AI Assistant
Transcript
00:08:390Paolo Guiotto: So, good morning.
00:12:320Paolo Guiotto: Let's face, there's something.
00:14:970Paolo Guiotto: Okay, the topic of today is basically, let's say, the first real probabilistic idea, which is the concept of independence. We start with the definition of independent events, so let's…
00:34:380Paolo Guiotto: Omega… F… P, the as usual probability space.
00:45:820Paolo Guiotto: two events…
00:51:240Paolo Guiotto: E and F.
00:55:110Paolo Guiotto: independent.
01:05:940Paolo Guiotto: If the probability
01:12:270Paolo Guiotto: of the intersection, E and F.
01:16:350Paolo Guiotto: So, this is the event.
01:19:830Paolo Guiotto: that both the two events here and there have happened.
01:26:370Paolo Guiotto: is equal to the product of the probabilities, P of E times E of F.
01:34:740Paolo Guiotto: This definition comes with the idea that if you, for example, toss a coin two times, the second toss
01:44:760Paolo Guiotto: is not affected by the first one. So the probability that you have two times head is the probability that the first coin is head, so one half times the probability that the second coin is head again. So another one half, so total one-fourth.
02:02:840Paolo Guiotto: Which is exactly what happens, because if you toss the coin two times, there can be…
02:08:630Paolo Guiotto: four possibilities. Two times head, two times tail, one times head, tail, and the other tail, head. So, 1 over 4 is the possibility that you are. So, this is the definition of independent events, and more in general.
02:28:490Paolo Guiotto: We say that, a number, N, of events, E1, EN.
02:39:160Paolo Guiotto: E. F. R.
02:41:490Paolo Guiotto: independent.
02:45:940Paolo Guiotto: If… similar definition. The probability of the intersection, E1, Intersection E2, intersection N,
02:57:520Paolo Guiotto: is the product of the probabilities. E1 times E2, Bones. Yeah.
03:10:710Paolo Guiotto: So, did you go to the hubs?
03:15:990Paolo Guiotto: So that's from next Tuesday.
03:18:970Paolo Guiotto: Not from today. I don't know if, shouldn't they say that Professor Marson told you that it's from today? Yes.
03:27:570Paolo Guiotto: I don't know, I hope that he is sure of what he says, because the official communication is that from the next Tuesday, we are, definitely there.
03:42:270Paolo Guiotto: I don't know for today. For today, we are supposed to be here.
03:47:690Paolo Guiotto: Okay, now this is the definition of independent events, and we have a definition of independent random variables, so let's continue here.
03:58:350Paolo Guiotto: If, We have now two random variables, if X, and Y.
04:07:120Paolo Guiotto: R.
04:08:350Paolo Guiotto: random variables.
04:10:750Paolo Guiotto: we say, that… They independent.
04:26:340Paolo Guiotto: if, well, the events associated, the natural events associated to random variables are things like probability that X belong to some set E,
04:38:630Paolo Guiotto: intersection with the prob… with the Y in some F.
04:46:520Paolo Guiotto: So this is the product of the two probabilities, that X belongs to E times the probability that Y belongs to F. This for every E and F borel sets.
05:01:550Paolo Guiotto: of, in this case, R, or RN, if we have arrays.
05:09:900Paolo Guiotto: And then we have, of course, a definition of N variables independent, so… A similar X1.
05:21:450Paolo Guiotto: XN. Independent.
05:25:710Paolo Guiotto: If, the probability of the intersection, let's say, XK belongs to some EK,
05:37:100Paolo Guiotto: When K goes from 1 to N, this is the product
05:42:220Paolo Guiotto: K goes from 1 to n of the probabilities.
05:50:850Paolo Guiotto: Now, these are definitions, and we now focus on,
05:56:420Paolo Guiotto: on this point, what does it mean that two or more variables are independent in terms of their characteristic entities? In particular, in terms of
06:09:860Paolo Guiotto: of the CDF density and characteristic function, and so on.
06:16:280Paolo Guiotto: A first remark that comes from… immediately from this definition, so let's focus on the case of two variables.
06:26:30Paolo Guiotto: can be easily extended to end variables, is the following. If you… if we rewrite this.
06:34:660Paolo Guiotto: Identities, so we have this remark.
06:39:630Paolo Guiotto: So, that identity means that probability that the pair XY well, let's say…
06:53:720Paolo Guiotto: Let's say that, yeah, that XY belongs to the Cartesian product, it cross F.
07:04:250Paolo Guiotto: Which is the, law of the,
07:08:950Paolo Guiotto: array mu XY evaluated on the set P times F.
07:15:490Paolo Guiotto: a way to write this, this is the integral on R… this case would be R2, because the law is a law of a two-dimensional array of the indicator of the set E types F.
07:32:40Paolo Guiotto: Or, we can also write this as an expected value of that indicator, E cross F.
07:40:440Paolo Guiotto: of the, pair, X, Y.
07:46:530Paolo Guiotto: So… Now, this indicator… is 1 if and only if, of course, XY belongs to,
07:57:380Paolo Guiotto: E cross F, so if and only if indicator of X
08:03:790Paolo Guiotto: is 1, indicator on F of Y is 1. So, if and only if this product is 1. So we can say that this, at the end, is the expected value of indicator E.
08:19:880Paolo Guiotto: indicator F, Y, And this is the value of that probability.
08:26:580Paolo Guiotto: Similarly, the product of the two probabilities, X belongs to me times probability that Y belongs to F,
08:38:640Paolo Guiotto: Each of them can be seen as the expected value of the indicator on set E of X times expected value indicator set FY.
08:51:500Paolo Guiotto: So, basically, the independence can be recasted in this form. So…
08:59:660Paolo Guiotto: We can say that, XY are independent.
09:07:300Paolo Guiotto: If, and only if, the expected value of indicator set in.
09:14:770Paolo Guiotto: of X times the indicator of set F, Y,
09:20:290Paolo Guiotto: So this is, the expectation of this product is the product of the expectations.
09:28:580Paolo Guiotto: So, expectation of E.
09:31:260Paolo Guiotto: indicator set E of X times the expectation, indicator set FY.
09:41:160Paolo Guiotto: Okay? Now, this identity, in fact, extends to a more general product, like a function of X times a function of Y,
09:52:320Paolo Guiotto: Whenever you have expected value of function of X times function of Y, this splits into the product of the expectations, which is something useful to know.
10:03:460Paolo Guiotto: proposition. We have this.
10:07:10Paolo Guiotto: that XY independent.
10:15:110Paolo Guiotto: It's a characterization if and only if the expected value of PX times CY,
10:26:410Paolo Guiotto: is the product of the two expectations, expectation of phi of XL,
10:33:220Paolo Guiotto: times the expectation of C of Y.
10:37:240Paolo Guiotto: Well, technically, for every phi and the C,
10:41:580Paolo Guiotto: such that phi of X… well, what is written makes sense. This is the idea. So you need that this expectation here is well-defined, and this is phi of X belongs to L1.
10:58:300Paolo Guiotto: omega, and the same C of Y for the second expectation, belongs to L1 omega.
11:09:900Paolo Guiotto: Just a remark before we,
11:14:390Paolo Guiotto: we look at the proof. Saying that phi of X belongs to L1 omega means that the expected value of the modulus of phi of x
11:29:440Paolo Guiotto: is filed, right?
11:33:290Paolo Guiotto: We know that this expectation can be written as an integral on r with respect to the law of x of the function modulus phi of x, where x is now a real variable, it's not a random variable.
11:50:780Paolo Guiotto: So, that expectation coincides with this integral, so that expectation is finite if and only if this integral is finite, and this means if and only if the function modulus of phi as a numerical function, here is a function of random variable of phi.
12:08:250Paolo Guiotto: of random variable X, so it's a random variable, this thing, okay? It's a function of X, capital X, random variable. Here, the function is a numerical function, phi of X,
12:23:320Paolo Guiotto: numerical function.
12:27:330Paolo Guiotto: of X, of the variable X. So this must be integral in L1, in the real line, respect to this measure mu X. So, respect to the law of X.
12:44:410Paolo Guiotto: So this is to say, how do I check if phi of X is in L1? Well, maybe you already have… you can do directly, or you can do checking that the numerical function phi is in L1, so let's take out the absolute value, is in L1 with respect to the law of x.
13:04:230Paolo Guiotto: Okay.
13:05:750Paolo Guiotto: Now, let's go back here. So, clearly, you see that, if, this property, let's give a name to this, we call, 2Star, this,
13:18:670Paolo Guiotto: Two-star is an extension of one star, which is this one, basically, because this is just a way to rewrite the definition.
13:27:420Paolo Guiotto: So, it is, evident that 2 star, contains 1 star, because you have just to take phi, empty, two indicators, and you are done, okay? So, proof.
13:44:450Paolo Guiotto: 2 star.
13:47:50Paolo Guiotto: Implies, easily, one star.
13:51:180Paolo Guiotto: Baker.
13:53:300Paolo Guiotto: as function ph, the indicator of set E, as function C, the indicator of set F,
14:00:690Paolo Guiotto: You can see, for example, that these two functions are clearly in any one with respect to the law of x, because they are bounded, they are indicators, so they take very single one, and this is the probability measure. Bounded functions are always integral with respect to probability measures.
14:21:710Paolo Guiotto: What is, let's say, less evident is that, in fact, one star implies two stars. So, let's say that one star, which is a particular case of two stars, can be extended to two stars.
14:36:30Paolo Guiotto: And this is done with the standard procedure we have already seen, in fact. So, the idea is that we have that one star is the two star with the indicators.
14:47:570Paolo Guiotto: So, by linearity, we can first extend this one-star poly linear combination of indicators, which are simple functions. So, step one…
15:01:950Paolo Guiotto: take C equal sum CJ indicator set J, and C equal sum DI indicator set F.
15:16:370Paolo Guiotto: Aye.
15:17:760Paolo Guiotto: So simple… functions.
15:24:240Paolo Guiotto: So, if you compute the expected value of phi of capital X times C of capital Y,
15:32:970Paolo Guiotto: Now, this is the sum of…
15:36:850Paolo Guiotto: It's a double sum, so sum over I and J, of C, CJD.
15:44:20Paolo Guiotto: I, indicator EJ of X, indicator FI of Y.
15:52:720Paolo Guiotto: Now, by linearity, you carry outside the sum, you carry outside the constant, so you have a sum.
15:59:650Paolo Guiotto: I, J… DICJ, and then you have the expectation of indicator EJ of X times indicator Fi of Y.
16:13:360Paolo Guiotto: But here is where you'll use the assumption. By the assumption, this splits into the product of the two expectations.
16:22:180Paolo Guiotto: EJ Exon.
16:25:180Paolo Guiotto: expectation indicator, FI.
16:29:770Paolo Guiotto: Why?
16:32:430Paolo Guiotto: And now you carry back into the expectations the sums. This is a double sum, it's a sum of an I, a sum of a J.
16:41:660Paolo Guiotto: So we can carry the sum of a J and the coefficient CJ into this first expectation here. So we get expected value of sum of a J, CJ indicator EJ,
16:56:770Paolo Guiotto: of capital X, and we do the same with the other expectation sum of I, DI indicator, FI of Y.
17:08:280Paolo Guiotto: And these are, our… C of capital X, and C of capital Y.
17:16:240Paolo Guiotto: So we can say that,
17:18:530Paolo Guiotto: If it holds for the indicators, it holds automatically for simple functions.
17:25:119Paolo Guiotto: Now, step two… We extend this to… positive,
17:34:490Paolo Guiotto: functions, phi and C, positive measurable functions, phi and C. So, let the phi, be positive.
17:43:160Paolo Guiotto: And the same for obscure.
17:46:820Paolo Guiotto: Now, we know that this is a general fact we proved in the first part of
17:51:950Paolo Guiotto: Because there exists a sequence, say, SN for the function phi, and another sequence SN for the function phi, so two sequences, I will denote with these symbols, of simple
18:12:470Paolo Guiotto: and, positive.
18:16:350Paolo Guiotto: functions.
18:20:280Paolo Guiotto: such that they converge pointwise X by X for every x, so SN phi of little x converges to phi of X for every X.
18:34:270Paolo Guiotto: And the same for the other one. S and C of X converges to C of X for every X. And moreover, this sequence is built in a way that it is an increasing sequence.
18:47:660Paolo Guiotto: And, SN plus 1 phi is larger than SN phi.
18:56:110Paolo Guiotto: And the same for the… other sequence, SN plus 1.
19:01:180Paolo Guiotto: C is larger fork width of SMC.
19:06:470Paolo Guiotto: Okay, so for DSM, the conclusion holds… by step one…
19:16:130Paolo Guiotto: We can say that the expected value of SN phi, of capital X, SNC of capital Y,
19:25:840Paolo Guiotto: It is equal to the product of the expectations of S and phi of X,
19:32:730Paolo Guiotto: times expected value for SNC of Y.
19:41:420Paolo Guiotto: And now we pass to the limit.
19:43:630Paolo Guiotto: We can do that because when we increase N, since both S and phi and S and C are increasing, this object inside increases with N.
19:59:360Paolo Guiotto: with N, and also this one, and this one.
20:03:800Paolo Guiotto: And moreover, they pointwise converge to CMC. So, by monotone convergence, Here, we're applying monotone.
20:14:950Paolo Guiotto: convergence. We will go to, at the left, to the expected value of phiXCX,
20:23:430Paolo Guiotto: TY, sorry. At right, we will go, respectively, to expected value of phi X,
20:30:650Paolo Guiotto: and here to the expected value of CY.
20:35:500Paolo Guiotto: And putting these together, this identity together, because of course they must have the same limit, we have the conclusion.
20:44:590Paolo Guiotto: So now we know that… so the conclusion is that true star holds… for… phi and C positive.
21:00:520Paolo Guiotto: And now we… we remove this restriction of positivity, step 3.
21:09:570Paolo Guiotto: It is the general case.
21:14:00Paolo Guiotto: of, phi of X,
21:16:690Paolo Guiotto: and the C of Y, generically in L1.
21:21:570Paolo Guiotto: Amiga.
21:23:780Paolo Guiotto: So what we do, we split each of them into the positive and negative part, so we write phi of X equal the positive part of phi evaluated at X minus the negative part of phi evaluated at X.
21:42:520Paolo Guiotto: And the same authority of Y.
21:47:800Paolo Guiotto: Then, for brevity, I will just write phi plus phi minus… I will not write the argument X. So, when you do the expected value of phi of X, let's do once C,
22:04:540Paolo Guiotto: Why?
22:06:300Paolo Guiotto: You are doing the expectation of phi plus minus, phi minus times C plus, minus mines.
22:18:530Paolo Guiotto: Then you do the product, it is phi plus, C++.
22:23:580Paolo Guiotto: Then I take also plus Ph, C-.
22:30:550Paolo Guiotto: This thing is one term, huh?
22:33:670Paolo Guiotto: And then we have, a minus…
22:37:400Paolo Guiotto: Now, that object, which is, phi plus… times C minus… plus phi minus… times C plus.
22:49:470Paolo Guiotto: And this is another object.
22:52:890Paolo Guiotto: Now, by linearity, the expectation will split into expected value of C++, C++
23:02:480Paolo Guiotto: plus expected value of C-, C-, and so on.
23:08:00Paolo Guiotto: minus the expected value of phi plus C minus.
23:13:850Paolo Guiotto: minus the expected value of phi minus C plus.
23:20:540Paolo Guiotto: But here, you're reminded that phi plus, phi minus, C+, C minus are all positive. Reminded also the negative part is positive, okay? So these are positive functions, so this is positive, this is positive.
23:38:740Paolo Guiotto: And so on. So this… to these expectations, we can apply the, the…
23:44:990Paolo Guiotto: Pack that the expectation of the product is the product of the expectations, so we get this.
23:53:500Paolo Guiotto: Q plus expectation of C minus times expectation of C minus.
24:02:570Paolo Guiotto: and so on, minus, etc. And now we have to recombine things. So, for example, maybe it's better to write the next term here, which is the expected value of phi plus.
24:15:710Paolo Guiotto: times the expected value of C-, and then there is another one.
24:21:10Paolo Guiotto: minus the last one, which is expected value of phi minus times expected value of C+.
24:29:830Paolo Guiotto: Now, we recombine, for example, take this one, And this one.
24:37:690Paolo Guiotto: and put together these two. So we'll factorize the expectation of phi plus.
24:46:880Paolo Guiotto: and it multiplies expectation of C plus minus the expectation of C-.
24:56:640Paolo Guiotto: But this one… there is another term I don't write because it's the same music. This one is also… you can recombine as a unique expectation of C plus minus C minus.
25:12:800Paolo Guiotto: And the positive part minus the negative part is the function, so this is pig. So this is the expectation of
25:21:890Paolo Guiotto: Sorry, proceed.
25:24:790Paolo Guiotto: And the same here in the other two terms. As you can see, the common factor is the expected value of phi minus. So, we will get something like plus expected… well, maybe it's better if I factorize a minus.
25:40:640Paolo Guiotto: Because if I factorize a minus, I have expected value of phi minus. As you can see, it is multiplied by the expected value of C+.
25:51:220Paolo Guiotto: And here, since I have plus, I'll put minus, minus the expected value of C minus.
25:57:590Paolo Guiotto: And this is, again, the expected value of C.
26:01:840Paolo Guiotto: So now you factorize the expected value of P, and you get expected value of phi plus.
26:09:300Paolo Guiotto: minus expected value of C minus.
26:14:100Paolo Guiotto: This times the expected value of C finally is the expected value of Phi times expected value of C.
26:25:700Paolo Guiotto: Which finishes the booth.
26:28:720Paolo Guiotto: Okay, so this is a very important factor that we will frequently use. So, in particular.
26:37:920Paolo Guiotto: So, this is important to keep in mind. Remark, a special case of this.
26:44:680Paolo Guiotto: So, in particular, if, X and the… Why? Independent.
26:59:580Paolo Guiotto: expected value of X times… Whee?
27:04:840Paolo Guiotto: Which is a quantity used to compute the covariance.
27:09:790Paolo Guiotto: is exactly the product of the expected values, e of x times e of y.
27:16:60Paolo Guiotto: This is the previous formula, where you see phi of X is just
27:21:970Paolo Guiotto: capital X and C of Y is just Y.
27:27:940Paolo Guiotto: It's exactly a special case of the previous formula.
27:39:190Paolo Guiotto: Yes, what I want to emphasize here, this is something which is a little bit curious, if you want, because what we… if we interpret this in the language of integrals under the probability P, this is saying that the integral on omega of X of omega
27:58:860Paolo Guiotto: Why offer me?
28:02:610Paolo Guiotto: DP?
28:03:860Paolo Guiotto: in the valuable omega. This is the product of the integrals, X of omega, DP omega.
28:15:30Paolo Guiotto: Oh, wow.
28:16:850Paolo Guiotto: Got it.
28:18:780Paolo Guiotto: EP.
28:20:00Paolo Guiotto: Omega.
28:21:820Paolo Guiotto: If you think in terms of ordinary function, no, think of the classical Riemann integral, you do the integral of the product, it's almost never the product of integrals.
28:32:380Paolo Guiotto: So, you see, there is something which is not usual with ordinary integrals. Concept is a proper concept.
28:43:430Paolo Guiotto: And again, in particular, so… If you computed the covariance between X and Y, to know, which is…
28:58:300Paolo Guiotto: This is the expected value of… Sometimes Y minus its expectation.
29:07:360Paolo Guiotto: So this is, by definition, what the covariance is.
29:12:210Paolo Guiotto: Now…
29:13:530Paolo Guiotto: that there is an equivalent representation of this, which is the expected value of X times Y minus the product of the expectations, X times expectation Y.
29:30:30Paolo Guiotto: I don't know if we have seen this formula, but this is just an algebraic derivation from this, because if you do the algebraic product in the expectation, you have expected value of X times Y minus its expectation.
29:49:810Paolo Guiotto: Then minus expectation of X,
29:54:510Paolo Guiotto: times Y minus expectation of Y, I'm doing just the algebra, into the expected value.
30:03:760Paolo Guiotto: Now, you notice that if you do this, this is X times Y minus X times the expected value of Y.
30:14:70Paolo Guiotto: When you do the expectation of this thing, you get exactly, by linearity, expectation of X, Y, minus…
30:24:710Paolo Guiotto: Since the expectation of Y is a constant, you can carry outside of this expectation, you get expectation of X times expectation of Y.
30:36:970Paolo Guiotto: While on the other factor, for the same reason, you carry outside this constant of the expectation, so you continue here, you have minus E of X,
30:48:540Paolo Guiotto: times the expectation of Y minus its expectation.
30:56:340Paolo Guiotto: But this one is zero.
30:58:880Paolo Guiotto: Because it is expectation of Y minus the expectation of expectation of Y, but expectation of Y is a constant, its expectation is the constant itself.
31:08:780Paolo Guiotto: And therefore, we get this. So, if the variables are independent, since the expectation of the product splits into the product of the expectations, we get that this is zero. So the covariance of two independent random variables is equal to zero.
31:29:900Paolo Guiotto: Which is something that… Important to be, you know… Okay.
31:38:690Paolo Guiotto: Now, let's come to the,
31:42:980Paolo Guiotto: I said, we want to understand what does it mean to be independent in terms of the important quantities associated to two random variables. So, in particular, to the CDF, the density, to the characteristic function.
32:00:930Paolo Guiotto: I skip a beat… The case of, the law, because, while there is some…
32:12:120Paolo Guiotto: that we miss about the measure theory, so… and in any case, it wouldn't be so important here. So let's start directly on this CDF. So let's talk about independence.
32:34:650Paolo Guiotto: And, CDF.
32:40:470Paolo Guiotto: Now, this idea that when you have independence, the joint thing
32:45:490Paolo Guiotto: The product is also, present, on the, on the CDF. We have this proposition.
32:59:170Paolo Guiotto: So, the following…
33:05:270Paolo Guiotto: statements.
33:10:510Paolo Guiotto: equivalent.
33:14:360Paolo Guiotto: So, number 1, X and Y independent.
33:21:560Paolo Guiotto: Well, number two… The joint CDF, so the CDF of the vector made by these two components, XY,
33:34:450Paolo Guiotto: Splits into the product of the CDF, precisely FXY at point little x y is equal to FX of little X times FY of little y for every XY real.
33:54:330Paolo Guiotto: Number 3… In the case that the, the pair, XY,
34:03:710Paolo Guiotto: So, is a p variet, random variable, is absolutely continuous.
34:11:90Paolo Guiotto: So this means, to exist A density for… The vector XY.
34:19:850Paolo Guiotto: Then… They are independent if and only if the joint density
34:29:130Paolo Guiotto: Splits into the product of the single densities.
34:36:239Paolo Guiotto: Well, this is not necessarily for every XY, but for almost every XY,
34:44:300Paolo Guiotto: in R2. Now, here, there is a little warning to do. That almost every means respect to what? Because here we have so many measures that it's not clear with respect to which measure we are talking about. Well, that means respect to the Lebeg…
35:02:440Paolo Guiotto: measure.
35:05:500Paolo Guiotto: It's important because for certain… for the rebell measure, for example, single points, they have measure zero. But for other measures, single points may have positive measure. Think to a delta, all the measure is concentrated in one single point, okay?
35:24:990Paolo Guiotto: Okay, so let's see this, the little proof of this.
35:32:580Paolo Guiotto: Basically, I'm doing these proofs because they help to get familiar with the definition of independence with these objects, the CDF, densities, and so on.
35:46:50Paolo Guiotto: Not a specific, and this one, it's a bit interesting.
35:52:920Paolo Guiotto: Let's start showing that suevalent. This is the idea, because we can always talk about the joint CDF, but we…
36:05:270Paolo Guiotto: We are not… so the third one is a special case. Okay, so, from 1 to 2 is easy, because…
36:14:390Paolo Guiotto: What is the joint CDF at point XY?
36:19:430Paolo Guiotto: By definition, this is the probability that capital X is less or equal than little x, and capital Y is less or equal than little y. So here, the hypothesis is XY are independent.
36:36:890Paolo Guiotto: Because 1 means XY are independent.
36:40:300Paolo Guiotto: Now, since they are independent, this is… you see a special probability X belongs to something, Y belongs to something else. This means X is in the set minus infinity to little x, and Y is in the set from minus infinity to little y.
36:58:810Paolo Guiotto: So it is, the probability of the intersection of these two events.
37:05:280Paolo Guiotto: And since the two variables are independent, this probability splits into the product of the two probabilities. So, the probability… well, let's just write X is less or equal than little x times the probability that Y is…
37:20:690Paolo Guiotto: would equal, then, little y. But these are, just by definition, the CDF of X at little x, and the CDF of Y at little y. And this happens for every X, Y,
37:35:440Paolo Guiotto: In Av2.
37:42:890Paolo Guiotto: A little bit more complicated is to prove that from 2 it follows 1.
37:53:370Paolo Guiotto: let's say that 2 that you have here, basically, so 2 is… assumption now is that the joint CDF
38:05:590Paolo Guiotto: splits into the product of the two single… of the two marginal CBR.
38:13:850Paolo Guiotto: So this is for every XY.
38:20:700Paolo Guiotto: So basically, it is, like if, we have that, this identity.
38:28:410Paolo Guiotto: This is equal to this, huh?
38:31:220Paolo Guiotto: But this means that this probability is equal to this probability. So it's basically a very special case of the independence of X and Y. For the independence of the two variables, I would need to prove
38:50:390Paolo Guiotto: Either that, for example, the expected value of a product phi is the product of the expectations. But here, I don't know how to do, because we have not introduced
39:06:270Paolo Guiotto: how the law of X, how it works, the Law of X, in terms of the CVF.
39:14:400Paolo Guiotto: of the single measure, so… or I should go back down to the definition that says that to prove that probability that X belongs to E and Y belongs to F, E is the product of the probabilities for E and F borel set.
39:31:650Paolo Guiotto: Basically, for very particular boreal sets.
39:34:840Paolo Guiotto: the intervals from minus infinity to a point, okay? You see that?
39:40:780Paolo Guiotto: So, the, the thesis is to prove,
39:47:790Paolo Guiotto: So let's say that thesis is, X, Y… Independent.
39:56:630Paolo Guiotto: That is the probability that X belongs to E, and Y belongs to F.
40:05:540Paolo Guiotto: is the product of the probabilities that X belongs to E times probability that Y belongs to F for every E and F borel set.
40:20:650Paolo Guiotto: What we know… by the hypothesis.
40:26:220Paolo Guiotto: We know that the product of the CDF… the joint CDF is the product of the CDF. That means, in fact, that the probability that X belongs to the offline minus infinity to X
40:42:360Paolo Guiotto: and Y belongs to the offline minus infinity to Y,
40:48:40Paolo Guiotto: This probability splits into the product of… Pitties.
40:53:320Paolo Guiotto: X belongs to minus infinity to X.
40:57:360Paolo Guiotto: times probability that Y belongs
41:00:710Paolo Guiotto: minus infinity to Y. So, what we know is a particular case of what we have to prove.
41:12:400Paolo Guiotto: How do we get that?
41:14:700Paolo Guiotto: Well, let's say that, That if we prove that for any…
41:21:280Paolo Guiotto: of E, and for any interval in place of F, the conclusion holds, basically, we are saying that it's for every set that generates the sigma algebra, so probably it will be true for every set of the sigma algebra. This will be a technical passage.
41:37:800Paolo Guiotto: So, what we do is, let's prove this, let's call it one star, when E and F are intervals. We know that it is true for special
41:48:690Paolo Guiotto: Once, you see the intervals from minus infinity to X, interval minus infinity to Y. So for this particular E and F, this is true, the one star is true. Now we want to extend a little bit to every type of interval, okay?
42:06:570Paolo Guiotto: So… Let's… prove, huh?
42:13:810Paolo Guiotto: That one star holds when…
42:29:510Paolo Guiotto: Yeah, when E is… well, for the moment, I will do for this type of interval. An interval AB, open at left, close at right, and F is a similar interval from C to D.
42:45:740Paolo Guiotto: So what we have to do is we have to compute the probability that X belongs to the interval AB,
42:55:130Paolo Guiotto: And Y belongs to the interval CD.
42:59:570Paolo Guiotto: Now, of course, I want to formulate this quantity in terms of the CDFs. That's what I want to try to do. So now, this means that
43:14:20Paolo Guiotto: If you want, it is the probability that X is greater than A, less or equal than B, and Y is greater than
43:25:550Paolo Guiotto: intellectually what… D.
43:28:530Paolo Guiotto: Now, the idea is I want to represent this as… somehow, with a set where this is a set where X is less or equal than X, Y less or equal than Y.
43:41:320Paolo Guiotto: So, it's something where, like, A and C are minus infinity. I want to do with A and C find it. So, what could I say here?
43:51:590Paolo Guiotto: Now, it's maybe convenient to do this kind of figure in Cartesian plane.
43:58:930Paolo Guiotto: So it means that the first coordinate is between A and B, and the second coordinate is between C and D. So the pair XY, the pair XY belongs to what? To a product, interval AB,
44:18:80Paolo Guiotto: times interval CD, which is a rectangle.
44:22:700Paolo Guiotto: And the particular rectangle, because you have… since A is not included, it means that we are not considering points with equal A, and since C is not included, we are not considering points where the ordinary is C.
44:38:720Paolo Guiotto: while we are considering points where abshesa is B, and points where ordinary is D. So we are taking, say.
44:46:830Paolo Guiotto: This particular rectangle without, say, half of the edge.
44:55:330Paolo Guiotto: Okay, now I want to transform this into a calculation of the… with the CDI.
45:01:10Paolo Guiotto: Now, what is the CDF XY at point, for example, BD?
45:09:130Paolo Guiotto: This is the probability that X is less or equal than B, and Y is less or equal than D.
45:21:290Paolo Guiotto: So, it means that XY belongs to a much larger region. Let's do with a second color.
45:29:450Paolo Guiotto: It is in blue here.
45:34:320Paolo Guiotto: Which is this infinite rectangle.
45:38:900Paolo Guiotto: Yeah.
45:41:270Paolo Guiotto: So now, the idea is that this FXYBD gives the probability that XY belongs to the blue region. I want the probability that XY belongs into the black region. So what I can do is, if you are in the blue region, sorry, if you are in the black region.
46:01:480Paolo Guiotto: you can think you are in the blue region, but you are not in these three… in these other sub… two sub-regions. So this one…
46:13:270Paolo Guiotto: Now, this is… this is the sub-region, the red one, is X less or equal than B, and Y less or equal than C, the red one.
46:30:900Paolo Guiotto: And let's do a second one, which is the green one, this one.
46:35:810Paolo Guiotto: Sorry, green one.
46:40:950Paolo Guiotto: This green one is the region where X is less or equal A,
46:47:790Paolo Guiotto: And Y is less or equal than D.
46:58:870Paolo Guiotto: But, okay, so I could say…
47:01:790Paolo Guiotto: So, you are in the black rectangle if you are in the blue rectangle, but not in the red union with green.
47:11:570Paolo Guiotto: You see? So I could say that the region where X is above A less or equal B,
47:20:420Paolo Guiotto: Sorry, END.
47:23:940Paolo Guiotto: Why?
47:26:770Paolo Guiotto: is above C, both, or equal D,
47:34:860Paolo Guiotto: Is equal to the red region, which is,
47:41:280Paolo Guiotto: The red region, which is X, less or equal B, and Y less or equal D.
47:50:940Paolo Guiotto: But too much,
47:55:120Paolo Guiotto: I have to eliminate B, well, no, sorry, that was in blue. That was… the colors was in blue, so X less or equal B and equals D.
48:15:80Paolo Guiotto: minus… Now, the red one…
48:20:30Paolo Guiotto: minus, let's say, the union of the red one, which is X less or equal B and Y less or equal C, X less or equal B, Y less or equal C,
48:34:690Paolo Guiotto: union with the green one. The green one is X less or equal A.
48:43:750Paolo Guiotto: And why lesser equality?
48:50:420Paolo Guiotto: And since both the red and the green are contained into the blue, I could say that their union is contained into the blue, so in particular, when I take the probability of all this, so probability of this, which is the quantity I want to compute, will be the probability of this.
49:08:550Paolo Guiotto: Now, since this part, this is contained in this…
49:17:710Paolo Guiotto: As you can see from the figure.
49:20:30Paolo Guiotto: So I can say that the probability of the difference applies, so I have probability of X less or equal B.
49:28:440Paolo Guiotto: Y less or equal D,
49:31:700Paolo Guiotto: Minus the probability for the moment of that union, so the union of X less or equal B, Y less or equal C,
49:42:530Paolo Guiotto: union with X less or equal AY less or equal D.
49:50:840Paolo Guiotto: Now, the point is that this union, as you can see from the figure, is not disjoint.
49:57:200Paolo Guiotto: Because there is a part, apart here… Which belongs to the two.
50:04:860Paolo Guiotto: So I cannot say that the probability of the union is the sum of the probabilities, because this would be wrong.
50:11:500Paolo Guiotto: In fact, I could say that it is the sum, provided, since I count two times the yellow region, I subtract the value of that probability. So it is… this is…
50:25:610Paolo Guiotto: So the probability that X is less or equal B.
50:30:90Paolo Guiotto: Y is less or equal d minus…
50:34:290Paolo Guiotto: Yeah, the probability of the sum of the two probabilities, so X less or equal B, Y less or equal C,
50:42:200Paolo Guiotto: plus probability that X is less or equal A, and Y is less or equal D. But in this way, I'm counting two times the yellow.
50:54:340Paolo Guiotto: probability, which is the probability that X is less than A and Y is less than C. So, I have to subtract the probability that X is less or equal A,
51:08:210Paolo Guiotto: And Y is less equal C. Now, I am with an exact formula.
51:13:130Paolo Guiotto: Now, the point is that all these are joint CDF, because this one…
51:20:410Paolo Guiotto: is the joint CDF at point BD.
51:26:760Paolo Guiotto: This one… is the joint CDF at point BC.
51:33:820Paolo Guiotto: And so on. This is the joint CDF at point AB, and this one, finally, is the joint CDF.
51:43:500Paolo Guiotto: at point AC. So we have that our probability, we are computing the probability that X belongs to an interval, Y belongs to another interval.
51:56:20Paolo Guiotto: ease.
51:57:370Paolo Guiotto: So let's copy down here the conclusion. So, the probability
52:04:840Paolo Guiotto: that X belongs to the interval A to B.
52:09:770Paolo Guiotto: Y belongs to the interval C to DE, is equal to…
52:17:10Paolo Guiotto: the joint CDF at BD,
52:22:550Paolo Guiotto: minus this sum, the joint CDF, at BC.
52:30:940Paolo Guiotto: plus the joint CBF, at AD, Minus the joints, yeah, F… at AC.
52:45:40Paolo Guiotto: Okay, now let's use the hypothesis. The hypothesis says that the joint CDF splits into the product of the two CDFs. So, this is the hypothesis. This is FX of B times FY of B,
53:00:730Paolo Guiotto: Minus. This is FX of B.
53:05:390Paolo Guiotto: times FY of C plus FX of A.
53:12:260Paolo Guiotto: times F, Y.
53:14:970Paolo Guiotto: D?
53:17:50Paolo Guiotto: minus FXA times FY.
53:22:610Paolo Guiotto: C.
53:24:490Paolo Guiotto: And now let's rework this thing. Let's put together these two.
53:30:40Paolo Guiotto: This is… you see that there is a common factor, which is FXB, that multiplies FY D…
53:42:270Paolo Guiotto: minus F, Y, C.
53:46:780Paolo Guiotto: And here, I prefer to factorize D minus, so minus F, the common factor is this one, FXA, that you have in both terms, so FXA times FY
54:05:850Paolo Guiotto: D… minus F… Why?
54:10:780Paolo Guiotto: D, this is a D. FYC.
54:15:630Paolo Guiotto: What is now this thing?
54:18:720Paolo Guiotto: This is the CDF of Y, so this is the probability that Y is less or equal than D minus the probability that Y is less or equal than C.
54:32:630Paolo Guiotto: I'll remind that to… D is about here, C is about here, is less than D.
54:39:550Paolo Guiotto: So the set where Y is less or equal than C is contained
54:46:520Paolo Guiotto: is included into this one. So you can transform this difference between the probabilities into the probability of the difference. So this is the probability that Y is less or equal than V minus Y less or equal than C.
55:02:840Paolo Guiotto: So, Y must be less or equal than D, but not less or equal than C, so it must be larger than C.
55:11:30Paolo Guiotto: And so this is the probability that Y belongs to CD. So, it's the probability that Y belongs to the interval CD.
55:21:830Paolo Guiotto: So this difference here is the probability that Y belongs to CD.
55:31:320Paolo Guiotto: And since this quantity here is the same, also this one is the probability that Y belongs to CDF.
55:40:20Paolo Guiotto: It's the same number, we can factorize, so we get the probability that Y is in the interval.
55:50:300Paolo Guiotto: CD…
55:54:720Paolo Guiotto: and times what remains here is FXB minus FXA.
56:03:230Paolo Guiotto: Which is, for the same reason, the probability that X belongs to the interval from A to B.
56:13:620Paolo Guiotto: And so now we have the conclusion.
56:16:560Paolo Guiotto: So, we proved them.
56:22:640Paolo Guiotto: that, the probability.
56:27:430Paolo Guiotto: that.
56:28:470Paolo Guiotto: X is in the interval, AB.
56:32:700Paolo Guiotto: And Y is in the interval.
56:35:410Paolo Guiotto: CD… splits into the product of the probabilities where X is in AB,
56:45:170Paolo Guiotto: times the probability that Y is in CD.
56:53:570Paolo Guiotto: So, with the few steps, we can extend this to any kind of intervals easily.
57:04:880Paolo Guiotto: This… expense… 2…
57:12:140Paolo Guiotto: the probability that X belongs to an interval I, and Y belongs to an interval J is the same of the product of probabilities that X belongs to I times the probability that Y belongs to J. This for every iJ
57:32:960Paolo Guiotto: intervals.
57:37:900Paolo Guiotto: And from this,
57:40:20Paolo Guiotto: it's not trivial, but it requires some work. The idea is that if something holds for the generators, more or less, of a sigma algebra, it holds for the sigma algebra. So, and from this.
58:02:420Paolo Guiotto: the probability that X belongs to E.
58:05:740Paolo Guiotto: and Y belongs to F, huh?
58:09:00Paolo Guiotto: splits into the product of the probability that X belongs to E, times probability that Y is in F.
58:17:970Paolo Guiotto: Now, for every E and F, Borel sets, so for every element of the Sigma algebra, and this means that X, Y are independent.
58:35:100Paolo Guiotto: Okay? So now, we know that independence can be seen as the CDF, the joint CDF, splits into the product of the CDFs, okay?
58:48:840Paolo Guiotto: Now, let's see the… the… the third property.
58:54:920Paolo Guiotto: which is what happens to independence in the case the variable is absolutely continuous, so when there is a density. And this, again, says that the density splits into the product of the densities.
59:12:90Paolo Guiotto: So, in other words, this is used in, say, two ways.
59:17:280Paolo Guiotto: Number one… How can I check if an absolutely continuous
59:27:200Paolo Guiotto: random variable is, solving. If two random variables are independent, if I know the joint distribution.
59:38:10Paolo Guiotto: And I see that the joint distribution splits perfectly in a function of X times a function of Y. This means that they are independent.
59:47:600Paolo Guiotto: So this is a concrete way, we will see in exercises, a concrete way to test the independence, because otherwise it remains a property. You know that they are independent, yes, but how do I check they are independent? If you have the joint density, and you see that this joint density can be split into a product, algebraic product of a function of
00:12:330Paolo Guiotto: The first variable times a function of the second variable, that means that they are independent.
00:18:180Paolo Guiotto: And in fact, it's a characterization.
00:20:690Paolo Guiotto: Well, if XY Easter.
00:29:660Paolo Guiotto: absolutely continuous.
00:33:590Paolo Guiotto: So, we know that, we know that, F, the joint density.
00:42:730Paolo Guiotto: Well, in principle, this would mean that the, the joint law, of a set, say, E,
00:54:700Paolo Guiotto: Hmm… Well, it is better if I bravely write this set,
01:03:70Paolo Guiotto: Well, let's take this. E is the integral on E of the joint density XY…
01:11:420Paolo Guiotto: EX DY for every E in the Borel class, in this case of R2, so for every Borel set.
01:23:500Paolo Guiotto: In particular.
01:28:730Paolo Guiotto: So, you have that. If you take the joint CDF, evaluated at point XY, this is what? This is the probability that X is less or equal than capital XY less or equal than capital Y, which is the same of saying that, or if you want, probability that the pair XY
01:50:800Paolo Guiotto: belongs to the set minus infinity X for the first component.
01:55:480Paolo Guiotto: Minus infinity Y for the second component.
01:59:280Paolo Guiotto: But that's… the probability that the pair XY belongs to some set is the low of… the joint law of that set.
02:11:570Paolo Guiotto: So…
02:16:240Paolo Guiotto: Which is, The integral on that set, minus infinity to X, times minus infinity toy.
02:27:830Paolo Guiotto: of the joint density, so let's call it FXY, the variables, let's call them UVE, DUDV, because the,
02:38:660Paolo Guiotto: the variable is supposed to be absolutely continuous, so the law writes in that way. So this splits into, by Fubini.
02:48:700Paolo Guiotto: theorem, this splits into the double integration from minus infinity to X minus infinity to Y of FXY UV
03:02:410Paolo Guiotto: Well, let's say that this is first in V, and then in U.
03:07:770Paolo Guiotto: This is the joint density.
03:15:980Paolo Guiotto: Okay, now, the goal is to prove that the joint density is the product of the two densities.
03:23:440Paolo Guiotto: Mmm… Well, on the other side, since we know that they are independent, so by independence.
03:35:670Paolo Guiotto: This is also the product of FXXFY.
03:40:950Paolo Guiotto: little y, which are respectively the integral from minus infinity to X of the density of
03:51:660Paolo Guiotto: X times integral minus infinity 2Y of the density.
03:58:380Paolo Guiotto: YD1.
03:59:970Paolo Guiotto: So now, these two are the same, and now we want to draw a conclusion about the identity between the two. Now, this one can be written also as the integral from minus infinity to X, integral minus infinity toy, of the product FXUFY
04:19:540Paolo Guiotto: U.
04:24:240Paolo Guiotto: Brilliant.
04:25:590Paolo Guiotto: I did the math with letters. This is a V.
04:29:990Paolo Guiotto: So this is V, DVTU.
04:41:60Paolo Guiotto: And now, now, of course.
04:46:770Paolo Guiotto: If you… from this, you can…
04:51:150Paolo Guiotto: Well, let's say this… let's for a second,
04:55:540Paolo Guiotto: eliminate this step here. So the joint density when XY is absolutely continuous is this one, while the product of the two CDF
05:07:530Paolo Guiotto: is this one.
05:10:340Paolo Guiotto: So, here I can say, so… IFA… I have that FXY
05:19:190Paolo Guiotto: X times Y is equal to the product FX of X, FY. Y, at least for almost every XY,
05:30:970Paolo Guiotto: So, you see that the two functions that you have here and here are the same almost everywhere. Their integral will be the same.
05:41:650Paolo Guiotto: So, I will conclude from this that the two integrals are… the first one is the joint CDF at point XY, the second one is the product of the two CDFs.
05:57:430Paolo Guiotto: This holds for every XY.
06:01:90Paolo Guiotto: And so this means that XY are independent.
06:10:40Paolo Guiotto: So, if this formula holds, if the joint density is the product of the marginals, we have that the two variables are independent. Vice versa.
06:27:820Paolo Guiotto: If XY are independent.
06:34:960Paolo Guiotto: then, this one, happens, so F, X, Y,
06:44:230Paolo Guiotto: the joint CDF is equal to the product of the CDFs.
06:53:240Paolo Guiotto: Now, since this answer for every XY, And the two are,
07:04:420Paolo Guiotto: We have that this is the double integral minus infinity to X minus infinity toy of the joint density. This is the double integral minus infinity to X minus infinity to Y of the product of the two densities.
07:20:230Paolo Guiotto: by computing the derivative, so since they are the same, so doing the derivative with respect to X, I will eliminate the integral in X, and then doing the derivative with respect to Y of the CDF,
07:36:470Paolo Guiotto: So, doing this derivative yields by the,
07:41:420Paolo Guiotto: The fundamental theorem of integral calculus, the joint density at point XY.
07:46:830Paolo Guiotto: while, when I do the derivative with respect to X of the derivative with respect to Y of this product.
07:56:50Paolo Guiotto: Again, I get FX of X, FY of Y.
08:02:660Paolo Guiotto: And so, since these two are the same, these two must be also the same, and this means that the density splits into the product of the two densities.
08:16:330Paolo Guiotto: Okay, so let's see, it's the moment we can see an exercise, and also to leave,
08:25:229Paolo Guiotto: some exercises to do. So, do exercises… 6… 4… 1… 2…
08:54:220Paolo Guiotto: So… 1, 2, 3…
09:03:450Paolo Guiotto: Maybe I, I will, I will do the number 4 now. 64.
09:09:779Paolo Guiotto: 4.
09:11:399Paolo Guiotto: So here it says we have two variables, X, Y,
09:15:569Paolo Guiotto: Which are both exponential with the parameter lambda.
09:21:130Paolo Guiotto: So I remind you that they are absolutely continuous with densities at
09:27:430Paolo Guiotto: Well, let's use the same formula, so a neutral letter.
09:33:460Paolo Guiotto: both FX and the FY, they have this density, which is lambda E minus lambda U indicator, 0 plus infinity.
09:46:680Paolo Guiotto: U.
09:51:920Paolo Guiotto: Okay, it says that.
09:58:50Paolo Guiotto: If… That's why.
10:01:50Paolo Guiotto: independent.
10:04:570Paolo Guiotto: Then, also, this pair, X plus Y, O value is two variables.
10:14:820Paolo Guiotto: independent.
10:22:540Paolo Guiotto: So, we have to check if these new variables, X plus Y and X over Y, are independent. So, let's call them Z equal X plus Y.
10:35:00Paolo Guiotto: and W equal X over Y. Now.
10:40:740Paolo Guiotto: since in this case, there is something which could be dangerous, dividing by Y, well, let's check. This is a general factor. Since Y is supposed to be absolutely continuous.
11:00:820Paolo Guiotto: We know that the probability that Y is… belongs to a single term, so in particular is a single value, is equal to zero, because this would be the probability that
11:13:450Paolo Guiotto: Y belongs to the singleton equals zero, and since we have a density, this is the integral on that set of FY.
11:24:60Paolo Guiotto: Y in DY, and that's equal to 0 because of the back match. So that's a general fact. So this means that, X over Y
11:34:460Paolo Guiotto: is wild.
11:37:460Paolo Guiotto: define the… Almost surely.
11:44:380Paolo Guiotto: Okay, now, to check if they are independent, what is the idea? I know that…
11:51:850Paolo Guiotto: I have the… the information is you have the single distribution of X and Y, and moreover, you have something else, which is this one. They are independent, so in practice, you have also the joint distribution of X and Y doing just the product of the two, okay?
12:10:700Paolo Guiotto: So, since now we have the joint distribution of XY, and this new pair is a pair which is obtained by doing a map of this XY,
12:21:390Paolo Guiotto: Let's do this. First, we computed the joint distribution of the new variables, and then we will see if the joint distribution of ZW is the product of two functions, one of Z and the other of W. So this is the
12:36:660Paolo Guiotto: the way we solve this, okay? So, we noticed that, since,
12:43:800Paolo Guiotto: So, in general, if you don't know anything, and you have only the single distributions of X and Y, you cannot draw anything about the joint distribution.
12:53:410Paolo Guiotto: We can say that in this case, the joint distribution is the product of the two, because we know they are independent. Since X and Y
13:02:00Paolo Guiotto: are independent.
13:06:270Paolo Guiotto: The joint distribution It's just the product of the two single distributions.
13:18:20Paolo Guiotto: So, in particular, this will be… the first one is lambda E minus lambda X indicator 0 plus infinity.
13:26:20Paolo Guiotto: X.
13:27:730Paolo Guiotto: The second one is lambda E minus… say, why?
13:32:520Paolo Guiotto: Indeed.
13:34:240Paolo Guiotto: 0 plus infinity of Y. So the joint distribution is, you may say, lambda squared P minus lambda X plus Y indicator. We can write in a uniform. It is, both coordinate must…
13:53:660Paolo Guiotto: Positive point, exploit.
13:57:320Paolo Guiotto: 0 plus infinity squared. So that's the joint distribution, okay?
14:02:710Paolo Guiotto: Now, D variables Z equal X plus Y W equal X over Y.
14:12:170Paolo Guiotto: can be seen as a map of X and Y.
14:21:270Paolo Guiotto: Well, the map is, of course, phi of XY equal X plus YX over Y.
14:31:540Paolo Guiotto: We want to derive from this the density of ZW.
14:40:320Paolo Guiotto: We have to understand if this map is invertible, and, with the derivative, with differentiable, because also we need the Jacobian of the inverse matrix of this.
14:55:820Paolo Guiotto: Now, maybe this map is not invertible as an map from R2 to R2. Here you can see, for example, if you look at the second component, if you change signs to both X and Y…
15:10:620Paolo Guiotto: So replace X with minus X. Why? With minus Y, the ratio is always X over Y. So this means that the point
15:19:30Paolo Guiotto: well, the first one would be different, so that would be the case. However, the second one could be dangerous. But this is not the case here, because if we look at where our variable XY is distributed, since we have this
15:38:740Paolo Guiotto: cutoff here that says that there is no probability for X and Y, at least one of the two negatives, so we can say we notice
15:48:290Paolo Guiotto: we notice… that, huh?
15:54:620Paolo Guiotto: the probability that X is positive, well, let's say strictly positive, even, so we eliminate the case Y equals 0.
16:05:630Paolo Guiotto: This would be the probability that we are integrating for X, let's say, little x, positive little y, positive our joint density, FXY,
16:20:350Paolo Guiotto: in BXDY.
16:25:440Paolo Guiotto: But that's exactly, when we look at that joint density, that's exactly the integral from 0 to plus infinity, 0 to plus infinity, lambda square E minus lambda X plus Y times the indicator is 1, dxty.
16:42:820Paolo Guiotto: And, easily, you have that this is equal to 1. So, our variable, XY,
16:49:730Paolo Guiotto: starts from not the interior Cartesian plane, but the point XY
16:56:50Paolo Guiotto: belongs to the first quarter, so we have to focus our map fee on the first quarter, because there is no XY out of there. So, let's say that we start from this domain, and we map through feed.
17:11:620Paolo Guiotto: somewhere else. Where?
17:14:140Paolo Guiotto: Well, if you look, since X and Y are both positive, X plus Y is positive, and X over Y is positive. So, necessarily, phi of E, the image, will be somewhere in this quarter.
17:30:280Paolo Guiotto: Okay, so this means that, so that in particular, the pair ZW belongs to 0 plus infinity.
17:40:810Paolo Guiotto: Square.
17:42:520Paolo Guiotto: So, now, Z equal X plus Y, so say ZW, let's see if this is invertible, ZW equal phi of XY means
17:57:330Paolo Guiotto: Z equal X plus Y, and W equal X over Y.
18:07:910Paolo Guiotto: So now, if we try to invert this, we have to express X and Y as functions of Z and W.
18:16:600Paolo Guiotto: So, we could do the following. For example, you multiply by Y in the second line, you get X equal YW. We plug this into the first line, we get Z equal YW plus Y.
18:35:360Paolo Guiotto: So, from this, we see that, Y…
18:39:280Paolo Guiotto: is equal to Z over W plus 1.
18:44:170Paolo Guiotto: And, X is,
18:47:950Paolo Guiotto: YW, so it means ZW divided W plus 1.
18:55:910Paolo Guiotto: So this is the, the inverse. We noticed that for,
19:04:520Paolo Guiotto: for XY positive, we get ZW positive, and vice versa. So this is the map P minus 1 of ZW, which is the map we needed to compute the density of ZW.
19:21:180Paolo Guiotto: So, the density of ZW is equal to the density of XY
19:30:220Paolo Guiotto: to be evaluated at phi minus 1ZW,
19:35:390Paolo Guiotto: times minus modulus of the determinant of the Jacobian matrix of the inverse in ZW.
19:46:60Paolo Guiotto: If you want, we can compute, or we can compute the Jacobian of the direct map, and then do the reciprocal. I noticed that the phi prime
19:57:40Paolo Guiotto: which is slightly easier, so the direct map. The phi prime as function of XY is the matrix
20:06:990Paolo Guiotto: So the direct map is X plus Y, X over Y. So the derivatives of X plus Y are 1, 1,
20:14:500Paolo Guiotto: The derivative of X over Y are 1 over Y, and minus X over Y squared.
20:22:340Paolo Guiotto: So when I do the determinant of this, metrics, I get…
20:29:410Paolo Guiotto: minus X over Y squared.
20:33:920Paolo Guiotto: minus 1 over Y.
20:37:180Paolo Guiotto: So, at the end…
20:44:760Paolo Guiotto: At the end, we have, 1 over Y.
20:51:290Paolo Guiotto: minus, one, plus X over Y.
20:58:340Paolo Guiotto: Right? So this is the determinant. So the modulus of the determinant of phi prime, this is the direct map, is one of… I should put the modulus, but since here X and Y are positive.
21:13:850Paolo Guiotto: Are in 0 plus infinity.
21:18:280Paolo Guiotto: Excluding 0, positive, I write just 1 over Y, 1 plus X over Y.
21:26:960Paolo Guiotto: So, the determinant of the inverse, the modulus of the determinant of phi minus 1
21:33:840Paolo Guiotto: now is function of ZW is 1 over the modulus of the determinant of phi prime, but phi prime is function of XY, so this will be evaluated minus 1 of ZW.
21:53:140Paolo Guiotto: So it is the reciprocal of that,
21:55:860Paolo Guiotto: which is 1 over, we can write with letters XY, then we will replace what are these quantities in terms of ZW. This guy is just W,
22:10:410Paolo Guiotto: And so we get Y over 1 plus W. What is Y? I can use this equation, Z over W plus 1.
22:22:580Paolo Guiotto: So this is Z over 1 plus W. So at the end, we get Z over 1 plus W squared.
22:31:650Paolo Guiotto: And this is the modulus of the determinant. If you do directly, maybe in this case the calculations were not so complicated, we could have done directly, and we would obtain this.
22:45:640Paolo Guiotto: So, the density of ZW is equal to…
22:53:70Paolo Guiotto: So we said we have to take the density of XY, which is, lambda square E minus lambda
23:01:960Paolo Guiotto: Then there is X plus Y. But keep in mind that X plus Y is… I don't need to do the calculation doing X plus Y here, because I know that X plus Y is Z.
23:12:720Paolo Guiotto: So this is, Z. Then we have the indicator, 0 plus infinity.
23:18:980Paolo Guiotto: square of, the two components, Z over W plus 1,
23:29:70Paolo Guiotto: which is DX, and ZW over W plus 1.
23:36:910Paolo Guiotto: And then we have the modulus of the determinant, which is Z over 1 plus W.
23:44:470Paolo Guiotto: square.
23:47:390Paolo Guiotto: Okay, now, what is this, indicator? Because, as you can see, there is a… this is starting to have a shape which is… which is a product of two things, okay? Because, you see, there is… this is,
24:02:690Paolo Guiotto: apart for factors, there is a function of Z, there is a function of Z, there is a function of W, we can split. Now, the point is, what is this indicator? Because, in principle, the indicator is an indicator of ZW, okay? It's not split, but…
24:19:450Paolo Guiotto: You notice that this indicator is equal to 1 if and only if Z over W plus 1 is greater or equal than 0, and at the same time, ZW, I prefer to write like that.
24:39:840Paolo Guiotto: 0 equals then 0, okay?
24:42:590Paolo Guiotto: Or if you want, we can even take it as strictly positive, because we know that this is irrelevant. Now, since this is positive, the second condition is possible if and only if W is greater or equal than 0.
24:57:180Paolo Guiotto: Okay, and now, if W is greater or equal than zero, you plug back into the first condition, this denominator is positive, so the first one is possible if and only if Z is greater or equal than 0. So at the end, I have that Z pair
25:13:860Paolo Guiotto: Z over W plus 1, and ZW over W plus 1.
25:19:30Paolo Guiotto: They are both positive if and only if Z and W are positive. So this means that this indicator is 1 if and only if the indicator of 0 plus infinity for Z times the indicator of 0 plus infinity for W,
25:37:60Paolo Guiotto: Are both equal to 1.
25:40:270Paolo Guiotto: And so you see that this splits. So finally, I have to, I have this. Lambda squared, E minus lambda Z,
25:50:110Paolo Guiotto: Let's put a Z also here, then I have the indicators, 0 plus infinity.
25:56:630Paolo Guiotto: Z. And then for W, I have 1 over 1 plus W squared times indicator 0 plus infinity.
26:08:70Paolo Guiotto: W. As you can see, I read the…
26:11:730Paolo Guiotto: Well, now, the point is where this constant goes, into one factor or into the other.
26:17:380Paolo Guiotto: Okay, it must go to the factor that makes a probability density, so probably it will go with the first factor, we will check in a second, but we see that here there is a function of Z, and here there is a function of W, and they are multiplied.
26:33:800Paolo Guiotto: So, this necessarily means that since this is the joint distribution, this must be the product of the two densities, this is the density of Z,
26:49:190Paolo Guiotto: And this will be the density of W.
26:52:970Paolo Guiotto: I repeat, maybe there could be some, something to decide about the constants. Let's see if this is the case.
27:01:760Paolo Guiotto: You have to check if they have probability densities. Let's take this one.
27:07:00Paolo Guiotto: The density, I need to verify it is positive, so it is the first requirement, and the second one, I have to verify that the integral from minus infinity to plus infinity is 1. So I have to compute this integral, 1 plus W squared indicator 0 plus infinity.
27:24:450Paolo Guiotto: W.
27:25:900Paolo Guiotto: in the W.
27:27:480Paolo Guiotto: So this becomes the integral from 0 plus infinity of 1 over 1 plus W squared DW.
27:37:190Paolo Guiotto: You see why, no? I can always put a constant here and one of a constant here, and the formula won't change, no? Maybe I have to rebalance to have the true densities, but it's just a matter of a multiplicative constant.
27:52:380Paolo Guiotto: Now, this is the derivative of minus 1 over 1 plus W,
27:58:220Paolo Guiotto: that we have to evaluate between W equals 0 and W equals plus infinity.
28:03:380Paolo Guiotto: At plus infinity, we get 0, at 0 we get minus 1, so the difference is plus 1. So, this means that the integral of that FW is 1,
28:13:330Paolo Guiotto: It is positive, so that's a probability density. I don't need any rebalance with constants. And this automatically will mean that also the other one will have integral equal to 1, because the product is a probability density.
28:27:150Paolo Guiotto: So, in particular, since the conclusion is that
28:36:730Paolo Guiotto: scenes…
28:39:260Paolo Guiotto: FZW, the joint density, is equal to the product of a function of Z times a function of W, it means that Z and W are
28:52:910Paolo Guiotto: Independent.
28:59:580Paolo Guiotto: And this finishes the program.
29:03:450Paolo Guiotto: And also finishes today's class.
29:07:80Paolo Guiotto: Also, I would suggest you to do… This problem, 646.
29:19:190Paolo Guiotto: and 7. So, let's say that next Monday, I will solve most of these problems I left to you, so please…
29:29:820Paolo Guiotto: Is it… No, Monday we do not have class.
29:34:60Paolo Guiotto: Okay.
29:35:130Paolo Guiotto: And we will be in the…
29:37:970Paolo Guiotto: forever in the innovative room at the Hub.
29:45:310Paolo Guiotto: Okay, let's stop recording.