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00:12:340Paolo Guiotto: Okay, so today we…
00:15:200Paolo Guiotto: We will see a few more facts about the conditional expectation, and do some exercises on this.
00:24:420Paolo Guiotto: So, we reminded that we introduced, We introduced the…
00:35:320Paolo Guiotto: the… conditional.
00:39:750Paolo Guiotto: It's the fishing.
00:47:410Paolo Guiotto: Which is denoted by these symbols. Expected
00:52:10Paolo Guiotto: Expected value of X given G.
00:55:750Paolo Guiotto: Where these are the ingredients.
01:00:200Paolo Guiotto: Where, X is a random variable.
01:05:190Paolo Guiotto: G, contained in F is…
01:11:670Paolo Guiotto: sigma algebra, and of course, there is an underlying probability space omega FB.
01:19:390Paolo Guiotto: Probability.
01:22:00Paolo Guiotto: space.
01:25:200Paolo Guiotto: We have seen that this operation is well-defined. We have seen first when X is an L2 random variable, and then we extend it to L1 belonging to L1 omega.
01:44:570Paolo Guiotto: And we have seen that this operation has these properties. So, properties…
01:57:540Paolo Guiotto: You have, linearity.
02:08:160Paolo Guiotto: That, basically, that's why…
02:11:560Paolo Guiotto: These properties are similar to properties of ordinary expectations, so if we have a linear combination of two variables, alpha x plus beta y.
02:23:290Paolo Guiotto: The conditional expectation of the linear combination is the linear combination of the conditional expectations.
02:37:890Paolo Guiotto: And we have, monotonicity…
02:44:450Paolo Guiotto: If X is less or equal than Y, almost surely, then the conditional expectation of X
02:54:880Paolo Guiotto: even G will be less or equal than the conditional expectation of Y, even G.
03:04:890Paolo Guiotto: From this, it follows, in particular the triangular inequality.
03:14:340Paolo Guiotto: The conditional expectation that says, The absolute value of the conditional expectation.
03:21:820Paolo Guiotto: Is less or equal than the conditional expectation of the absolute value.
03:27:700Paolo Guiotto: So we can carry inside the modules.
03:31:600Paolo Guiotto: In the usual way.
03:35:50Paolo Guiotto: Then I do not remind exactly the order of properties, so when we have that a variable X is already G measurable, and we do the conditional expectation that
03:49:410Paolo Guiotto: intuitively is still the best prediction you can do on X, known the information G, well, this is X itself. While if X is independent.
04:04:170Paolo Guiotto: of… the events of the Sigma Algebra G.
04:09:30Paolo Guiotto: So a random variable independent of a sigma algebra means that the set, the events associated to X, so X belongs to some set, that's an event, are independent of the events of G.
04:22:760Paolo Guiotto: Then, the conditional expectation indicates that the best you can say about X is its expected value, so it's a constant.
04:32:680Paolo Guiotto: Then we have the property called the sub-conditioning.
04:39:290Paolo Guiotto: which says that if we have a second sigma algebra, which is smaller than G, which is more than F,
04:47:630Paolo Guiotto: And we do a nested conditioning. So we take X, we condition with respect to G. Then, this is a random variable, it is G measurable, and we do the conditional expectation of this with respect to the second sigma algebra age.
05:05:660Paolo Guiotto: Well, this is the same of directly doing the conditional expectation,
05:11:690Paolo Guiotto: Given… of X, given H.
05:16:950Paolo Guiotto: Number 7, this is a property we added. It's important to know that if you do the expectation of
05:26:560Paolo Guiotto: the conditional expectation of X, even G, well, this is the expectation, the same of the expectation of X. So, the expected value of the conditional expectation is the same of the expected value of X.
05:43:120Paolo Guiotto: All these properties follows by a characterization that is something we should always keep in mind, because whenever we have to prove anything about the conditional expectation, we refer always to this. All these properties…
06:08:570Paolo Guiotto: follow.
06:12:330Paolo Guiotto: Bye.
06:13:730Paolo Guiotto: the…
06:14:930Paolo Guiotto: which is, again, the same of the orthogonality condition, but here we are not in an inner product setup, L1 is not in inner product space, so we call that the duality.
06:32:660Paolo Guiotto: characterization.
06:37:570Paolo Guiotto: of the conditional expectation of X given G.
06:43:520Paolo Guiotto: Which is the following.
06:46:590Paolo Guiotto: If you take the expected value, Well, let's say, why?
06:55:820Paolo Guiotto: Let's say, let's write this way. Why?
06:59:660Paolo Guiotto: Jeep.
07:00:890Paolo Guiotto: immeasurable.
07:04:190Paolo Guiotto: is… the conditional expectation of X given G, if and only if.
07:13:650Paolo Guiotto: whenever we take the expected value of X,
07:18:780Paolo Guiotto: times a G measurable, genetic G-measurable random variable. This is the same of the expected value of Y
07:28:470Paolo Guiotto: times Z for every Z. Now, 2.
07:33:350Paolo Guiotto: Technically, since that product must have an expectation, and here X is in L1 as well as Y, this is why we have to put L infinity. This must be bounded and G measurable, so omega GP.
07:51:510Paolo Guiotto: In other words, if you want, you see the orthogonality condition, because if you do X minus the expected value of X given G,
08:02:430Paolo Guiotto: multiplied by Z, this would be the scalar product, no?
08:06:980Paolo Guiotto: This reminds of the product between X minus the conditional expectation of X given G with Z.
08:17:670Paolo Guiotto: In L2, right? It's formally this, no? The L2 product is the integral of the product for this case. The unique point is that we formally cannot interpret this as a scalar product, because this guy is in L1,
08:34:760Paolo Guiotto: And this is in L infinity. If L infinity is contained in all LPs, also in L2,
08:42:240Paolo Guiotto: So we could interpret Z as a vector of L2. We cannot do the same for the other one. So that's why we need to keep separate this with this scarlet product notation. But at the end, it's the same relation. This equals 0.
08:58:170Paolo Guiotto: for every Z in L infinity.
09:02:230Paolo Guiotto: Now, it's useful to know that this condition, let's say the orthogonality condition.
09:10:270Paolo Guiotto: is in fact equivalent to, also to this. In particular, you see that if I take,
09:18:890Paolo Guiotto: If Z is an indicator of an element G of the sigma algebra, script G…
09:30:140Paolo Guiotto: This is a G measurable function, so I can use… and it is also bounded, because it is value… the values are 0 and 1, so this is NL infinity, omega.
09:44:40Paolo Guiotto: G measurable, function, so it's one of these, Z, this random variable.
09:52:440Paolo Guiotto: that I can put into that formula. So this, in this case, O,
09:59:920Paolo Guiotto: implies that the expected value of X times the indicator of this set G
10:07:910Paolo Guiotto: is equal to the expected value of Y,
10:12:220Paolo Guiotto: Which is the expected value of X given G times the indicator.
10:19:930Paolo Guiotto: Oh, jeez.
10:26:900Paolo Guiotto: So, this is actually another characterization, because once, you have this, this, this is, O prime.
10:40:650Paolo Guiotto: For every G in the family, Script G.
10:46:940Paolo Guiotto: This O' is actually another characterization.
10:52:720Paolo Guiotto: is, equivalent
10:59:460Paolo Guiotto: 2.
11:00:470Paolo Guiotto: O.
11:01:720Paolo Guiotto: I don't do the formal steps, but the idea is that if it holds for the indicators by linearity, it holds for linear combinations of indicators, no? So, whenever you take a sum of CJ indicator GJ, you can take out the sum, take out the constant.
11:24:230Paolo Guiotto: And so you can get that if this all indicators, it holds also supports simple functions.
11:30:540Paolo Guiotto: If it holds for simple functions, since you can always approximate by an increasing sequence of simple functions.
11:37:870Paolo Guiotto: any G-measurable random variable, you get that from this, you have the first one, okay?
11:48:160Paolo Guiotto: So this is another characterization. Okay. So, some remarks. In general, so let's say the first remark.
12:00:880Paolo Guiotto: Is that, in general, it is not easy to determine explicitly the conditional expectation given a generic sigma algebra.
12:11:920Paolo Guiotto: Okay? So, in general.
12:19:330Paolo Guiotto: It is… very… Difficult.
12:30:500Paolo Guiotto: guests.
12:32:870Paolo Guiotto: explicit,
12:38:190Paolo Guiotto: representations…
12:46:50Paolo Guiotto: of the conditional expectation of Excel.
12:49:750Paolo Guiotto: even… gee…
12:51:740Paolo Guiotto: There are some nice cases, which is important to know, because in certain applications, they are interesting. A remarkable case is the following.
13:07:900Paolo Guiotto: is when…
13:10:340Paolo Guiotto: The sigma algebra G is basically a finite sigma algebra, and it is generated by a finite family of sets, so G is the sigma algebra generated by a family, let's say, E1,
13:27:630Paolo Guiotto: E. N.
13:30:240Paolo Guiotto: with this set that, to write the formula I'm going to write, that, fulfills this condition with
13:40:670Paolo Guiotto: This sets, E, E, EJ, J from 1… to N.
13:48:290Paolo Guiotto: et partition…
13:57:550Paolo Guiotto: off… Omega.
14:03:260Paolo Guiotto: with, well, let's say, none.
14:11:640Paolo Guiotto: degenerate…
14:16:70Paolo Guiotto: coefficient of omega. What does it mean, this?
14:19:450Paolo Guiotto: It means that they must be a coefficient, so you expect that they divide omega into parts. So, formally, omega is the union of this set EJ, J from 1 to N.
14:34:600Paolo Guiotto: But, moreover, this is a disjoint union, so… and EI in the section EJ is empty when I is different from J, so this becomes actually a disjoint union of these sets.
14:55:650Paolo Guiotto: And the second condition that explains this term, non-degenerate, we ask that all the probabilities of the EJ are strictly positive. Let's say, otherwise, I could eliminate, basically.
15:11:510Paolo Guiotto: From this partition. So, for every J, equal 1, 2, N.
15:21:10Paolo Guiotto: Okay, now, in this case, it is possible to prove water, Well, the idea is that.
15:29:620Paolo Guiotto: In this case.
15:36:190Paolo Guiotto: whatever is X, huh?
15:39:160Paolo Guiotto: if X is a genetic L1 omega F, measurable random variable.
15:47:990Paolo Guiotto: the conditional expectation of X given G,
15:54:610Paolo Guiotto: is, first of all, a Gmeasurable random variable.
16:02:680Paolo Guiotto: But now, since the sigma algebra
16:07:140Paolo Guiotto: is generated by this set. We may imagine what are the sets of this sigma algebra, because
16:13:730Paolo Guiotto: Our space, the probability space, omega, is divided into a certain, numbers of sets E1, etc, generic, EJ. There are a finite number of these sets.
16:28:400Paolo Guiotto: So, imagine what could be the sigma algebra G. The Sigma algebra G, we say that it's not easy in general, to know what is the sigma algebra generated by a certain family, because this could be a very complicated
16:44:220Paolo Guiotto: a complicated family, we cannot characterize all the elements of a certain syntabra. But in this case, we can do that, because
16:54:310Paolo Guiotto: Of course, since G is the sigma algebra generated by these sets, these sets will be in the sigma algebra, because this is, let's say, the first requirement, the sigma algebra G generated by these sets is the minimal.
17:11:160Paolo Guiotto: Sigma algebra that contains those sets, okay?
17:15:210Paolo Guiotto: So, we know that these sets are in the sigma algebra, and also we know that G is a sigma algebra, so it must contain also unions, complementaries, and all possible set operations. But let's think about what happens if we do,
17:33:270Paolo Guiotto: let's say a finite union, let's say a union of two is not one of them, because if I do the union of these two, I get a new set, which is this one, okay? So, clearly, in this family G, there are all possible finite unions, okay?
17:52:920Paolo Guiotto: But, actually, I claim that these are all possible sets, because imagine, you have one of these sets, EJ, there will be also its complementary. In the figure, the complementary is in green.
18:07:420Paolo Guiotto: But the complementary is actually because they have a disjoint partition of omega, is the union of all the other sets, no? So it's, again, a finite union of these sets. And whenever you do the intersection of two finite unions, you get one finite union of them. So, in other words, the sigma algebra of… the sigma algebra g
18:31:960Paolo Guiotto: In this case.
18:37:890Paolo Guiotto: G is made, basically, of all possible finite unions.
18:44:270Paolo Guiotto: let's say, of the sets EJ, where J belongs to a set of indexes capital J, where J is all possible subsets of the indexes from 1 to N. So.
19:01:180Paolo Guiotto: J is, whatever contained in the sets 1N. So, it's a finite.
19:09:120Paolo Guiotto: sigma algebra is made by a finite number of sets. We include in this sigma algebra also the empty set that could be obtained by doing this union with J equals empties, or if you want, you add.
19:25:890Paolo Guiotto: So, union, empty set, while you don't need to add the full space, because the full space is already a disjoint union of them. So, basically, G is made of these sets.
19:40:680Paolo Guiotto: So, when you think to now a G measurable function.
19:46:660Paolo Guiotto: Of course, examples of Gmeasurable functions are the indicators of sets of this family, okay?
19:57:50Paolo Guiotto: But in general, how do you expect that it is a G measurable function? So if Y is G measurable, is G measurable, then you have that the anti-images of sets like Y minus 1 of, or let's say, Y belongs to.
20:16:240Paolo Guiotto: Let's write in the… form Y belongs to a set F, which is a Borel set of R.
20:29:420Paolo Guiotto: Now, this must be a set of this type, so it will be a finite union of these sets. So, in other words, you have that to be a G measurable, this Y must be a combination of indicators.
20:48:160Paolo Guiotto: CJ indicator, EJ. So, the unique possible measurable functions are the simple function, because this is only a finite number of sets.
20:58:850Paolo Guiotto: And, and so this is what happens. So, our conditional expectation will be a function of this type. So…
21:09:500Paolo Guiotto: Also, the conditional expectation of X given G will be a function of type some CJ indicator of EJ.
21:21:360Paolo Guiotto: J from 1 to N.
21:24:680Paolo Guiotto: Now, how can we determine these, these, values, CJ?
21:31:510Paolo Guiotto: Well, you can… you can see that if we,
21:40:30Paolo Guiotto: If we take the expectation, both sides, in particular, from this, it follows that the expectation
21:56:550Paolo Guiotto: So if I want to isolate, one… so if I take the expectation of the…
22:03:20Paolo Guiotto: conditional expectation, for example, I get that on one side, I know that doing the expectation of the conditional expectation, I get the expectation of X. On the other side, I get some J from 1 to N,
22:18:610Paolo Guiotto: of,
22:20:940Paolo Guiotto: CJ, and the expectation of the indicator of EJ, this is the probability of EJ, so we get some J12N of CJ probability of EJ.
22:35:270Paolo Guiotto: which is not yet sufficient to determine the single CJ.
22:40:330Paolo Guiotto: Okay, but I can do better this, and…
22:45:40Paolo Guiotto: If we do, the expectation of
22:48:310Paolo Guiotto: The conditional expectation of X given G times an indicator of these sets, EJ, so let's say of the indicator of EI, we compute this.
23:02:90Paolo Guiotto: Now, we remind that this is a G-measurable random variable, right? And we have seen above that there is this equivalent characterization.
23:12:140Paolo Guiotto: Whenever you compute the expectation of the conditional expectation times a G measurable indicator, it is the same of computing the expected value of X times the same indicator.
23:24:930Paolo Guiotto: So I can say that by the characterization.
23:27:970Paolo Guiotto: comes equal to expected value of X, indicator of EI.
23:36:130Paolo Guiotto: on one side. And on the other side, since this is also the sum of J of CJ indicator EJ, I have that into this expectation, there is a sum
23:51:820Paolo Guiotto: of a JCJ indicator, EJ, times indicator EI.
24:00:330Paolo Guiotto: But we remind that these sets form a disjoint partition of the space omega.
24:07:670Paolo Guiotto: So, what happens? This product…
24:11:880Paolo Guiotto: this product, omega by omega, is 0 when i is different from J, because the indicator EJ times indicator EIE
24:23:540Paolo Guiotto: is equal to 1 if and only if the 2 are equal to 1. So the indicator of EI is equal to 1, and also the indicator of J is equal to 1.
24:34:800Paolo Guiotto: But this happens if and only if the omega, where you evaluate the indicators, belongs to EI, and at the same time.
24:44:510Paolo Guiotto: belongs to EJ, so it belongs to the intersection. Omega belongs to EI intersection EJ, which is empty.
24:54:850Paolo Guiotto: when i is different from J. So, this means that there is no omega for which that product is 1 when i is different from J. So.
25:07:130Paolo Guiotto: indicator EI times indicator EJ is identical, is 0, when i is different from J. Because to be 1, that would be at least an omega into the intersection, but that's empty.
25:21:790Paolo Guiotto: And therefore, in that sum, it survives only one term, which is the term with J equal to I, so we have expectation of CI indicator EI,
25:36:690Paolo Guiotto: which is CI times the expectation of the indicator, EI, so the probability of EI.
25:47:260Paolo Guiotto: And therefore, we get this formula.
25:50:340Paolo Guiotto: So…
25:53:580Paolo Guiotto: we obtained that the expectation of X times the indicator of EI is equal to CI, probability EI, so that's right here, expectation of X times indicator EI,
26:12:10Paolo Guiotto: is equal to CI probability.
26:15:410Paolo Guiotto: EI.
26:16:890Paolo Guiotto: And since we suppose that these are all positive numbers, we finally get to the formula for CI.
26:23:540Paolo Guiotto: CI is equal 1 over the probability of EI, times the expectation of X.
26:32:310Paolo Guiotto: indicator of EI. So we get a formula for the conditional expectation, which is the following. This CI
26:42:430Paolo Guiotto: I remind you that these CI are the coefficients of these indicators that provides the conditional expectation of X given G. So we have found this formula. Conditional expectation of X given G
26:58:480Paolo Guiotto: is equal to sum…
27:01:230Paolo Guiotto: Let's restore the letter J from 1 to N of CJ, which is now 1 over the probability of EJ.
27:11:210Paolo Guiotto: times the expectation of X indicator EJ. This is the coefficient CJ.
27:20:110Paolo Guiotto: times the indicator out of the expectation of the STEJ.
27:25:620Paolo Guiotto: Now, this is a formula. As you will see in a second, this is, at the same time, natural, and we could say it is also well known to us, because it's the same formula we would have with theild space structure of writing a vector into a basis.
27:46:990Paolo Guiotto: So this formula is nice.
27:49:900Paolo Guiotto: this formula.
27:55:350Paolo Guiotto: Says… that.
27:59:990Paolo Guiotto: You see that, because of this form, because DPJ are a partition of the space omega.
28:11:160Paolo Guiotto: Okay? Only one of these indicators is equal to 1, and all the other are equal to 0 for every omega. So we could say that, if we evaluate.
28:24:770Paolo Guiotto: the conditional expectation of X.
28:28:190Paolo Guiotto: given G. Now, remind that this is a random variable, so it's a function of omega. You will never see this in probabilistic notations, because normally we never write this as function of omega. It's not a very nice notation, even if you think a second, no?
28:46:880Paolo Guiotto: But let's say, if omega belongs to the set PIE,
28:52:730Paolo Guiotto: to one of them, so only one of these terms will be different from zero, all the others will be zero, and it will be that one with the index j equal i. So we get that the value is one of the probability
29:07:830Paolo Guiotto: of EI times the expected value of X, Indicator, EI.
29:14:990Paolo Guiotto: And of course, I do not write the indicator EI at 2 omega, because that is equal to 1.
29:22:360Paolo Guiotto: So you see that the conditional expectation, so if you want to return to the generic… to the general picture, this is the sample space omega that we partition in a number of sets, E1, E2, let's say the generic EJ,
29:39:800Paolo Guiotto: Okay? The conditional expectation is a function, it's a random variable, it's a function from omega to r, let's say.
29:49:490Paolo Guiotto: So this is the conditional expectation.
29:53:830Paolo Guiotto: Now, on the set EJ, here, the conditional expectation of X given G
30:04:490Paolo Guiotto: is constantly equal to this value here, you see? Because that's a constant.
30:11:820Paolo Guiotto: No? I say, when omega is… well, let's say, let's keep the same letter, so let's write this as EI.
30:19:410Paolo Guiotto: When omega is in this set, and I evaluate the conditional expectation for this particular situation, okay, where the sigma algebra g is generated by this finite number of sets that are a partition of omega.
30:33:850Paolo Guiotto: On that, on each element of the partition, the conditional expectation is constant, and it is constant equal to this value, 1 over…
30:44:450Paolo Guiotto: One of the probability of EIE.
30:47:970Paolo Guiotto: times the expected value of X
30:50:830Paolo Guiotto: times, indicator of EI. Or, in another language, I don't know, perhaps it's my… is my… is my way to… I was… I… I studied analysis first, and probably later, so I am used to see these things as integrals.
31:11:100Paolo Guiotto: And if you see as integrals, you see this. It is 1 over the probability of
31:15:350Paolo Guiotto: EI, the integral on EI of X in DP.
31:21:140Paolo Guiotto: Now, this way, it says that it's like if you are doing an average of X, but only on the subset PIE, and you have a… this one order is scaling factor, it is just because it says you are doing an average.
31:36:310Paolo Guiotto: No? There's such a way that it's like the sum of these is equal to 1. If you put X equal to 1, you get 1.
31:44:210Paolo Guiotto: Okay, so you have this scaling factor.
31:48:240Paolo Guiotto: So, in other words, in this case, the conditional expectation works in this way. On the set EI, you have to take the average of X on that set.
32:01:300Paolo Guiotto: the average respect to the probability. So again, it's a sort of mean value, but it's not the expectation of X. You see? It's the expected value of X restricted to EI, averaged by the probability of EI, to make this still an expectation, okay?
32:19:800Paolo Guiotto: If you put X equal 1, you get 1, as it should be.
32:23:920Paolo Guiotto: Okay, second… This formula is also natural, because if you…
32:30:400Paolo Guiotto: force the language of Hilbert spaces. If you look at this, you can… You… can.
32:41:270Paolo Guiotto: also notice… That, huh?
32:47:170Paolo Guiotto: this thing, 1 over the probability of EJ, expected value of X indicator EJ times EJ,
33:00:120Paolo Guiotto: In a very, let's say, is a… is a forced, let's say, notation.
33:07:30Paolo Guiotto: Now, this is what? Well, in the… and actually, it would be a correct notation if X is in L2, okay? This would be, let's say, between commas, the product between X and the indicator of EJ times the indicator of EJ
33:26:680Paolo Guiotto: Which is here rescaled by the probability of EJ.
33:32:750Paolo Guiotto: Now, what is this probability?
33:35:460Paolo Guiotto: Well, this should remind something like, say, something like, reminds X scala vector E times E, or if you want EJ,
33:49:780Paolo Guiotto: When, we do the decomposition.
33:53:780Paolo Guiotto: in a… in an orthonormal basis of the space L2.
33:58:140Paolo Guiotto: It reminds something like that, where… who is EJ? Well, if you split this probability into this, you give a root this and a root of the probability here, here, to the inner factor.
34:13:40Paolo Guiotto: Now, you may notice that if you call EJ this element, this function, indicator of EJ divided by the root of the probability of EJ,
34:25:900Paolo Guiotto: Now, this is a unit vector in the L2 norm.
34:30:250Paolo Guiotto: Here we have that the L2 norm of EJ, well, let's do 2 power 2, would be the integral on omega of EJ squared, right? DP.
34:42:929Paolo Guiotto: But EJ square, it would be the integral on omega of the indicator of EJ square, which is itself, because that function is 0, 1 function, so when you do the square, you get the same thing, divided by the probability of EJ
35:00:420Paolo Guiotto: the root of the probability to power 2, so this becomes just the probability of J in DP.
35:07:660Paolo Guiotto: But, since this is exactly equal to the indicator of EJ, and you carry outside that probability, so you get 1 over the probability of EJ,
35:18:580Paolo Guiotto: times the integral over omega, or the expected value of the indicator of set EJ, this is the probability of EJ, so you get, in fact, at the end, equal 1, this ratio.
35:32:20Paolo Guiotto: Okay? So, in other words, this formula that we have here for this conditional expectation is nothing but the way we would write the vector X with respect to the orthonormal basis made of this vector CJ. You can check also that they are orthonormal, because if you do EI, it's color EJ,
35:56:190Paolo Guiotto: This would mean that we say that EJ is indicator divided the root of the probability, so we would have 1 over the root of the probability of PI.
36:08:110Paolo Guiotto: times the probability of PJ, EEJ.
36:12:00Paolo Guiotto: And then we would have the scalar product between the two indicators, indicator of EI, scalar product with the indicator of EJ. But that's the integral on omega of the product, indicator EI, indicator EJ.
36:27:330Paolo Guiotto: DP, and that product is 1 if and only if
36:32:540Paolo Guiotto: there is an omega in the intersection, but that's never true, you see? So when i is different from J,
36:40:850Paolo Guiotto: you see that this is always equal to zero, and you get 0. So they are orthogonal.
36:47:440Paolo Guiotto: They are unit vectors, so they are an orthonormal basis. So basically, this formula, the conditional expectation, so the conditional expectation of X given G, would be just the classical orthonormal expression of X
37:06:130Paolo Guiotto: in the orthonormal basis made of the vectors EJ.
37:10:450Paolo Guiotto: J from 1 to N.
37:13:700Paolo Guiotto: So that's the shape… at the end, what is the shape of this formula. Now, this formula is exactly this when X is in L2. Otherwise, with the different notations, but in practice the same thing, it is equal to…
37:31:280Paolo Guiotto: this when X is in L1, okay? The only thing is that I am not authorized to write. This is the scalar product of X between EJ, because formally speaking, X is not in L1, okay? So that's… but that's just a formal
37:48:190Paolo Guiotto: point. It's exactly the same formula. It's not a different one that looks the same. No, it's the same formula.
37:55:660Paolo Guiotto: And so, remind of this formula, because this might be helpful whenever you do condition… you condition on, basically, a finite sigma algebra. Because if you… if you condition on a finite sigma algebra.
38:09:930Paolo Guiotto: there will be generators at the end, no? So if the sigma algebra is made by a finite number of sets, you take the intersections of these sets, at the end, you have a set which generates all the other sets.
38:23:270Paolo Guiotto: You see the point, no? Imagine that your space omega, and you have a finite sigma algebra, so there is only a finite number of parts of omega. Maybe they overlap here and there, there are many intersections, it doesn't matter, no?
38:39:340Paolo Guiotto: But what happens is that you see that, taking all these sets, since also the intersections are in the sigma algebra, you can see that, for example, this is a set of the sigma algebra. This one is another set of the sigma algebra, this one is another set of the sigma algebra, and so on. At the end, any finite sigma algebra will be generated by a finite partition.
39:02:790Paolo Guiotto: So the problem is just determining what is the partition. Once you have the partition, you can use this formula. So when… when… when…
39:12:690Paolo Guiotto: You have a sigma algebra generated by a certain number, finite number of sets.
39:19:690Paolo Guiotto: So, which have the condition that omega must be a disjoint union of the EJ, and we need also that the probability of each of these sets be positive.
39:33:270Paolo Guiotto: Well, if one of them is a probability zero, is an event which is negligible for the probability, you can eliminate, basically, because you understand that on that set, which is a probability zero set, it doesn't matter what is the conditional expectation.
39:48:530Paolo Guiotto: Okay? Because it's like to say, you have a variable which is defined on a measure zero subset, who cares what happens?
39:56:550Paolo Guiotto: Any random variable is always defined modular, let's say, almost surely, or almost everywhere in the measure language. So this is the formula that you have.
40:10:890Paolo Guiotto: We may say that this formula is the basic formula for all concrete, real-world applications, because in real world.
40:23:150Paolo Guiotto: in real-world application, we have always finite approximations of everything, so we can always say that for a real-world problem, we use a finite sigma algebra, and therefore there will be generators, and therefore we can compute the conditional expectation by using a formula of this type. So, it's not just an example, it's a concrete
40:47:450Paolo Guiotto: Tool that you use whenever you do finite approximations.
40:53:970Paolo Guiotto: And notice that, a little particular case,
40:59:620Paolo Guiotto: That, if you want, we get back,
41:02:890Paolo Guiotto: From this, the, the… This final property here.
41:16:30Paolo Guiotto: Oh, well, let's, let's, let's, let's, let's, let's keep a… Okay
41:24:280Paolo Guiotto: I wanted… I added mine something else.
41:27:20Paolo Guiotto: Okay, A second factor I want to point out here…
41:36:370Paolo Guiotto: Is another general property of the conditional expectation, which is an extension of the triangular inequality.
41:49:440Paolo Guiotto: Now, if we look at the triangular inequality, We see this, huh?
41:56:920Paolo Guiotto: So… If… Weird.
42:00:590Paolo Guiotto: Locke.
42:04:560Paolo Guiotto: the triangular… inequality.
42:11:390Paolo Guiotto: So models of the conditional expectation.
42:17:250Paolo Guiotto: Is less or equal than conditional expectation, of the modulus
42:26:230Paolo Guiotto: We see, we can recast this property in this way.
42:34:110Paolo Guiotto: We have that, huh?
42:42:380Paolo Guiotto: If we call, Fee.
42:46:20Paolo Guiotto: of, let's say, intellects. The absolute value.
42:51:870Paolo Guiotto: This formula says this nice,
42:54:930Paolo Guiotto: Property, that if you do fee of the conditional expectation.
43:02:830Paolo Guiotto: This is less or equal than conditional expectation of B of X.
43:08:970Paolo Guiotto: In some sense, you can flip the order between expectation carrying inside the function. This happens for the models, but it turns out that it happens for a more generalized class, and this is called the Janssen inequality.
43:32:150Paolo Guiotto: which is, at least in a simplified version, very easy to prove. It says that if
43:40:430Paolo Guiotto: phi from R to R. Actually, it holds in a very…
43:46:700Paolo Guiotto: In a little bit more… in a little… in a more general case, is convex.
43:59:170Paolo Guiotto: Now, convex, for a generic function, would mean this.
44:05:550Paolo Guiotto: So phi convex, huh?
44:09:580Paolo Guiotto: If a… what happens?
44:16:150Paolo Guiotto: Is there any problem? Okay. Geometrically, convexity means that
44:22:450Paolo Guiotto: This is a… we are talking about a function
44:26:10Paolo Guiotto: real… of real valuable, real value. The convexity means something like this. So, the more general formulation of this property is the following. If you take any two points.
44:39:780Paolo Guiotto: on the graph of the function, and you join by a segment, this one. As you can see, the segment is above the graph of the function on, between the initial and the final point.
44:55:450Paolo Guiotto: How can we express this? So, we can say that if this is point X and this is point Y, the abscesses of the initial and final point.
45:07:440Paolo Guiotto: Any point between these two.
45:10:30Paolo Guiotto: can be represented in this way. We can say that this is X plus TY minus X, where T ranges from 0 to 1.
45:24:100Paolo Guiotto: You see, when T is 0, you get X, when T is 1, you get Y, and this moves linearly between X and Y. Another way to rewrite this, if you expand this notation, you see that we can say 1 minus T.
45:39:10Paolo Guiotto: X plus TY.
45:43:220Paolo Guiotto: This is not just a linear combination of X and Y, but it is called convex combination, because the two coefficients are not arbitrary, A and B, so you have not AX plus BY for A and B generic, but
45:59:580Paolo Guiotto: The A and B must be related in such a way that their sum is 1 and they are between 0 and 1. This is…
46:05:500Paolo Guiotto: Basically, what is it?
46:07:810Paolo Guiotto: Now, this point, the condition, the convexity condition, is that the ordinate here is greater than the ordinate here.
46:19:270Paolo Guiotto: Now, the originate here is the value of the function at this point, so let's start writing this. So phi F, normally the condition is written with this notation, okay? So that's why I write it with that notation. 1 minus phi.
46:33:310Paolo Guiotto: X plus TY.
46:36:910Paolo Guiotto: And this is the value of the function, so it is this number here.
46:42:280Paolo Guiotto: So this number here is phi of 1 minus TX plus TY.
46:50:820Paolo Guiotto: And what about this one? Well, this is the Y that… the Y… well, actually, we have used Y, so let's call Z the ordinates.
47:02:240Paolo Guiotto: Okay, this is the Z of the points that is on that red segment, when the abshi is that value, no? Is 1 minus TX TY. What is the Y here? Well, this is a straight line joining two points. This is the point X.
47:22:260Paolo Guiotto: P of X,
47:24:330Paolo Guiotto: And this is the point, Y.
47:27:60Paolo Guiotto: P of Y, right?
47:30:90Paolo Guiotto: So now there is a standard equation for the line that joins two points. You have done… you have seen this from… when you have seen the derivative. So this, the straight line here has form Z equal
47:45:640Paolo Guiotto: there is an angular coefficient, and to have the angular coefficient, you get… you do… you have two points, you do the ratio between the difference of ordinates and the difference of abscissas, so I do phi of Y.
47:58:290Paolo Guiotto: minus phi of X divided by Y minus X.
48:03:410Paolo Guiotto: Of course, Y and X are not the same point, otherwise this argument degenerates. So this is the angular coefficient times, say, if I am at some point here.
48:17:300Paolo Guiotto: W, let's call it W, minus, X.
48:23:710Paolo Guiotto: So, when W is equal to X, that quantity is 0, so I need to add the value, which is phi of X.
48:34:110Paolo Guiotto: And you can see that when W is X, you get Z equals phi of X, so this line passed through this point. And when Z… when W is Y,
48:46:700Paolo Guiotto: You see that, that ratio becomes Y is when X divided by minus X, 1, so field y minus P of X plus P of X, P of Y.
48:59:660Paolo Guiotto: So, the Z when W is Y is field Y. This means that this is the equation of the line, straight line, joining, passing through these two points.
49:18:660Paolo Guiotto: Now, the Z at the midpoint, so the Z here, when W is… when W is 1 minus T…
49:29:890Paolo Guiotto: X plus TY.
49:33:610Paolo Guiotto: The Z corresponding to this would be P of Y minus phi of X
49:40:810Paolo Guiotto: divided Y minus X, and there is W minus X, maybe it's better if we rewrite under this form, X plus TY minus X is the same.
49:52:760Paolo Guiotto: So when you do W minus X, you see, you carry the X in the other side, what remains is T times Y minus X.
50:02:730Paolo Guiotto: plus phi of X.
50:04:940Paolo Guiotto: Now, if we do the other, we simplify this with this, we get that this is T times phi Y.
50:12:270Paolo Guiotto: minus phi X, plus fax, or…
50:17:920Paolo Guiotto: Reorganizing this formula, we have 1 minus T phi X,
50:24:730Paolo Guiotto: plus T phi Y. So this is,
50:29:540Paolo Guiotto: Equivalently, the Z that corresponds to that W. So this Z is 1 minus T.
50:39:240Paolo Guiotto: phi of X.
50:40:720Paolo Guiotto: plus T field Y.
50:43:750Paolo Guiotto: So we want to say that this quarter
50:47:690Paolo Guiotto: So the quote on the graph of the function is smaller than this quote, which is the quote on the segment. And that's why we have this condition, 1 minus T.
50:58:770Paolo Guiotto: P of X,
51:00:930Paolo Guiotto: plus P of Y.
51:04:630Paolo Guiotto: for every T between 0 and 1.
51:09:890Paolo Guiotto: Now, this is the standard… the standard, the more general definition of a convex function.
51:19:120Paolo Guiotto: you know, because in first-year calculus, you mostly deal with regular functions, so functions who have a derivative, or maybe a second derivative.
51:33:690Paolo Guiotto: Phi is differentiable.
51:38:90Paolo Guiotto: Phi is convex, If and only if the derivative
51:44:810Paolo Guiotto: Now, the derivative is the angular… the angular coefficient of the tangent. If you see the tangent line to this function, when you move the variable left to right, you see that this is negative, that this is still negative, this is positive.
52:03:770Paolo Guiotto: the point is that the value of the angular coefficient is increasing with the variable, so we have this equivalent characterization, this is increasing, and if
52:18:430Paolo Guiotto: phi prime is differentiable, so if there is the second derivative, we have also this characterization that the second derivative must be greater or equal than zero, okay?
52:32:830Paolo Guiotto: Okay, now, the theorem, let's go, return on the theorem. So, if phi is a convex function.
52:40:290Paolo Guiotto: And X, let's add here. And the capital X is an L1 random variable.
52:48:320Paolo Guiotto: 1 omega, F… P.
52:54:460Paolo Guiotto: So let's close these parentheses. 10…
52:59:480Paolo Guiotto: We have this inequality, that, phi
53:03:740Paolo Guiotto: of the conditional expectation of X given G,
53:09:930Paolo Guiotto: is less or equal than the conditional expectation of P of X given G. And this is the Janssen inequality.
53:21:890Paolo Guiotto: Now, intuitively, this is exactly an extension of the definition of convexity.
53:29:780Paolo Guiotto: Because the idea is that, what is an expectation?
53:34:670Paolo Guiotto: An expectation is a sum of values of X, no? Informally, I'm not saying that this is what is an expectation.
53:43:210Paolo Guiotto: But expectation is an integral. It's an integral with respect to a probability, so it's something like the sum of X times DP.
53:52:570Paolo Guiotto: But the sum of the DP is 1.
53:55:450Paolo Guiotto: Because the probability has total sum equals 1. So, it is like to have…
54:01:870Paolo Guiotto: P of this, you see, these are the value of X, and these are the values of the probabilities, the total value is 1.
54:09:570Paolo Guiotto: So, this is phi of the expectation less or equal than the expectation of phi.
54:16:690Paolo Guiotto: So, in some sense, this is natural because of this. The proof is a bit more complicated, because, of course, the expectation is not the sum, etc. But
54:29:390Paolo Guiotto: It is easy if we assume that the function phi is regular, is differentiable. We assume… for simplicity.
54:47:900Paolo Guiotto: be differentiable.
54:49:960Paolo Guiotto: Because, in this case.
54:57:360Paolo Guiotto: P is convex, huh?
55:01:530Paolo Guiotto: If and only if convexity is a very beautiful property, because it's full of geometrical insight. So we said here, we're reminded that phi is convex if and only if the derivative is increasing, because we see this on the graph, but there is another remarkable way
55:20:890Paolo Guiotto: to connect convexity to the derivative. If this is a convex function, and you take a generic point.
55:30:80Paolo Guiotto: say, a point X.
55:32:610Paolo Guiotto: and you trace the tangent to the graph, you see that the tangent is all time below the graph, or the graph is above the tangent. So we can say that the function phi, so phi at any point Y,
55:48:670Paolo Guiotto: phi of Y is above the tangent to the graph through the point x.
55:55:900Paolo Guiotto: So what is the equation of the tangent to the graph at that point? The equation is the derivative of the function at point x is the angular coefficient, so times y minus X.
56:09:360Paolo Guiotto: Now, this in function of Y is a linear function, which is equal to 0 when a Y is equal to X, so we have to add the phi of X, and this solves for every X and for every y.
56:22:910Paolo Guiotto: Okay? So this is the way we formally say the function is above each of the tangents. Each, because if you variety point X, you variety the tangent, so you have to look at this inequality in this way. Fix an x, so for X, fixed the…
56:40:510Paolo Guiotto: This blue line has this equation in variable 1, okay?
56:46:280Paolo Guiotto: a point Y equal X, this vanishes, so you get Y equals this equal, so Z equal phi of X. So, for X fixed and Y variable, this says function phi of Y above blue line tangent.
57:02:580Paolo Guiotto: Now, move X, so you have another tangent, but still the same thing, function above the new tangent, and so on.
57:10:670Paolo Guiotto: Okay, and now… So… if we apply this inequality, replacing Y with capital X, so…
57:24:600Paolo Guiotto: phi of capital X,
57:27:430Paolo Guiotto: will be greater or equal than phi prime of little x, whatever little x is, we will fix in a moment, capital X minus little x plus phi of little x.
57:40:960Paolo Guiotto: Now, we choose as little x the conditional expectation of X given G.
57:48:290Paolo Guiotto: So this means that we get capital phi of X will be larger than phi prime of the conditional expectation of X given G,
58:00:290Paolo Guiotto: times X minus the conditional expectation of X.
58:05:70Paolo Guiotto: given G… plus fee, Of the conditional expectation, of X, given G.
58:18:70Paolo Guiotto: Okay? So this now is the consequence of convexity.
58:25:180Paolo Guiotto: Okay? When I put in that formula, as Y, the capital X, and as X, the conditional expectation, the value that comes from the conditional expectation.
58:37:140Paolo Guiotto: Of X given G. Now, what we do, we take the conditional expectation into this inequality. So, apply expectation of whatever it is given G. What we get?
58:51:500Paolo Guiotto: So look here, we have… expectation of phi of X, Given. G.
59:00:170Paolo Guiotto: Now, since this is almost surely greater than the other one, and you reminded that the conditional expectation is monotonic, so when someone is greater than someone else, the conditional expectation of someone will be greater than the conditional expectation of someone else. We have that this is monotonicity.
59:20:720Paolo Guiotto: greater than the conditional expectation of all this, so phi prime, conditional expectation of X given G,
59:30:620Paolo Guiotto: times X minus its conditional expectation.
59:35:460Paolo Guiotto: Given G… Last…
59:39:40Paolo Guiotto: phi of the conditional expectation of X given G. This is the random variable. We are conditioning respect to G in this big conditional expectation.
59:52:220Paolo Guiotto: Now, let's apply all the other properties we know on the conditional expectation. The conditional expectation, this is the conditional expectation of this block here, given G. You see? It doesn't matter how complicated it is. It is the conditional expectation of something given G. Now, we use linearity.
00:14:160Paolo Guiotto: And this is the sum of the conditional expectations, so phi prime EX given G.
00:23:710Paolo Guiotto: times X minus conditional expectation of X given G, all this given G,
00:33:610Paolo Guiotto: plus, second line, expectation of phi, of the conditional expectation of X, given G, condition at respect to G.
00:47:900Paolo Guiotto: But now, let's start with the second conditional expectation.
00:52:440Paolo Guiotto: This quantity, I'm doing the conditional expectation is a function of this random variable, which is a conditional expectation. So, in particular, this is already G measurable.
01:06:630Paolo Guiotto: No? Because the conditional expectation yields a G measurable random variable. So this is a function of a Gmeasurable random variable. It is G measurable.
01:16:970Paolo Guiotto: this… Quantity inside this conditional expectation.
01:22:00Paolo Guiotto: here.
01:24:30Paolo Guiotto: is G measurable.
01:27:250Paolo Guiotto: Well, what happens when you do the conditional expectation given G of a G measurable random variable? You get the random variable itself. There is no effect of the conditional expectation, no? This is the property…
01:43:450Paolo Guiotto: of this list, no? When…
01:46:50Paolo Guiotto: you already have something which is G-measurable, the conditional expectation gives the same random variable. So, here we have that this part, this, this last here.
01:59:730Paolo Guiotto: is equal to… what we have inside, which is P of the conditional expectation of X given G.
02:10:370Paolo Guiotto: Let's now work on the first one.
02:13:800Paolo Guiotto: The first one is made in part of something which is G measurable, which is this one. It is still a function, now this time it's phi prime, but the sense is the same. This is a function of the G measurable stuff, so it's G measurable.
02:28:760Paolo Guiotto: This one, yes, it's made the part is G measurable, but part is not necessarily G measurable. X is supposed to be F measurable.
02:37:620Paolo Guiotto: So I cannot say that this round parenthesis is immeasurable, because there is one piece which is not.
02:43:400Paolo Guiotto: So I can say that I have a product where this is G major, and this is F measure.
02:49:300Paolo Guiotto: Now, what is the property? And you know that when you have a factor which is already measurable with respect to G, you can treat it as a constant. You can carry outside of the conditional expectation, and this was… this…
03:04:50Paolo Guiotto: other remarkable property.
03:08:900Paolo Guiotto: that I forgot to write, but… well, we haven't seen yesterday, you reminded this far, no? So, this is…
03:20:370Paolo Guiotto: This, so this can be written outside, huh?
03:24:910Paolo Guiotto: So let's see what remains. So we have equal to…
03:29:190Paolo Guiotto: Conditional expectation of phi prime of condition… sorry, sorry, not… there is no conditional expectation. Phi prime of…
03:38:340Paolo Guiotto: conditional expectation of X given G, this is the factor that comes outside. Then we have expected value of X minus the conditional expectation of X given G,
03:50:300Paolo Guiotto: Given Gia gain.
03:53:340Paolo Guiotto: And then there is the other term, which is plus phi of expectation of X given G.
04:01:540Paolo Guiotto: Now, can you tell me what is this quantity?
04:06:320Paolo Guiotto: Lucy?
04:17:420Paolo Guiotto: You don't see… so let's split linearity. This is expectation of X given G,
04:24:20Paolo Guiotto: minus the expectation of the expectation of X, given G, again, given G.
04:33:410Paolo Guiotto: Now, the second one is the double conditioning, you see?
04:37:880Paolo Guiotto: It's too simple, they are the same, in fact. Oh, if you want, I'm conditioning this guy, which is already G measured all with respect to G, so I will get the same quantity. So the second term is equal, in fact, to the conditional expectation of X given G, but the first term.
04:57:480Paolo Guiotto: It's the same, so the difference will be equal to zero.
05:00:890Paolo Guiotto: So, all this is equal to zero.
05:04:370Paolo Guiotto: Multiplied by the other…
05:06:620Paolo Guiotto: the 1 gives 0, so at the end I get equal to phi of expected value of X given G.
05:15:650Paolo Guiotto: But then I have the conclusion, because,
05:19:410Paolo Guiotto: We started from this, the conditional expectation of phi X given G, so the conditional expectation of phi X
05:29:830Paolo Guiotto: even G.
05:31:610Paolo Guiotto: has been proved to be greater or equal than phi of the conditional expectation of X given G, which is exactly
05:39:770Paolo Guiotto: the conclusion.
05:41:800Paolo Guiotto: Of course, if the function is concave.
05:46:680Paolo Guiotto: the inequality is inverted, okay? So, remark.
05:54:550Paolo Guiotto: If phi is concave, Well, concave, you can write the condition, but means that minus phi is convex.
06:07:330Paolo Guiotto: So you get easily that the conditional expectation, of phi of X,
06:16:200Paolo Guiotto: given G, in this case, will be less or equal than phi of the conditional expectation of X given G.
06:25:680Paolo Guiotto: Okay?
06:30:740Paolo Guiotto: Okay.
06:38:430Paolo Guiotto: Let's return again to the question, how do we compute the conditional expectation? So, we have seen a case…
06:47:520Paolo Guiotto: that says if the sigma algebra, G, on which we are conditioning is generated by a finite number of events, basically, if we assume that the events are already a partition, we have a direct formula, otherwise we have to find a partition.
07:05:290Paolo Guiotto: For the, for the sigma algebra.
07:09:210Paolo Guiotto: We have a formula, okay?
07:11:940Paolo Guiotto: Another important case, which is not necessarily of this type, so G in general could be infinite as a sigma algebra, is the following, is…
07:25:670Paolo Guiotto: Another important case…
07:33:930Paolo Guiotto: Important.
07:37:410Paolo Guiotto: case… of conditional expectation of X in G.
07:44:610Paolo Guiotto: that.
07:47:400Paolo Guiotto: 10.
07:49:700Paolo Guiotto: Hmm.
07:51:170Paolo Guiotto: computed.
07:55:360Paolo Guiotto: is when… G is the sigma algebra generated by another random variable.
08:07:840Paolo Guiotto: But Y is a random variable.
08:10:870Paolo Guiotto: So this is the sigma algebra made of the events, so it's the family of the events.
08:18:550Paolo Guiotto: If the family of the events Y belongs to E when E is a Borel set.
08:25:850Paolo Guiotto: In, so in general, unless Y is, is,
08:33:660Paolo Guiotto: we say has finite rank, so takes a finite number of values. This is not a finite sigma algebra. This would be… if Y is a random variable with the density, for example, this is basically the Borel sigma algebra again, okay?
08:53:630Paolo Guiotto: Well, we do. The interesting case is, moreover, when there is a density, a joint density of X and Y. So there is a formula that yields the calculation of the conditional expectation of X
09:13:300Paolo Guiotto: given this family generated by Y,
09:16:609Paolo Guiotto: in terms of the density, the density, the joint density between F and Y, so we have to assume, moreover, that the two are absolutely continuous, okay?
09:26:260Paolo Guiotto: So, but before to say that, in this case, let's say that in this case.
09:36:479Paolo Guiotto: we use… the… notation… Which is an improper notation, clearly.
09:45:109Paolo Guiotto: We say that the conditional expectation of X given Y,
09:53:200Paolo Guiotto: So, it's improper because we define this operation, the conditional expectation, when we condition respect to a sigma algebra, not respect to a random value.
10:04:380Paolo Guiotto: This means that if you want, by definition, this is the conditional expectation of X given the sigma algebra generated by Y. But there is a good, very good reason to keep that notation, as you will see in a moment.
10:19:960Paolo Guiotto: We have this proposition that yields a formula.
10:25:710Paolo Guiotto: for the conditional expectation. So, let,
10:29:960Paolo Guiotto: XY, the pair X, Y, B, absolutely continuous.
10:39:540Paolo Guiotto: Got you.
10:42:480Paolo Guiotto: with density.
10:49:870Paolo Guiotto: F… X, Y.
10:54:450Paolo Guiotto: Ben…
10:58:160Paolo Guiotto: So, I will do, formally what is the formula, and then I will write you the informal notation, which is,
11:09:640Paolo Guiotto: a more immediate notation. Then, the expected, the conditional expectation of X given Y
11:19:160Paolo Guiotto: So now this is a function which is measurable with respect to a sigma, the sigma algebra generated by Y, so it turns out to be a function of Y.
11:31:940Paolo Guiotto: a numerical function of Y.
11:35:00Paolo Guiotto: Okay?
11:36:150Paolo Guiotto: Where this function of y is computed from this formula. So phi of little y, now little y is the real number, okay?
11:46:410Paolo Guiotto: is the integral on R over X,
11:52:370Paolo Guiotto: times this function that is written with this symbol, or if you want, times this function, F of XY, the joint density at XY, divided by FY of Y.
12:08:130Paolo Guiotto: in DX.
12:10:290Paolo Guiotto: This ratio here is also denoted with this symbol, F of X given Y, with little x given little y. It is used in this X annotation.
12:26:470Paolo Guiotto: And, as you can see.
12:30:590Paolo Guiotto: If you look at these relations, so this says that this conditional expectation Peace.
12:37:800Paolo Guiotto: a function of Y where this numerical function is obtained by doing these integrals, so this is literally what it means. Now, we can give an interpretation, because this is the integral of X with respect to something
12:52:960Paolo Guiotto: This reminds of, what?
12:56:60Paolo Guiotto: So when we have a density, Let's see where we… Yeah, we introduced densities.
13:07:290Paolo Guiotto: When we have a density.
13:13:290Paolo Guiotto: You remind that…
13:19:00Paolo Guiotto: So this is the density, let's see if we have a formula.
13:23:410Paolo Guiotto: Bobby.
13:24:290Paolo Guiotto: So the density is, that,
13:28:970Paolo Guiotto: That function that says that the new X, the low of X, is F times the X. So, when you compute the integral of X times the density, you have the expected value of the random varietal, no? The formula is this one. So, here.
13:47:350Paolo Guiotto: This is nice, because it's saying that the conditional expectation of X given Y is this function of Y. Well, you do the expectation, the mean value, this is the integral of X, with respect to this function that we call conditional density for this reason.
14:09:890Paolo Guiotto: conditional…
14:16:220Paolo Guiotto: Density.
14:19:50Paolo Guiotto: Very informally, since these are densities, you know that they are not probability. Now, a density is not a probability, but it is probability per unit of length, no? It's the distribution of probability along the x-axis, no? But informally, we may say that this quantity
14:37:740Paolo Guiotto: X given Y, X… slash Y is, so we may think that it is the probability
14:47:340Paolo Guiotto: that, X, why?
14:53:630Paolo Guiotto: between, let's say, X and X plus DX,
15:02:170Paolo Guiotto: And Y is… no, well, actually, there is no condition for Y.
15:09:640Paolo Guiotto: Divided by the probability that Y is between…
15:15:730Paolo Guiotto: what, let's say Y and Y plus DY.
15:21:240Paolo Guiotto: So it's a sort of ratio of probabilities, this thing. However, this is an informal notation, has not any meaning what I'm writing.
15:30:540Paolo Guiotto: Now, for this reason, for this reason, since that function phi of the random variable y that gives the conditional expectation is this thing, so it is also used this notation, so…
15:46:850Paolo Guiotto: We also write.
15:53:400Paolo Guiotto: That if we do the conditional expectation of X given Y,
15:57:730Paolo Guiotto: By knowing that Y is little y, which is, of course, just a notation, it has no meaning, because now.
16:06:650Paolo Guiotto: Y equal to little Y with PL event couldn't be any more as in algebra. But it's just a notation to say that… you see that the conditional expectation of X given Y is the function of capital Y,
16:22:70Paolo Guiotto: Well, when capital Y is this value, little y, this is the value of this function.
16:27:740Paolo Guiotto: So it is like, if we can say that the conditional expectation of X given Y equal little y is that function phi of little y, so it is the integral on r of x respect to the conditional density, X given Y.
16:46:430Paolo Guiotto: in the acts.
16:49:410Paolo Guiotto: At the end, the right-hand side, we compute the function of capital Y that yields the conditional expectation.
16:58:120Paolo Guiotto: So, to prove this, We have to verify that the characteristic property holds. So,
17:10:80Paolo Guiotto: We define… so, let…
17:14:250Paolo Guiotto: Let's start defining phi of y as that integral on r of X times the conditional density.
17:24:460Paolo Guiotto: which is, in other words, it is the integral on R, X, F, X, Y,
17:34:710Paolo Guiotto: X comma Y divided FY of Y. Notice that we write that quantity FY of Y inside for convenience.
17:45:180Paolo Guiotto: But it is not integrated, it's a constant, so we could also write it out here. FYY integral in R of XFX.
17:56:820Paolo Guiotto: Why?
17:58:10Paolo Guiotto: X, Y.
17:59:380Paolo Guiotto: the X. Now, there is a little care to take to define this, because you see that there might be problems when this quantity is zero.
18:09:900Paolo Guiotto: Okay, so we have to be a little bit careful on this. What does it mean when that density is zero? How do we define? Well, the idea could be, if the density is zero, when the density is zero?
18:24:00Paolo Guiotto: there is no distribution of probability of Y, so if I say, where, in the range of values, Y,
18:31:960Paolo Guiotto: where the density is zero, it means that the probability that capital Y is there is null, is 0. So I could say, okay, for those Y, who cares, we put the function equal to zero, and for the others, Y, we put this. So we do this definition for Y different from 0.
18:51:240Paolo Guiotto: I'm sorry, 4F…
18:54:150Paolo Guiotto: Y of Y different from 0? When… while… when F, Y is equal to 0, we just set the definition equal to 0. So this is formally how it is defined. Now, there is no problem with the definition. We have a well-defined function.
19:13:500Paolo Guiotto: We have to verify.
19:15:560Paolo Guiotto: We have… to check… that if we do phi of Y,
19:23:480Paolo Guiotto: This is the conditional expectation of X given Y, right?
19:30:370Paolo Guiotto: So this is what we have to check.
19:32:830Paolo Guiotto: thatism.
19:35:930Paolo Guiotto: Now, when, how do we check this? We use the orthogonality, or duality condition.
19:45:680Paolo Guiotto: that characterizes any conditional expectation. So, we use this form here.
19:53:20Paolo Guiotto: So, if we are able to prove that expected value of X times Z equal expected value of… now there will be phi of Y times Z, because Y here in this notation is
20:08:40Paolo Guiotto: the conditional expectation, so sorry for the notations, but G is Y here, so we should say another letter here, so here there will be phi of Y, I write, and I cancel immediately, because otherwise, at the end.
20:22:410Paolo Guiotto: will be. Okay, so let me write it down here.
20:27:970Paolo Guiotto: we have to check if and only if, D.
20:34:260Paolo Guiotto: conditional… the expectation of X times Z is equal to the expectation of the conditional expectation, the candidate, phi of Y times Z, for every Z, which is an L infinity
20:51:130Paolo Guiotto: G… measurable function, where G is the sigma algebra generated by Y.
20:59:50Paolo Guiotto: So, in other words, we can say that Z will be a function, say, C, of Y, where C is a bounded function.
21:15:600Paolo Guiotto: So, we have to show that this identity holds. Let's start from this.
21:19:740Paolo Guiotto: So we compute expectation of phi y times a generic function CY,
21:30:670Paolo Guiotto: Now, since as you can see, everything here depends on y, we transform with the density into the integral on r of phi of little y, c of little y times the density of capital Y, DY in DY.
21:50:290Paolo Guiotto: Now, let's plug the, definition of phi.
21:55:100Paolo Guiotto: And you see that when FY of Y is 0,
21:59:00Paolo Guiotto: Okay, so we could split this integral into the integral on Y is for which FY is equal to 0, plus the integral on Y is for which FY is different from 0.
22:09:840Paolo Guiotto: But the integral when FY is 0, whatever is between here is 0. And by the way, this will be zero, this is something, but it doesn't matter, this is also zero. So we have that everything here is zero.
22:22:970Paolo Guiotto: So, clearly, this integral… this first integral is equal to zero.
22:28:400Paolo Guiotto: In the second integral, f of y is different from 0, so we plug that formula. So we have that this is…
22:37:620Paolo Guiotto: 1 over F, Y, Y, integral on R of X, the joint density, FXY, DX.
22:50:550Paolo Guiotto: And then, out here, we have still the C of Y.
22:56:70Paolo Guiotto: the density, F, Y, of Y, DY.
23:01:420Paolo Guiotto: Well, as you can see, we can simplify these two densities.
23:06:870Paolo Guiotto: Now, we put everything under a unique integral, so because you have… let's restore now, since,
23:17:700Paolo Guiotto: We have… this is, the integral.
23:21:990Paolo Guiotto: of integral in R, we can carry this inside, it's constant for the first integration. We have X, sorry, C,
23:31:860Paolo Guiotto: of Y.
23:34:230Paolo Guiotto: F, X, Y.
23:37:670Paolo Guiotto: XY, DX, and then DY.
23:41:760Paolo Guiotto: And this is the double integral, so integral on R2 of XC, of Y, F, X, Y,
23:53:60Paolo Guiotto: X, Y, DX, DY.
23:56:690Paolo Guiotto: Now, since this is the joint density of the pair XY, and you have the integral of this function with respect to this joint vent density, this is the expectation of
24:12:130Paolo Guiotto: The function where you replace the little x, the capital X, and the little y, the capital Y, so we got this.
24:20:230Paolo Guiotto: And as you can see, we have the conclusion, because we started from the expected value of phi of Y, C of Y,
24:28:490Paolo Guiotto: And we obtained that this is equal to the expected value of X times C of Y, which is exactly what we need to prove, no? Expected value of X times Z, where Z is a function of Y,
24:42:430Paolo Guiotto: So, if you see, it's exactly the identity we needed. So this is the conclusion of the platform.
24:51:790Paolo Guiotto: We have only 5 minutes, so we can,
24:56:420Paolo Guiotto: maybe do an example, I don't know if I… I… Finish.
25:01:50Paolo Guiotto: So this is the example 724. It says, suppose that X is normal, mean m variance sigma squared, and Y is still normal, mean 0, variance sigma squared.
25:18:90Paolo Guiotto: And they are independent.
25:21:920Paolo Guiotto: So, it's asked to determine The conditional expectation of X plus Y given X, so… and of X, so…
25:35:660Paolo Guiotto: given X plus Y.
25:41:00Paolo Guiotto: Now, the first one is, easy, because it can be…
25:45:990Paolo Guiotto: solved immediately by just using the properties of the conditional expectation. Because if you do the conditional expectation of X plus Y,
25:57:40Paolo Guiotto: given X, By linearity, you can split into the expected value of X
26:04:570Paolo Guiotto: given X plus the expected value of Y given X.
26:11:960Paolo Guiotto: But the first one is the expected value of X, given the sigma algebra generated by X. So, since X is measurable with respect to the sigma algebra that it generates, you get what here? This is equal to…
26:26:570Paolo Guiotto: X.
26:28:230Paolo Guiotto: About this second one, we know that Y is independent of X, so Y will be independent of the sigma algebra generated by X, and when a variable is independent of a sigma algebra, the conditional expectation is just the expectation. Why?
26:44:650Paolo Guiotto: So the expectation of Y is 0.
26:47:650Paolo Guiotto: So we get that at the end, this conditional expectation is just X.
26:52:800Paolo Guiotto: For the other one, it's more complicated, because,
26:56:670Paolo Guiotto: Here, we want to compute the conditional expectation of
27:00:140Paolo Guiotto: X given X plus Y.
27:03:500Paolo Guiotto: Now, one may think, okay, let's add and subtract Y, so the conditional expectation of X plus Y is easy, because it's X plus Y given X plus Y, but then we have to compute Y given X plus Y, so we basically get the same problem.
27:17:460Paolo Guiotto: So, how can we compute this? Well, here, we can use this formula. So, if you want to compute the conditional expectation, now here, unfortunately, the letters are XY, but let's say, of a variable with respect to another.
27:34:400Paolo Guiotto: Okay, provided you know that the pair is absolutely continuous, you can do through this formula.
27:42:490Paolo Guiotto: So the steps are, we need as pair… so the two variables are X and X plus Y, okay? So let's take X and X plus Y.
27:54:220Paolo Guiotto: So, we apply.
27:59:580Paolo Guiotto: The… Conditional.
28:04:780Paolo Guiotto: Density.
28:08:750Paolo Guiotto: formula.
28:11:580Paolo Guiotto: To this sale, we have to… Determine faster. We have… to determine… First.
28:23:720Paolo Guiotto: If the pair, which is the pair is XX plus Y,
28:29:310Paolo Guiotto: is absolutely conditioned. It is absolutely continuous.
28:34:920Paolo Guiotto: Why this pair? I repeat, because I'm doing the conditioning with respect to the second variable, so I have to take these two.
28:41:860Paolo Guiotto: And I need to understand if this is absolutely continuous. So now let's treat this as a new random B variant random variable, and we need to determine whether or not we have a density.
28:55:500Paolo Guiotto: Now, since this is a transformation, is a map phi of the original pair, XY, for which we know that there is a joint density, no?
29:07:540Paolo Guiotto: We know that the joint density of FXY is, and this is important because they are independent, it is said in the assumption, so by independence.
29:20:70Paolo Guiotto: By independence, I know that this is the product of the two densities, so this is at point XY, this is FX of X times FY of Y. So I know that the pair XY is absolutely continuous, and the minute pair, XX plus Y is a function of this one.
29:37:780Paolo Guiotto: Now, what kind of function? The function is easy, because the function is, of course, phi of XY is the map X plus Y is, say, a linear map, if we call this ZW.
29:54:150Paolo Guiotto: This is equivalent of saying that Z is X and W is X plus Y, so we can easily see that it is invertible, because we have X is Z,
30:07:110Paolo Guiotto: And, Y is W minus X, so W minus Z. So this is the phi minus 1.
30:14:910Paolo Guiotto: So it's clearly a big jack, and it's a linear map, so there is no problem.
30:20:610Paolo Guiotto: So we know that the density of the pair X plus Y, which is the pair ZW with these notations, so perhaps it is better to use these notations, ZW.
30:35:550Paolo Guiotto: For the letters here. This is the original density, FXY, evaluated at phi minus 1ZW.
30:45:710Paolo Guiotto: times the modulus of the determinant of the Jacobian matrix of the inverse map.
30:54:150Paolo Guiotto: Which is easy, because the determinant of phi minus 1 prime is the determinant of… here we have a 2x2 matrix.
31:06:310Paolo Guiotto: So, the phi minus 1, phi minus 1 of ZW is ZW minus Z.
31:16:950Paolo Guiotto: the derivatives of respect to the variables of phi minus 1, so ZW. So, the first component is this, derivative with respect to Z is 1, derivative with respect to W is 0, the second component is this, derivative with respect to Z is minus 1, respect to W is 1.
31:33:170Paolo Guiotto: And the determinant of this is just a constant equal to 1.
31:37:620Paolo Guiotto: So, we have that the joint density for the pair XX plus Z
31:44:560Paolo Guiotto: Sorry, it's not X plus Z, but X plus Y.
31:54:370Paolo Guiotto: at point ZW is the initial density.
32:00:610Paolo Guiotto: evaluated at phi minus 1, so ZW minus Z times 1, that comes from the modulus of the determinant, etc. Now, this is FX at Z times F
32:15:680Paolo Guiotto: Y at W minus Z.
32:21:50Paolo Guiotto: we can write, but if we don't need to write… so we can write because we know that they are Gaussian, so we have this formula. But let's see where do we need this. Now we have the joint density of XX plus Y, so we have…
32:40:150Paolo Guiotto: the ingredient that we need in this formula, but we need also the density of the second variable, which is X plus Y, okay?
32:49:850Paolo Guiotto: Now, the density of X plus Y
32:54:380Paolo Guiotto: is obtained by the, let's say, as function of the variable W, is obtained by the joint density integrating in, in the first variable. So, it is the integral on R of FXX plus Y,
33:12:330Paolo Guiotto: Zed.
33:13:790Paolo Guiotto: W.
33:15:190Paolo Guiotto: in DZ.
33:16:800Paolo Guiotto: So we have to compute this integral.
33:20:700Paolo Guiotto: F, X, Z, F, Y, W, minus Z in DZ.
33:29:210Paolo Guiotto: Now, this integral is a convolution, you see?
33:33:340Paolo Guiotto: So I should compute the convolution of FX star FY at point W.
33:42:250Paolo Guiotto: Now, to compute this convolution, we could do, in principle, we have the ingredients, FX is the Gaussian, so if you want FX of X is, what is it, is 1 over root of 2 pi…
33:58:790Paolo Guiotto: sigma squared, E minus its mean M, X minus m square divided 2 sigma squared, and we have also FYY, which is the same with mean 0, 1 over root of 2 pi.
34:13:670Paolo Guiotto: sigma square, E minus Y squared over 2 sigma squared.
34:21:50Paolo Guiotto: Well, time is over…
34:23:150Paolo Guiotto: Well, let me just indicate how to proceed. You could compute, but here, the best thing to do to compute that convolution is to use the Fourier transform, because if you do the Fourier transform of the convolution.
34:42:270Paolo Guiotto: This is the product of the Fourier transforms, sir.
34:46:660Paolo Guiotto: of the two, which is easy.
34:51:340Paolo Guiotto: And then you should easily recognize that it is the Fourier transform of something. And therefore, this something is what you need to put here.
34:59:890Paolo Guiotto: Okay.
35:00:850Paolo Guiotto: So do this, maybe I will, I will,
35:05:780Paolo Guiotto: Because there is too much time to mandate, I could write the solution later.
35:13:40Paolo Guiotto: Once you have done this, you have the, if you want, I can…
35:18:420Paolo Guiotto: Tell you what is the result.
35:24:820Paolo Guiotto: What is it?
35:27:870Paolo Guiotto: I've not written yet.
35:29:790Paolo Guiotto: So once you have done this, you have the density of X plus Y, you have the joint density, so you have now the ingredients to write the conditional density. So the conditional density of X given X plus Y
35:44:880Paolo Guiotto: let's say, let's still use the letters Z given W, otherwise we should write X given X plus Y. This is the joint density of the two, XX plus Y,
36:01:460Paolo Guiotto: ZW, and we have this, we computed here, is this guy here.
36:08:430Paolo Guiotto: divided by… The density of…
36:15:130Paolo Guiotto: X plus Y, which is this convolution, so you have to compute the convolution, so the density of X plus Y
36:24:260Paolo Guiotto: at point W.
36:26:930Paolo Guiotto: And, now you should, once you have this, so 1 to this, 2 to this, and 3, the conditional expectation of X given, X plus Y
36:43:850Paolo Guiotto: equal W will be the integral on R of X times the conditional density, X given X plus Y.
36:56:30Paolo Guiotto: X… given W, DX.
37:01:320Paolo Guiotto: So once you have done these three, you have the result.
37:05:910Paolo Guiotto: So maybe I will, it will take about 5 minutes yet to do this, to finish with the details. I will add the calculation later.
37:19:870Paolo Guiotto: And let me just leave… exercises to do. Do the seven… Three. One, two… 3…
37:36:610Paolo Guiotto: 4…
37:42:60Paolo Guiotto: Five.
37:43:730Paolo Guiotto: Then there is… the remaining are theoretical exercises, so I would suggest you to do
37:52:120Paolo Guiotto: Maybe the 7… 3… 7… And they ate.
38:01:680Paolo Guiotto: Okay.
38:04:290Paolo Guiotto: Okay, sorry for… I think being a little bit longer.
38:10:230Paolo Guiotto: Let's stop.