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00:02:760Paolo Guiotto: You want your exam?
00:11:230Paolo Guiotto: Oh, thank you.
00:12:380Paolo Guiotto: Right, you can click the exam until the end of the class, and you give that back to me, okay?
00:19:440Paolo Guiotto: So we are recording… Great.
00:25:430Paolo Guiotto: Today, we start With some of the exercises left on multivariate.
00:33:480Paolo Guiotto: Random variables, then we will go.
00:36:380Paolo Guiotto: Let's see… We just closed this part on multivarity. So let's take this exercise, the 431.
00:50:890Paolo Guiotto: I started to publish on model solutions of the lecture notes exercises, so check regularly.
00:59:260Paolo Guiotto: And please do the exercises before I publish the solution, so you can… Use that to check.
01:08:120Paolo Guiotto: So here we have, XY… He's a… mmm… well… In this case, when…
01:17:530Paolo Guiotto: this vector as two components, we say that it is a B variant, not just multivariate, B variant.
01:27:600Paolo Guiotto: random variable.
01:29:970Paolo Guiotto: with density.
01:36:290Paolo Guiotto: Do you want to see your exam?
01:39:120Paolo Guiotto: Your test?
01:41:590Paolo Guiotto: What's your last name?
01:48:30Paolo Guiotto: events, right?
01:49:970Paolo Guiotto: Should be… Yes.
01:53:240Paolo Guiotto: Okay, so with density F… XY…
01:59:810Paolo Guiotto: of little x, little y equal 4… XY indicates a function of 01 square at point X1.
02:14:370Paolo Guiotto: So, number… first question is to determine the… Joint CDF, capital FXY.
02:23:990Paolo Guiotto: The second question is compute.
02:29:200Paolo Guiotto: the probability… that X plus Y be less than 1.
02:38:200Paolo Guiotto: Oh, this is… Not particularly complicated.
02:43:520Paolo Guiotto: It's just to… Take some confidence with these, definitions, etc.
02:51:40Paolo Guiotto: Now, first of all, we have to compute the, this, joint CDF with a mind of the definition, FXY,
03:01:980Paolo Guiotto: at point little x, little y is the probability that X is less or equal than little x, and y is less or equal than little y.
03:15:170Paolo Guiotto: Equivalently, this is the probability that the pair, XY belongs to this set, minus infinity to X.
03:28:120Paolo Guiotto: Cartesian product minus infinity to Y.
03:34:210Paolo Guiotto: So, this means that it is the mu X,
03:38:800Paolo Guiotto: I'm doing for the first time, so I write all these things, but…
03:42:960Paolo Guiotto: You can proceed directly, of course.
03:47:830Paolo Guiotto: And this is, since we have a density, the integral on that domain here, so minus infinity X
03:57:390Paolo Guiotto: cross minus infinity Y of the joint density, FXY, so let's use the other letters UV.
04:07:20Paolo Guiotto: DUDV. So at the end, we have to compute this integral.
04:11:570Paolo Guiotto: So this, means that, by using reduction formula or Fubini theorem.
04:18:670Paolo Guiotto: that this is the integral for U from minus infinity to X, for V from minus infinity to Y of the density FXY.
04:29:350Paolo Guiotto: UV, DUDV. So, this is DV, and then DU.
04:36:120Paolo Guiotto: Now, the density is a 4 UV indicator of the square, 01 square UV.
04:47:500Paolo Guiotto: Now, because of that indicator that we can also see as a…
04:52:350Paolo Guiotto: This is equal to 1 if and only if,
04:57:400Paolo Guiotto: The point UV belongs to the square zero one.
05:01:530Paolo Guiotto: So it means that U must be between 01 and V, between 0, 1, so we can also write this in this way.
05:08:300Paolo Guiotto: Indicator 01 of U.
05:11:360Paolo Guiotto: times indicator 01 of V.
05:14:920Paolo Guiotto: Because this is 1 exactly if, and only if, both U and V are in the interval 01. So, in particular, we can split into a product this density.
05:28:510Paolo Guiotto: 4UB times this. So since the first integration here is in V, we can carry outside the, 4. The 4 is outside everything, integral minus infinity to X, then U indicator 0,
05:44:860Paolo Guiotto: One of you.
05:47:570Paolo Guiotto: integral minus infinity to Y, V indicator 0, 1, V… DV.
05:57:890Paolo Guiotto: DU. As you can see, the innermost integral is independent of X, so it's a constant, for the other one, so basically we can split this into the product. Integral minus D2X of U indicator 0, 1,
06:14:450Paolo Guiotto: U. D. U.
06:17:520Paolo Guiotto: times integral minus infinity 2Y, the same thing. V indicator 0, 1, V, DVD.
06:27:880Paolo Guiotto: Now, since the indicator is 1 only when the variable is in the U, integration variable, is between 0, 1, we see that if X
06:37:710Paolo Guiotto: is less than zero, that you are integrating on an interval where the function is zero, and the same for the other integral, so we may say that this is equal to
06:51:20Paolo Guiotto: 0. Let's say, if one of XY is less than 0, okay, so if you look in the Cartesian play, we can do a little figure here, where there is point XY,
07:03:00Paolo Guiotto: It means that if one of the two coordinates is negative, so if X is less or equal, also equal, because here.
07:13:270Paolo Guiotto: the Lebec measure has no relevance, points have no relevance for the Lebec measure, or Y less or equal 0. Now, X less or equal 0 means the left half plane, so here we have X less or equal 0, and here.
07:31:610Paolo Guiotto: Y less than or equal 0 is the half plane below the x-axis, so Y is equal… less than or equal 0 here and here. So it means that that density, no, sorry, that value, the CDF at the end, this is the CDF,
07:48:590Paolo Guiotto: the value of capital FXY at point XY. Here, we get 0.
07:54:800Paolo Guiotto: Okay? So, if X is negative or positive. Now, if XY
08:00:650Paolo Guiotto: Belongs to the first quarter.
08:03:910Paolo Guiotto: That we will have something.
08:06:320Paolo Guiotto: So, you understand that clearly this depends on where is this X with respect to the interval 0, 1, because now, if X, for example, is positive, if X is less than 1, so if…
08:23:850Paolo Guiotto: X is less than 1. The first integral becomes 4 integrals minus infinity to X, because we are in the integral 01 of UDU.
08:36:90Paolo Guiotto: And the second interval is similar, so if also Y is less or equal than 1, we have the integral from minus infinity to Y of VDV.
08:48:940Paolo Guiotto: If one of the two is greater than 1, it happens just that the corresponding integral is the integral between 0 and 1. So actually, sorry, sorry, sorry, I…
08:59:920Paolo Guiotto: There is a mistake here. It's not minus infinite, but zero here, because there is the indicator that cuts off everything. So, if, for example, X is greater than 1, and Y is less than 1,
09:16:160Paolo Guiotto: maybe it is better if I continue down here. If X is greater than 1, Y is less than 1,
09:27:20Paolo Guiotto: We will have 4 integr, times integral 0y of VDV.
09:35:460Paolo Guiotto: Similar, when x is less than 1 and Y is greater than 1, so we have integral 0 to X UDU, times integral 0 to 1 of V dB.
09:49:490Paolo Guiotto: And when both are greater or equal than 1,
09:55:60Paolo Guiotto: We have 4 integrals 0 to 1 of u, the u, integral 0 to 1VDV.
10:03:120Paolo Guiotto: So basically, we have to compute all these things. It's not such a…
10:07:210Paolo Guiotto: Great job. So, let's say equal…
10:11:450Paolo Guiotto: 0 for X less or equal than zero, or Y less or equal than 0.
10:21:650Paolo Guiotto: Then, when both are less or equal than 1, so positive, X between 0 and 1, Y between 0 and 1,
10:32:530Paolo Guiotto: We have, 4, the integral, you see they are the same integral, of the variable, so the integrals, integral
10:42:60Paolo Guiotto: say, integral from 0 to X of u to u is u squared divided 2, evaluated between 0 and X, so it is X squared divided 2.
10:54:80Paolo Guiotto: So, when we have this case, we have 4, X squared divided 2, Y squared divided 2, so it means X squared, Y squared divided 4 times 4 is just X squared, Y squared. This is the value.
11:10:630Paolo Guiotto: When x is greater than 1 and Y is less or equal than 1, then it is the integral from 0 to 1 of U, which is equal to 1 half. So this one is equal to 1 half
11:26:490Paolo Guiotto: times 4 times this one, which is Y squared over 2. So, it comes, Y squared.
11:34:40Paolo Guiotto: And similarly, when exit is less or equal than 1, of course, positive.
11:40:220Paolo Guiotto: and positive, and Y greater than 1, we will have X squared here. And when they are both
11:47:630Paolo Guiotto: Great or equal, or greater than 1,
11:51:310Paolo Guiotto: So we have… the two integrals are both equal to 1 half, so everything makes, no.
11:59:430Paolo Guiotto: Yeah, makes one. So this… is 1.
12:05:330Paolo Guiotto: So, like, we write everything here. So, FXY at the end, XY comes out to be.
12:13:230Paolo Guiotto: 0, if X is less than zero, or Y is less than 0. It is X square, Y squared when both are between 0 and 1.
12:27:320Paolo Guiotto: It is Y squared when X is greater than 1Y between 0 and 1.
12:37:300Paolo Guiotto: It is X squared when it is X to be between 0
12:42:990Paolo Guiotto: And 1 and Y greater than 1. And finally, it is 1 when both are greater than 1.
12:51:700Paolo Guiotto: So if we want, we can divide the plane, XY,
12:58:520Paolo Guiotto: In these ways. So here, this CDF, it is equal to zero.
13:04:210Paolo Guiotto: Then when we are for the 2 between 0 and 1, so it means we are in this square.
13:10:980Paolo Guiotto: The value is X squared, Y squared.
13:15:210Paolo Guiotto: This case here means X greater than 1, so X right of 1, and Y between 0, 1, so it means when XY is here, the CDF is Y squared.
13:28:510Paolo Guiotto: This case here, it is Y greater than 1, so above the original line, Y equals 1, and X between 0, 1, so it is here, so we got X squared. And finally, here, the CDF is 1. So this is the
13:43:270Paolo Guiotto: the shape of, the CDF is not the graph of the CDF, I… I should put it…
13:49:510Paolo Guiotto: The figure in three dimensions, but it's not necessary.
13:54:850Paolo Guiotto: Okay, question two, we have to compute the probability that X plus Y is less than 1.
14:06:420Paolo Guiotto: Now, we can always transform this, into something like probability that the point XY belongs to some set E of this, in this case, of the Cartesian plane. What is E? But simply.
14:24:700Paolo Guiotto: It is the set of points in the plane
14:27:880Paolo Guiotto: XY, well, X plus Y must be less than 1, so if you want Y less than 1 minus X,
14:36:730Paolo Guiotto: 1 minus X is a straight line, more or less like that, Y equals 1 minus X. So this is Y equal 1, this is X equal 1. So basically, what we are computing is the probability that the pair, XY belongs to this
14:54:640Paolo Guiotto: satta, in the Caucasian prey, okay?
14:59:530Paolo Guiotto: Now, in fact, the probability that, for example, XY is here, in this part, In this red region.
15:16:980Paolo Guiotto: Will be equal to zero.
15:19:390Paolo Guiotto: No? Because, I can… I can say that, that, that's the probability. So, the probability that the pair XY belongs to E,
15:32:210Paolo Guiotto: If you call the red set, A,
15:35:750Paolo Guiotto: And this one, let's say, green set, B.
15:42:740Paolo Guiotto: Is the probability that, the pair, X, Y,
15:48:80Paolo Guiotto: belongs to A, union B. Now, if, for example, you exclude the intersections, you can do, for example, you give to the green set this
16:01:80Paolo Guiotto: these points on the axis, so on A, you don't have those points in the axis, so this becomes a disjoint union. You can say that this is the probability that
16:13:500Paolo Guiotto: XY belongs to A, plus the probability that XY belongs to B.
16:25:530Paolo Guiotto: Now, the probability that XY belongs to A is 0, you know, because that's the probability that X is less or equal than zero, or…
16:36:270Paolo Guiotto: Y is less or equal than 0. If you want, exaggerating a bit, this is less or equal than the probability that X is less or equal than zero, plus the probability that Y is less or equal than 0.
16:54:860Paolo Guiotto: But this is, if you want, the CDFFXY.
16:59:980Paolo Guiotto: evaluated 0 and plus infinity for the second coordinate, because there is no condition on Y. However, when the first variable is less or equal than zero, it doesn't matter what is the second variable. The value of the sphere is 0.
17:15:900Paolo Guiotto: In other words, the density of this vector is concentrated into the square 0, 1, so if you look for points.
17:25:790Paolo Guiotto: Where this pair, XY, falls outside of that square, you will have nothing. So this is 0, and similarly, this is 0.
17:36:220Paolo Guiotto: So basically, we have that probability.
17:40:280Paolo Guiotto: the probability that XY belongs to B.
17:45:150Paolo Guiotto: And, this is, the mu, the loop… the vector XY.
17:51:580Paolo Guiotto: on the set B, so it is the integral on B of the joint density FXY.
17:59:410Paolo Guiotto: the X, the Y.
18:02:850Paolo Guiotto: Now, the set B is the set of points X, Y,
18:10:30Paolo Guiotto: of R2, such that it's that green triangle, so the axis between 0, 1. The Y is between 0, 1,
18:20:700Paolo Guiotto: But I have to impose a further condition, which is Y is less than 1 minus X, otherwise I take out the full square. So, X is between 0, 1, also Y is between 0, 1, and Y is less or equal than 1 minus X.
18:38:330Paolo Guiotto: So at the end, we have to compute the integral on that domain, XY, both between 0 and 1.
18:46:760Paolo Guiotto: and Y less than 1 minus X, of the density, which is 4XY. Now, I can write the indicator, but in this case, because I'm already with XY,
19:02:870Paolo Guiotto: In that square. Now, the square is this one.
19:09:930Paolo Guiotto: So the square is all this.
19:12:880Paolo Guiotto: this quantity is identical equal to 1 for these points, because the two coordinates are between 0 and 1. So I have to do the integral for XY between 0 and 1,
19:26:330Paolo Guiotto: Y less than 1 minus X of 4XY, The X, the Y.
19:35:330Paolo Guiotto: Now, I apply the reduction formula, or Fubini theorem, as you prefer.
19:41:590Paolo Guiotto: And, example, I can say, let's integrate 4X between 0, 1, and then fory, I have the condition Y between 0 and 1 minus X. What? For XY? This is integration in Y first, and then in X.
20:00:50Paolo Guiotto: Now, this 4X comes outside.
20:04:390Paolo Guiotto: of this integral. We have an integral between 0, 1 for X, then the integral of Y is Y square over 2 to be evaluated between y equals 0 and y equals 1 minus X.
20:21:230Paolo Guiotto: all this integrated in Access, so we get…
20:29:110Paolo Guiotto: So let's just write 2.
20:32:290Paolo Guiotto: times 1 minus X squared dx.
20:39:720Paolo Guiotto: Now, we can see that this is more or less a derivative.
20:45:390Paolo Guiotto: It should be a derivative of something… a power of 1 minus X. You know that when you differentiate powers, you decrease the exponent by 1 unit, so it should be the derivative, more or less, of this.
21:00:30Paolo Guiotto: In fact, if we compute the derivative of this, we get 3, 1… no, it's not this lengthy, because they have the factor 2X. So, okay. However, no panic, we just,
21:17:510Paolo Guiotto: Yeah, well, we can develop the square, or we add and subtract 1 to this X, and we create 1 minus X. It's the same. So let's do…
21:28:550Paolo Guiotto: It's not this is the point of the exercise. So let's write the squares. We have integral 01 of 2x minus 4x squared.
21:39:440Paolo Guiotto: Plus, 2X cubed DX.
21:44:230Paolo Guiotto: And now these are integrals 0, 1, sorry, this is, 2.
21:50:510Paolo Guiotto: X squared over 2, between 0 and 1, minus 4, X cubed over 3 between 0 and 1, plus 2, X to the power of 4 divided 4 between 0 and 1. And we get the
22:08:410Paolo Guiotto: Desired quantities, so we get 2 times 1 half minus 4 times 1 third, Plus 2 times 1 fourth.
22:18:80Paolo Guiotto: Well, since this is a probability, let's at least check if it comes a number between 0 and 1, because if it is negative or greater than 1, it's certainly wrong, this. This is 1 minus 4 thirds…
22:33:490Paolo Guiotto: plus 1 half.
22:35:760Paolo Guiotto: So… it is, 3 half…
22:42:90Paolo Guiotto: 3 half minus 4 thirds, what is this?
22:46:870Paolo Guiotto: 6… 9 minus 8, 1 6.
22:52:720Paolo Guiotto: So it is positive less than 1. It is acceptable, it doesn't mean that it is correct.
22:57:720Paolo Guiotto: But it is, let's say, it is acceptable. Okay.
23:12:260Paolo Guiotto: Oh, God.
23:13:720Paolo Guiotto: The next exercise, the 42, is the… basically the same kind of thing, so…
23:19:810Paolo Guiotto: If you have not yet done, you can do by your own. Let's do the 4-3 trip.
23:26:300Paolo Guiotto: which is, still, B variate, random variable.
23:32:910Paolo Guiotto: with the density, so it is saying, basically, it's an absolutely continuous random variable, with this density e to minus X plus Y,
23:44:310Paolo Guiotto: Then we have the indicator of 0 plus infinity, X squared XY.
23:56:710Paolo Guiotto: So, it says, determine… Determined.
24:02:290Paolo Guiotto: Deep.
24:04:130Paolo Guiotto: Sensitivity.
24:08:220Paolo Guiotto: of X… over why.
24:13:330Paolo Guiotto: Now, this is not evident, So…
24:18:170Paolo Guiotto: who says that there is a density. It is not the density of X that is a marginal for this.
24:24:570Paolo Guiotto: So, we should check that there is a density factor.
24:28:630Paolo Guiotto: Not only… just to complete, it's not an evident fact at this.
24:33:100Paolo Guiotto: Okay, so…
24:37:990Paolo Guiotto: The exercise, gives this hint to start computing the CDF. That function is always defined. So, let's,
24:52:90Paolo Guiotto: Startups… computing… the CDR… Oh.
25:04:300Paolo Guiotto: X over Y.
25:08:850Paolo Guiotto: Well, actually, there is a little, there could be, potentially, a problem, because we are doing a division, so what if we divide by zero?
25:18:350Paolo Guiotto: Now, the point is that this makes sense, because let's see what is the probability that Y is 0, okay? Notice that…
25:33:390Paolo Guiotto: The probability that Y is equal to 0, so the probability that we cannot make the division.
25:39:510Paolo Guiotto: is, how do I check this? Well, I don't have the density of Y, so let's first see what… I don't have the density. In this case, there is the density because I have a joint density, so there is a
25:54:610Paolo Guiotto: If the vector is absolutely continuous, each of the components are absolutely continuous, and the density of the components are obtained by integrating the joint density. So, here…
26:09:520Paolo Guiotto: Why?
26:10:800Paolo Guiotto: is absolutely continuous, and this should be just sufficient to respond to this question, because we know that there is a density with
26:21:400Paolo Guiotto: density.
26:24:860Paolo Guiotto: Let's do this calculation just for the sake of doing this, which is, since it is a marginal, it is the integral from minus infinity to plus infinity of the joint density
26:39:530Paolo Guiotto: in the other variable. In this case, there is only one other variable.
26:44:730Paolo Guiotto: So this is the integral from minus infinity to plus infinity of that function, E minus X plus Y indicator 0.
26:56:70Paolo Guiotto: plus infinity squared XY DX, DY.
27:02:670Paolo Guiotto: as I noticed above, we can always represent this indicator as the product of two indicators, the indicator for X, because this means XY must be in that
27:14:540Paolo Guiotto: in that rectangle, no? This is saying XY is in this region here. So this is 0 plus infinity square is this quarter here.
27:27:640Paolo Guiotto: But this is equivalent of saying that X must be positive and Y must be positive.
27:32:220Paolo Guiotto: So you… it means that this indicator must be 1, and also this other indicator must be 1.
27:40:940Paolo Guiotto: But vice versa, you see that this indicator is one exactly if and not if the two are together positive, so they are the same.
27:49:390Paolo Guiotto: So this makes possible to split this integer… I'm sorry, this is, I'm sorry.
27:56:290Paolo Guiotto: This is the integral only in DX, not in Y.
28:00:210Paolo Guiotto: It's the marginal density. So we… this makes possible to extract everything, depending on y, so we got E2 minus Y indicator, 0 to plus infinity of y out of the integral, then we have integral from minus infinity to plus infinity of E minus X indicator, 0 plus infinity.
28:22:120Paolo Guiotto: X, DX.
28:27:40Paolo Guiotto: Now, this integral is the integral from 0 to plus infinity of e to minus x dx, which is easily equal to 1.
28:36:360Paolo Guiotto: So at the end, FY of Y is E minus Y indicator 0.
28:45:130Paolo Guiotto: plus infinity to Y. So, Y is an exponential with the parameter 1.
28:55:120Paolo Guiotto: PCs.
28:58:870Paolo Guiotto: Now, the fact, however, the fact that Y is absolutely continuous is a general fact, so we have seen in general that if the vector XY is, or more in general, if a vector, a multivariate vector is absolutely continuous.
29:17:110Paolo Guiotto: So there is a density. Each of the components is also absolutely continuous, so it has a density, and you compute the density of the components by doing the integral of the joint density from all the… all the… the range for the variables.
29:33:600Paolo Guiotto: Okay? With respect to the remaining variables, so this is the general rule. In particular, since this is absolutely continuous, this probability will always be equal to…
29:51:770Paolo Guiotto: Boop.
29:53:120Paolo Guiotto: So this is the integral on… this is, if you want, this is the low.
29:58:600Paolo Guiotto: or, let's say, it is the probability that Y belongs to the singleton made by 0, right?
30:06:410Paolo Guiotto: That's the new Y.
30:08:310Paolo Guiotto: Of the singleton.
30:10:980Paolo Guiotto: which is equal to the integral on that domain, the single term of the density, with respect to the LeBague measure.
30:19:530Paolo Guiotto: And that's now where you can say that this integral is…
30:22:640Paolo Guiotto: is 0. That's not true in general, because if the measure… if the density is not absolutely continuous, there is no reason. Imagine the delta is… the low is a direct delta centered at that point, you got 1 as well, you know?
30:38:700Paolo Guiotto: But, for this, you get 0. So, since the probability of Y equals 0 is equal to 0, we conclude that the variable X over Y
30:49:110Paolo Guiotto: is well defined.
30:53:990Paolo Guiotto: Almost everywhere, so it makes sense as random variable.
31:00:490Paolo Guiotto: And it is the ratio of measurable functions, so it's a random variable.
31:05:490Paolo Guiotto: Now, we can now rejoin to the point. The point is computing the CDF of X over Y.
31:14:120Paolo Guiotto: that once we have a random variable, there is always a CDF, okay? Maybe there is not a density, but the CDF always exists. And what is the CDF of this variable? Let's use a letter… I don't know, U, it is better if we do not use X.
31:32:270Paolo Guiotto: This means the probability that the ratio X over Y is less or equal than U.
31:42:310Paolo Guiotto: Now, I want to reduce this to some calculation on the joint density, because that's the unique ingredient I have, no?
31:53:510Paolo Guiotto: So I want to see this as probability that the pair XY belongs to some set E, in such a way that if I transform this into this, I will have that this is the integral of that E of the joint density
32:08:640Paolo Guiotto: FXYXYBXDY.
32:12:800Paolo Guiotto: So the point is, how can we see in this way? Now.
32:17:390Paolo Guiotto: The temptation could be, okay, XY, X over Y, less or equal than U. Well, first of all, if you think about…
32:30:820Paolo Guiotto: For this, particular case.
32:35:250Paolo Guiotto: Well, let's say what I would like to do, and then we try to make it precise, I would like to rewrite this in line, multiply by Y.
32:43:700Paolo Guiotto: Now, the point is, can I say that this is equivalent of saying that X is less than UY? Because in this case, if I continue, this becomes a set, no? XY belongs to the set E,
32:59:940Paolo Guiotto: What is E is the set of points in the Cartesian plane, little x, little y, in R2, such that X is less or equal than UY.
33:12:340Paolo Guiotto: And then I will,
33:14:240Paolo Guiotto: I will move to this integral here, and then I will have to compute a double integral here.
33:21:820Paolo Guiotto: Now, the point is that, of course, when you multiply something on an inequality, you have to be careful, because if you multiply by positive, you do not change. If you multiply by negative, you change the sign of the inequality.
33:35:280Paolo Guiotto: So the problem is, what can be said about this? Should I split the two cases? Of course, I could always say that
33:42:120Paolo Guiotto: probability that X over Y is less or equal than U could be splitted in this
33:48:690Paolo Guiotto: In these two. Let's do an argument which is slightly more general and useless here.
33:55:370Paolo Guiotto: I could say, take X over Y less than U and Y positive.
34:02:760Paolo Guiotto: And then, union with X over Y, less or equal than U, and Y negative. So in this case, I know
34:11:510Paolo Guiotto: better. In the first set, I can carry Y on the other side.
34:16:980Paolo Guiotto: and I do not change the inequality. In the other case, I will have to change the inequality. There would be a third alternative, which is X over Y less to equal than U,
34:26:190Paolo Guiotto: and Y equals 0, which is a nonsense. However, this is a measure zero set, because we are in the subset where Y is equal to zero, so we discard it. It's not important.
34:38:340Paolo Guiotto: Now, this is clearly a disjoint union, because Y positive cannot be at the same time Y negative. So this allows to split this probability as the probability that X over Y is less or equal than U and Y is positive.
34:56:690Paolo Guiotto: plus the probability where X over Y
35:00:940Paolo Guiotto: is less or equal than U, and Y is negative.
35:05:250Paolo Guiotto: Now, the key point is, well, in this case, this will be the probability that X is less or equal than Uy, and Y is positive.
35:18:870Paolo Guiotto: While this one will be the probability that X is greater or equal than UY. You see that I inverted the inequality, because Y here is negative.
35:31:110Paolo Guiotto: Okay. Now, this is general. It does not,
35:35:730Paolo Guiotto: Involve in any way this particular case with this particular distribution.
35:42:290Paolo Guiotto: Now, what can be said? We say that above that, for example, the probability that Y is 0 is 0. What is the probability that Y is positive or negative?
35:52:400Paolo Guiotto: Let's see, because one of these two is an empty case. You can understand, because the density, the joint density.
36:03:110Paolo Guiotto: It's concentrated only on points where the two coordinates are positive.
36:09:150Paolo Guiotto: So this is, so what is the probability that Y is negative, for example? Probability that Y is less than 0.
36:18:80Paolo Guiotto: Well, I can see this as a joint probability. I take the pair XY, where X is where? There is no condition, so X is any real, and Y is in the offline minus infinity to zero, okay?
36:36:440Paolo Guiotto: So this is… the pair X, Y is in this set, means Y negative. Now, this is the integral on that domain R cross minus infinity to zero of the joint density
36:51:250Paolo Guiotto: in the X, the Y.
36:55:110Paolo Guiotto: But…
36:56:710Paolo Guiotto: We know that the joint density is apart for the exponential E minus X plus Y. There is the indicator 0 plus infinity for X,
37:06:510Paolo Guiotto: times indicator 0 plus infinity for Y.
37:11:120Paolo Guiotto: And so, in that, Y is in minus infinity to zero, so the…
37:18:350Paolo Guiotto: constantly equal to 0. Therefore, you get zero here.
37:24:100Paolo Guiotto: Okay, so the probability that Y is negative is 0, and that is a subset of Y negative, this probability will be 0. So at that, you have only this.
37:35:900Paolo Guiotto: Okay?
37:36:900Paolo Guiotto: So, this probability. So this…
37:41:50Paolo Guiotto: compute the probability that X over West and U, which is the CDF of the ratio X over Y at 2.
37:51:180Paolo Guiotto: this step.
37:52:470Paolo Guiotto: Okay, excellent.
37:58:270Paolo Guiotto: Because the police.
38:01:700Paolo Guiotto: What remains is just that term.
38:05:540Paolo Guiotto: X… It's less or equal than UI,
38:10:910Paolo Guiotto: and U… NY is positive. So now let's turn this to an integral for the density variables for the density, okay? So it means that X is less or equal than UY, and Y is positive.
38:25:560Paolo Guiotto: of the density, which is E minus X plus Y, indicator for X positive.
38:33:520Paolo Guiotto: indicator for Y positive.
38:38:330Paolo Guiotto: DX, DY.
38:40:100Paolo Guiotto: Now, on this integration domain, we already know that Y is positive.
38:46:550Paolo Guiotto: Euro will be equal to 1.
38:51:10Paolo Guiotto: Okay, will constantly be equal to 1.
38:53:540Paolo Guiotto: Now, what is that domain?
38:56:810Paolo Guiotto: The domain is 1, so we are above here. X less than UY now depends on what is the value of U, okay?
39:11:560Paolo Guiotto: What makes the difference?
39:23:830Paolo Guiotto: Why is positive, no? So this guy is positive.
39:27:890Paolo Guiotto: If this guy is negative, if U is negative, the right-hand side would be negative.
39:35:100Paolo Guiotto: And so this would be more than X less than 0. The indicator would be 0, we would get a 0. So basically, we have two cases. The case one is when U is less or equal to 0.
39:50:480Paolo Guiotto: In that case, this number would be negative, and we are integrating on a domain, but X is negative.
40:00:210Paolo Guiotto: For the indicator would,
40:02:880Paolo Guiotto: be zero, so we would be integrating, all the X, DY, and we would get, of course, equal to 0.
40:12:970Paolo Guiotto: USC or not?
40:16:100Paolo Guiotto: For why you octave, We may have something. Now, what is X less?
40:23:980Paolo Guiotto: and UY. This is, X equal to UY is a straight line where you have to see Y, Y…
40:35:330Paolo Guiotto: Right? So that's a positive… for a positive U. With a positive Y, we have something like, yes, it's something like this, still with positive slope, that passes to the origin, so we have actually
40:54:480Paolo Guiotto: Let's say… This is X equal U.
41:00:820Paolo Guiotto: And we have to take… UI.
41:05:930Paolo Guiotto: So, it's the, left or the right region.
41:14:280Paolo Guiotto: It's the left, okay? It's this part here. So, it would be this half plane. Then we will discard the X negative in a moment, because, correct?
41:23:570Paolo Guiotto: the indicator, kills, everything. So, let's say that this becomes, for you, positive.
41:30:440Paolo Guiotto: So it would be reasonable.
41:33:690Paolo Guiotto: this variable as the last integration variable, so this is Y from 0 to plus infinity, and then 4X, so the last integration in Y, the first in X. 4X, we have from… yes, from minus, if you want, from minus infinity to UY,
41:52:530Paolo Guiotto: function, which is E minus X plus pi.
41:55:820Paolo Guiotto: still solved the indicator, 080x, so this cuts off this, half of that integral, so this makes this integral into the integral from 0 to
42:11:520Paolo Guiotto: X plus 1.
42:13:980Paolo Guiotto: Now, this is in VR.
42:19:70Paolo Guiotto: Indicator is 0 for negative. So, 0, there is nothing, you are integrating 0.
42:26:740Paolo Guiotto: So this, the Andy, is that… Why?
42:35:90Paolo Guiotto: minus X plus Y. This is integrated the first, then, in Y.
42:41:220Paolo Guiotto: Now, the thing is…
42:44:980Paolo Guiotto: a D value of the ratio U positive. So this simple calculation, let's carry… let's split this into E minus X, E minus Y, we carry the exponential minus y outside integral plus EP, minus Y integral 0 to UY, e to minus dx, and DY.
43:05:480Paolo Guiotto: Now, e to minus X is the derivative with respect to X of minus E to minus X.
43:11:280Paolo Guiotto: So this integral here will come equal to minus e to max to be evaluated x equals 0 to
43:21:00Paolo Guiotto: Why?
43:23:380Paolo Guiotto: So, carrying outside the minus, so minus E2 minus UY minus E2 minus 0, so E201.
43:36:280Paolo Guiotto: Or, if you want to reduce the number of signs, 1 minus E2 minus UY.
43:43:370Paolo Guiotto: This is the value of the innermost integral. So, returning to the population of… the double integral.
43:53:160Paolo Guiotto: from zero to glassing feet of E to minus green.
43:59:490Paolo Guiotto: times 1 minus E2 minus UY DY.
44:07:820Paolo Guiotto: Now, this is, to E my… E minus… Why?
44:19:380Paolo Guiotto: That we have to integrate from 0 to plus infinity.
44:23:00Paolo Guiotto: So for the first evaluation, this is the derivative with respect to Y of minus E to minus y, so we have the evaluation of,
44:33:440Paolo Guiotto: minus… E to minus Y from Y equals 0 to y equals plus infinity.
44:43:510Paolo Guiotto: And maybe we keep the minus together with this. This is the derivative with respect to Y of E minus U plus 1.
44:52:660Paolo Guiotto: Y, but here we have to divide by U plus 1.
44:56:710Paolo Guiotto: Because when we do the derivative, we get the exponential, then we have the derivative of the exponent with respect to Y,
45:04:360Paolo Guiotto: And this is the factor minus U plus 1.
45:07:720Paolo Guiotto: So we kill the U plus 1 with that denominator, and we are done. So this is plus… this is plus because I gave the minus to the exponential to reproduce the derivative.
45:21:400Paolo Guiotto: is E minus U plus… And why?
45:26:400Paolo Guiotto: divided by U plus 1 to be evaluated, and 0 plus 1.
45:34:970Paolo Guiotto: Okay, now the value of the exponential at plus infinity here is 0, minus. The value at y equals 0 is minus equals 0, so…
45:44:70Paolo Guiotto: So the difference gives plus 1, plus, if you want, we've put outside this factor, 1 over U plus 1.
45:53:50Paolo Guiotto: Then we have to do the evaluation at plus infinity of the other exponential, E minus U plus 1 plus times Y. Remind that here, U, we are doing the calculation for U positive.
46:06:810Paolo Guiotto: So, in particular, that U plus 1 here, this number is a positive number, because U is positive.
46:14:960Paolo Guiotto: So when y is plus infinity, the exponent is going to minus infinity, the exponential is going to 0. So at plus infinity, we get 0. Minus, at Y equals 0, we get 1.
46:27:460Paolo Guiotto: So, at the end, the value of the calculation yields 1 minus 1 over u… want.
46:37:110Paolo Guiotto: So we can now put the summary, so…
46:41:170Paolo Guiotto: We have F over Y at U is equal to… we obtain that it's equal to 0 for U negative.
46:50:220Paolo Guiotto: for U positive, it is equal to 1 minus 1 over U plus 1.
46:58:700Paolo Guiotto: You see that it is a function… well, it is, of course, the CDF of something, so it has the properties of the CDF, okay, because it is a CDF.
47:09:230Paolo Guiotto: If you want, we can, have,
47:13:530Paolo Guiotto: Here, how it's made this. For U negative, we got 0.
47:19:460Paolo Guiotto: at 0, U is zero, so actually it's continuous there. And then, when you range from 0 to plus infinity.
47:31:680Paolo Guiotto: Since the quantity 1 over U plus 1 is decreasing down to 0, minus will be increasing up to 0, so 1 minus that quantity is increasing to 1.
47:42:720Paolo Guiotto: So, this is one. We have some like this, more or less, there.
47:51:10Paolo Guiotto: So in particular, it is continuous, F, X.
47:55:400Paolo Guiotto: over why.
47:57:730Paolo Guiotto: continuous, but this is not sufficient to say that the variable X over Y is absolutely continuous, which is the goal. We want to have a density. However, it is also differentiable, and
48:12:320Paolo Guiotto: at least a part for u equals 0, maybe, one single point, there existed the derivative with respect to U of this FX over Y.
48:25:670Paolo Guiotto: at U, which is equal to 0 for U-
48:30:300Paolo Guiotto: And for U positive, it is plus 1 over U plus 1 square for U positive. You see that at U equals 0, there is a discontinuity, so the derivative would be 1.
48:44:270Paolo Guiotto: It doesn't matter, the derivatively exists almost everywhere, okay? So there exists the derivative, with respect to its variable,
48:56:120Paolo Guiotto: of the CDF of X over Y.
48:59:310Paolo Guiotto: at almost every U, and of course, it is an L1 function, and therefore, this… we can now baptize this as the density of X over Y.
49:13:180Paolo Guiotto: so this shows that X of the white.
49:19:300Paolo Guiotto: Is absolutely continuous with density.
49:27:530Paolo Guiotto: this function, F, Little F, X.
49:32:120Paolo Guiotto: But finally, we obtain the E.
49:36:460Paolo Guiotto: Answer to this problem.
49:40:750Paolo Guiotto: Okay?
49:43:500Paolo Guiotto: Do you have any questions?
49:54:310Paolo Guiotto: Okay.
49:57:240Paolo Guiotto: It is important, because it may…
50:00:770Paolo Guiotto: problem to be able to see how, CDF and, in particular, the joint density.
50:12:90Paolo Guiotto: change under a transformation of the multivariate random variable. So, let's say that, mapping
50:26:740Paolo Guiotto: multivariate.
50:28:620Paolo Guiotto: It's something about the…
50:31:710Paolo Guiotto: scene for one-dimensional random variables. Here, it is the same, but a little bit more technical. So, let's say that letter X
50:46:270Paolo Guiotto: be, on omega with values into some Rn.
50:52:490Paolo Guiotto: be a multivariate.
51:00:980Paolo Guiotto: Random variable.
51:04:900Paolo Guiotto: on some probability space, omega F… B.
51:11:60Paolo Guiotto: So, assume that, suppose… that this X be absolutely continuous.
51:24:430Paolo Guiotto: So, register a density, F, XM.
51:30:40Paolo Guiotto: density.
51:35:960Paolo Guiotto: Excellent. And I remind you that this means that, So, such that,
51:46:860Paolo Guiotto: the probability that the vector X the second E, where PRE,
51:54:630Paolo Guiotto: since the values of X are vectors of Rn, that set will be N, so it's a multi-dimensional set, and it must be technically a Borel set.
52:06:910Paolo Guiotto: of Rn. Now, this is, by definition, the low of X on the set E. This law can be computed by doing the integral over E of the density X
52:21:250Paolo Guiotto: XVX. Well, this is a multidimensional, is an n-dimensional integral, okay, LeBague integral.
52:29:630Paolo Guiotto: This guy here is, N… Let that go.
52:36:710Paolo Guiotto: You can figure it out.
52:40:970Paolo Guiotto: Now, I told, sometimes it may happen that, we define a new random variable, a new random…
52:59:140Paolo Guiotto: capital fee be a transformation. Here, I will assume that the transformation is defined on the full space.
53:09:210Paolo Guiotto: it is not needed because, for example, imagine that our random variable X is positive, so it means that all entries are positive. The transformation of X is not
53:23:800Paolo Guiotto: it's… it is not required that it's defined for every vector, just for positive vectors, okay? But to avoid…
53:31:20Paolo Guiotto: Not complications. For simplicity, I will write as if this phi is a map of Rn in itself.
53:41:680Paolo Guiotto: Let t be this, and define a new variable, Y,
53:51:800Paolo Guiotto: is the image through this map, fee of… Yes.
53:58:690Paolo Guiotto: Now, in general, you will not obtain a random variable, because you will need some minimal requirement on this function phi.
54:09:70Paolo Guiotto: But since the requirements we needed to use here
54:15:110Paolo Guiotto: strong, because we want to discuss this problem. So, problem is… And… We… conditions… Is…
54:37:180Paolo Guiotto: Absolutely.
54:38:330Paolo Guiotto: Continos.
54:43:780Paolo Guiotto: And here's… the UI and the desktop browser.
54:55:500Paolo Guiotto: 18…
55:01:280Paolo Guiotto: Bye.
55:03:50Paolo Guiotto: X.
55:05:710Paolo Guiotto: Now, we understand, basically, this is related to the variable in multiple integrals. It's a problem related to that, because, I…
55:16:960Paolo Guiotto: approval with an informal, like, argument that on a pre…
55:23:740Paolo Guiotto: However, I will not take the technical details. Imagine that you take your Y equals P , and you want to say if this Y is absolute. Well, what you have to do is you have to show that
55:43:230Paolo Guiotto: of, Y, and then represented to an integral certain, which will be the density of Y, no? Now, Y is absolutely tin's
55:57:930Paolo Guiotto: Even though it… "… If the law of Y evaluated On a SE, SE… E, or… an FY insight.
56:13:490Paolo Guiotto: Why?
56:14:660Paolo Guiotto: for function F,
56:22:430Paolo Guiotto: positive, the characteristic of the density must be that the function FY must be positive, and probability distribution, so that integral
56:34:420Paolo Guiotto: And I'll have… DY must be equal to 1.
56:42:640Paolo Guiotto: So, the goal should be Please stop. Oh.
56:47:920Paolo Guiotto: What they about to mean?
56:50:620Paolo Guiotto: What?
56:51:950Paolo Guiotto: definition, the prob… Why?
56:56:250Paolo Guiotto: Bill, E.
56:58:860Paolo Guiotto: Y phi of X means the probability that phr belongs to the sector E.
57:05:900Paolo Guiotto: Literally, this means, this notation, phi of X, belongs to E, is what? Since you understand? We are doing a probability, so this will be
57:19:250Paolo Guiotto: a set of omegas, so we are in a sample space, so this is the set of omegas in capital omega, such that XP, sorry, of omega below that set P.
57:35:890Paolo Guiotto: Hello!
57:37:800Paolo Guiotto: functions. This is… came off saying that…
57:46:910Paolo Guiotto: Oh.
57:49:110Paolo Guiotto: must be the anti-ishman of SAP.
57:56:710Paolo Guiotto: Is that we have a short note… Is this… For the chapter, the event, It's a beautiful… Use from.
58:11:220Paolo Guiotto: When we probate, we have that muy
58:19:810Paolo Guiotto: Feel max.
58:21:330Paolo Guiotto: belongs.
58:26:160Paolo Guiotto: bet.
00:07:980Paolo Guiotto: mute.
00:09:90Paolo Guiotto: Are you still there?
00:10:930Paolo Guiotto: Can you give me a feedback?
00:13:100LEONARDO RIZZOTTO: Yes.
00:13:530Paolo Guiotto: Seeing the messages?
00:15:260Paolo Guiotto: Maybe you are sleeping?
00:39:920Paolo Guiotto: Okay, so we were saying that muy of,
00:48:630Paolo Guiotto: the Boris set E, which is the probability that phi of X belongs to E, because this is
00:54:460Paolo Guiotto: our Y is the probability that X belongs to this other set, that for a moment, we will assume that it's a Borel set. At the end, it will be true, because the assumption we have to do on this map fee are quite strong.
01:10:600Paolo Guiotto: So, since this is now the low of X on the set phi minus 1E, for this one, since we know that
01:22:310Paolo Guiotto: X is supposed to be absolutely continuous, change the
01:27:590Paolo Guiotto: set E into phi minus 1E. This becomes… so, X on that set.
01:54:740Paolo Guiotto: Shannon, very good.
01:59:00Paolo Guiotto: Okay.
02:00:980Paolo Guiotto: Let's see… So, this is the integral of the density of the vector X.
02:12:500Paolo Guiotto: Boom?
02:14:660Paolo Guiotto: So let's see what we obtained. What we wanted was to show that there exists a function. This function.
04:59:690Paolo Guiotto: Okay, let's see…
05:08:480Paolo Guiotto: So it is recording. Let's continue.
05:13:500Paolo Guiotto: So we wanted to find, and this is the last time, if it happens once more, I will just continue the class.
05:21:800Paolo Guiotto: So, we had to find such function, FY, for which new Y of E is the integral of that density. Apparently, we obtained something similar, because we proved that, so let's say that we want
05:42:530Paolo Guiotto: muy of E equal to the integral of E of some density FY.
05:51:310Paolo Guiotto: In the Bible.
05:52:650Paolo Guiotto: Why? We have what we have…
05:56:810Paolo Guiotto: is this formula, muy of E,
06:01:470Paolo Guiotto: It looks like this one, but it's slightly different, because we obtain the integral of phi minus 1 of E,
06:10:660Paolo Guiotto: of the density of X in the X. Well, the letter is not the problem, because I could always call y X equal Y, but the problem is that that function is not the same.
06:23:910Paolo Guiotto: So now the point is, how can I transform that, and you see also the domain, here you have domain E, and here you have domain
06:35:910Paolo Guiotto: C minus 1 of E. So here, the Y belongs to the set E, and here, the X belongs to the set P-1E.
06:46:670Paolo Guiotto: But if X belongs to phi minus 1E, it means that the quantity phi of X
06:54:370Paolo Guiotto: is where? Well, if X is in phi minus 1E, you map back without true phi, you go in E. So this quantity, Y equals phi of X, would be in E. So, if we now do the change of variables, so set y equal phi of X,
07:14:600Paolo Guiotto: So did we change variable in the integral. In the new variable Y,
07:21:10Paolo Guiotto: DY will be in the right set, in the set E,
07:26:270Paolo Guiotto: The function FX has to be evaluated here is an X, but X is what with respect to Y? If Y is phi of X, and this phi is good enough to be invertible in such a way that 2NX there is a Y, but also the vice versa, so I can say that to a Y there is an X, that will be the phi minus 1
07:49:470Paolo Guiotto: Why? So here we need the first requirement, which is phi invertible.
07:59:600Paolo Guiotto: So far.
08:00:920Paolo Guiotto: I… I… I… to write phi minus 1 of a set, it's the anti-image of a set, this is defined whatever is the map phi. You do not require… you do not need that this is invertible, but it is here whether you use the change of variable that you need. The map fee must be invertible, so in this way, you have that this X belongs phi minus 1,
08:24:279Paolo Guiotto: of Y. And since we are using a change variable in the integral, you know that there is something here to be written to make this formula correct, and this something requires a lot more, which is also this. So, phi invertible and phi, phi minus 1, both differentiable.
08:50:720Paolo Guiotto: So, in fact, this means that this map is a change of variable that must be a regular function. So, in particular, it will be continuous, and this explains why the anti-image of a Borel set is a Borel set. So, let's just remark this in particular.
09:10:620Paolo Guiotto: phi and phi-1 are continuous, and this is sufficient to show that phi-1 of E belongs to the Borel class of N if E is a Borel set.
09:26:620Paolo Guiotto: you may remind that at some point, I just said a word here. I was saying, provided this is a Borel set. What is the condition? Well, if the condition to make that a Borel set, well, is a Borel set is very weak.
09:43:380Paolo Guiotto: But here we have a very strong assumption that implies that weak assumption. Now, with these assumptions, the quantity you have to put here is the… the… is the… is what, no? The idea is that Y is…
09:57:310Paolo Guiotto: phi of X, so X is phi minus 1 of Y, so literally, let's say that dx becomes what times DY? Well, if we are in one variable, this is just the absolute value of the derivative of the inverse function.
10:14:680Paolo Guiotto: But if we are in, as in this case, with a map from Rn to Rn, the situation is more involved, we need to put the determinant of the Jacobian matrix of phi minus 1, so determinant.
10:28:790Paolo Guiotto: Of the derivative, phi minus 1 prime.
10:32:860Paolo Guiotto: Why? Well, this is the Jacobian metrics,
10:41:870Paolo Guiotto: Of the map, phi minus 1.
10:44:980Paolo Guiotto: Okay.
10:46:370Paolo Guiotto: So…
10:47:380Paolo Guiotto: I hope you remind what is the Jacobian matrix. These maps, phi and phi minus 1, are maps from Rn to Rn, no? So just let's do a quick recap, no?
11:00:700Paolo Guiotto: Imagine, since here we have to compute the Jacobian matrix of phi minus 1, let's talk about phi minus 1. So, phi minus 1 is a function phi minus 1 of vector Y.
11:11:680Paolo Guiotto: This is a function from Rn to Rn, so this will be a vector with n components. Each of them is a function of vector Y. So let's call these components
11:30:340Paolo Guiotto: well, let's call them a C-1 first component at Y, C minus 1 second component at Y, and so on.
11:39:830Paolo Guiotto: phi minus 1 nth component at Y. Now, this function here
11:45:240Paolo Guiotto: P minus 1, the j component, is a function that goes from Rn, because the variable, the Y, is a vector, and they are components of a vector, they are numerical functions, so they are real-valued stuff.
12:00:120Paolo Guiotto: Now, the Jacobian matrix of this map…
12:05:570Paolo Guiotto: is a matrix. In this case, since the map is a map from Rn to RN, this is an N times N matrix.
12:20:460Paolo Guiotto: Well, a quick way to remind of this is that you take the components, which are these ones.
12:27:180Paolo Guiotto: components of the map, phi-1, and in each of the lines of this matrix, you put the gradient of this component. So here you will have the gradient of the first component of phi-1. Here you have the gradient of the second component of phi-1 until you finish the components. You have end lines, the end component of phi-1.
12:50:110Paolo Guiotto: And each queue identity is an array with all the partial derivatives, okay?
12:55:570Paolo Guiotto: So this is the Jacobian matrix.
12:58:590Paolo Guiotto: Okay, now… So, with this, we obtained what we wanted, because… so.
13:06:660Paolo Guiotto: If, Fee.
13:09:790Paolo Guiotto: And phi minus 1.
13:12:100Paolo Guiotto: R.
13:14:830Paolo Guiotto: differentiable.
13:18:70Paolo Guiotto: we have that… we found this on one side, well, we found that muy of E, at the end turns out to be the integral… well, let's put it in red… muy of E
13:33:930Paolo Guiotto: is the integral on E of this function, the density of the initial vector X, evaluated, let's say, in the X, that corresponds to Y through the change of variable. This is phi minus 1 of Y.
13:50:670Paolo Guiotto: modules of determinant of the Jacobian matrix of phi minus 1 at point Y in DY.
14:00:370Paolo Guiotto: And we wanted that muy of E be equal to a function of Y in DY, integral of a function of Y in DY on the domain E. And that's exactly the stuff we have inside this integral here.
14:16:820Paolo Guiotto: So this means that… So… In particular…
14:25:60Paolo Guiotto: In these assumptions, y equals P of X Is absolutely continuous.
14:34:930Paolo Guiotto: and the density… of, Y,
14:41:820Paolo Guiotto: is given by this formula. It is the density of X,
14:46:300Paolo Guiotto: with the variable changes, not the letter, but just the variable, because there is a change of variable between X and Y, times the modulus of the determinant of the inverse
14:58:20Paolo Guiotto: of the Jacobians.
15:02:120Paolo Guiotto: the Jacobian metric of the inverse function, and that's the formula.
15:06:460Paolo Guiotto: For the change of density.
15:11:780Paolo Guiotto: Let's see an example. For example, we take… The exercise 435.
15:29:260Paolo Guiotto: So, here it says, XY…
15:33:830Paolo Guiotto: So I'm sorry now for the notations, because for the general, let's say, formulas, I use the X and Y as if they are arrays, okay? So in the formula you see here, X and Y are multivariate.
15:49:510Paolo Guiotto: Random variable. Both, okay? Now, here, in this exercise, X and Y are the components of a multivariate. We have
15:58:640Paolo Guiotto: We need to be a little bit flexible on this. So this says B, B variant.
16:05:880Paolo Guiotto: Random variable with density.
16:13:670Paolo Guiotto: FXY is this function.
16:18:830Paolo Guiotto: is 1 fourth, it's similar to the previous example, X plus Y over 2 indicator, 0 plus infinity squared.
16:29:810Paolo Guiotto: Oh, that's white.
16:31:80Paolo Guiotto: The point is, however, is the…
16:34:540Paolo Guiotto: the application of the formula. Now, we define a new vector, let's ZW,
16:41:400Paolo Guiotto: The vector defined in this way, X, sorry.
16:45:980Paolo Guiotto: W. X… plus Y.
16:51:210Paolo Guiotto: No, it is minus Y.
16:57:290Paolo Guiotto: of 2, and Y.
17:04:60Paolo Guiotto: determine the density of ZW, so the joint density, and then the densities of Z and W, the marginal densities of these.
17:18:90Paolo Guiotto: So it's just a calculation. It's not particularly interesting, but it shows how to use this formula here.
17:27:980Paolo Guiotto: So, in this… in the notation of this formula, you see that we start with a variable, this would be the X we have here, and we transform into this other variable. This is the Y we have there.
17:44:920Paolo Guiotto: Okay? Because you see that this ZW is a function of the pair XY. So here… ZW…
17:56:860Paolo Guiotto: is a function phi of the pair XY, where the function phi
18:05:920Paolo Guiotto: This phi is defined on R2 and is R2 vectors.
18:12:510Paolo Guiotto: And the definition is, of course, you transform the uppercase letters into the lowercase letters. So, phi XY is… the first component is X minus Y over 2, the second component is just Y.
18:29:120Paolo Guiotto: So this is the map.
18:31:300Paolo Guiotto: Okay?
18:34:30Paolo Guiotto: Therefore, we can say that if this map is invaluable with both phi and phi-1 differentiable.
18:42:770Paolo Guiotto: the, the arrival variable, this ZW, will be absolutely continuous with density
18:50:150Paolo Guiotto: given by this formula with the appropriate quantities, okay? So that's what we want to do.
18:56:520Paolo Guiotto: Now, we notice that the fee is clearly
19:03:830Paolo Guiotto: differentiable.
19:05:760Paolo Guiotto: The components are first-degree polynomials, so there is no question.
19:10:700Paolo Guiotto: And about phi minus 1, well, just for the sake of doing this once, because maybe we need the phi is…
19:20:920Paolo Guiotto: Invertible.
19:26:350Paolo Guiotto: Because…
19:29:590Paolo Guiotto: Well, to show that it is invertible, you have to take an arrival vector and show that it comes from some XY, so you have to invert the transformation. So, let's say that we call… we use the same letters, ZW is…
19:46:430Paolo Guiotto: phi of XY, if and all if, if you want to have a system, Z must be X minus Y over 2, and W must be Y.
19:57:580Paolo Guiotto: From this, we see that now we have to invert this relation to express X and Y as function of ZW. So the second equation yields Y equals W.
20:09:670Paolo Guiotto: So we plug into the first one, so we can write Z equals X minus W divided 2.
20:18:30Paolo Guiotto: So, equivalently, you carry the 2 on the other side, so we have X equals 2Z plus W, so that's the inverse function. Of course, it's a linear function, the inverse will be a linear function.
20:30:730Paolo Guiotto: X is equal to Z plus W, Y is W.
20:36:450Paolo Guiotto: This function yields XY in function of ZW, so this is the inverse of the map phi. So if you want to see the phi minus 1 as a function of ZW, is this function, is 2Z plus W,
20:54:10Paolo Guiotto: W. And clearly, it is differentiable.
20:58:200Paolo Guiotto: As well as the direct map fee.
21:02:620Paolo Guiotto: Okay, so this in particular says that,
21:06:240Paolo Guiotto: without computing the density, that the arrival variable, so the capital Zw, is absolutely continuous.
21:17:490Paolo Guiotto: And if we want the density, that we will need to compute the… it's required, and also to compute the other two densities, the density, FZW, let's say in the little z, little w.
21:32:530Paolo Guiotto: is equal to what? The original density, that will be the density between… of XY,
21:39:770Paolo Guiotto: at point… the pointer is the XY that corresponds to ZW, so that's P minus 1.
21:46:590Paolo Guiotto: ZW.
21:49:380Paolo Guiotto: times the modulus of determinant of the Jacobian matrix of phi minus 1. This phi minus 1 is a function of ZW, so this quantity, in principle, would be a function of ZW.
22:02:650Paolo Guiotto: And here we have the transformation formula. So this means that this is FXY, let's put the ingredient into this. So phi minus 1 is 2Z plus W,
22:16:30Paolo Guiotto: Double.
22:17:530Paolo Guiotto: Then, what is the phi-1 prime? You have phi minus 1,
22:22:600Paolo Guiotto: And phi minus 1 prime is the Jacobian matrix of phi minus 1. The phi minus 1 is here.
22:30:500Paolo Guiotto: So you have to take, in first line, the gradient of the first component. So it's a matrix where you have here the gradient of the first component is 2Z plus W.
22:43:140Paolo Guiotto: 2Z plus W. So the gradient means that there will be the partial derivatives with respect to the two variables that for this function are Z and W, okay? Because this is a function of ZW. And here we have the gradient of the second component, which is just W.
23:00:760Paolo Guiotto: So remind that the gradient is an array made of the partial derivatives with respect to the variables you are considering. For this function, the derivative with respect to Z, the derivative with respect to W.
23:15:170Paolo Guiotto: So when we do the derivative with respect to Z, we get 2, derivative with respect to W, we get 1, derivative with respect to Z, 0, derivative with respect to W1. That's the Jacobian matrix for this transformation. So the determinant of phi minus 1 prime
23:32:660Paolo Guiotto: which is the quantity I have put here in absolute value, is now easily equal to 2. So now this is the quantity we have to put here, modulus of 2.
23:47:970Paolo Guiotto: Okay, so we have.
23:49:740Paolo Guiotto: equal. Let's continue here, after S star.
23:57:270Paolo Guiotto: 2. Now, we plug the density of XY, we have this, is, this function, 1 fourth the exponential, etc. So, 1 fourth…
24:11:580Paolo Guiotto: E to minus… There is, here, X plus Y.
24:17:440Paolo Guiotto: Right? Where X is the first argument of this density, Y is the second argument.
24:23:460Paolo Guiotto: So X plus Y will be, in this case, since the X is 2Z plus W, the Y is W, X plus Y is 2Z plus W plus W, so 2Z plus 2W. So, 2Z plus W,
24:39:150Paolo Guiotto: Then we have the indicator, yeah. It is the indicator of the pair XY, so again, indicator 0 plus infinity square.
24:49:670Paolo Guiotto: of, what is X is 2Z plus W.
24:54:930Paolo Guiotto: And Y is W.
24:57:250Paolo Guiotto: So, simplifying a bit, we have 1 half e to minus 2…
25:03:250Paolo Guiotto: Z plus W times this indicator. We can, of course… an indicator of a square is always a product of indicators, because it says the first component must be positive, the second component must be positive, so indicator 0 plus infinity of 2Z plus W,
25:22:940Paolo Guiotto: Indicators 0 plus infinity.
25:26:50Paolo Guiotto: of W.
25:28:60Paolo Guiotto: And this is the density of the pair ZW, okay? So this quantity here is the joint density of ZW.
25:42:80Paolo Guiotto: To finish the exercise, we need to compute the single densities of Z and W, but they are marginals, so since…
25:51:860Paolo Guiotto: the pair, ZW is… absolutely continuous.
25:59:80Paolo Guiotto: Also the marginals, Z and W's, are… absolutely continuous.
26:12:430Paolo Guiotto: And, let's start with FZ. FZZ, little z, is the integral of the joint density, FZWZW, in DW, in the other variable, from minus infinity to plus infinity.
26:32:790Paolo Guiotto: Now, we plug that formula into this integral, let's see what happens.
26:37:380Paolo Guiotto: We have the integral from minus infinity plus infinity.
26:41:60Paolo Guiotto: So, first of all, I have 1 half this exponential.
26:45:130Paolo Guiotto: Well, maybe we already split the exponential, because some part won't enter in the integral, for example, the factor E2 minus 2Z, so 1 half E minus 2Z E minus 2W,
26:59:40Paolo Guiotto: Then we have two indicators. One is the indicator 0 plus infinity.
27:05:460Paolo Guiotto: of 2Z plus W, and the other is the indicator 0 plus infinity of W.
27:14:160Paolo Guiotto: in the W.
27:16:360Paolo Guiotto: Okay, so first of all, let's, let's put outside the quantities that do not depend on
27:23:210Paolo Guiotto: on W. For example, this one goes outside. We cannot put outside this one, because you see, there is still W inside the indicator, so let's keep there. For the last one, we can cut the integral, because it is 0 when W is negative, so we can say
27:41:380Paolo Guiotto: First of all, 1 half e to minus 2W, integral from 0 to plus infinity, and from 0 to plus infinity, the last indicator is 1.
27:50:740Paolo Guiotto: Then we have E2 minus 2W, indicator 0 plus infinity of… 2Z plus W.
27:59:800Paolo Guiotto: in DW. Now, let's see what this indicator tells.
28:03:970Paolo Guiotto: This indicator is 1 if and only if equal 1 if and only if the quantity 2Z plus W is positive, right?
28:16:650Paolo Guiotto: Now, we can read this as a condition for W, which is the integration variable. This says that, equivalently, W must be greater or equal than minus 2Z. So it means if W is less than minus 2Z, that quantity is 0, so you are integrating nothing, okay?
28:37:340Paolo Guiotto: And, it is equal to 1 when W is greater than minus 2Z, so it means that I can cut the integral from minus inf… no, from 0 to,
28:50:200Paolo Guiotto: To plus infinity, okay, okay.
28:53:110Paolo Guiotto: Well, let's write this way. Integral 0 plus infinity E minus 2W. So let's transform into an indicator for W.
29:01:900Paolo Guiotto: So this could be written as indicator for the interval from minus 2Z to plus infinity.
29:09:110Paolo Guiotto: of W in the W, you see?
29:12:590Paolo Guiotto: Because this is 1 exactly when W is greater than minus to Z, which is the same of this condition, which is the same of that one, okay?
29:22:630Paolo Guiotto: Now, I want to, of course, put this with this integral, because this is saying you should integrate from minus to Z plus infinity, but this depends when is this minus to Z, because what we see… I'm already on slow plus infinity, so the difference would be made by Z upon Z negative.
29:41:810Paolo Guiotto: So, let's see what is the situation. So for…
29:47:00Paolo Guiotto: Z, say, negative, let's start from this.
29:51:760Paolo Guiotto: For Z negative, minus to Z is positive.
29:55:750Paolo Guiotto: So, I… I have a farther restriction, because now that indicator is 0 from 0 to minus 2Z, and then it is 1. So this becomes the 1 half, the E minus 2Z, the integral from minus 2Z to plus infinity of E minus 2W.
30:15:00Paolo Guiotto: DW.
30:18:450Paolo Guiotto: Well, let's leave some space. We will complete the calculation here directly.
30:26:190Paolo Guiotto: While for Z positive, what happens? For Z positive, minus 2Z is negative. So that indicator is now red in that, because you are saying, taking 1 when W is between that negative number and positive 3, but W is already between 0 plus 3.
30:44:10Paolo Guiotto: So you are sure that that indicator is 1. And therefore, this reduces to 1 half
30:49:960Paolo Guiotto: E minus 2Z, integral now from 0 to plus infinity, E minus 2WDW. This 4Z positive.
31:01:320Paolo Guiotto: Now, in the first case, they are similar, the same function. You see that in both cases, this is the derivative with respect to W of minus 1 half E minus 2W, right?
31:16:200Paolo Guiotto: Because if you do the derivative, you get minus 1 half the exponential, E minus 2W, times the derivative of the exponent, which is minus 2, that kills the minus 1 half.
31:26:100Paolo Guiotto: Okay, so we get that this is 1 half E minus 2Z evaluation of minus 1 half E minus 2W,
31:37:30Paolo Guiotto: from W equals minus 2Z, that's to W equal plus infinity.
31:43:940Paolo Guiotto: At plus infinity, we get 0, because it's a negative exponential, minus the value at minus 2Z, so if I'm not wrong, it becomes plus… so 1 half, 1 half.
31:58:260Paolo Guiotto: E minus 2Z, this becomes plus 1 half E minus, minus plus 4… for… Z, right?
32:15:310Paolo Guiotto: So if I'm not wrong, it comes 1 fourth… E to 2Z.
32:23:330Paolo Guiotto: For Z positive, we have 1 half E minus 2Z. Now, the calculation is the same, minus 1 half
32:32:270Paolo Guiotto: E minus 2W from W equals 0 to W equals plus infinity. Again, at plus infinity, we get 0. At 0, we get minus 1 half times 1, so minus 1 half.
32:46:10Paolo Guiotto: It is negative, so the difference, we get plus 1 half, so 1 fourth… E minus 2Z.
32:54:500Paolo Guiotto: So, the conclusion is that the density of F… this is the density of FZ,
33:04:820Paolo Guiotto: Yes. The density… of FZ at point Z is equal to…
33:15:840Paolo Guiotto: So we have two phases. Is that negative? Is that positive?
33:19:990Paolo Guiotto: Was that negative?
33:22:590Paolo Guiotto: we obtained 1 fourth E22Z, 1 fourth E to 2Z. Y, for Z positive.
33:35:920Paolo Guiotto: It is 142 minus 2Z, so we could unify.
33:40:190Paolo Guiotto: 1 fourth.
33:41:600Paolo Guiotto: E2 minus 2Z.
33:44:510Paolo Guiotto: we can write in a unique way. 1 fourth… E to minus 2… absolute value of Z, right?
33:56:570Paolo Guiotto: Should be this. So this is the density respect to,
34:01:380Paolo Guiotto: Z, and now it remains the density with respect to W, which is, similar.
34:07:570Paolo Guiotto: So FW of little w is, again, the integral for minus infinity plus infinity of the joint density ZW.
34:21:30Paolo Guiotto: This time is integrated in Z. So, you have the same calculation, I don't know what comes out, but we can finish, I think.
34:29:80Paolo Guiotto: Okay, so, you do the exercises, 4… 3… 6… 7… 8 and 9.
34:48:90Paolo Guiotto: All these type of exercises, with questions related to transformations of random biomes.
34:54:950Paolo Guiotto: Okay.
34:56:260Paolo Guiotto: Let's stop here.