AI Assistant
Transcript
00:01:950Paolo Guiotto: Okay.
00:03:460Paolo Guiotto: We can start.
00:08:540Paolo Guiotto: I agree.
00:11:930Paolo Guiotto: Let's start with some of the exercises on the independence. I do the 646, which is nice.
00:24:290Paolo Guiotto: it says we have, A and B, independent… random variables.
00:34:130Paolo Guiotto: uniformly distributed in 01, so both A and D have uniform distribution.
00:41:770Paolo Guiotto: In the interval 01.
00:45:360Paolo Guiotto: And it asks, this. What is the probability
00:58:940Paolo Guiotto: bets.
01:01:630Paolo Guiotto: The second degree equation, X squared plus 2AX plus B equals 0.
01:09:700Paolo Guiotto: pause.
01:12:310Paolo Guiotto: real solution.
01:21:520Paolo Guiotto: Okay, so, in general, for a second, the equation, to have real solution, we need that, to…
01:33:470Paolo Guiotto: Reality.
01:35:190Paolo Guiotto: solutions… We didn't end up… that, the… discriminant,
01:47:420Paolo Guiotto: what's the… this square… so this one square, two-way… Square.
01:53:520Paolo Guiotto: Minus 4 times the coefficient of the X squared 1.
01:58:860Paolo Guiotto: Times the coefficient of degree zero, this quantity, so 4… A square minus 4B.
02:08:160Paolo Guiotto: P equator or equal than zero.
02:11:510Paolo Guiotto: So, the probability… that this equation, X squared plus 2AX plus B equals zero, pause.
02:26:680Paolo Guiotto: Rio?
02:28:460Paolo Guiotto: solutions… It's the same of the probability that, that quantity so…
02:37:840Paolo Guiotto: we can factorize the 4, so basically A square minus B.
02:43:840Paolo Guiotto: be greater or equal than zero. This is the problem. So we have to compute.
02:51:970Paolo Guiotto: Or, if you want, the probability that A square is greater than B.
02:55:960Paolo Guiotto: Now, we can say that this probability, which is,
03:01:510Paolo Guiotto: probability that the point AB belongs to a set E, well, set E is the set of points XY, such that X squared minus Y is greater than or equal than zero, can be written in terms of
03:23:280Paolo Guiotto: But if we have the density, which is the case, as we see in a moment, it can be written as the integral on E of the joint density FAB,
03:35:390Paolo Guiotto: XY… Yeah, so… BY.
03:41:590Paolo Guiotto: Well… The joint density exists because NDT is equal to the product of the Single densities, because Bye.
04:00:00Paolo Guiotto: assumptions.
04:02:280Paolo Guiotto: The variables A and B are independent.
04:15:840Paolo Guiotto: And the… absolutely continuous.
04:21:60Paolo Guiotto: It's because they have a density, no? Both densities, FA, say, U, is the same of FB.
04:30:800Paolo Guiotto: U, they are both uniformly distributed in the interval 0, 1, so the density is just the indicator of the interval 0, 1.
04:39:640Paolo Guiotto: Yo.
04:41:920Paolo Guiotto: Now, you remind that, if, a variable, X,
04:47:270Paolo Guiotto: is uniformly distributed in AB, it is absolutely continuous with density, which is 1 over B minus A, the indicator of the interval AB.
04:59:610Paolo Guiotto: In this case, the interval AB is 0, 1, and therefore we have this.
05:04:160Paolo Guiotto: So, the, joint density Is just the indicator of 01.
05:16:350Paolo Guiotto: variable X times the indicator… 0, 1, variable Y.
05:25:10Paolo Guiotto: Okay, so this means that that integral, which is the probability, We are looking for…
05:38:220Paolo Guiotto: can be written as the integral,
05:41:900Paolo Guiotto: on domain E, the domain E is the domain where X squared minus Y is greater or equal than 0.
05:53:510Paolo Guiotto: of the density is this, 01 for the variable X, 01 for the variable Y.
06:03:190Paolo Guiotto: the XDY.
06:05:580Paolo Guiotto: Now, the domain is this in the Cartesian plane, XY.
06:12:300Paolo Guiotto: is, Y less or equal than X squared? Y equal X squared is the parabola.
06:22:670Paolo Guiotto: Why? Less or equal is below the parabola.
06:26:90Paolo Guiotto: So, is this… region here.
06:32:860Paolo Guiotto: This is Y less or equal than X squared off.
06:37:220Paolo Guiotto: And because of the indicators, we, basically, we have to take X between 0 and 1.
06:46:860Paolo Guiotto: Y, between 0 and 1.
06:51:180Paolo Guiotto: And, of course, Y less or equal than X squared of 1.
06:56:670Paolo Guiotto: DX, D, Y.
06:58:630Paolo Guiotto: So we have, to compute, basically, the area, if this is 1,
07:06:110Paolo Guiotto: the area of this plane region. This is the region described here.
07:14:40Paolo Guiotto: So we can, easily do the integral. We have the integral 0, 1, say, for X, then for Y, we have from 0 to X squared.
07:24:640Paolo Guiotto: The function is 1.
07:27:910Paolo Guiotto: So we get integral 1 of X pair DX, which is X cubed over 3, between 0 and 1, so it is equal to 1 third.
07:41:980Paolo Guiotto: So that's the probability we are looking for.
07:59:140Paolo Guiotto: Let's see this, the next one, this 647.
08:08:880Paolo Guiotto: We have two variables, X and Y, which are both uniform.
08:14:820Paolo Guiotto: In the interval 01.
08:18:340Paolo Guiotto: And they are independent.
08:27:820Paolo Guiotto: Then it says you define capital U as the minimum between X and Y.
08:38:340Paolo Guiotto: And capital V as the maximum.
08:43:890Paolo Guiotto: between X and Y.
08:52:210Paolo Guiotto: well, it asks to determine the expected value of U, the expected value of V… And the covariance…
09:04:210Paolo Guiotto: between U and B.
09:16:270Paolo Guiotto: Yeah, we can do this, but I wasn't thinking if we can also compute the…
09:23:710Paolo Guiotto: to see if there is a density for U, for V.
09:29:830Paolo Guiotto: Could be interesting.
09:32:220Paolo Guiotto: Let's see, but let's add the… I don't know what is, we have to verify. I will add these questions. Is there a density for U, for V?
09:45:890Paolo Guiotto: Mmm…
09:47:910Paolo Guiotto: we don't need the densities necessarily to compute this expectation, because, for example, we could say the expected value of U
09:59:90Paolo Guiotto: So how could we, read this, well…
10:10:10Paolo Guiotto: But let's use a…
10:11:910Paolo Guiotto: let's use the… we have that the two are independent, so we have the joint density, so we can say that it is the integral on R2 of the function minimum between X and Y.
10:27:670Paolo Guiotto: the density of the variable X.
10:31:600Paolo Guiotto: times the density of the variable Y,
10:35:110Paolo Guiotto: PXDY. The two are uniformly distributed in 01. This is the joint density, because of independence.
10:43:740Paolo Guiotto: So, in principle, I should write the joint density, but because they are independent, it is this one. Now, since these are the indicators of intervals 0, 1 each.
10:58:30Paolo Guiotto: So, basically, it means that we are integrating for X between 0 and 1, for Y between 0 and 1.
11:09:670Paolo Guiotto: the minimum.
11:11:570Paolo Guiotto: between X and Y.
11:18:690Paolo Guiotto: Bandai is, well… we could say, let's split into two iterated integrations. So, for example, let's…
11:28:60Paolo Guiotto: integrate last in X, and the first in Y.
11:34:350Paolo Guiotto: So, formally, I should say, from 0-1 minimum between X…
11:39:610Paolo Guiotto: and Y. The minimum is X. If X is less than Y, it is Y if it is Y less than X. So I could say that this integral here could be splitted into the integral from 0 to X plus the integral from X to 1.
12:00:10Paolo Guiotto: When Y is between 0 and X,
12:05:810Paolo Guiotto: It means that it is smaller than X, so the minimum between X and Y will be Y.
12:11:590Paolo Guiotto: When Y is between X and 1, it thinks that, in particular, Y is larger than X, so the minimum will be X.
12:20:710Paolo Guiotto: So this becomes the integral 0 to 1.
12:24:520Paolo Guiotto: This first integral is… this is the integral of Y, so it is Y squared divided 2 between 0 and X, plus this is instead a constant for the integration variable, so it is X times the integral from X to 1 of 1, it is the length of the interval 1 minus X.
12:45:10Paolo Guiotto: This is in the axe.
12:47:160Paolo Guiotto: This evaluation yields X squared over 2, so plus X minus X squared.
12:54:420Paolo Guiotto: So all this at the end, it's X minus X squared over 2.
13:00:970Paolo Guiotto: That we have to integrate between 0 and 1.
13:09:130Paolo Guiotto: So this will… between 0 and 1, minus 1 half X cubed over 3, between 0 and 1.
13:23:30Paolo Guiotto: At zero, you get 0, so the unique values are at 1, so at 1A, we have 1 half minus 1 half times 1 third, so 1 half minus 1 sixth,
13:35:840Paolo Guiotto: What is it? 2 6th, 1 3rd.
13:39:330Paolo Guiotto: So this is the expected value of U.
13:42:950Paolo Guiotto: The expected value of V will be the same.
13:48:390Paolo Guiotto: Since the exercise asks for the covariance, we need to compute also this value.
13:55:290Paolo Guiotto: So, this is just the same thing, integral, for, XY, both between 0 and 1. Now, the function is the maximum between
14:08:250Paolo Guiotto: X and Y.
14:18:100Paolo Guiotto: It should be the same, more or less. So we have integral 0, 1,
14:23:350Paolo Guiotto: integral 01 of the maximum. So, again, we split between 0 and X, plus the integral from X to 1.
14:33:500Paolo Guiotto: when Y is between 0 and X, now the maximum is X, so this is this.
14:42:20Paolo Guiotto: When Y is between X and 1, the maximum is Y, so we have this.
14:49:480Paolo Guiotto: DX, huh?
14:54:150Paolo Guiotto: Okay, so this is X times the integral 0 to X of 1, so X squared, plus here we have Y squared over 2 between x and 1, so 1 half minus X squared over 2.
15:18:200Paolo Guiotto: So we have integral between 0, 1 of, 1 half… Plus… X squared over 2, dx.
15:31:140Paolo Guiotto: So, 1 half for the first integral, plus 1 half X cubed over 3.
15:39:70Paolo Guiotto: Between 0 and 1, and this yields 1 half plus 1 sixth.
15:45:700Paolo Guiotto: 1 half is 3 sixths, so 4 sixths, so 2 thirds.
15:52:00Paolo Guiotto: Probably the… there was something,
15:59:770Paolo Guiotto: I don't know, faster to get this. So finally, the covariance between these variables, U and V, Now.
16:13:40Paolo Guiotto: We observed that if they are independent, the covariance would be zero.
16:19:250Paolo Guiotto: We have not, signed this.
16:23:40Paolo Guiotto: Probably they won't be, let's see what happens to the covariance. The covariance is the expected value of U times V
16:31:150Paolo Guiotto: Minus the… can be also written in this way, since we have the two expectations, minus…
16:38:570Paolo Guiotto: the… the product of the expectations. So, the expectation of U is 1 3rd.
16:45:900Paolo Guiotto: the expectation of V is 2 thirds, so this is 2… Mine.
16:53:780Paolo Guiotto: Now, the expectation of U times V…
16:58:590Paolo Guiotto: is, again, the integral for XY between 0 and 1.
17:07:569Paolo Guiotto: the minimum between X and Y
17:11:720Paolo Guiotto: times the maximum between X and Y.
17:15:599Paolo Guiotto: DX, DY.
17:20:810Paolo Guiotto: So we have the integral from 0 to 1, then 4Y,
17:26:359Paolo Guiotto: as above, we split the integration into 0 to X. From 0 to X, the minimum Y is smaller than X, so the minimum is Y, while the maximum is X.
17:41:650Paolo Guiotto: plus the integral from X to 1. Y now is larger than X, so the minimum is X, and the maximum is Y.
17:50:120Paolo Guiotto: So, basically, this, in DY, this is the integral from 0 to 1 of XY.
17:57:770Paolo Guiotto: DY in the integral 0 to 1 of the X.
18:05:530Paolo Guiotto: Peace.
18:06:650Paolo Guiotto: So this comes out, we have… this actually splits into the product 0 to 1 of… because the two integrals are independent, say.
18:21:750Paolo Guiotto: And this is, the expected value of X, so it's 1 half times, One half.
18:30:480Paolo Guiotto: So, it's 1-4.
18:33:140Paolo Guiotto: So the covariance at the end seems to be equal, so they are not independent.
18:40:530Paolo Guiotto: Equal to 1 fourth minus 2 over 9, whatever it is. This is the numerical graph.
18:51:540Paolo Guiotto: We said, let's add these questions about the…
18:57:100Paolo Guiotto: density of U, density of V.
19:01:30Paolo Guiotto: So, we could…
19:04:480Paolo Guiotto: we could, discuss this by seeing U and V as a map of X and Y, and then,
19:12:920Paolo Guiotto: computing,
19:15:360Paolo Guiotto: the, the joint density and reducing the marginal by doing the integral of the joint density. This is a possibility.
19:26:420Paolo Guiotto: Otherwise, we can proceed directly, but in this case, we have to start from the CDF.
19:35:140Paolo Guiotto: to compute,
19:40:270Paolo Guiotto: But let's see what happens. To compute the… Fu.
19:45:470Paolo Guiotto: If any.
19:48:460Paolo Guiotto: We start from the CDF, we compute
19:56:190Paolo Guiotto: D.
19:57:620Paolo Guiotto: CDF.
20:00:170Paolo Guiotto: Because this always exists,
20:03:150Paolo Guiotto: This would be the probability that U is less or equal than little u. And u is what is the minimum
20:11:920Paolo Guiotto: between X and Y.
20:16:140Paolo Guiotto: Is less or equal than 0.
20:19:30Paolo Guiotto: Now, this minimum can be only X or Y, so we could say that this is…
20:31:20Paolo Guiotto: Yeah, we will see, that this is a… this type of variable, the minimum, or the maximums, or variables,
20:40:300Paolo Guiotto: more than 2 is a frequent, important problem. So, how can we represent this event? We could say, for example, that the minimum is X,
20:55:500Paolo Guiotto: Index must be less than U.
20:58:730Paolo Guiotto: So, this means that X is less or equal U, and Y is greater than U. In this case, the minimum is X, and X is less than U.
21:12:340Paolo Guiotto: Or… X is greater than U, and it is Y to be less or equal than
21:25:110Paolo Guiotto: What if they are both?
21:33:190Paolo Guiotto: I need also this one.
21:36:300Paolo Guiotto: I want to… because I wanted to split into the… this joint thing.
21:48:970Paolo Guiotto: Well, let's say… no. This is, because in principle, there is also this case. X less or equal U, and Y less or equal U.
22:00:90Paolo Guiotto: But I want to write this as disjoint union.
22:05:570Paolo Guiotto: But… no, no, no.
22:07:500Paolo Guiotto: But in this case, no, this,
22:13:910Paolo Guiotto: No.
22:19:510Paolo Guiotto: No, this is not correct. So, the minimum…
22:25:560Paolo Guiotto: is less or equal than U. If X is less than U,
22:44:170Paolo Guiotto: Can't realize this.
22:48:300Paolo Guiotto: Because if we have this type of events, then we can use the fact that the two are independent, so the probabilities will factorize, etc.
23:01:550Paolo Guiotto: the new ones.
23:05:00Paolo Guiotto: If X is less than U, Y, greater than U, the minimum is X.
23:09:440Paolo Guiotto: And, if X is larger than U, and Y is less than U being more. Is Y?
23:17:740Paolo Guiotto: I should, have also required…
23:22:310Paolo Guiotto: If they are both greater than U.
23:27:20Paolo Guiotto: The minimum cannot be less than U.
23:36:500Paolo Guiotto: Well, let's say that, let's do this. Minimum XY is less than U, is, formally…
23:47:20Paolo Guiotto: the same of X is less than U, or Y is less than U, right? One of the two must be… at least one of the two must be less than U.
23:58:850Paolo Guiotto: Now… This is not necessarily a disjoint union.
24:05:690Paolo Guiotto: So I can, like, design tier now.
24:18:540Paolo Guiotto: Yeah, I can make a disjoint unit this way. X is less than U, Union. X.
24:26:960Paolo Guiotto: is larger than U.
24:31:100Paolo Guiotto: and Y is less than U.
24:34:800Paolo Guiotto: It's the same, because… If you are, of course,
24:40:940Paolo Guiotto: if you are here, this is clearly contained in this one, no? Because either you are less than U, or if you are in the other side, it is Y to be less than U, so you are here.
24:54:80Paolo Guiotto: But if you are here, it will be in at least one of these two.
25:04:00Paolo Guiotto: Okay? If it is X less or equal than U, you are here. If you are here.
25:09:550Paolo Guiotto: then you have two possibilities. Either X is less than U, so you would be here.
25:17:620Paolo Guiotto: or X should be greater than U, and you should be here. So I can always split this into Y less or equal U, and X less or equal U union
25:33:150Paolo Guiotto: Why? Less or equal you?
25:36:420Paolo Guiotto: and, X greater than U.
25:39:620Paolo Guiotto: This… this is not a disjoint union, because there are parts which are in common. But I'm saying the two sides are the same, because if you… clearly, if you are in this union, either X is less or equal than U, and therefore you are here, or Y is less or equal to U, if you are in this path, and you are here.
25:57:830Paolo Guiotto: The vice versa, if you are in one of these two, if you are in this one, you are in this. If you are in this.
26:04:480Paolo Guiotto: Okay? It can be that X is strictly less than U, but is less or equal to U also, and therefore you are there. Or X is greater than U, and you are here. But now the point is that this union is disjoint.
26:20:980Paolo Guiotto: Because, of course, you cannot be less than U and strictly greater than U. So, it means that the probability
26:30:120Paolo Guiotto: that the minimum between XY is less or equal than U,
26:38:90Paolo Guiotto: which is the quantity we are computing, probability that capital U is less or equal than U. This is the probability that X is less or equal U, plus the probability that X is larger than U, and Y is less or equal U.
26:56:800Paolo Guiotto: Now, the first one, since this is the CDF of capital X evaluated at U. The second one, since the two are independent.
27:09:400Paolo Guiotto: by independence.
27:14:240Paolo Guiotto: This splits into the product probability that X is greater than U times the probability that Y is less or equal U.
27:23:820Paolo Guiotto: The second one is the CDF of Y.
27:28:430Paolo Guiotto: at U, while the first one, I can see this probability as 1 minus the probability that X is less or equal U.
27:42:300Paolo Guiotto: So this is, again, 1 minus FX of U. So at the end, I get this formula.
27:50:860Paolo Guiotto: the, probability.
27:54:60Paolo Guiotto: That's the minimum.
27:56:420Paolo Guiotto: Between X and Y.
27:59:130Paolo Guiotto: is less or equal than U, it is equal to FX.
28:05:90Paolo Guiotto: of you.
28:06:920Paolo Guiotto: plus 1 minus FXU, times F, Y, You.
28:18:250Paolo Guiotto: And this is the probability that the capital U, which is this variable, is less than U, so the CDFFU at point little u.
28:29:740Paolo Guiotto: Now, these effects,
28:32:210Paolo Guiotto: are the CDF of the uniform. They are the same thing in this case. In our case, Okay, zip.
28:43:870Paolo Guiotto: FX is equal to FY, because they are supposed to have the same distribution, and they have both the derivative, the derivative of FX, the derivative of FY, is the indicator of the… is the density, indicator of interval 01.
29:03:530Paolo Guiotto: So we can say that, so there exists the derivative.
29:09:100Paolo Guiotto: This, for, let's say, let's put the letter U here, for every U except the two points where we have the
29:18:890Paolo Guiotto: the CDF would be not differentiable, because the CDF is made like that. It's 0 here, then from 0 to 1 goes up linearly, and then it remains equal to 1. This is the CDF.
29:34:150Paolo Guiotto: F, X, F, Y are both made like this. So they are all everywhere differentiable except at two points, so they are almost everywhere differentiable. There exist a
29:46:340Paolo Guiotto: the derivative, consequently, there exists a derivative with respect to U of F capital U,
29:55:500Paolo Guiotto: Of you, except for 0 and 1.
30:01:540Paolo Guiotto: And it is equal to… Well, the derivative of FX, which is the indicator 01,
30:12:920Paolo Guiotto: Time, plus, here we have a product, so we will have the derivative of the first, which is, minus the indicator 0, 1,
30:23:820Paolo Guiotto: You?
30:25:80Paolo Guiotto: times the second, F, Y… You.
30:32:670Paolo Guiotto: plus 1.
30:35:210Paolo Guiotto: well, let's call the common, the common CDF just F, so this is the…
30:41:420Paolo Guiotto: function here, 1 minus FU times the derivative, which is again the indicator, 0, 1.
30:49:10Paolo Guiotto: You?
30:52:440Paolo Guiotto: So, we have Indicator 01, You…
31:00:150Paolo Guiotto: Plus another indicator, so 2 times the indicator.
31:05:940Paolo Guiotto: There are two indicators. One is here, and another comes out from this times this.
31:14:170Paolo Guiotto: Then we have minus 2 times F of U, indicator 01 of U. So this is…
31:23:990Paolo Guiotto: Or if you want, that's right better, 2, indicator 01.
31:29:500Paolo Guiotto: U.
31:31:340Paolo Guiotto: 1 minus F… of fuel.
31:35:780Paolo Guiotto: This is for U different from 0 and 1.
31:40:570Paolo Guiotto: And this is the density of,
31:45:310Paolo Guiotto: So we can say that U is absolutely continuous, and this is the density of FU.
31:52:230Paolo Guiotto: And similarly, for FV, we can… Procedure, 4.
32:00:610Paolo Guiotto: V equal the maximum.
32:03:380Paolo Guiotto: between X and Y.
32:09:310Paolo Guiotto: we can say that the probability that V is less or equal than little d
32:17:230Paolo Guiotto: Is the probability that the maximum Between X and Y.
32:23:420Paolo Guiotto: is less or equal than V.
32:26:840Paolo Guiotto: So this means that X, in this case it is easier, because to have the maximum less or equal than something, it means both must be less than something. So X less than V and Y less or equal V. This is exactly the same thing.
32:44:960Paolo Guiotto: Now, because of the independence, this can be written as the probability independence
32:52:290Paolo Guiotto: the probability that X is less than V times the probability that Y is less or equal V.
33:01:400Paolo Guiotto: And these are the CDF of X at V times the CDF of Y at V, which is the same function at the end, so it is F squared of V. But F is the CDF of the uniform.
33:17:450Paolo Guiotto: So, if you want, this is the CDF of,
33:22:720Paolo Guiotto: V at point V. So for the maximum, the CDF is just the CDF of F squared, so you have that here we get the parabola, like this, and then we have 1.
33:36:220Paolo Guiotto: So this is F squared.
33:39:780Paolo Guiotto: So, also, for this case, we have that, for… except, for the values B equals 0 and 1, where… well, for B equals 0, probably it is differentiable.
33:51:570Paolo Guiotto: So we have that the derivative with respect to V of the CDF,
33:58:600Paolo Guiotto: exists, and it is equal to 2F of V times F prime of V, so derivative with respect to V of F, and this one is the indicator of 01.
34:12:699Paolo Guiotto: So, for the maximum, the… density is this one. Two…
34:19:440Paolo Guiotto: times F of B, where this is the CDF of the uniform
34:26:170Paolo Guiotto: Distribution times the indicator of 01.
34:31:270Paolo Guiotto: this. This is the formula for the… That's it.
34:51:139Paolo Guiotto: Okay, mmm… I want to add something now, so we said that, so, if,
35:04:970Paolo Guiotto: X and Y are two random variables.
35:16:160Paolo Guiotto: We have seen that XY are independent.
35:25:810Paolo Guiotto: if and only if, well, with the CDF,
35:31:220Paolo Guiotto: It happens that this joint CDF
35:35:850Paolo Guiotto: Splits into the product of the two.
35:39:610Paolo Guiotto: single CDF, huh?
35:42:400Paolo Guiotto: this, at every point, XY.
35:48:150Paolo Guiotto: And this is equivalent.
35:50:900Paolo Guiotto: if the vector XY is absolutely continuous, if XY…
35:59:50Paolo Guiotto: is absolutely continuous. So, in other words, If there exists a density, Okay, joint density.
36:15:340Paolo Guiotto: So it is equivalent to say that D joint density splits into the product, of… The marginals.
36:26:740Paolo Guiotto: This is not necessarily for every XY, but for almost for almost every XY, so…
36:38:230Paolo Guiotto: almost every XY. This, almost every, is referred to the Lebec measure, so the DXDY.
36:47:670Paolo Guiotto: with respect… to… Le Baguet Measure.
36:55:330Paolo Guiotto: Now, there is a further property, which is known as the Katz Theorem, this was… an important Polish mathematician.
37:08:240Paolo Guiotto: Which is the following.
37:17:570Paolo Guiotto: But the two variables are independent if and only if also the characteristic function splits.
37:24:500Paolo Guiotto: So, XY… are independent.
37:31:510Paolo Guiotto: If and only if, when you take the characteristic function of the pair, X, Y.
37:39:270Paolo Guiotto: This is a function of two variables, let's denote them Xi and eta. This splits into the product of the two characteristic functions.
37:52:790Paolo Guiotto: Where I read.
37:54:280Paolo Guiotto: Per si ITa.
37:56:520Paolo Guiotto: in R2.
38:00:680Paolo Guiotto: Now, the one way is easy.
38:04:210Paolo Guiotto: The other way is a bit sophisticated, which is the interesting part for this statement. However, this one is easy, because if you take
38:13:290Paolo Guiotto: the, joint… the characteristic function of the, joint vector, C eta. This is, by definition, the expected value of e to i, the vector, c eta, scalar product with vector XY.
38:33:880Paolo Guiotto: Now, this color product is C times capital X plus eta times capital Y.
38:41:660Paolo Guiotto: And therefore, in this exponential, you have e to iCX plus eta y.
38:51:370Paolo Guiotto: Of course, you can split this into the product, e to iCX times e to i eta y. So, the joint characteristic function is the expectation of this, expectation of e to iCX
39:10:30Paolo Guiotto: times e to i eta y. Now, as you can see, these are a function
39:16:540Paolo Guiotto: of capital X times a function of capital Y. And we know that by independence, this thing splits into the product of the expectations. So this becomes
39:29:520Paolo Guiotto: Here is where we use independence. It becomes E2I X, the expectation of this, and times the expectation of e to i eta y.
39:42:510Paolo Guiotto: And now these are exactly the two characteristic functions of, respectively, capital X and capital Y.
39:51:970Paolo Guiotto: So, from the independence, we get that also the characteristic function splits into the
39:59:270Paolo Guiotto: the joint characteristic function splits into the product of the single characteristic functions. For the vice versa, we will do… the little proof becomes very easy in the case when the pair XY is absolutely continuous.
40:14:660Paolo Guiotto: If, XY.
40:18:920Paolo Guiotto: is absolutely continuous.
40:22:580Paolo Guiotto: Otherwise, it is a bit complicated, depends on the fact that the product measure is the measure of the product, but with the density, it's easier. So, suppose that,
40:34:630Paolo Guiotto: So there exists the joint density, FXY. I want to show that this is the product of the single densities.
40:47:190Paolo Guiotto: There's also… there are also the single
40:50:130Paolo Guiotto: marginals, FXFY. So, the thesis is to prove That, the…
40:58:590Paolo Guiotto: Joint density is exactly the product of the marginals.
41:04:350Paolo Guiotto: Okay? So if we prove that this is true almost everywhere, we have the conclusion.
41:10:370Paolo Guiotto: Of course, what is the connection with the characteristic function is that the characteristic function, when there is density, is the ordinary Fourier transform. So what we have here is that we notice
41:25:390Paolo Guiotto: we noticed… that… If we take the characteristic function.
41:32:800Paolo Guiotto: of the joint vector XY, so this is the expectation of EI, again, XCX dot XY.
41:47:890Paolo Guiotto: Now, this can be written as an integral.
41:51:210Paolo Guiotto: Now we go down to the low, and since we have the density, we can write an integral on R2 of,
42:01:240Paolo Guiotto: the function that you have here, e to iC eta dot XY,
42:09:860Paolo Guiotto: integrated versus the function, the joint density, DXDY.
42:17:260Paolo Guiotto: But this thing is, apart for the minus that we don't have in the exponent, is the Fourier's form of the function FXY, and exactly is the hat of XFXY evaluated at minus C minus eta.
42:35:440Paolo Guiotto: But since this joint characteristic function is also, by assumptions.
42:42:180Paolo Guiotto: By hypothesis, so here we are assuming that
42:46:340Paolo Guiotto: we are in the vice versa. We are assuming that Visa,
42:52:200Paolo Guiotto: characteristic function for the joint vector splits into the product of the two characteristic functions. So, by assumption, this is also PX of Xi times phiy of theta.
43:05:450Paolo Guiotto: That is, they are the single, one variable Fourier transform, so this is the integral in R of EI,
43:13:520Paolo Guiotto: XC, say, X, F, X, X in DX, times integral on R of EI eta y F, Y,
43:26:600Paolo Guiotto: Y theta for the same reason, so this is the transformation of the expectation into the integral with respect to the law of x, the law of Y, no? So these are
43:38:00Paolo Guiotto: These expectations were written, PIXC capital X, and EI eta, capital Y,
43:48:280Paolo Guiotto: when we rewrite this as integrals in the… on the image of the variable, so with respect to the law of x, so this would be D mu X, and this would be D muy. Since there are densities, these are the integrals
44:03:560Paolo Guiotto: dx becomes the function of X times the bag measure.
44:09:610Paolo Guiotto: And now, recombining these two integrals into a unique integral, so carrying one inside to the other, and apply the Fubini theorem.
44:18:440Paolo Guiotto: This integral, this product becomes an integral in R2. We can also recombine the two exponentials. We have e to i, if you want CX plus I eta y.
44:32:780Paolo Guiotto: then fx little x times FY little y in dx dy.
44:41:170Paolo Guiotto: And this is, of course, a unique exponential e to i vector C.
44:47:90Paolo Guiotto: Ita, Scala product XY.
44:52:210Paolo Guiotto: Well, now you see that this is the Fourier transform of this function here, apart from the sine here. So this is the hat of FX
45:05:20Paolo Guiotto: of one variable, be careful, because this is function of X, this is function of Y, so let's, say, sharp this and ballet the other to specify that these are two different variables. Evaluated again at minus C minus eta.
45:24:330Paolo Guiotto: So, by the assumption, so… from…
45:30:730Paolo Guiotto: The fact that the joint characteristic function splits into the product of the characteristics It comes that,
45:40:390Paolo Guiotto: We have that deflated form of the joint density.
45:45:240Paolo Guiotto: evaluated at minus C minus theta, is the same of the written form of this product, of the algebraic product, of the one variable densities.
45:57:890Paolo Guiotto: still at minus C minus eta.
46:02:290Paolo Guiotto: And since the L1 Fourier transform is injective, So, injectivity…
46:14:280Paolo Guiotto: of L1.
46:16:660Paolo Guiotto: Pourier transform.
46:19:60Paolo Guiotto: And the two functions are two densities, so they are L1 functions, they have the same Fourier transform, they must be the same function. So we get that FXY
46:29:760Paolo Guiotto: XY must be equal to FXXFY. Why? Now, the injectivity holds in the L1 sensor, so it means almost every point XY.
46:44:360Paolo Guiotto: in R2.
46:46:570Paolo Guiotto: And this means exactly that.
46:49:430Paolo Guiotto: X, Y.
46:51:530Paolo Guiotto: R.
46:52:750Paolo Guiotto: independent.
46:57:840Paolo Guiotto: Now, the fact that, the, The, joint,
47:06:580Paolo Guiotto: characteristic function splits into the products of the characteristics as some remarkable application, mainly with the Gaussian distributions, but not only. For example, we have this result, this nice result.
47:24:760Paolo Guiotto: That says… that if X, Y, R, independent.
47:37:400Paolo Guiotto: And absolutely continuous.
47:43:930Paolo Guiotto: random variables.
47:51:160Paolo Guiotto: Then, also the sum of the two variables is absolutely continuous, and this formula holds… the density for the sum is exactly the convolution of the densities of X and Y.
48:07:380Paolo Guiotto: So it's another place where you see the interesting in the convolution, because it provides a formula for the density of the sum. Otherwise, if they are not independent, the density of the sum is not an easy object to get. But in this case, we can see this.
48:27:790Paolo Guiotto: Because of… If we take, the,
48:36:570Paolo Guiotto: Let's take the characteristic function of the sum.
48:40:90Paolo Guiotto: So, the characteristic function of the sum…
48:45:320Paolo Guiotto: at point C is equal to the expected value of EIC times X plus Y.
48:54:980Paolo Guiotto: This is the definition of characteristic function for the sum.
49:00:270Paolo Guiotto: Now, of course, this splits into the exponential of EIX times EICY.
49:09:660Paolo Guiotto: And because of the independence, my independence.
49:17:200Paolo Guiotto: This expectation is the expectation of a function of X times a function of Y splits into the product of the expectations.
49:26:80Paolo Guiotto: E to ICX.
49:30:640Paolo Guiotto: times E2… I. C. Y.
49:38:110Paolo Guiotto: And these are exactly the characteristic function of X, evaluated at XC, times the characteristic function of Y, evaluated at the same point C.
49:51:740Paolo Guiotto: But… You may remind that, if there is a density, And since,
50:03:960Paolo Guiotto: the characteristic function of X, and the same for Y.
50:09:140Paolo Guiotto: is also the integral on R of EIX times the density. So this comes from the change of variable formula when we
50:22:160Paolo Guiotto: transform the expectation into an integral with respect to the load. This one is the Fourier transform of FX, but evaluated at minus C.
50:38:10Paolo Guiotto: And maybe you may remind that this is the same of the transformer.
50:43:930Paolo Guiotto: of F changing the sign of the argument, evaluated at X. This is…
50:51:170Paolo Guiotto: a simple property of the Fourier transform, because if you want, you can see here, if you change values, if you emphasize the minus here, you have to put the minus, for example, in the argument minus X, in such a way that the two minus
51:06:140Paolo Guiotto: Simplify into a plus. Now, if you change variable, you call y equal minus X, this becomes the integral. Of course, R remains R, because it's just a reflection.
51:19:130Paolo Guiotto: This is E minus XY. F of X becomes F of minus Y and DY. So this is exactly the clear transform of FX with the argument with the opposite algorithm.
51:35:760Paolo Guiotto: And the same is for FY, for the characteristic function of Y, so phi y Xi is the hat of FY evaluated at minus the argument.
51:47:190Paolo Guiotto: point C. So, returning back to this calculation, we have that the characteristic function of the sum is the product of the characteristic functions.
52:02:440Paolo Guiotto: which is the product of the heads of these two functions, so FX at minus
52:09:570Paolo Guiotto: the argument FY minus the argument point C.
52:15:50Paolo Guiotto: But we remind of a remarkable property of the Pietan spawn. This product of a hat is the hat.
52:24:290Paolo Guiotto: of something, and this something is the convolution product of the two functions. So, the convolution product of this with the convolution product of
52:34:510Paolo Guiotto: FY minus.
52:39:400Paolo Guiotto: Okay?
52:40:540Paolo Guiotto: And this says, going back, that, so the X plus Y
52:47:840Paolo Guiotto: has characteristic function, which is, this, let's call, for a moment, F, this function, F of argument, this function.
52:59:120Paolo Guiotto: Now, this function, if you want F of X equal, it should be integral, because for the sine is FX at minus, so if we write the convolution as X minus YF
53:17:30Paolo Guiotto: the other function of Y. We have to change the sign of everything, so this will become Y minus X, FY of minus Y DY. This would be the convolution, but now if you flip back
53:33:920Paolo Guiotto: the, the… But it is better if we denote this with F minus something.
53:42:850Paolo Guiotto: So, if we put the minus X, this becomes plus X. If we now change variable y equal minus U, this is integral on R of FX of X minus U, FYUDU.
54:01:810Paolo Guiotto: So it is the convolution between FX and FY at point X.
54:10:680Paolo Guiotto: So… At the end, it doesn't change anything here.
54:20:870Paolo Guiotto: So it says that the characteristic function of the… of the sum is the Fourier transform of this, and this means that… so this F of minus X, which is the convolution of FX with FY,
54:37:160Paolo Guiotto: F, X, with FY… is D.
54:43:210Paolo Guiotto: density.
54:46:890Paolo Guiotto: of X plus Y.
54:49:630Paolo Guiotto: Because its characteristic function, its Fourier transform is the characteristic function of X plus 1.
55:04:40Paolo Guiotto: Alright.
55:05:490Paolo Guiotto: Let's think about this exercise.
55:10:20Paolo Guiotto: 642.
55:12:590Paolo Guiotto: So, XY are independent.
55:20:830Paolo Guiotto: The first question is, number one, check that raw XY is equal to zero, so the correlation is equal to zero. Remind you that
55:34:230Paolo Guiotto: Raw X, what?
55:35:790Paolo Guiotto: is equal to… well, the covariance of XY…
55:43:620Paolo Guiotto: divided by the, the standard deviations, the product of the standard deviations, so the variance of X
55:54:160Paolo Guiotto: to exponent, 1 half variant of Y to exponent.
55:58:180Paolo Guiotto: 1 half. Well, let's put that, independent and, non… constant.
56:07:890Paolo Guiotto: Otherwise, this would be trivial. The variance are zero, and it's… the correlation is not fine. Well, this is easy, because, in fact, if you look at the covariance between X and Y,
56:24:100Paolo Guiotto: This is the… we already said this, probably. It is the expected value, one of the form of the covariance is this one, X times Y minus the product of the expectations, so expectation of X times expectation of Y.
56:42:190Paolo Guiotto: And the two are the same, because, you see, we have… we know that for independent random variables, the product… the expectation of the product splits into the product of the expectations, so these two are the same, the difference is zero, so ri is 0.
56:57:670Paolo Guiotto: Now, the nice thing is, number two, If the vector XY is Gaussian.
57:10:110Paolo Guiotto: Ben, XY are independent, if and all if, actually, XY.
57:18:470Paolo Guiotto: independent, if and only if the correlation is equal to zero.
57:27:850Paolo Guiotto: Which is not true in general, but for Gaussian is true.
57:34:270Paolo Guiotto: And,
57:40:410Paolo Guiotto: Why? Let's see. Probably we can check this directly,
57:55:840Paolo Guiotto: So, if we have to do directly.
58:01:280Paolo Guiotto: We could say… so, we have to verify the only if, no?
58:05:710Paolo Guiotto: So, if raw XY is equal to 0, means that the covariance is equal to 0.
58:30:800Paolo Guiotto: Yeah.
58:32:980Paolo Guiotto: If, XY is Gaussian, So it's normal with mean M,
58:40:410Paolo Guiotto: and covariance matrix C, which is, in this case, a 2x2 matrix.
58:48:380Paolo Guiotto: Yes, this case is easy, also directly, because
58:53:190Paolo Guiotto: You remind, I don't… I don't think we have done the calculation I told you to verify. We have done for the mean value, but…
59:02:740Paolo Guiotto: the entries of this matrix C, which is, in this case, a 2x2 matrix, so there is C11, C12, C21, C22.
59:16:810Paolo Guiotto: The entries of this metrics are the… So… C.
59:24:320Paolo Guiotto: Well, this is… let's say it's better if we write with,
59:30:490Paolo Guiotto: Instead of writing XY, I write X1, X2 for a seconder.
59:35:960Paolo Guiotto: So, if you have an array, in this case of two components.
59:40:980Paolo Guiotto: Which is normal. CIJ, the entry line i, column J, is just the covariance between Xi and Y… and XJ.
59:53:780Paolo Guiotto: So, in particular, C12 is the covariance
00:00:220Paolo Guiotto: If X1, X2 is our array XY,
00:05:460Paolo Guiotto: is the covariance between X and Y.
00:10:310Paolo Guiotto: Now, if we assume that the covariance between X and Y is 0, this number is 0, as well as C21, which is the covariance between Y and X, which is the same.
00:22:630Paolo Guiotto: Equals 0. So it means that the covariance matrix C is actually a diagonal matrix, C1100C22.
00:34:270Paolo Guiotto: So in this case, it is easy to verify that the two variables are independent, because if you take the joint density, FXY, which is 1 over root of 2 pi, in this case, we have two variables, so square, determinant of matrix C.
00:52:440Paolo Guiotto: E minus 1 half. The C minus 1 matrix is easy, because it is the matrix.
01:00:220Paolo Guiotto: with the diagonal matrix, with the elements on the diagonal, which are the reciprocal of those numbers. So there is 1 over C1100, 1 over C1, C22 here.
01:13:100Paolo Guiotto: So this is the matrix C minus 1, and then we have, say…
01:22:700Paolo Guiotto: It says here we should have the vector notation, and… So, what, could the…
01:32:70Paolo Guiotto: I don't want to use the letter, so let's write this, it's a little bit heavy way, XY minus M1M2…
01:42:710Paolo Guiotto: Scala product with it itself, XY… minus M1M2.
01:51:920Paolo Guiotto: However, since the matrix C-1 is diagonal, what happens when you do C minus 1 times this vector scales this vector? Because it is diagonal, we get this 1 over C11 times the first component of this squared, so this evaluation here
02:11:590Paolo Guiotto: is 1 over C11 times X minus M1 squared plus 1 over C22XY minus M2 squared.
02:25:850Paolo Guiotto: Because the matrix is diagonal.
02:28:480Paolo Guiotto: When you multiply by the vector, it takes exactly the components and multiplies by…
02:36:180Paolo Guiotto: the elements on the diagonal. So, in particular, this means that… and also the terminant of C is equal to C11 times C22, so this splits into 1 over root of 2 pi.
02:51:100Paolo Guiotto: C11E minus 1 over 2C11X minus M1 squared times 1 over root of 2 pi C22E minus 1 over 2C22
03:10:440Paolo Guiotto: Y minus M2 square.
03:13:770Paolo Guiotto: So you see that the joint density splits into the product of two functions. One is function of X, the other is function of Y, that are both densities. They are Gaussian densities, so this must necessarily be the density of X, and this is the density of Y.
03:32:940Paolo Guiotto: And since the joint density splits into the product, this means that the two variables are independent, so XY
03:43:740Paolo Guiotto: independent.
03:51:180Paolo Guiotto: We could have done this calculation
03:57:190Paolo Guiotto: You know, how much is better.
04:00:290Paolo Guiotto: equivalently with the characteristic functions, but let's say that this is enough. There is a third question.
04:07:910Paolo Guiotto: But we have to think about, huh?
04:11:930Paolo Guiotto: Show that, huh?
04:16:260Paolo Guiotto: Xy, the correlation could be equal to zero.
04:23:130Paolo Guiotto: But… X, Y… Not.
04:30:260Paolo Guiotto: Independent.
04:37:600Paolo Guiotto: Okay, let's see. Now, we need to build an example here, where we get correlation equals zero, and the two variables are not independent.
04:51:200Paolo Guiotto: So… Basically, we needed to, to…
04:56:860Paolo Guiotto: determine two variables such that the covariance between X and Y is 0.
05:03:850Paolo Guiotto: And, since the covariance is the same of the covariance.
05:11:240Paolo Guiotto: of, X minus, expected value of X.
05:16:790Paolo Guiotto: let's say Y, because when you, the covariance is linear in each of the two variables. So when you split this, this becomes covariance XY minus covariance of a constant times Y, but the covariance of a constant is 0.
05:32:590Paolo Guiotto: If one of the two factors is constant, it's zero. And the same can be done for the second factor, so I can center, basically, this in such a way that we can just… so since these are expected value equals zero variables.
05:52:60Paolo Guiotto: And the covariancy is the expectation of the product, minus the product of the expectation, so we can limit to prove that, expected value of XY
06:04:670Paolo Guiotto: He's zero.
06:06:450Paolo Guiotto: with the… we… We did the two, X and Y… We'd mean zero.
06:19:630Paolo Guiotto: So we need to build, in other words, two variables, X and Y, with mean 0,
06:26:720Paolo Guiotto: Such that the product… the expectation of their product is zero.
06:31:730Paolo Guiotto: But, such that, They are not independent.
06:36:980Paolo Guiotto: In other words, this is also saying that, In other words,
06:48:400Paolo Guiotto: this… is also… it count… their example.
06:58:320Paolo Guiotto: tool.
06:59:320Paolo Guiotto: So, we know that if XY are independent, Then, the expected value.
07:09:700Paolo Guiotto: of X times Y splits into the product of the expected values, so expected values of X, expected value of Y. And therefore, if they are equal 0, the expected value of the product is 0 as well.
07:25:420Paolo Guiotto: But the vice versa is not true.
07:30:520Paolo Guiotto: So we know that, what is true is that,
07:34:160Paolo Guiotto: This is equivalent of saying that whenever you have expectation of function of X times function of Y,
07:42:380Paolo Guiotto: This splits into the product of expected value function of X times expected value function of Y. But this, for every function ph.
07:56:330Paolo Guiotto: Okay? Which is much stronger.
07:59:740Paolo Guiotto: This one contains the fact that the expectation of the product splits into the product of the expectations, but since it must hold for every phi and t, it is much more stronger.
08:12:760Paolo Guiotto: So it's not equivalent to this. So, this one implies this, but the vice versa is not true.
08:25:649Paolo Guiotto: Unless the variables are Gaussian. As we have demonstrated, if they are Gaussian.
08:32:140Paolo Guiotto: covariance equals zero, or expected value… expectation of the product equal product of the expectation says… tells that they are independent. But just in this case.
08:44:620Paolo Guiotto: Now, the point is, we want… so at the end, we want to find two variables, X and Y, with mean 0, okay, such that the expectation of the product is zero, but they are not independent.
09:01:20Paolo Guiotto: So, how can we build this example?
09:26:500Paolo Guiotto: Well, let's… let's see if we do this.
09:30:399Paolo Guiotto: So, let's say that normally, to build this kind of examples, we could take as probability space, for example, the interval 01 with the Lebec measure, so it's a probability space, and take just functions, no, ordinary functions, as random variables.
09:48:800Paolo Guiotto: And, so we need the…
09:51:710Paolo Guiotto: For example, if I want to construct an example on omega equals 01,
09:58:360Paolo Guiotto: with the probability P equal the Lebec measure.
10:02:590Paolo Guiotto: So, it means that X of omegas are functions, of omega.
10:11:290Paolo Guiotto: X and Y omega are functions of omega, such that… so the expected value becomes the integral. What I want is that… I want to find two functions such that when I integrate from 0 to 1 their product, X omega.
10:29:110Paolo Guiotto: times Y omega in D omega, I get 0.
10:33:730Paolo Guiotto: when I integrate each of them, I get 0.
10:43:390Paolo Guiotto: But I want to add that they are not independent.
10:48:360Paolo Guiotto: So, but… XY… Not independent, though.
11:02:620Paolo Guiotto: Well, the opposite of… Independence could be what?
11:08:340Paolo Guiotto: could be, for example, they are totally dependent. One is a function of the other, for example, something like this.
11:17:490Paolo Guiotto: So… But they are the same, for example, the same variable, no?
11:23:670Paolo Guiotto: But if they are the same, I get exposed, so it's not possible to have 0 unless X is 0. But I could do… let's see if we do this…
11:43:340Paolo Guiotto: Okay, maybe we modify the…
12:00:470Paolo Guiotto: I'm thinking to this. Instead of taking 0, 1, let's take a semantic interval, like minus 1, so we have to divide by 2 base, no, to make a probability measure, so it's not, important, here. So let's put the integrals from minus 1, 1.
12:21:260Paolo Guiotto: If we take…
12:28:330Paolo Guiotto: So I have to take, something that, in the product… so let's start with X of omega equals omega.
12:37:450Paolo Guiotto: In this way, the integral from minus 1 1 of X omega is clearly equal to 0.
12:52:580Paolo Guiotto: what we can take for Y omega.
13:16:180Paolo Guiotto: Let's see, if we take something like sine omega.
13:20:590Paolo Guiotto: No, this is not good because… It's even… No, we take cosine omega.
13:32:510Paolo Guiotto: So, the integral minus 1 1 of Y is 0, and also the integral minus 1, 1 of X times Y is 0, because it is…
13:45:320Paolo Guiotto: Because it is, it is an even function. If you do omega, no XY,
13:54:580Paolo Guiotto: X omega, Y omega.
13:58:410Paolo Guiotto: Is equal to omega cosine omega.
14:01:850Paolo Guiotto: And this is, odd, no? Because,
14:05:250Paolo Guiotto: to change omega with minus omega, and this remains the same, and this changed sine. So when I do the integral of X times Y, I'm doing the integral from minus 1 to 1 of omega times cosine of omega, so I get 6. The problem is, are these variables independent?
14:22:320Paolo Guiotto: So,
14:28:80Paolo Guiotto: how can we check if these are independent? So, probability that,
14:37:840Paolo Guiotto: X, how to check independence.
14:46:160Paolo Guiotto: So, if I say, mmm…
15:06:630Paolo Guiotto: let's say, probability that X is greater than 0, and Y…
15:16:500Paolo Guiotto: Let's see what happens to this. So, this is the probability that
15:22:10Paolo Guiotto: X is greater or equal than zero, and Y is greater or equal than zero. So the first says omega greater or equal than zero, the second cosine of omega greater or equal than zero.
15:42:640Paolo Guiotto: Now, since we have omega is… our probability space is the interval between minus 1 and 1.
15:51:240Paolo Guiotto: So, the cosign…
15:57:170Paolo Guiotto: Now, this one…
16:00:200Paolo Guiotto: This is just omega between 0, this would be 1 half, right? Because cosine is 1 here, a pi half is where it goes to zero, so it is positive here. So this is the interval
16:13:870Paolo Guiotto: as you don't…
16:24:440Paolo Guiotto: Okay, yeah. This is for omega positive, and cosine of omega positive… well, cosine is always positive, so this is the probability where omega is greater than zero, and this is 1 half.
16:40:620Paolo Guiotto: No, yeah, that's the product of the two probabilities, because the probability where X is greater or equal than 0 times Y greater or equal than zero
16:50:130Paolo Guiotto: Now, this is one… Fa…
16:54:670Paolo Guiotto: This is equal to 1 half, and this is equal to 1, because the cosine is always positive, so we get the same.
17:01:410Paolo Guiotto: We have to take something different.
17:07:410Paolo Guiotto: So… This is not a good example.
17:18:550Paolo Guiotto: Hmm.
17:46:470Paolo Guiotto: So, we need to find a second variable.
17:50:810Paolo Guiotto: Why?
17:52:240Paolo Guiotto: So let's say that we take X of omega equals omega, We look for Y omega.
18:00:380Paolo Guiotto: In such a way that integral B is 0.
18:04:790Paolo Guiotto: What I would like to take is, but that's… it's not working with the…
18:11:320Paolo Guiotto: If I take also Y omega equal…
18:15:790Paolo Guiotto: Equal… equal X squared, X omega square.
18:25:630LEONARDO RIZZOTTO: Excuse me, Professor.
18:27:390Paolo Guiotto: Yeah.
18:28:830LEONARDO RIZZOTTO: We weren't looking for a two dependent random variable.
18:35:590Paolo Guiotto: Yes, what I'm looking for, that's why I'm…
18:39:700LEONARDO RIZZOTTO: Omega and cosine of omega, epsilon of omega, that is, cosine of omega.
18:46:350LEONARDO RIZZOTTO: it's like epsilon of X omega, because it's cosine of omega, and X omega is omega, so.
18:54:570Paolo Guiotto: is cosine of x omega also, yes.
18:58:300LEONARDO RIZZOTTO: Yes, so… they are dependent.
19:03:360LEONARDO RIZZOTTO: So…
19:04:90Paolo Guiotto: I want to… I want to find variables for which the correlation is zero, but they are not independent.
19:12:960Paolo Guiotto: So, mmm…
19:15:350LEONARDO RIZZOTTO: Isn't it? So isn't that good?
19:18:30LEONARDO RIZZOTTO: A good example.
19:20:90Paolo Guiotto: No, because these variables, they look to be independent.
19:25:620Paolo Guiotto: So here, down here, I was checking if it is true that the probability
19:32:80Paolo Guiotto: Of the product is not the product of the probabilities, to be independent.
19:38:230Paolo Guiotto: So, I did with this set when the two are positive, when X is in 0 plus infinity, say, and Y is in 0 plus infinity. If they are independent, this probability
19:54:150Paolo Guiotto: Shouldn't be equal to the product of the probabilities, okay?
20:00:150Paolo Guiotto: But in this case, it turns out this is true, because, you see, the probability that the two are both greater or equal than zero. The point is that Y is always greater or equal than zero, so that's a condition…
20:20:670Paolo Guiotto: A condition that,
20:26:530Paolo Guiotto: Yeah, but actually, I'm stupid also, because this is not true, that integral of cosine is zero.
20:37:160Paolo Guiotto: So we should take, now, this also does not verify this.
20:46:160Paolo Guiotto: And that's a problem.
20:50:270Paolo Guiotto: Okay, let's try again. So, X omega…
20:54:450Paolo Guiotto: Let's see what happens. X omega equal omega.
20:58:700Paolo Guiotto: Let's take y omega equal omega square, for example.
21:05:690Paolo Guiotto: Which does not mean equal to 0, because integral from minus 1 to 1 of y omega
21:12:240Paolo Guiotto: in this case, is not zero. It is the integral for minus 1 1 of omega squared, so it is omega cubed over a 3, so…
21:23:920Paolo Guiotto: what is 2 thirds? 2 3rd is the value. But if we center, now it becomes a mean zero variable. So let's take this.
21:34:980Paolo Guiotto: Now, this, the expected value of X is the integral from minus 1 to 1 of omega in the omega, so this is it.
21:45:260Paolo Guiotto: The expected value of Y is 0, because it is the integral from minus 1 to 1 of omega square minus 2 thirds.
21:54:20Paolo Guiotto: In the omega, and this is equal to zero.
21:58:560Paolo Guiotto: by, construction.
22:01:540Paolo Guiotto: What is the expected value of X times Y?
22:07:90Paolo Guiotto: Now, this is equal to the integral from minus 1 to 1, x is omega times omega square minus 2 thirds.
22:16:670Paolo Guiotto: that this splits into two integrals, integral 4 minus 1, 1 of omega cubed minus integral minus 1, 1, 2 third omega. But both these integrals are zero because the functions are, odd.
22:34:810Paolo Guiotto: So the expected value of the product is zero. Now,
22:46:920Paolo Guiotto: So their point is that they shouldn't be independent, because one is function of the other.
22:52:690Paolo Guiotto: So, but if I want to convince you, I have to convince you that some probability does not split into the product of the probabilities. So…
23:12:660Paolo Guiotto: Well, let's see.
23:16:410Paolo Guiotto: If I say, take probability where X is greater or equal than 0 and Y,
23:21:990Paolo Guiotto: Greater or equal than zero.
23:26:350Paolo Guiotto: So this is,
23:28:430Paolo Guiotto: the probability that omega is greater or equal than zero, and omega square minus 2 thirds is greater or equal than zero. So this is omega square greater or equal than 2 thirds…
23:42:510Paolo Guiotto: So remind that our measure is Telebeck measure, so we are taking, in the domain omega, which is the interval minus 1,
23:51:910Paolo Guiotto: The omegas were… which are positive, and omega square greater than 2 thirds, so root of… 2 thirds…
24:02:580Paolo Guiotto: And the minus root of 2 thirds.
24:05:970Paolo Guiotto: So this is this interval here, the interval for the omegas between this value and this value. So the probability here is 1 minus root of 2 thirds.
24:18:190Paolo Guiotto: Now, what is the product of the two probabilities when X is positive and Y is positive?
24:27:590Paolo Guiotto: The probability that X is positive is the probability that omega is greater or equal than zero.
24:33:250Paolo Guiotto: And the other one is the probability that omega square is greater or equal than 2 thirds.
24:39:320Paolo Guiotto: Now, the probability that omega is greater or equal than 0 is 1 half, and the probability that omega squared is greater or equal than 2 thirds means the probability that omega is less than minus root of 2 thirds, or omega greater than root of 2 thirds. So this is 2 times…
24:58:680Paolo Guiotto: No. 1 minus the root of 2 thirds, it doesn't come.
25:04:520Paolo Guiotto: Because it's, again, the same value.
25:08:960Paolo Guiotto: Oh.
25:13:280Paolo Guiotto: They shouldn't be.
25:17:00Paolo Guiotto: So what led I should,
25:36:10Paolo Guiotto: It does not mean yet that they are not independent, but… because the point is that, you see, since this says that Y is X squared minus 2 thirds, so 1 is the function of the other.
25:51:530Paolo Guiotto: And so, we have to get… to take the right choice here for the sets for which we do not have that the product splits… the probability of the intersection splits into the product of the probability.
26:05:750Paolo Guiotto: Hmm… Okay, let me think about, but it should work with something like this, so… Oh.
26:15:270Paolo Guiotto: We have only 5 minutes left.
26:20:60Paolo Guiotto: Okay, so… Try to… try to figure out, to… Except… this,
26:31:760Paolo Guiotto: to make… to check.
26:40:540Paolo Guiotto: bet.
26:42:660Paolo Guiotto: X.
26:44:380Paolo Guiotto: Why? Not.
26:48:300Paolo Guiotto: Independent.
26:57:970Paolo Guiotto: Also, you see that since one is function of the other.
27:17:900Paolo Guiotto: So, certainly, these are not absolutely continuous, because since in this example we have that Y is X squared minus 2 thirds, so it means that if you look in the plane, XY at the point
27:35:520Paolo Guiotto: with coordinates capital X, capital Y. This point belongs to this parabola.
27:41:880Paolo Guiotto: So, it means that our distribution is concentrated here.
27:53:40Paolo Guiotto: You see this, no? The probability that Y is equal to X squared minus 2 thirds is 1, right?
28:03:480Paolo Guiotto: No? If Y is X squared minus 2, this probability is 1. So it means that the probability that point XY belongs to that set, parabola.
28:16:50Paolo Guiotto: is 1.
28:19:290Paolo Guiotto: And this is, this is somehow stranger with the fact that, the 2XY are absolutely continuous. They have densities, you know, because, so, X,
28:35:850Paolo Guiotto: And R… absolutely continuous.
28:42:810Paolo Guiotto: before… because, for example, if you look for X, the probability that capital X is less than U is what? Is the probability of the set of omegas in the interval minus 1,
28:59:560Paolo Guiotto: such that omega is less or equal than U, but this is an interval. It's the interval from minus 1 to U.
29:08:710Paolo Guiotto: And therefore, the probability is the LeBague measure. This is U minus minus 1, so it is U plus 1.
29:15:620Paolo Guiotto: is this one. It's a very nice function. This for U between minus 1 and 1. For U less than minus 1, there is no X, and for U greater than 1, there is no X. So, this means that this is the CDF of capital X at U, and therefore, the density
29:39:520Paolo Guiotto: F, X.
29:41:360Paolo Guiotto: U is the derivative of this, which is 1, when U is between minus 1, 1.
29:51:320Paolo Guiotto: Yes, sorry, the measure was the Lebec measure, so was the Lebec measure divided by 2, because the probability of the interval… the interval is minus 1, so it must have measure 1, so this is divided by 2.
30:07:490Paolo Guiotto: So, the derivative is just 1 half, there exists the derivative, there exists the density, so X…
30:15:150Paolo Guiotto: is absolutely continuous.
30:19:90Paolo Guiotto: And since Y is a nice function of X, also Y is absolutely continuous.
30:28:400Paolo Guiotto: But now, if… so this is an indirect argument, because I've not yet found an example.
30:36:40Paolo Guiotto: of a set where the probability of the intersection is not the product of the probabilities. But with this, I can say that the two are absolutely continuous, so there are density. If they were independent.
30:50:280Paolo Guiotto: If, they… Independent.
30:57:470Paolo Guiotto: There would be a density, the joint density.
31:01:120Paolo Guiotto: FXY would be the product of the two densities.
31:06:790Paolo Guiotto: But look at what happens. We have that these variables are concentrated on that black line.
31:13:670Paolo Guiotto: So we would have that… so the probability that you are elsewhere, so in the full plane, except the parabola, this probability would be zero, right? So we would have that.
31:27:740Paolo Guiotto: Zero would be the probability… the probability that, XY.
31:34:510Paolo Guiotto: Does not belong to the parabola.
31:40:980Paolo Guiotto: But this would be the integral on the set where… of points XY, where Y is not X squared minus 2 thirds of the density, joint density.
32:01:730Paolo Guiotto: Or, sorry, let's use that one with the parabola, which is better, because we have 1 would be the probability that point XY belongs to the parabola, so this…
32:14:540Paolo Guiotto: is equal.
32:17:270Paolo Guiotto: But this is impossible.
32:19:00Paolo Guiotto: Because this set, for the Lebec measure, is a measure zero set. You have a line in the Cartesian plane. But this…
32:28:380Paolo Guiotto: is a Lebeg… measure.
32:33:970Paolo Guiotto: zero set. So whatever is this thing.
32:38:640Paolo Guiotto: The integral will be equal to zero.
32:41:640Paolo Guiotto: And this is a contradiction. So I don't know how to find now, I don't see,
32:50:530Paolo Guiotto: an event where I can show you directly that the probability of the intersection is not the product of the probabilities, but this… with this indirect argument, and as you can see, this is a sort of general idea, no? Whenever you have that a random variable is a function of another.
33:09:970Paolo Guiotto: This is somehow important because,
33:15:90Paolo Guiotto: Lots of probability is based on the idea of independence, and this is important because independence happens in many random phenomena.
33:23:880Paolo Guiotto: You know, when you… when noises are made by superposition of independence, independent events that affect a signal, for example, and you have this biased signal.
33:38:580Paolo Guiotto: But it's also important to know when there is a dependence between quantities, because maybe somehow a big data analysis is based on this, on searching for dependencies, maybe complicated dependencies between data, no? And what happens here is that this is a case of dependence, extremely strong dependence. One variable is a function of
34:02:490Paolo Guiotto: another.
34:03:520Paolo Guiotto: And this means that if you look at the pair, XY, this will be concentrated along the set of points that verify that condition, the relation between the two, Y equal function of X,
34:17:10Paolo Guiotto: identifies if the function is good, like this one, identifies a line in the Cartesian plane.
34:24:730Paolo Guiotto: Now, this means that if this happens, all points of the pair, XY falls in this line.
34:33:380Paolo Guiotto: So the probability is somehow concentrated on a set, which is a measures your set for the plane measure. And this means that that density, that pair cannot have a density. Otherwise, as you can see here, the
34:50:530Paolo Guiotto: If you compute the probability that XY falls on the parabola, we would have 1. But at the same type, if there is a density, it should be the integral of the density on that set.
35:00:300Paolo Guiotto: And whatever is the density, the Lebec measure, kills this, because this is a measure zero set for the Lebec measure. So this integral, whatever is this density, will be zero, and this will be the contradiction.
35:14:830Paolo Guiotto: Okay, so, assuming that they are independent, since we know that each of them is absolutely continuous, it has a density, we would have that the pair would have a joint density that would be the product of the two. So, in this case, we would get a contradiction, and therefore they cannot be
35:34:120Paolo Guiotto: in the time, then times.
35:38:50Paolo Guiotto: Okay, it's an indirect check, but it seems to be correct.
35:42:800Paolo Guiotto: Okay, let's stop here. Just one thing. So, tomorrow, we say that we do class, online class directly, okay?
35:52:800Paolo Guiotto: So that's only for tomorrow.