is, a trouble, because we have,
44:13:700Paolo Guiotto: That this integral is this one.
44:17:380Paolo Guiotto: since here Z is positive, means for X greater than root of Z, there is not the other case x less than minus the root, because X is positive.
44:27:120Paolo Guiotto: And this is X less than root of Z. So we should split, the integration for X,
44:40:950Paolo Guiotto: That goes from 0 to plus infinity into 0 to
44:47:620Paolo Guiotto: that root of Z, and we write this thing into the integral. And from root of Z to plus infinity, we write this other thing.
44:56:500Paolo Guiotto: Okay, so it's a bit… it's a bit messy. So FZ of Z for this case would be integral 0 to plus infinity of, we say, the E2 minus X,
45:10:310Paolo Guiotto: Then we have to put the first line for Z greater than root of X. So, let's say that we split this already in 0 to
45:23:260Paolo Guiotto: root of Z plus root of Z plus infinity, E minus Z.
45:30:40Paolo Guiotto: Then…
45:31:330Paolo Guiotto: When x is less than the root of Z, we have this formula, 1 minus 1 half, etc. So, 1 minus 1 half E2, what is it?
45:44:820Paolo Guiotto: minus 2Z plus 2X squared.
45:54:170Paolo Guiotto: And this is an integration in Excel. And then the second one.
45:59:110Paolo Guiotto: From root of Z to plus infinity, we have 1 half e to 2Z minus 2x squared.
46:05:110Paolo Guiotto: 1 kaf e to 2Z, E to minus 2X squared UX.
46:14:500Paolo Guiotto: So let's see what to… yeah, let's see a…
46:20:520Paolo Guiotto: So we have integral from 0 to root of Z of E minus X DX.
46:29:130Paolo Guiotto: minus 1 half integral from 0 to root of Z. Here we have E2 minus X.
46:37:220Paolo Guiotto: E to minus X.
46:40:360Paolo Guiotto: Plus 2X square… And then, de facto e to minus 2Z can be… Written outside.
46:54:740Paolo Guiotto: And then, finally, we have 1 half e to 2Z integral from root of z to plus infinity. This is a Z.
47:05:540Paolo Guiotto: E to minus X minus 2x squared.
47:12:570Paolo Guiotto: It's very ugly, this…
47:17:990Paolo Guiotto: So because these integrals are not explicitly… we cannot compute explicitly. The first one we can do, it is minus C2 minus X to be evaluated from 0 to root of Z, so the evaluation at final point will yield E minus root of Z negative.
47:35:280Paolo Guiotto: And the valuation at 0 will produce 1.
47:39:600Paolo Guiotto: So it should be this with the right sign. Minus 1 half E2 minus 2Z.
47:50:600Paolo Guiotto: And how do we compute that integral?
48:01:890Paolo Guiotto: I don't think we can compute the… because,
48:13:900Paolo Guiotto: But I mean, I mean, I'm really stupid, because who cares? We have to do the… I know, but there is the product here, and that we will,
48:23:100Paolo Guiotto: Because this is the CDF. Then we have to do the derivative to get the density.
48:30:190Paolo Guiotto: So, for this case, it's easy, because it's constant E22Z. This one is not easy, because…
48:42:660Paolo Guiotto: And we can't… we can't do anything better than this. We can compute the derivative, this becomes explicit for this part, but then we will have a factor, because this… the derivative of this time this, we cannot compute that quantity.
49:01:730Paolo Guiotto: So we… we have to leave,
49:05:760Paolo Guiotto: indicated, I don't know how to do… Better than this.
49:13:330Paolo Guiotto: And also, for this one, we can't do anything.
49:22:20Paolo Guiotto: I don't see how to simplify. However.
49:27:310Paolo Guiotto: Let's leave this exercise to its destination.
49:34:990Paolo Guiotto: I don't… I don't know if, doing it directly with the density… I took this way because it looked to be easier, but at the end, you… you see that it becomes much harder. I don't know if integrating this thing
49:53:830Paolo Guiotto: Respect to W.
50:05:150Paolo Guiotto: I don't know. I don't know. I don't know if you… if you try to compute this… yeah. It works.
50:17:270Paolo Guiotto: explore any reason.
50:20:230Paolo Guiotto: So, it's… it's wild.
50:23:770Paolo Guiotto: Okay, well, let's leave with the exercise here.
50:29:180Paolo Guiotto: Maybe we do another one, hoping to be a little bit more lucky.
50:36:380Paolo Guiotto: Let's do the 437, let's see, because…
50:40:490Paolo Guiotto: I've not completely checked these exercises, so sometimes it happens to find something.
50:49:320Paolo Guiotto: So maybe I just added the question that was… So, XY here has a density FXY equals E minus…
50:59:780Paolo Guiotto: Y indicator 0, 1, X indicator 0 plus infinity, Why?
51:12:350Paolo Guiotto: Number one, determine… the density… Oh.
51:21:640Paolo Guiotto: Z equal X times Y.
51:26:350Paolo Guiotto: and W equal… X over Y.
51:33:710Paolo Guiotto: Question 2, compute the probability that Z times W be greater than 1.
51:48:160Paolo Guiotto: Okay, so again, we have a map here. ZW is phi of XY.
51:58:550Paolo Guiotto: Where the map phi takes a point XY and yields X times Y and X over Y.
52:07:820Paolo Guiotto: Of course, this map is not defined when Y is 0, so we should, we should verify,
52:16:520Paolo Guiotto: when I put a capital Y here, if this capital Y can be 0. So, basically, we have to compute what is the probability that capital Y is 0. Here, I don't need to give… to do the calculation. I already know that this is 0,
52:32:840Paolo Guiotto: This because, since, the pair XY Is absolutely continuous.
52:42:510Paolo Guiotto: This will be also for the marginals, so particular Y is absolutely continuous.
52:50:480Paolo Guiotto: So this means that there exists a density that I don't need to compute. I know in general that it exists, and therefore, when I do the probability that Y belongs to… is equal to 0, so it means it belongs to a singleton.
53:06:940Paolo Guiotto: This probability is always zero, because this is the integral on the singleton of the density.
53:15:570Paolo Guiotto: in the wild.
53:17:50Paolo Guiotto: And, since here means that you are integrating with respect to the Lebang measure, singletons have measure 0, and therefore the integral will be zero, so…
53:28:100Paolo Guiotto: So we know that,
53:31:730Paolo Guiotto: with probability 1, we can compute phi of XY, okay?
53:39:400Paolo Guiotto: Moreover, what else can be said? We noticed that, because of these,
53:47:550Paolo Guiotto: indicators, the probability that X is between 0 and 1, and the probability that… and Y
53:57:290Paolo Guiotto: is greater or equal than zero, or if you want strictly greater than zero. This probability is the integral when x is between 0, 1, and Y.
54:09:70Paolo Guiotto: It's positive of the, density.
54:14:70Paolo Guiotto: FX. Why?
54:17:590Paolo Guiotto: the X, DY.
54:19:310Paolo Guiotto: But on that set, the two indicators are one.
54:24:230Paolo Guiotto: So what remains is just the exponential. So we have integral for X between 0 and 1,
54:31:220Paolo Guiotto: Y positive of E2 minus Y is E2 minus y, right?
54:38:40Paolo Guiotto: DXDY.
54:39:780Paolo Guiotto: And this integral is equal to 1, because we use the reduction formula. We integrate, for example, first in X, that is, from 0 to 1,
54:50:590Paolo Guiotto: that we put outside E2 minus y, which is independent of X, and then we have to integrate this from 0 to plus infinity. This integral is equal to 1, and the integral from 0 to plus infinity of e to minus y, dy, is exactly equal to 1.
55:06:680Paolo Guiotto: So, this means that the pair XY belongs to this domain, the interval 0, 1,
55:16:220Paolo Guiotto: times, the offline 0 plus infinity, with the probability Wow.
55:25:620Paolo Guiotto: So this means that this is the domain we have to consider for the map feed here, for the transformation of XY, because at the point
55:36:820Paolo Guiotto: Capital X, capital Y, wherever it is, it is on this set, okay? So this means that the domain in Cartesian plane XY
55:47:120Paolo Guiotto: we should be interested. For this map is X between 01 and Y between 0 plus infinity, so a strip like this.
55:57:490Paolo Guiotto: This is the domain.
56:00:00Paolo Guiotto: Well, we should consider the map.
56:02:760Paolo Guiotto: feed.
56:04:130Paolo Guiotto: It's because maybe this map, it doesn't look to be a big action. If you change, for example, both signs to X and Y,
56:15:160Paolo Guiotto: you see that the product X times Y don't change, and also the ratio X over Y is the same. So, certainly, this is not injective, no? But you see that in this domain, I cannot change the coordinate assigned to the two coordinates.
56:31:250Paolo Guiotto: Because I leave the domain. So, probably in that domain, it is injective. So this map phi will map this Cartesian plane, or better, that domain, into, let's say, a ZW plane.
56:45:830Paolo Guiotto: Well, we have that…
56:48:270Paolo Guiotto: If you want, if you look at the coordinate, if X and Y are both positive, as for this domain, X times Y and X over Y are both positive, so we are in the first quarter, for sure.
57:10:280Paolo Guiotto: I would say that, let's see if this is invertible. So, let's say that if ZW is,
57:18:830Paolo Guiotto: fee of XY, This means that it is X times Y, X over Y, so we get that,
57:29:30Paolo Guiotto: Z is X times Y, W is X over Y. Now, if I want to extract from this X and Y,
57:39:730Paolo Guiotto: I could, for example, multiply the two equations, and I could get that…
57:47:750Paolo Guiotto: X squared, because if you multiply the first times the second, the Y disappears, is equal to ZW.
57:58:310Paolo Guiotto: And, if you do the ratio, right.
58:04:890Paolo Guiotto: The first over the second.
58:08:470Paolo Guiotto: you get Y squared equal to, Z over W.
58:18:730Paolo Guiotto: Now, here, if you want to be 100% precise, you should check whether this is really a dejection. Notice that, in particular, you see that we say that the Y equals zero
58:36:270Paolo Guiotto: as probability equal to 0. So, we can discard y equals 0 from the domain, taking only Y positive, because this is a set of omega which is irrelevant.
58:49:530Paolo Guiotto: Y equals 0 goes where with this map? The map is here. If I put Y equals 0, the second component wouldn't be defined.
59:01:930Paolo Guiotto: On the other side, W equals 0 would be a problem for the denominator here. W is X over Y, this is W. This can be 0 if x is 0.
59:16:260Paolo Guiotto: Okay, so I should exclude also X equals 0, which is this axis.
59:22:460Paolo Guiotto: Again, I could identify that the probability that X is equal to 0 for the same reason, this is equal to 0. So we are not, the set of omega, where the X is on the y-axis.
59:36:870Paolo Guiotto: where the point XY is on the y-axis is an event with probability equals 0. So, in fact, to be precise, I could say, well, the domain B should be…
59:49:700Paolo Guiotto: We should exclude the 0 for the X and 0 for the Y.
59:54:60Paolo Guiotto: In this way, the two coordinates are strictly positive, so it means that X times Y is a positive number, and X over Y is a positive number.
00:06:770Paolo Guiotto: Okay?
00:08:40Paolo Guiotto: And, now, if, in the… so, it means that, in this, space, ZY, my point
00:20:660Paolo Guiotto: Belongs to this, quarter.
00:23:950Paolo Guiotto: And now I claim that probably this is the image of the map piece, so it is a big action between these two sets.
00:33:660Paolo Guiotto: Because if XY belongs to the domain D, which is, let's say, the red domain without the true axis, this is okay.
00:45:530Paolo Guiotto: The fee of X belongs to the blue satin.
00:50:550Paolo Guiotto: And I would say vice versa, because from ZW equal XY,
00:58:860Paolo Guiotto: Index over Y, so these ones.
01:02:600Paolo Guiotto: I can multiply and divide, and multiply divided by numbers which are different from zero, so you can always do that. I get this, and now from this, since X, in any case, is positive, and Y is positive, you get that X is the root of ZW
01:23:830Paolo Guiotto: and Y is the root of Z over W.
01:28:400Paolo Guiotto: So…
01:33:770Paolo Guiotto: No, yeah, that's not the entire domain, because if we want that X be between 0 and 1, we must have that root of ZW be between 0 and 1.
01:47:460Paolo Guiotto: greater or equal than 0 is always true, but less than 1 not. So we have root of ZW, less or equal 1, this means ZW less or equal than
01:58:760Paolo Guiotto: 1, and so this means that if we are in the first quarter, W must be with ZW positive.
02:07:370Paolo Guiotto: W must be less than 1 over Z. Okay, so it means that we are not taking all points here, but the points which are below the hyperbola W equal 1 over Z. So, this is the image, the true image of the Z.
02:26:190Paolo Guiotto: So now, the map fee is a big action between these two sets.
02:31:710Paolo Guiotto: And this is the inverse map.
02:35:820Paolo Guiotto: This is the feet minus 1.
02:38:840Paolo Guiotto: So we can write the formula that says that the density for ZW
02:44:950Paolo Guiotto: is equal to the density of XY at point of phi minus 1.
02:51:10Paolo Guiotto: of ZW.
02:53:500Paolo Guiotto: times the modulus of the determinant of phi minus 1 prime ZW.
03:02:250Paolo Guiotto: We need the determinant of phi minus 1, let's compute directly for this example. So, determinant of phi minus 1 prime is equal to the determinant of the Jacobian matrix made by the two gradients.
03:17:570Paolo Guiotto: of the first component of phi minus 1, which is root of ZW,
03:25:90Paolo Guiotto: And the second component, which is root of Z over W.
03:29:750Paolo Guiotto: So we get the determinant. Here, derivatives are with respect to Z and W, so derivative with respect to Z is like root of Z root of W, so the derivative is 1 over 2 root of Z, so we get root of W over Z with 1 half, and here is the same, but
03:51:810Paolo Guiotto: flip the components, so it is root of Z of a W exponent.
03:57:260Paolo Guiotto: Times 1 alpha.
03:59:520Paolo Guiotto: For denominator, when we do the derivative with respect to Z, we get 1 over 2 root of Z, the root of W, so root of ZW. When we do the derivative with respect to W, it is Z to minus 1 half… sorry.
04:16:60Paolo Guiotto: This is the root of Z times,
04:21:410Paolo Guiotto: It would have been better if we've righted powers.
04:25:330Paolo Guiotto: So let's write powers.
04:29:290Paolo Guiotto: So this is Z1 half W1 half. This is Z1 half W minus 1 half. So you see that…
04:39:830Paolo Guiotto: Normally, if the inverse… see, if the direct function are nice, the inverse are bad. So that's why it is sometimes better to compute on the direct metrics, and then you do the inverse, the reciprocal.
04:54:830Paolo Guiotto: So here we get 1 half Z minus 1 half W to 1 half.
05:01:720Paolo Guiotto: This is 1 half Z, 1 half
05:05:340Paolo Guiotto: W to minus 1 half. This is 1 half Z minus 1 half W1 minus 1 half.
05:16:440Paolo Guiotto: And this is minus 1 half Z1 half. Minus 1 half minus 1 is minus 3 half.
05:26:570Paolo Guiotto: So when we do the determinant, we get 1 half, one half, well, these factors, one half makes a 1 fourth in front of
05:34:860Paolo Guiotto: everything, then we have Z minus 1 half W, 1 half times Z1 half, so you see that these two, they kill, and we have a W 1 half minus 3 half, it makes W2 minus 1.
05:52:370Paolo Guiotto: Minus, when we do the cross product here, the Z disappears, and again.
06:01:430Paolo Guiotto: Sorry, there is a minus here because of…
06:04:150Paolo Guiotto: Assigned here, okay, minus another W is minus 1. So at the end, it is minus 2 fourths, so minus 1 half W to minus 1. 1 over 2W.
06:15:960Paolo Guiotto: So the modulus of the determinant of phi minus 1 prime is 1 over 2 modulus of W, and since our W in the domain of ZW is positive, this is 1 over 2W.
06:33:630Paolo Guiotto: So, finally, the density of ZW at point ZW.
06:40:110Paolo Guiotto: is equal to, the density of F,
06:46:290Paolo Guiotto: Well, it's better if I look directly here.
06:50:660Paolo Guiotto: This was E2 minus Y, so E2 minus Y, Y is… Ease, ease, is,
07:00:650Paolo Guiotto: I have to pick why… root of Z over W.
07:08:580Paolo Guiotto: Then I have the indicator, interval 01 of X. X is root of ZW, right? Yes, yes.
07:18:20Paolo Guiotto: root of ZW.
07:20:760Paolo Guiotto: Indicator, 0 plus infinity, of Y, which is the root of Z over W, that's constantly equal to 1,
07:32:350Paolo Guiotto: Times the modulus of the determinant, so this 1 over 2… value.
07:39:440Paolo Guiotto: That's the density, so maybe we can rearrange 1 over 2W, exponential minus root of Z over W,
07:50:580Paolo Guiotto: indicator of, well, let's keep in this…
07:57:550Paolo Guiotto: We could say it is the same as, say, ZW between 0 and 1.
08:05:700Paolo Guiotto: So this is the… joint density.
08:11:430Paolo Guiotto: Then, it asks to compute the probability that Z
08:17:569Paolo Guiotto: W is greater or equal than… is greater than 1.
08:26:880Paolo Guiotto: Well, since ZW is… I don't know, ZW is what? Remind that Z…
08:34:149Paolo Guiotto: was XY, and WX over Y, ZW is X squared. So that's the probability, then.
08:47:729Paolo Guiotto: X squared is greater than 1.
08:51:240Paolo Guiotto: So we can use the joint density.
08:55:430Paolo Guiotto: So, saying that this is the same of X, since X is positive for us, so X greater than 1, and Y whatever, real. Well, actually, also Y is positive.
09:12:670Paolo Guiotto: No, no, no.
09:17:810Paolo Guiotto: But that's zero.
09:20:310Paolo Guiotto: Sorry.
09:21:680Paolo Guiotto: I will just… strange question, because,
09:28:510Paolo Guiotto: If you think about… I can see this as the probability that X is greater than 1, and Y is whatever.
09:38:60Paolo Guiotto: So say Y real. So this means that this is the integral for X greater than 1 and Y any real of the density, which is E minus Y indicator 01 for X. That's where you see that this is 0.
09:55:400Paolo Guiotto: indicator of 0 plus infinity for X.
10:00:460Paolo Guiotto: But why?
10:02:430Paolo Guiotto: So, you are integrating 0, you get zero, so it's a strange question, so probably…
10:08:140Paolo Guiotto: I don't know. What was the spirit of this exercise has to be reviewed.
10:13:370Paolo Guiotto: Okay, however… I think we have understood more or less how it works this mapping with random variables.
10:21:490Paolo Guiotto: Okay, we have still 20 minutes.
10:24:510Paolo Guiotto: Let's, move a bit forward, introducing the next topic.
10:33:810Paolo Guiotto: So, we have seen that, for a random variable, for… a random variable X.
10:46:570Paolo Guiotto: Well, here I'm not precise if it is color or multivariate. We… associated…
10:59:610Paolo Guiotto: A probability measure.
11:06:680Paolo Guiotto: the law, of X, huh?
11:12:140Paolo Guiotto: Which is the probability measure on the class of boron sets of R, or the Borrel sets of Rn for the multivariate case.
11:24:420Paolo Guiotto: Then, we associate the function Which is the CDF.
11:31:10Paolo Guiotto: function.
11:36:100Paolo Guiotto: the CDF… of… Excellent.
11:42:870Paolo Guiotto: Which is a… An object defined on the real line with values in 01,
11:50:570Paolo Guiotto: or on our end, for the multivariate case, still with values in 01, with certain properties.
11:59:550Paolo Guiotto: And if the X is absolutely continuous, it…
12:06:600Paolo Guiotto: Density, which is still a function.
12:13:510Paolo Guiotto: if… That is absolutely… Continos.
12:20:530Paolo Guiotto: which is a function FX, defined on real, Real, positive, valued.
12:29:520Paolo Guiotto: or RN… Still positive value, though.
12:34:450Paolo Guiotto: With the integral equal to 1.
12:38:330Paolo Guiotto: Now, of course, a probability measure is an object which is…
12:43:610Paolo Guiotto: relatively difficult to handle. A function is much more concrete object, okay? Now, we associate now another object, which is basically the Fourier transform of, of something we see.
13:03:660Paolo Guiotto: We.
13:05:360Paolo Guiotto: Let's start from the particular case, if we have a density.
13:10:540Paolo Guiotto: if X… let's say that the idea is that if this object exists, they characterize the random variable.
13:19:570Paolo Guiotto: Because, basically, you want to compute the probability that X belongs to a certain set E. If you know the law, this means you compute mu X of E.
13:30:570Paolo Guiotto: If you have the CDF, well, you don't have a formula for this. Well, that would be a formula, but we have not seen what is the integral with respect to a CDF. If we have a density, this is an integral of the density
13:43:980Paolo Guiotto: with respect to delay back measure. There would be a formula also here, but it's something we have not done. Let's say that for special cases, we can do, no, for example, here we can say that the probability that X is between
14:03:00Paolo Guiotto: A and the B in this way.
14:05:870Paolo Guiotto: I write this way just to have that this is the difference of the CDF at point B minus the CDF at point A.
14:17:560Paolo Guiotto: So let's say that somehow you can compute probabilities with these tools. Now, effects is absolutely continuous.
14:35:790Paolo Guiotto: And so we have a density, FX,
14:40:110Paolo Guiotto: Now, this function, the density, is in particular NL1 in the real line, or Rn, if we are in the multivariate case function, with the integral equal to 1. So, in particular, D.
14:57:830Paolo Guiotto: L1.
14:59:880Paolo Guiotto: Puy a transforma.
15:01:670Paolo Guiotto: of FX is well.
15:06:350Paolo Guiotto: defined.
15:09:640Paolo Guiotto: Well, what's up, then?
15:15:780Paolo Guiotto: It is well defined.
15:19:100Paolo Guiotto: And as we know, it is a function that we denoted with this symbol, FX hat C. It is formally, the integral in r of the function density FXXE minus ICFDX.
15:42:490Paolo Guiotto: Now, reminder that, remind…
15:50:390Paolo Guiotto: Better.
15:52:860Paolo Guiotto: The integral of a generic function phi of x times the density.
16:01:810Paolo Guiotto: This is the integral on R of that function.
16:07:590Paolo Guiotto: in, the measure, which is the law of X,
16:12:220Paolo Guiotto: So we get this starting with phi and indicator, and then you extend with the simple function and limits of simple function to NF. And this is the expected, also the expected value of phi of capital X.
16:31:850Paolo Guiotto: So, in particular, since this integral here has exactly this form with the phi of x, this function.
16:41:840Paolo Guiotto: We get that the Fourier transform of the density
16:47:150Paolo Guiotto: is the integral on R of this function, E minus iCX d mu X.
16:56:920Paolo Guiotto: Which is also, equal to the expected value of E
17:03:510Paolo Guiotto: to minus IC capital X. So these quantities are the same quantity written under different representations.
17:15:220Paolo Guiotto: Now, the key point is that if we look at this.
17:19:440Paolo Guiotto: Now, whatever is the random variable of X,
17:24:850Paolo Guiotto: With or without the density is going to make sense.
17:29:540Paolo Guiotto: And also, each one makes sense if there is the random Bible accent.
17:34:560Paolo Guiotto: Now, this yields this definition.
17:38:860Paolo Guiotto: definition.
17:43:110Paolo Guiotto: If, mu is A.
17:48:580Paolo Guiotto: Probability.
17:51:500Paolo Guiotto: measure.
17:53:960Paolo Guiotto: on this particular space, R, with the Borel sets of R, we… Cool.
18:05:480Paolo Guiotto: Fourier transform.
18:15:220Paolo Guiotto: of the measure mu, this quantity, mu hat, C…
18:22:260Paolo Guiotto: which is, by definition, the integral on R of E minus ICX.
18:29:820Paolo Guiotto: Integrated with respect to the measure mu.
18:35:770Paolo Guiotto: In general, If, mu.
18:41:320Paolo Guiotto: is a measure Probability.
18:48:230Paolo Guiotto: Measure.
18:50:660Paolo Guiotto: on… RN… with the Borel sets of RN,
18:58:880Paolo Guiotto: The Fourier transform of mu will be a function of vector variable Xi, so here Xi is a numerical variable. Here is a vector in our N.
19:12:720Paolo Guiotto: And the definition is similar. We do the integral on our N.
19:17:50Paolo Guiotto: E2 minus i, the product XCX is replaced by this color product between C and X d mu X.
19:32:350Paolo Guiotto: And this is called the Fourier transform of the measure mu.
19:39:30Paolo Guiotto: Now, the remark is that,
19:43:400Paolo Guiotto: This is a well-posed definition, so we may notice that the mu hat is always defined…
19:56:860Paolo Guiotto: We don't need to have a density to define the Fourier transformer, because
20:06:560Paolo Guiotto: We have to show that the function, this exponential, is integral with respect to the measure mu, because what we know about the E minus iCX is that it is a continuous function in the real line.
20:24:610Paolo Guiotto: And the…
20:26:720Paolo Guiotto: So it will be measurable. And the integral on R of the absolute value of E minus iCX
20:35:110Paolo Guiotto: In the measure mu.
20:37:360Paolo Guiotto: Now, this quantity is constantly equal to 1, because it is the absolute value, the modulus of a unitary complex number, so it is the integral on R of 1 d mu, but since mu is a probability measure, this number is equal to 1, so it is finite.
20:56:480Paolo Guiotto: So, for a probability measure, this quantity is well-defined and defines a function that we call Fourier transform.
21:06:180Paolo Guiotto: So, Muat… C is… Well… define the… For every seat.
21:17:190Paolo Guiotto: Riyadh, or RN for the general case.
21:23:570Paolo Guiotto: Second remark… IFA.
21:29:940Paolo Guiotto: Mu is the law of random variable X.
21:33:750Paolo Guiotto: And that's is… absolutely continuous.
21:39:230Paolo Guiotto: So this means that,
21:42:40Paolo Guiotto: In fact, the mu X, is the density times the Lebec measure, the density FX times the Lebec measure. When you plug there.
21:53:940Paolo Guiotto: the measure mu X, which is FX times DX, you get back this integral, so the Fourier transform of the density.
22:03:230Paolo Guiotto: So in that case, the Fourier transform of the measure, mu X, is nothing but the Fourier transform of the L1 function, density of FX.
22:18:440Paolo Guiotto: So this is the L1 Fourier transform of FX.
22:24:800Paolo Guiotto: The object we studied, let's say, in the first part of the course.
22:30:700Paolo Guiotto: So, in particular, for example.
22:34:240Paolo Guiotto: An important example is, at least for probability, if X is a Gaussian normal with mean m and variance sigma squared, so here we are with the scalar case, so M is a real number.
22:53:160Paolo Guiotto: And sigma square is a real positive number.
22:58:530Paolo Guiotto: Well, the Fourier transform of,
23:03:130Paolo Guiotto: If we had some form of this,
23:06:70Paolo Guiotto: Of the law of this distribution.
23:08:950Paolo Guiotto: is the Fourier transform of the density of the Gaussian
23:14:610Paolo Guiotto: Which is, we know, the Gaussian has density 1 over root of 2 pi sigma squared E minus the square minus mean squared over 2 sigma squared.
23:30:860Paolo Guiotto: So it is the Fourier transform of this thing.
23:34:180Paolo Guiotto: Now, I don't know if you remember, we did some times ago, but when you do the translation in the argument, in the free transform, you get a product by a character, so this becomes E2, I, C, M.
23:52:720Paolo Guiotto: times the Fourier transform of 1 over root of 2 pi sigma square, E minus, now it is centered, the square over 2 sigma squared.
24:05:390Paolo Guiotto: And, if you remind of the formula for the Fourier transform of this, we get e to ICM
24:13:80Paolo Guiotto: E to minus 1 half sigma squared c squared.
24:18:580Paolo Guiotto: So we get that, if this remarkable formula that you should remind, if X is normal, mean m variance sigma squared.
24:29:960Paolo Guiotto: the, law of X has fully transformed this function
24:35:850Paolo Guiotto: E to ICM, or better we normally put the mean. In fact, EIMC minus 1 half sigma square C squared. This is the formula for the FLR transformer
24:52:670Paolo Guiotto: of the…
24:58:360Paolo Guiotto: Of the, no, there should be, probably, there is a minus here.
25:04:660Paolo Guiotto: With the translation, there should be a minus with this factor.
25:12:270Paolo Guiotto: Okay.
25:14:70Paolo Guiotto: If we take a multivariate Gaussian.
25:19:00Paolo Guiotto: So, another example, important example to know, is the case when X is normal with mean m and covariance C, where now M is a vector of Rn, and C is a positive definite, symmetric.
25:41:190Paolo Guiotto: In this case, the, characteristic, the Fourier transform of, this, of the law of this,
25:52:180Paolo Guiotto: Of this variable is similar with the suitable,
25:57:710Paolo Guiotto: adjustments, so it is… M is now a vector, C is now also a vector of the same
26:05:640Paolo Guiotto: space RN, where we have the… the array. So this is this color product between M and C, minus 1 half. So now.
26:19:400Paolo Guiotto: you see that this sigma square is here, is exactly this, the role of C, and in fact, here we have a CC dot C for the general case.
26:37:120Paolo Guiotto: We can, do this calculation,
26:41:920Paolo Guiotto: Actually, we have never seen the calculation of the multi-dimensional Fourier transformer, so, in this case, mu X,
26:52:280Paolo Guiotto: Let's do other sort of exercise. It is the multidimensional Fourier transform.
27:00:310Paolo Guiotto: So E minus I C dot…
27:04:510Paolo Guiotto: X mu… mu X DX. So, since we have the density, this is the integral on Rn of E minus IC dot X.
27:19:10Paolo Guiotto: E minus 1 half the density is C minus 1X minus m dot X minus M.
27:30:10Paolo Guiotto: However, the scaling factor is root of 2 pi to the n determinant of the matrix C, DX.
27:41:70Paolo Guiotto: Now, to compute this integral.
27:45:690Paolo Guiotto: We first do a change of variable to center the integral, so we call y this.
27:51:890Paolo Guiotto: So, setting Y equal X minus M, nothing changed for the…
27:58:600Paolo Guiotto: integration domain, because it's just a translation. We have E2 minus iC dot X is Y plus M, so I have Y plus M.
28:12:700Paolo Guiotto: Here, and about the exponential E minus 1 half C minus 1Y dot Y, and the scaling factor is the same, root of 2 pi to the n determinant of matrix C, DY.
28:30:640Paolo Guiotto: the modulus of the Jacobian of the change of variables is 1, because it's just a translation.
28:36:130Paolo Guiotto: Now, as you can see here, we get E minus IC.
28:40:370Paolo Guiotto: dot Y, and then there is an E minus ITC dot M that comes outside, so you can say E minus ICM, so you see that there is the minus, as I said here, for the one-dimensional case.
28:57:710Paolo Guiotto: And then we have to integrate on Rn E minus IC dot y E minus 1 half C minus 1Y dot Y over root of 2 pi
29:13:340Paolo Guiotto: to the N determinant C, DY.
29:18:00Paolo Guiotto: Now, to compute this integral.
29:21:530Paolo Guiotto: The trick is to diagonalize the matrix C. So we know that there exists a T orthogonal.
29:32:500Paolo Guiotto: of terminal metrics.
29:36:160Paolo Guiotto: Such that, TCT transposed is equal to a diagonal matrix, D. Diagonal.
29:48:410Paolo Guiotto: Well, on the diagonal, we call the eigenvalues, since it is positive, the infinite, sigma 1 square, etc, sigma n squared.
29:59:840Paolo Guiotto: So now we have to recreate for C minus 1 what can be said there, so this means that T, T transpose is the identity, this means orthogonal matrix, so that the inverse of T is the transpose of T.
30:16:950Paolo Guiotto: So, from this formula, inverting the formula and then isolating T, multiplying by T transpose this.
30:23:980Paolo Guiotto: So we get that C minus 1… Is equal to, well…
30:32:890Paolo Guiotto: If I do the inverse, this is TC-1T transposed.
30:40:300Paolo Guiotto: Right, because the inverse inverts the, order. Yes, it is this one, this is D minus 1, and therefore multiplying by t transpose and by t here.
30:53:390Paolo Guiotto: So D transpose the T here, I have that C minus 1 is T transpose the D minus 1 T.
31:00:600Paolo Guiotto: So if you now plug this into the expression over there, you get E minus iC dot m integral on Rn.
31:10:60Paolo Guiotto: E minus IC dot Y.
31:15:130Paolo Guiotto: T minus 1 half C minus… well, C minus 1 is replaced by T transpose the D minus 1T
31:25:790Paolo Guiotto: Y dot Y.
31:29:200Paolo Guiotto: Over the scaling factor, you don't copy.
31:33:310Paolo Guiotto: Okay, now, put this matrix this side, so this becomes a TY, and change again variable, call, Z equal TY.
31:48:150Paolo Guiotto: This becomes C minus, like, C dot M.
31:51:860Paolo Guiotto: The transformation is a linear transformation of an invertible of Rn, so the domain of integration remains Rn. E minus IC dot
32:06:30Paolo Guiotto: Y would be T minus 1Z, but this T inverse of T is the transpose of T, so basically we have T transposed
32:18:690Paolo Guiotto: Z here, E minus 1 half DZ dot Z over the scaling factor DZ.
32:30:590Paolo Guiotto: the determinant is 1, because we know that modulus of determinant of T is 1, so the factor, due to the change of variable would be, again, equal to 1.
32:43:960Paolo Guiotto: But now the point is that this is a diagonal product, so this can be written as e to minus 1 half the sum of the sigma j squared ZJ squared.
32:58:980Paolo Guiotto: So, in other words, we can transform also into a product of a J1 from N of this E minus ZJ squared over 2 sigma j squared.
33:11:250Paolo Guiotto: And since also this one transforms into the product, because we can carry these metrics on the other factor by doing the transposed there.
33:21:260Paolo Guiotto: we get E minus IC dot m integral on Rn, so this becomes E minus iTC
33:32:350Paolo Guiotto: dot Z, so I now split this into product of E minus i, the J component of Txi.
33:43:210Paolo Guiotto: times, the J component of Z,
33:48:430Paolo Guiotto: And, here we have E minus ZJ squared divided to sigma J.
33:55:370Paolo Guiotto: You'll remind that there is a scaling factor, which is the root of 2 pi to the n, so I can give to each factor a 2 pi.
34:04:330Paolo Guiotto: And then there is the determinant of the matrix C, which is the determinant of the matrix D, which is the product of the eigenvalues, so sigma 1 square, sigma n squared, so I can put… split also this into a product.
34:20:630Paolo Guiotto: And therefore, when I do the product J from 1 to N, I get that.
34:26:80Paolo Guiotto: this integral. Now, this integral splits into the product of integral, because it is a product of functions depending each on one single variable.
34:37:560Paolo Guiotto: So this one becomes the product for J going from 1 to n of single integral in one single variable.
34:46:450Paolo Guiotto: So the integral is on R, E minus i, the jade component of this vector TC,
34:55:770Paolo Guiotto: ZJ E minus ZJ squared over 2 sigma J,
35:03:20Paolo Guiotto: root of 2 pi sigma j squared.
35:09:430Paolo Guiotto: DZJ.
35:11:140Paolo Guiotto: And these are the Fourier transforms of the standard Gaussians.
35:16:840Paolo Guiotto: At this point here.
35:20:310Paolo Guiotto: So this is the hat of E minus square over 2 sigma j squared, rescaled by the factor root of 2 pi.
35:32:730Paolo Guiotto: sigma j squared, evaluated at this point the jade component of tick C.
35:42:420Paolo Guiotto: And these are, because of the free transformer, of the standard Gaussian, e to minus 1 half sigma j squared this quantity, the J component of TC.
35:58:350Paolo Guiotto: L squared.
36:01:640Paolo Guiotto: So, the total will be the product of these things, so we have a product here.
36:09:660Paolo Guiotto: But putting together these factors, we have E minus 1 half. The product of the exponentials is the exponential of the sum.
36:19:350Paolo Guiotto: So, we have a sum of a J of sigma j squared Txi, J component squared.
36:28:400Paolo Guiotto: Now, you can easily see that this is exactly the action, this part here, is diagonal matrix D applied on the vector tick C, scalar product with tick C.
36:43:650Paolo Guiotto: That is, we finally have E minus 1 half.
36:47:660Paolo Guiotto: carrying back this T. On the other side, we have T transpose the DT,
36:52:660Paolo Guiotto: C, Scala C, but that's the matrix C.
36:58:450Paolo Guiotto: So we have E to minus 1 half 6C dot C.
37:04:460Paolo Guiotto: This is for the integral. Remind that we have a factor in front of everything, so we can conclude that the mu X hat C, for this case, is e to minus iC dot.
37:20:990Paolo Guiotto: M times this exponential, E minus 1 half
37:25:330Paolo Guiotto: CC dot C. And here we have D.
37:29:710Paolo Guiotto: formula for the Fourier transform of the,
37:36:300Paolo Guiotto: multivariate Gaussian distribution, or better for the filter form of the law of this.
37:43:550Paolo Guiotto: Okay, we stop here.