AI Assistant
Transcript
00:32:460Paolo Guiotto: Okay, good morning. Yesterday, we computed the density for a lognormal distribution.
00:42:220Paolo Guiotto: Log normal means a random variable whose logarithm is a Gaussian, is a normal random variable, or equivalently, the random variable is the exponential of a Gaussian random variable. Of course, we understand that this is a stylized notation, because
01:01:250Paolo Guiotto: And m sigma squared means a random variable distributed with that.
01:06:760Paolo Guiotto: distribution. Okay, now this is the, the densities, so let's see what can be computed from this. So, just to do some calculation to see… take a bit of confidence with these, concepts. For example.
01:23:380Paolo Guiotto: CDF here.
01:27:360Paolo Guiotto: The CDF, in this case, is not complicated, because we should… we should start… it's better to start from the definition, because this is the probability that capital X is less or equal than little X.
01:44:450Paolo Guiotto: We don't need to pass… we may say that, in principle, it should be the integral… since I have a density, I could write the CDF as integral from minus infinity to X of the density, say, in a variable YDY.
02:00:640Paolo Guiotto: But instead of computing this integral, it is much easier… well, much easier, we can do, but at the end, we cannot compute, as you will see.
02:09:800Paolo Guiotto: If we just rewrite what is this? This X is e to y, where Y is a Gaussian, less or equal than X.
02:21:90Paolo Guiotto: So now, since logarithm is a strictly increasing function, I may observe that e to y is less or equal than x if and only if applying logarithm
02:34:550Paolo Guiotto: both sides, so log of E to Y is less or equal than log of X.
02:41:590Paolo Guiotto: And this means that Y must be less or equal than log of X.
02:47:570Paolo Guiotto: So at the end, this turns out to be the probability that Y is less or equal than log of X, where Y is Gaussian, normal, mean m, and variance sigma squared.
03:03:690Paolo Guiotto: And this is the CDF, the cumulative distribution function, of Y evaluated at point log of X.
03:15:430Paolo Guiotto: Now, we know that for a Gaussian, we do not have, let's say, an explicit expression for the CDF.
03:25:780Paolo Guiotto: Unless we consider the function, which is the CDF of the standard Gaussian, the mean 0 minus 1 Gaussian, which is the function we…
03:37:710Paolo Guiotto: Introduced, here.
03:40:80Paolo Guiotto: the function capital phi as an elementary function, and in fact, it is considered as an elementary function. So, this that you see here is the CDF of a random variable X, which is distributed as a Gaussian, mean m variable sigma squared. So we just take this formula.
04:01:70Paolo Guiotto: and replace X with log X. So we get, down here, this.
04:08:130Paolo Guiotto: This is phi, capital phi, of log X minus m divided by sigma.
04:17:390Paolo Guiotto: That's the, CDF of X.
04:23:160Paolo Guiotto: Is it the phi of log X?
04:28:20Paolo Guiotto: minus N divided sigma. This 4X are positive.
04:33:530Paolo Guiotto: Because for X negative, the CDF will be 0. Well, let's say, roughly, when X is 0, log of x is minus infinity, the argument of that capital phi would be minus infinity, and since capital phi of
04:48:440Paolo Guiotto: U is the integral from minus infinity to u of the standard motion.
04:54:960Paolo Guiotto: P minus T squared half divided root of 2 pi dt, the value at 0 is 0, and the value for x negative is 0, okay? Because,
05:05:700Paolo Guiotto: the probability that X, capital X, is less than little x negative, it's zero, no? E2Y cannot be negative, so it cannot be less than any negative number. So we may say that this is 4x positive, and 0 for X less than or equal than 0.
05:24:140Paolo Guiotto: That's the CDF of this, distribution.
05:30:50Paolo Guiotto: We may consider, in some sense, this as explicit.
05:35:530Paolo Guiotto: Let's do some other calculation, for example. Suppose we want to compute the expected value of X for this,
05:45:250Paolo Guiotto: random variable, which is, let's see what… which form should be convenient. Now, we know that if we have a density, this is the integral on R of X times the density.
05:59:370Paolo Guiotto: in the Xer.
06:01:280Paolo Guiotto: Now, the density is this function, which is 0 for x less frequent than 0, and that expression for X positive. So, I can say the integral restricts to the interval from 0 to plus infinity. I have X times
06:17:890Paolo Guiotto: Now I copy this density. Since I have the root of X squared, I carry outside it, right, just a modulus of X for X positive.
06:27:280Paolo Guiotto: is just X.
06:29:390Paolo Guiotto: Turned on, yeah.
06:33:180Paolo Guiotto: Okay.
06:34:430Paolo Guiotto: So, I haven't… This is 1 over X, root of 2 pi sigma squared.
06:44:420Paolo Guiotto: E minus… we have the log of X minus m squared divided to sigma squared dx. This is the expected value.
06:56:940Paolo Guiotto: Now, you'll see that we can… Simplify a little bit this.
07:02:600Paolo Guiotto: And now, to compute that integral, we can change a variable, setting y equal log of X.
07:12:240Paolo Guiotto: So this means that the X is, of course, E to Y, so DX is E2YDY.
07:21:210Paolo Guiotto: Now, since y is log of X, the range for X 0 to plus infinity becomes the range for y minus infinity to plus infinity, 1 over root of 2 pi.
07:34:300Paolo Guiotto: sigma squared, then we have E minus log of X becomes Y, so Y
07:42:330Paolo Guiotto: Y minus M squared divided by 2 sigma squared.
07:47:900Paolo Guiotto: times, then, DX becomes E2YBY.
07:54:420Paolo Guiotto: So we have to compute this integral, which is basically a Gaussian integral. We have, to…
08:01:900Paolo Guiotto: to reorder this in… to write in a way that it becomes a Gaussian integral. So I'm just showing this kind of calculations, not because they are interesting, but because
08:13:20Paolo Guiotto: Sometimes you have to do this kind of manipulations, and so it's better to get familiar with this kind of…
08:22:510Paolo Guiotto: Thanks.
08:24:90Paolo Guiotto: So they are, let's say, a hard worker, not particularly soft worker. We write integral from minus infinity plus infinity, we put in a unique exponential.
08:34:870Paolo Guiotto: So, what I do is, let's emphasize the fact of 1 over 2 sigma squared with a minus, and then we have that Y minus M squared
08:47:400Paolo Guiotto: I… I will expand the D squared, so I have Y squared minus 2MY plus M squared.
08:58:480Paolo Guiotto: And then I have this hit Y, that since I am, I have not the factor minus 1 over 2 sigma squared, I have to put minus 2 sigma squared Y here.
09:12:240Paolo Guiotto: To make this correct.
09:15:680Paolo Guiotto: Now, we have to rework this expression to recognize a square in Y. So let's work on this. This is Y squared, we may say minus 2M plus sigma square y.
09:34:650Paolo Guiotto: So this is the double product.
09:37:140Paolo Guiotto: So I want to recognize a square, I need the square of M plus sigma squared.
09:44:70Paolo Guiotto: Which I've not, so I have to add and subtract M plus sigma squared squared.
09:49:990Paolo Guiotto: And then there is this term, plus M squared, which is, however, a constant.
09:55:730Paolo Guiotto: So this part here is now Y minus
10:02:90Paolo Guiotto: Plus sigma squared, all this squared.
10:07:260Paolo Guiotto: So, the integral becomes 1 over root of 2 pi sigma squared, integral from minus infinity plus infinity.
10:16:950Paolo Guiotto: E minus 1 over 2 sigma squared. I have this, Y minus M plus sigma squared, or this squared.
10:28:410Paolo Guiotto: And then, what remains is something that is independent of Y. It's a constant, so I write directly outside.
10:39:360Paolo Guiotto: And what is this factor? It is exponential m square minus M squared M plus sigma squared
10:51:660Paolo Guiotto: Squared, divided by… divided by, or multiplied by, this thing, minus 1 over 2 sigma squared.
11:03:850Paolo Guiotto: Now, the integral… is easy, because if you redo the change of variable, pull all this quantity Z,
11:14:440Paolo Guiotto: So that's just a translation of the argument, because you set Z equal Y minus something. So this will become 1 over root of 2 pi sigma square integral from minus infinity plus infinity, E minus z squared
11:31:620Paolo Guiotto: This is a…
11:35:710Paolo Guiotto: divided by 2 sigma squared, DZ. DZ is equal to DY, because it's just a translation.
11:45:430Paolo Guiotto: And all this quantity is equal to 1.
11:50:270Paolo Guiotto: And, outside we have this… well, perhaps it's better if you write us…
11:55:870Paolo Guiotto: E2 fraction, 2 sigma squared at denominator, then extending the denominator and changing sine here, here.
12:04:810Paolo Guiotto: So we have M squared that cancels M squared.
12:08:770Paolo Guiotto: Now, we have the double product, 2M.
12:11:690Paolo Guiotto: Sigma squared plus sigma squared squared, so sigma to power 4.
12:18:570Paolo Guiotto: divided by that 2 sigma squared. At the end, we have that the value is e to m plus sigma squared divided 2.
12:31:660Paolo Guiotto: Okay, so the formula is, if X is e to a Gaussian of mean m and variance sigma squared, the expected value of X is finite.
12:46:860Paolo Guiotto: And it is equal to E2M plus sigma squared divided 2.
12:54:110Paolo Guiotto: There is this formula.
12:57:250Paolo Guiotto: So in particular, it is not E2M.
13:00:970Paolo Guiotto: even if m is the mean value of the exponent, the mean value of the exponential is not e to the mean value of the exponent. It's a bit…
13:10:120Paolo Guiotto: different?
13:12:270Paolo Guiotto: And similarly, you could compute the variance.
13:17:10Paolo Guiotto: It's a similar calculation, and you will obtain that the variance of this Random variable is,
13:26:30Paolo Guiotto: E to sigma square minus 1.
13:29:40Paolo Guiotto: E to 2M plus sigma squared, so it's the square of the… number.
13:37:120Paolo Guiotto: So these are the parameters for this distribution.
13:42:800Paolo Guiotto: Okay, now, I, published this morning basically half of the notes.
13:50:220Paolo Guiotto: So you have also exercises, I brought you to do some of the exercises, now we do some of these, but not all.
14:00:670Paolo Guiotto: So I now change a page here.
14:06:400Paolo Guiotto: I got the new one.
14:08:410Paolo Guiotto: So, for example,
14:10:840Paolo Guiotto: Let's take the exercise… these are exercises on these few things we have seen, CDF, densities, and things like this, transformations of random variables. So, the exercise 342.
14:27:500Paolo Guiotto: So, the question is, for… the… following… F.
14:41:990Paolo Guiotto: they… Yes, but… they… Cdf.
14:56:860Paolo Guiotto: So, for example, let's take the number 1.
15:01:380Paolo Guiotto: Here we have this function. F of X is equal to…
15:06:830Paolo Guiotto: 1 minus E to minus X divided by 3 times the indicator interval 0 plus infinity of X.
15:17:840Paolo Guiotto: Now, we remind that a CDF is basically characterized by its characteristic properties, no? So, to be… Cdf…
15:35:960Paolo Guiotto: masks.
15:38:180Paolo Guiotto: Verify… D.
15:43:250Paolo Guiotto: characteristic… Properties.
15:53:560Paolo Guiotto: that are. So we say, the number one, the function F must be increasing, not necessarily strictly increasing.
16:04:410Paolo Guiotto: So let me refresh quickly. Number two, the value at minus infinity must be 0, the value at plus infinity must be 1. By the way, these two together means that… implies that the function f must be always between 0 and 1, no?
16:23:250Paolo Guiotto: Number 3, the F must be right continuous.
16:32:130Paolo Guiotto: Continos.
16:34:570Paolo Guiotto: And left limit, well, if it is increasing, there is always the left limit, and it is less rippled than the function at that point.
16:43:20Paolo Guiotto: So these are basically the three characteristic features. Now, let's see if this function verifies. Well, in this case, we could also
16:51:740Paolo Guiotto: do a plot of this function to see what is it, no? So, in testing, we noticed that the function is 0 for x negative because of the indicator that cuts off everything. So, we may say that the function f of x
17:09:60Paolo Guiotto: can be also written in this way. It is 0 for X negative, it is 1 minus E to minus X divided 3 for X positive.
17:19:30Paolo Guiotto: Now, for X negative, it is 0, like this. For X positive, it is 1 minus E2 minus X. For X equals 0,
17:29:390Paolo Guiotto: You see that the value e to 0 is 1, so 1 minus 1 third is 2 thirds, so we have, let's say, 1 here. We have value 2 thirds here, and that's the value of the function here.
17:45:00Paolo Guiotto: Then E2 minus X would be decreasing, clearly, so with the minus in front becomes increasing, so this function clearly is increasing.
18:00:990Paolo Guiotto: And, if we compute the limit at plus infinity.
18:07:120Paolo Guiotto: So this is decreasing, and therefore F is increasing, also because the value, of course, at 0 is above the limit from the left.
18:17:640Paolo Guiotto: And, so the value at minus infinity is clearly equal to 0, the value at plus infinity, when we send X to plus infinity, the exponential goes to zero, so this is the limit for X going to plus infinity of that quantity, 1 minus
18:35:560Paolo Guiotto: E2 minus X divided by 3, but this guy goes to 0, so the limit is 1, as it should be.
18:44:180Paolo Guiotto: So the function is increasing, it goes to 1 at infinity, it goes to 0 at minus infinity, so 1 and 2 are verified, and it is right continuous. This is for X greater or equal than zero, and F is clearly continuous.
19:03:430Paolo Guiotto: R, except .0.
19:07:240Paolo Guiotto: And the limit for X going to 0 from the right of the function is exactly the value of the function at 0. So the right continuity is at the unique… so F…
19:23:780Paolo Guiotto: Capital F is right.
19:28:30Paolo Guiotto: continuous… on, and so the answer is, yes.
19:37:470Paolo Guiotto: F… is it.
19:40:190Paolo Guiotto: CDF.
19:44:10Paolo Guiotto: So this is the kind of check, we should do on the other,
19:51:20Paolo Guiotto: Examples, so do… there are two others, do 2 and 3.
19:59:840Paolo Guiotto: Well, let's do the next one, which is similar, exercise 3-4-3.
20:07:50Paolo Guiotto: Here, the problem says, let F of X be this function.
20:14:390Paolo Guiotto: be defined as 0 for X negative.
20:18:440Paolo Guiotto: And for X positive, you take 1 minus 3 to exponent minus the integer part of X.
20:26:260Paolo Guiotto: Or that's greater equivalency, though.
20:29:290Paolo Guiotto: Okay, so this thing is the integer.
20:37:340Paolo Guiotto: But… of number XL.
20:44:190Paolo Guiotto: which I remind you, the integer part of a number, a positive number, X, is a natural number.
20:54:670Paolo Guiotto: Such that,
20:56:530Paolo Guiotto: The number X is greater or equal than its integer part, but less, strictly less than integer part of X plus 1.
21:06:530Paolo Guiotto: So, you know, any real will be between two consecutive naturals, and the left, the smallest between these two, is the integer part of… of…
21:18:650Paolo Guiotto: X Now, the question one is the following, is capital F, CDF?
21:30:740Paolo Guiotto: And question two, if yes… Compute the probability
21:42:820Paolo Guiotto: what I say, to complete it, to write properly the question.
21:47:350Paolo Guiotto: Capital F is the F of some… Capital X.
21:53:870Paolo Guiotto: If yes, compute the probability that capital X be greater than 3,
22:00:100Paolo Guiotto: and the probability that capital X be equal to 2.
22:08:530Paolo Guiotto: Okay.
22:09:550Paolo Guiotto: So, to respond to the first question, again, we have to check that DSF fulfills the characteristic properties of any
22:18:850Paolo Guiotto: CDF. In this case, let's see how he's made this function, first of all, before we respond. So, the function, capital F, is clearly
22:32:240Paolo Guiotto: 0 for X negative.
22:38:160Paolo Guiotto: So at least up to zero, then there is… for x strictly negative, it is 0. For X greater or equal than zero, it is 1 minus 3 to
22:46:970Paolo Guiotto: minus integer part of X. Now, you know that the integer part is the same for all numbers between two integers, no? So this would make reasonable to say that, for example, I could say the integer part of X is equal to… when x is greater or equal than zero, but less than 1, the integer part is 0.
23:09:890Paolo Guiotto: When X is between 1 included and 2 excluded, this is 1.
23:15:510Paolo Guiotto: When X is between 2 and 3,
23:18:810Paolo Guiotto: this is 2, and so on, no? So, this quantity is, in fact, a constant on these intervals, and this suggests that maybe we should divide everything here into intervals with endpoints, the integers.
23:35:310Paolo Guiotto: Because now, for x between 0, 1, since the integer part is 0, what is the value of f of x for this X? F of X is always 1 minus 3 to minus the integer part of x when x is greater or equal than 0. So we may say that
23:54:940Paolo Guiotto: If X is greater than 0, less than 1,
23:58:520Paolo Guiotto: This is 1 minus 3 to… the integer path is 0, so 3 to minus 0. So, 1, it's like 3 to 0, 3 to 0 is 1, 1 minus 1, 0. So the function is 0, again, between 0 and 1, so we continue in this way.
24:18:230Paolo Guiotto: Up to this point.
24:19:900Paolo Guiotto: 1x is between 1
24:23:340Paolo Guiotto: Great or equal 1, less than 2. Now, the integer part, we said it is equal to 1, so this is 1 minus 3 to minus 1, so 1 minus 1 third, it is 2 thirds.
24:37:150Paolo Guiotto: So, it's, say, let's say that Rick said, yes, this is one, two-thirds about here.
24:43:510Paolo Guiotto: So… The function is constantly equal to 2 thirds up to x equal 2.
24:51:820Paolo Guiotto: Now, for X greater or equal than 2, but less than 3, the integer part will be equal to 2, so we will have 1 minus 3 to minus 2, so…
25:06:280Paolo Guiotto: 1 minus 1 9th, so 8 ninths, okay?
25:12:610Paolo Guiotto: And so on. So, we will have that, between, 3 and 4.
25:17:910Paolo Guiotto: You see that perhaps it is convenient, since we have powers of 3, basically, here, no? So this is 2 thirds, this is 3 square, and down upstairs, we have 3 square minus 1.
25:34:960Paolo Guiotto: So the next one, and this is, 3 minus 1 divided 3.
25:41:90Paolo Guiotto: So you see there is a similar formula. So, in general, we will have… this will be probably 3 to 3 minus 1 divided by 3 tubed.
25:50:830Paolo Guiotto: So, in general, When X is between n and n plus 1,
25:58:150Paolo Guiotto: The value will be 3 to N.
26:00:780Paolo Guiotto: or if you prefer, 1 minus 1 over 3 to n.
26:04:980Paolo Guiotto: When it's a…
26:07:720Paolo Guiotto: Okay, so, this means that this, the next interval, this will be, 1 minus, 1 9th, so it will be very close to 1, but below.
26:21:330Paolo Guiotto: And then… and so on.
26:23:690Paolo Guiotto: Of course, we cannot throw too many.
26:27:10Paolo Guiotto: Pieces of this.
26:29:970Paolo Guiotto: Okay, so this is the F, huh?
26:33:10Paolo Guiotto: we can see that it looks like an increasing. It must… it is not needed that it is strictly increasing. It is increasing. At minus infinity, it goes to 0. At plus infinity, it seems it goes to 1, and it seems to be right to continue. So, we can cite all these things.
26:54:680Paolo Guiotto: If you want, we can say, so F for… is constant equal to 0 for… X negative.
27:05:50Paolo Guiotto: For X positive, F of X is equal to 1 minus 3 to minus integer part of X for X greater than equal to 0.
27:14:680Paolo Guiotto: Now, what is the behavior of the integer part of X when you increase X? When you increase X, you increase the integer part of X, no? As it is natural. So this quantity is increasing with the minus is decreasing
27:29:50Paolo Guiotto: 3 to… the exponential with base 3 is an increasing function, so if the argument increases, it increases. If the argument decreases, it decreases. So, if X is increasing, the integer part increases.
27:43:410Paolo Guiotto: Minus the integer part decreases, so the exponent decreases.
27:47:860Paolo Guiotto: The exponential will be decreasing, because it is an exponential with the base greater than 1, with the minus will be again increasing. So, all the function is definitely an increase in function.
28:04:940Paolo Guiotto: So, F…
28:06:190Paolo Guiotto: is increasing. F at minus infinity is clearly equal to 0. F at plus infinity is what? Well, is the limit when x goes to plus infinity of this 1 minus 3 to minus the integer part of X. But when x goes to plus infinity, this quantity goes to plus infinity, so the
28:30:150Paolo Guiotto: exponential goes to 0, and therefore the argument of the limit will go to 1.
28:36:130Paolo Guiotto: And finally, F is right continuous, because as we can see, this is piecewise constant function.
28:46:390Paolo Guiotto: So it's a constant function, everywhere, except, let's say, on each interval, so from minus infinity to 1, then from 1 to 2, excluding the end points, and from 2 to 3. So at the end, we can say, clearly.
29:04:140Paolo Guiotto: This F is a continuous function everywhere, except the points 1, 2, 3, etc.
29:15:640Paolo Guiotto: At this point, however, when you do the limit.
29:20:290Paolo Guiotto: well, it is evident from the figure. If you want at this stage, we could say it is clear from the figure that this function is continuous also at these points. But if I do the limit when x goes to some natural n of my function f of x, this is what? This is the limit
29:37:770Paolo Guiotto: So N is supposed to be greater or equal than 1 natural here.
29:43:470Paolo Guiotto: So I'm doing the limit at the points where I don't know if… I'm sure the function is not continuous.
29:51:180Paolo Guiotto: and I want to show that it is right continuous, so I do the right limit, and the idea is that you are moving X to n from this side, so what is the limit? The limit is of 1 minus 3 to minus the integer part of X.
30:08:300Paolo Guiotto: But, since you are moving, this is the interval from n to n plus 1, you are moving your X here, going to n from the right.
30:18:630Paolo Guiotto: It is clear that the integer part of this X, of an X, which is between n and n plus 1 at right of n, is exactly equal to n. So that quantity is constantly equal to number n.
30:33:700Paolo Guiotto: So the quantity you have inside here is constantly equal to 1 minus 3 to minus n. It's constantly with respect to the variable x, which is moving to n from the right.
30:47:260Paolo Guiotto: So, N+.
30:50:920Paolo Guiotto: Since we are doing limit of a constant, the limit will be the value of that constant, Y minus 3 to minus N, and this is exactly the value of F at point n. So this shows the right continuity. So the conclusion is that, yes.
31:12:420Paolo Guiotto: F is… the CDF… of some…
31:20:180Paolo Guiotto: random variable capital X. It's because we proved that
31:25:330Paolo Guiotto: If you have a function that verifies this characteristic function, there is always some probability space, some random variable for which that F is the CDF, okay? So the CDF basically identifies
31:39:500Paolo Guiotto: identifies the random variable. Now, the second question is to compute the probability that that random variable is greater than 3, and the probability that it is equal to 2.
31:52:620Paolo Guiotto: Now, how do we compute the probability that X is greater than 3? It's CF, because the CDF is probability that X is less or equal than something.
32:05:350Paolo Guiotto: However, we can transform this to make a CDF value, no?
32:12:620Paolo Guiotto: Ow.
32:15:410Paolo Guiotto: Yes, exactly, the complementary. So, this is the complementary of the set where X is less or equal than 3.
32:26:850Paolo Guiotto: So the probability of the complementary is 1 minus the probability of the set, so probability that X is less ripple than 3, and so we have 1 minus F of 3.
32:38:50Paolo Guiotto: So if we want, this is 1 minus, there is 1 minus 3, 2, what is it, minus 3, so we get, 1 over 3 to 3, 1 over 27. That's the probability.
32:53:790Paolo Guiotto: About this one, probably the tax is equal to 2.
33:08:560Paolo Guiotto: You see?
33:14:670Paolo Guiotto: Now, in general, if there is a density, so if the variable is absolutely continuous, which is not the case here, because to be absolutely continuous, the CDF must be continuous.
33:27:740Paolo Guiotto: And not only continuous must have a derivative at least almost everywhere, but here it is not even continuous, the function, so this is not absolutely continuous, for sure.
33:40:330Paolo Guiotto: If it is absolutely continuous, we will see in a moment, these quantities are always zero, no? Because the idea is that the variable is… has a distribution along all the real line, and so points will not have any particular relevance. But for this.
33:59:70Paolo Guiotto: We may say that this is the same of the event where X is less or equal than 2.
34:06:770Paolo Guiotto: Well, I take out the omegas for which X is strictly less than 2.
34:15:850Paolo Guiotto: You see?
34:17:449Paolo Guiotto: Now, this set here, acts strictly less than 2, is clearly contained into this one.
34:26:719Paolo Guiotto: So, yeah, the probability of big set minus a set which is included, no? So it's like A minus B with B contained into A.
34:37:770Paolo Guiotto: And this, since the probability is a finite measure, we have that the probability of a set difference in this circumstance is the difference of the probabilities. So this is P of A minus P of B,
34:55:409Paolo Guiotto: This because,
35:00:420Paolo Guiotto: P is a finite measure.
35:03:420Paolo Guiotto: So, we, we…
35:06:450Paolo Guiotto: We have seen that for a generic measure, in general, this is not true, you cannot do the difference, the measure of difference.
35:13:300Paolo Guiotto: Even if they are, of course, included, you cannot say it is the difference of the measures, but if the measure of the big set… well, actually, if the measure of the small set is we find it, you can do that, okay?
35:27:00Paolo Guiotto: In particular, if, as in the case here, we have a probability measure, everything is fine, and we can do this… this thing.
35:36:130Paolo Guiotto: Okay, so now this is the probability that X is less or equal than 2, and that's a CDF, so it will be F of 2.
35:44:990Paolo Guiotto: Minus the probability that tax is strictly less than 2.
35:49:390Paolo Guiotto: And this is similar, no, it's not similar. So, how…
35:58:720Paolo Guiotto: Well, how we compute this probability?
36:09:180Paolo Guiotto: Yeah, but does that become greater equal?
36:14:400Paolo Guiotto: Just doing more than one.
36:18:670Paolo Guiotto: No, in general, say.
36:21:540Paolo Guiotto: So we know that the probability of a random N… this is generic, this argument. X less or equal than little x, this is what we call, by definition, the CDF, okay?
36:34:610Paolo Guiotto: Now, let's see, in general, these… for example, probability that X is equal to the value little x can be written as we have done here. Probability that X is less or equal than little x.
36:50:860Paolo Guiotto: minus… Excellent.
36:53:730Paolo Guiotto: Strictly less than little x, right?
36:58:230Paolo Guiotto: So, I can use the same argument as above. This set here, this is contained into this one. The measure is finite, so the probability of the difference is the difference of the probabilities. So, this is this minus this.
37:15:800Paolo Guiotto: So we're doing, in general, in such a way that you can…
37:22:680Paolo Guiotto: Remind of these kind of situations.
37:26:490Paolo Guiotto: Now, this is the CDF, FX of X.
37:31:580Paolo Guiotto: And what is this?
37:36:860Paolo Guiotto: in terms of cost of the CDF,
37:40:30Paolo Guiotto: Yeah, that's the limit. So this is the limit when Y goes to X from left of the CDF.
37:51:220Paolo Guiotto: Let's see why.
37:52:890Paolo Guiotto: This is, again, a consequence of the continuity of a measure, because…
38:02:330Paolo Guiotto: So, this is the important fact. Probability that X is less than some value is the limit when Y goes to X from left of the CDF, FX of Y, which is not, in general, the same of
38:20:740Paolo Guiotto: the value of the CDF at point X.
38:25:190Paolo Guiotto: This because,
38:27:530Paolo Guiotto: So, let's start from the right. So, if we do the limit, when Y goes to X minus of FXY,
38:38:650Paolo Guiotto: We are doing the limit when y goes to X minus, so to X from the left, of the probability that X is less or equal than Y.
38:50:840Paolo Guiotto: So, we are doing a limit of probabilities, no, of certain sets. If you call the set EY is the set where X is less or equal than Y, we have here the probability of this EY.
39:08:760Paolo Guiotto: Now, what happens when we move this Y, and we move to the right?
39:17:10Paolo Guiotto: from the left to the value X minus, to the value X.
39:21:630Paolo Guiotto: X minus is not a value. So, we move… X minus means we move Y to X from the left, so with smaller values of X. Now, you understand that, for example, if I take two Ys, say, I take a Y1,
39:42:880Paolo Guiotto: which is smaller than a Y2,
39:46:860Paolo Guiotto: What is the relation between this set? Well, if you are less or equal, then Y1,
39:54:840Paolo Guiotto: you will be less or equal than Y2, no, because Y2 is bigger. So this means that if omega is such that X of omega is less than value Y1, it will be also X of omega less or equal than Y2.
40:11:900Paolo Guiotto: So this family of sets is an increasing family of sets. So this family, EY, is an increasing family of sets.
40:23:140Paolo Guiotto: Now, there is the usual K, because this Y here is a real variable, so it's not a discrete variable. And the continuity property of measure treats with a discrete sequence of sets.
40:40:680Paolo Guiotto: But we could arrange things in this way. Now, if you take a sequence of Y,
40:45:670Paolo Guiotto: that moves from right to left to X, so imagine that you have a sequence of Ys, so Y1, Y2, Y3, etc, at a certain point, YN, YN plus 1,
41:01:620Paolo Guiotto: And so on.
41:03:730Paolo Guiotto: and you define this set, EY1, EY2, EY3, etc, EYN, you can see that this is now a discrete sequence of sets.
41:15:690Paolo Guiotto: They are each included in the next one, so they form an increasing sequence of sets, and therefore, when you do the limit of their probabilities, of the probabilities of the sets YN,
41:30:890Paolo Guiotto: Sending N, sorry, N… to infinity.
41:38:170Paolo Guiotto: You will get what? This is the case of the continuity from below, because we have an increasing sequence of sets, so it doesn't matter if P is a probability or not, so continuity from
41:55:560Paolo Guiotto: Below… This will go to the probability of the union of these sets, EYN.
42:08:150Paolo Guiotto: But what is that?
42:10:570Paolo Guiotto: It is big.
42:13:780Paolo Guiotto: And probably it doesn't like that I have this…
42:23:500Paolo Guiotto: No.
42:25:490Paolo Guiotto: It was as long.
42:27:790Paolo Guiotto: Whatever.
42:33:550Paolo Guiotto: Okay.
42:34:970Paolo Guiotto: But what is the union? Now, the claim is that this union is exactly the set where X is strictly less than little x. Now, so we are doing the union of sets where X is less or equal than YN,
42:51:940Paolo Guiotto: with this YM, a discrete sequence that moves from left to right to point X. And I claim that the union of these sets is exactly the event where X is strictly less than capital X. That's because… so…
43:08:440Paolo Guiotto: X strict less than little x is exactly the union of these, events, X less or equal than YN union in N.
43:21:690Paolo Guiotto: Why? Because…
43:24:60Paolo Guiotto: If you pick an omega, which is in this side, X less than betel x, it means that when you take the random variable X and you evaluate on omega, you have a number which is less than this number little x.
43:42:320Paolo Guiotto: And so, I don't know where is it, but it will be somewhere at last of X, so you have a value here. So sooner or later, there will be a YN which is greater than that X of omega.
43:56:530Paolo Guiotto: No? So there will be… There will be… And then…
44:03:410Paolo Guiotto: Which is depending on omega, but who cares?
44:06:750Paolo Guiotto: It's an N that depends on omega, such that X of omega will be less or equal than that, the corresponding YN.
44:17:700Paolo Guiotto: Because this sequence is going to X, so sooner or later, we'll cross this capital X of omega. But this means that there exists an N
44:28:380Paolo Guiotto: for which our omega belongs to the set where capital X is less or equal than YN.
44:36:110Paolo Guiotto: So, I proved that if I take an omega in the event capital X less than, strictly less than X,
44:43:460Paolo Guiotto: that omega will belong in some of these sets, X less or equal than YN, and therefore, omega belongs to the union of these sets, X less or equal than
44:56:420Paolo Guiotto: YN.
44:57:500Paolo Guiotto: So this is, let's say, the difficult part. It shows that the set where capital X is strictly less than little x is contained into the union of sets where x is less or equal than YN.
45:13:570Paolo Guiotto: The vice versa is actually trivial, because if you have an omega on this distribution, so in one of these, X of omega is less than sum of the YN, which are less than x.
45:27:650Paolo Guiotto: So, we are also in the other part. This part here is, say, trivial.
45:35:620Paolo Guiotto: So the two are the same,
45:38:70Paolo Guiotto: And this means that,
45:40:300Paolo Guiotto: The limit set, is exactly that set. So we can say that the probability
45:47:830Paolo Guiotto: of the set where x is strictly less than little x is the limit
45:53:860Paolo Guiotto: in N of the probabilities of X less or equal than this YN, but this is the CDF, so limit in N of the CDF of X at point YN.
46:08:720Paolo Guiotto: And since YN is going to X from the left, this coincides with the limit when Y goes to X from the left of the CDFX at point Y.
46:22:820Paolo Guiotto: So this explains why that probability is the left limit of the CDF. Of course, if the CDF is continuous, that value would be the same of, let's say, remark.
46:43:500Paolo Guiotto: if the FX is Continuous. At point.
46:51:630Paolo Guiotto: X, we would have that. The probability that X is less or equal than little x, which is the value of the CDF at point x.
47:04:10Paolo Guiotto: If this, is continuous, this coincides with the left limit. This would be the limit when Y goes to X from the left of FX of Y.
47:17:540Paolo Guiotto: But this, as we just proved, is the probability that X is strictly less than X.
47:24:160Paolo Guiotto: So there would be no difference between the probability that X is less or equal than little x, and x is strictly less. And this, in particular, as a consequence that when you take the probability that
47:38:430Paolo Guiotto: X is exactly equal to the value little x, you get 0, because this would be the probability that X is less or equal than little x minus X strictly less than X.
47:55:780Paolo Guiotto: Right?
47:57:180Paolo Guiotto: But since these two, you would split into the difference, probability that X is less or equal than little x minus the probability that capital X is less than x.
48:09:450Paolo Guiotto: Since this is equal to this one, the difference would be zero.
48:14:360Paolo Guiotto: So we would get, in this case, that if the CDF is continuous at point X, then the probability that X, capital X, is equal to that value, little x, would be equal to 0.
48:30:620Paolo Guiotto: And actually, you can see that it is an if and only if, in fact, because if you do the other way, if you add that this is zero.
48:40:330Paolo Guiotto: It means that these two probabilities coincide, but this would mean that this, which is the left limit, is equal to the value of the CDF at point S. So this would say FX is continuous from the left.
48:55:470Paolo Guiotto: at contacts.
48:56:910Paolo Guiotto: We already know that, that CB is always continuous from the right, so the conclusion would be the atmosphere is continuous.
49:04:930Paolo Guiotto: at point X, so it's in fact an if and only if. So, the CDF is continuous if and only if the random variable, the probability that the random variable takes a specific value is zero.
49:21:630Paolo Guiotto: Okay, so, returning back to the exercise, all this
49:27:820Paolo Guiotto: was generated by respond… the attempt of responding to this question, what is the probability that that capital X is less than 2? So, so…
49:41:790Paolo Guiotto: We saved.
49:43:590Paolo Guiotto: Well, let's re… re-work. We started from… we had to compute probability that X is equal to 2. We cannot say that this is 0.
49:54:560Paolo Guiotto: Because that's a point where the F is discontinuous.
49:58:790Paolo Guiotto: So we cannot say that it is 0. We can say that this is the probability that X is less or equal than 2, minus the probability that X is strictly less than 2. The first one is the CDF at 0.2. The second one is the left limit.
50:15:890Paolo Guiotto: of the CDF at 0.2.
50:20:750Paolo Guiotto: So now we have to compute this limit.
50:24:540Paolo Guiotto: But this is, so FX2 is, what is 1 minus 3 to minus the integer part of 2, so minus 2, minus… this one is the limit when y goes to 2 from the left.
50:41:240Paolo Guiotto: of the formula is 1 minus 3 to minus integer part of Y.
50:47:300Paolo Guiotto: What is the integer part of Y?
50:50:130Paolo Guiotto: Y is moving to 2 from the left, okay? So Y is here, and it is moving to 2 in this way. So, let's say that it will be at the right of 1,
51:04:620Paolo Guiotto: Because it is approaching 2, and therefore the integer part will be…
51:10:740Paolo Guiotto: 1. So this is a constant equal to 1, so that quantity in the limit is 1 minus 3 to minus 1, and so at the end, we get 1 minus 3 to minus 2 to 1.
51:25:990Paolo Guiotto: Oh, no.
51:27:00Paolo Guiotto: Cheers.
51:27:920Paolo Guiotto: Where? Minus 1 minus 1 over 3.
51:32:870Paolo Guiotto: So they once disappear, so one-third…
51:36:210Paolo Guiotto: Minus 1 9th, so it's, so…
51:45:910Paolo Guiotto: And that's the value of the probability.
51:50:640Paolo Guiotto: Okay.
51:54:950Paolo Guiotto: And that's all about the 343.
51:58:990Paolo Guiotto: Okay, another thing I want you to point your attention is that to compute the CDF, sorry, the density for the log normal, we met this problem.
52:14:760Paolo Guiotto: So, what if we have the density of a random variable, say, Here it was Y.
52:22:960Paolo Guiotto: And we have an X, another random variable, which is a function of Y.
52:28:980Paolo Guiotto: So this says that, well, I have not been too much precise on this formula. This says that if basically phi is a change of variable, so change of variable because we used the change of variable in the integral.
52:44:60Paolo Guiotto: Now, what does it mean to be a change of variable? It means that there should be a function phi, but also its inverse. So, it means that this must be a big section, okay? And since you need the derivatives, it must be that both the function phi and its inverse are differentiable.
53:04:70Paolo Guiotto: And if you are within the context of Riemann integrals, this would also require that these functions are continuous. Otherwise.
53:14:730Paolo Guiotto: it's not needed. We can have mild assumptions. But let's say that… let's return a second on this.
53:23:980Paolo Guiotto: So… We proved that.
53:33:200Paolo Guiotto: this… Factor.
53:36:670Paolo Guiotto: So, which is a general rule.
53:41:200Paolo Guiotto: Suppose that I… we have, suppose that Y, B, Absolutely.
53:47:980Paolo Guiotto: Continos.
53:49:860Paolo Guiotto: with density.
53:55:910Paolo Guiotto: Say, FY. So, absolutely continuous means that there is a density.
54:02:450Paolo Guiotto: And let… X be a function of Y.
54:11:430Paolo Guiotto: If, fee, is… First, be objective.
54:21:740Paolo Guiotto: So, there exists also each inverse, females 1,
54:27:570Paolo Guiotto: And the second, both phi and phi minus 1 are differentiable.
54:35:410Paolo Guiotto: Let's say that maybe with continuous derivatives, or a little bit less is with the integral derivatives, then
54:47:500Paolo Guiotto: The variable x equals phi of Y, is… also.
54:54:260Paolo Guiotto: absolutely continuous.
54:57:680Paolo Guiotto: And we have a relation within the density, or we have a recipe to compute the density of X from the density of Y.
55:07:70Paolo Guiotto: And the density of X is given by this formula. It is equal to the density of Y
55:14:560Paolo Guiotto: well, you don't need to memorize, in fact. You have just to remind that you take the density of Y, the density of Y is a function of Y, no? Let's say, no, of its variable. So, if the change of variable is X equals phi of Y,
55:33:370Paolo Guiotto: It means that here I should put a Y, I cannot put phi of Y, because that's an X, so I need to use the formula that expresses y as phi over X, so in this case, phi minus 1 of X, and that's the quantity that goes here, phi minus 1 of Y.
55:52:370Paolo Guiotto: Hmm, sorry, I'm… 2 minus 1 of X, sorry.
55:59:770Paolo Guiotto: Because this gives DY, that corresponds to DX, no?
56:05:170Paolo Guiotto: for 2X, there is a Y, which is phi minus 1, and 2Y, there is an X. For each Y, there is an X, which is phi of Y, no? So this is the map, the change of variable here.
56:19:430Paolo Guiotto: And then there is the modulus of the derivative.
56:24:740Paolo Guiotto: of the function that gives Y function of X. So this will be the phi minus 1… phi minus 1 prime Y.
56:33:20Paolo Guiotto: And that's the formula for the change of density that we can use in a certain number of cases.
56:41:200Paolo Guiotto: But not always.
56:43:630Paolo Guiotto: For example,
56:54:810Paolo Guiotto: Let's do this… this one. There are a couple of exercises in general, like 3-4-4, 3-4-5. Let's do the 3-4-4.
57:05:450Paolo Guiotto: And you do the 3-4-5,
57:09:630Paolo Guiotto: This is a general question. It says it's not related to a density, but as you will see, it has to do with this context. Let X be a random variable.
57:23:220Paolo Guiotto: Random variable with… CDF.
57:29:470Paolo Guiotto: Capital FX.
57:32:140Paolo Guiotto: Then, the, the question is, what is… the… CDF… of absolute value of accent.
57:45:960Paolo Guiotto: So, for the CDF, there is no problem, because the CDF always exists. The density might not exist, because that's, is, exists, it's a stricter requirement for
58:01:300Paolo Guiotto: For the random variable, no. So, here we have that Y is a modulus of X, so strictly speaking, if this is our function phi of X, this function phi wouldn't be a function of this… that verifies this condition, because it's not rejected. The modulus is a function made like that, no?
58:24:240Paolo Guiotto: So, it is not subjective, it is not injective in this type of function.
58:30:570Paolo Guiotto: However, for the CDF, we can actually do directly.
58:35:650Paolo Guiotto: If we take any variable, Y, FY of little y is the probability that Y is less or equal than little y, right? This is by definition.
58:49:290Paolo Guiotto: Now, since Y is…
58:52:790Paolo Guiotto: the absolute value of X, so this is the probability that the absolute value of X is less than Y.
59:01:160Paolo Guiotto: In terms of X, this means the probability that X is between minus Y and Y, this when Y is positive, because when Y is negative, of course.
59:15:890Paolo Guiotto: there is no omega for which modulus of X omega is less than a negative number, no? Suppose that this is minus 1, modulus X. Minus 1 is an impossible effect, so the probability is 0. So we can say that for Y negative.
59:34:290Paolo Guiotto: This will be zero.
59:37:270Paolo Guiotto: for Y, positive.
59:39:540Paolo Guiotto: We will add this.
59:41:320Paolo Guiotto: Now, let's try to express this in terms of the CDF of exon.
59:46:570Paolo Guiotto: How can we do that?
59:48:420Paolo Guiotto: the probability that X is between minus Y and Y Can be seen as
59:58:380Paolo Guiotto: think that we have to represent this in terms of set X less or equal than something, right?
00:05:70Paolo Guiotto: So, it is X less or equal than Y,
00:10:920Paolo Guiotto: Since I want also that X be greater or equal than minus, it is literally n the X greater or equal than minus Y.
00:22:70Paolo Guiotto: In this case, I cannot split this, this probability.
00:29:190Paolo Guiotto: But I could say that, okay, it's like if I take X less or equal than Y, and die.
00:38:200Paolo Guiotto: subtract the X, which are strictly less than minus Y.
00:45:180Paolo Guiotto: So the omegas for which X of omega is less than minus y.
00:49:950Paolo Guiotto: You see? Because if you take an omega here, it means if omega is in this set, it means that X of omega is less or equal than Y.
01:02:60Paolo Guiotto: And it is not in that set, so omega is not in the set where X is less than minus 1.
01:12:490Paolo Guiotto: So this means that omega must be in the complementary, which is the set where X is greater or equal than minus Y.
01:20:270Paolo Guiotto: And this means that for an omega that belongs to this set, X of omega is, at the same time less or equal than Y and greater or equal than minus y. And that's what we want.
01:33:00Paolo Guiotto: So, in this way, I have that this probability is the probability that X is less or equal than Y, minus the probability that X is less than minus Y, reminded here Y is positive, or 0.
01:48:350Paolo Guiotto: So, it's the probability of a difference.
01:51:740Paolo Guiotto: Now, what is the relation between these two sets? Because in general, I cannot say that it is the difference of the probabilities. But, if they are included, I can do that.
02:02:810Paolo Guiotto: So, is this set contained into this one?
02:11:510Paolo Guiotto: I would say yes, no?
02:13:210Paolo Guiotto: Y is positive, minus Y is negative. If you are less than minus 1, you are clearly negative. You will be less than Y.
02:21:970Paolo Guiotto: Okay? So, yes, so I can transform this into a difference. This is probability that X is less frequent than Y, minus the probability that X is strictly less than minus Y.
02:34:630Paolo Guiotto: The first one is now the CDF, FX evaluated at Y.
02:42:850Paolo Guiotto: And the second one, we already encountered this. This is the left limit, so it is the limit when, let's use a different letter Z goes to minus Y from the left of FXZ.
03:03:840Paolo Guiotto: Okay?
03:05:340Paolo Guiotto: So, this… I… I… I wouldn't say that…
03:10:410Paolo Guiotto: with the conditions that are given X be a random variable with CDFX. Nothing else is given here, so we cannot do any
03:20:520Paolo Guiotto: You cannot say anything more than this, so this is exactly the CDF of this variable Y.
03:34:580Paolo Guiotto: So there exists… there is a 3-4-5, an exercise that has to deal with the density.
03:41:730Paolo Guiotto: And, the function is, X squared, so think about… I will do the 3, 4, 6,
03:49:450Paolo Guiotto: Which is, similar.
03:54:620Paolo Guiotto: So here we say that we have a Gaussian, Mean M variant sigma squared.
04:03:460Paolo Guiotto: There are two questions, I drew the number one.
04:06:280Paolo Guiotto: Number one asks, determine, If Y defined as the root of the modulus of X, is… Absolutely.
04:21:840Paolo Guiotto: continuous.
04:24:690Paolo Guiotto: And, in this case, If yes, determine… the density of Y.
04:34:720Paolo Guiotto: So I do this exercise because this is, sort of…
04:41:10Paolo Guiotto: If you can question difficult. If this transformation is a change of variable, we have the general recipe here. So, it's basically writing this formula.
04:53:870Paolo Guiotto: In this case, we do not have exactly that, because Y is a function of excerpt, but that function, the root of models of Xer, is not a projection.
05:05:630Paolo Guiotto: So, we have to be a little bit careful. We don't just apply that formula in a straight way, so we have to work a bit on this before we can say what is the answer.
05:20:330Paolo Guiotto: Okay, so, we know…
05:24:820Paolo Guiotto: Let's see what we can do. Here, we know that the density of X is well known. It is E2 minus X minus m squared divided 2 sigma squared over the scaling factor 2 pi sigma squared.
05:43:180Paolo Guiotto: for XRL.
05:45:570Paolo Guiotto: So that's the density.
05:47:870Paolo Guiotto: we have Y is the root of modulus of X. Of course, we do not have the expression of X explicitly, but it's not important, because at the end, what we need to do is just its distribution.
06:03:750Paolo Guiotto: Now, since this is phi of X, where phi is the numerical function, phi of x is the root of modulus of x, this function is not invatible, is not injective, because X and minus X
06:19:460Paolo Guiotto: ascending to the same value, no? It becomes subjective if you restrict to positive values, of course, so you can make a… we can make this a subjective, but definitely not an injective function. So it's not a one-to-one, and that's important because
06:36:710Paolo Guiotto: To use this formula, we need to compute phi minus 1, so what phi minus 1 should be.
06:43:110Paolo Guiotto: If you try to invert, you realize that you can't do that, because if you call this Y equal root of modulus of X, what should be the X? Well, you know, there are two X, because you see, you do the square, you get modules of X equal Y squared, and this means that X is either plus or minus Y squared, but plus or minus.
07:05:380Paolo Guiotto: So it's not embeddable, okay? So this fee… P is not… Invertible.
07:16:830Paolo Guiotto: So, the… Change.
07:25:60Paolo Guiotto: of density.
07:31:490Paolo Guiotto: Formula, which is the formula.
07:34:110Paolo Guiotto: We broke above, does not apply.
07:42:340Paolo Guiotto: But does this mean that the variable is not absolutely continuous? Well, from the formal point of view, this is a sufficient condition.
07:52:720Paolo Guiotto: it says.
07:54:100Paolo Guiotto: If X is phi of Y, and phi is a projection with the function phi and its inversely differentiable, then…
08:02:920Paolo Guiotto: X, if Y is absolutely continuous, the letters are inverted, but if one is absolutely continuous, also the other is absolutely continuous, and basically, and vice versa.
08:13:770Paolo Guiotto: One is absolutely continuous if and or if, the other is absolutely continuous. And you have that formula that tells you how the density of one variable becomes the density for the other variable.
08:25:680Paolo Guiotto: Okay?
08:27:149Paolo Guiotto: How to get one density from the other. But if that condition is not verified, does not mean that the conclusion is false.
08:34:939Paolo Guiotto: Let's see in this case if it is both. What can we do? Well, we know that absolutely continuous means that the CDF has certain properties, which is differentiable, etc. So, let's try to understand what is the CDF, and see if the CDF is differentiable.
08:53:490Paolo Guiotto: So let's compute the… let's compute… that…
09:00:720Paolo Guiotto: CDF. This guy always exists. The CDF always exists.
09:06:149Paolo Guiotto: Maybe there is no density, but the CDF, there's always a CDF. So I take the CDF of my Y.
09:13:950Paolo Guiotto: So this is,
09:16:870Paolo Guiotto: the probability that Y is less or equal than little y. Y is the root of the absolute value of X, right? This must be less or equal than Y.
09:28:479Paolo Guiotto: So this is the probability we have to compute.
09:32:240Paolo Guiotto: So what do we notice here?
09:41:40Paolo Guiotto: What should we notice?
09:45:220Paolo Guiotto: Well, of course, since I know X, I know everything about X, I want to transform this into probably that X is less than something, let's say, or X is in some range, because I know what is the distribution of X.
10:00:480Paolo Guiotto: So I want to go to transform this state into something that involves X diagonally.
10:06:530Paolo Guiotto: So, I should extract this X, so I should say, okay, this is less than this X squared, I try to extract X.
10:15:230Paolo Guiotto: Okay? But before we do that, I may notice something here.
10:20:300Paolo Guiotto: For setting Y, this number is 3.
10:24:750Paolo Guiotto: If Y is negative, we have the problem that the root is a positive, it's always the positive. If the root is negative, that's impossible. So we can say, that quantity is certainly equal to zero.
10:42:140Paolo Guiotto: If Y is negative.
10:46:20Paolo Guiotto: If Y is positive.
10:48:360Paolo Guiotto: Now, Y is positive? I can say that root of modulus of X is less than or equal to Y. It's equivalent, if y is positive.
10:59:210Paolo Guiotto: No? I can square, and the inequality remains… it's equivalent. So, it's equivalent to say that modulus of X is less or equal than Y squared. So this becomes the probability of the event modulus of X less or equal than Y squared.
11:15:730Paolo Guiotto: Which is not so bad, because this is the probability that X is similar to the problem we did.
11:25:80Paolo Guiotto: 2 minutes ago, is between minus Y squared and plus Y squared.
11:30:520Paolo Guiotto: Now, I know what is this probability.
11:34:90Paolo Guiotto: Well, you can do the same argument we have seen just a minute here. So, or, since we have a density, this is the integral, we have the density, no? You remind that probability that X belongs to E,
11:52:570Paolo Guiotto: In general, for any random variable is the integral on… on…
12:02:550Paolo Guiotto: Yeah, it's the… it would be the muax, the low…
12:06:760Paolo Guiotto: of X of the setting E, or if you want it in an exaggerated way, it's the integral on E of 1 in the mu X. I write it with the integral, because if there is a density, this is the integral on E of the density.
12:24:00Paolo Guiotto: In the Lebesgue, no? So, I'm using this to say that this is because now the integral between minus Y squared and Y squared of the density of X that I have, let's say here, I can write X, the X.
12:41:400Paolo Guiotto: Okay.
12:42:720Paolo Guiotto: So now, this is what? This is the FY at point Y. So, in conclusion, we have the F
12:52:580Paolo Guiotto: Y at point Y is the integral from minus y squared to plus Y squared of that density, which is the Gaussian density, 1 over root of 2 pi.
13:05:190Paolo Guiotto: Sigma square.
13:07:990Paolo Guiotto: E minus X minus M squared divided 2 sigma squared dx.
13:15:630Paolo Guiotto: Okay, now the question is, the question was, is Y absolutely continuous?
13:23:660Paolo Guiotto: Now, what can we say from FY? We said that, remind that Y is absolutely Continos.
13:37:120Paolo Guiotto: If and only if the function FY is continuous.
13:45:220Paolo Guiotto: There were three properties. Number two, there existed derivative with respect to its variable, which is Y here.
13:54:290Paolo Guiotto: And this must be an L1 function.
13:59:840Paolo Guiotto: In the real liner. And that's exactly, at the end, the density of Y will be the density of the function Y.
14:12:180Paolo Guiotto: Now, this function is continuous, because if you see this FY, it's basically an integral function. You remind what is an integral function?
14:24:330Paolo Guiotto: No? Fy… of Y is… Basically.
14:34:280Paolo Guiotto: And… An object which is known as integral function.
14:42:420Paolo Guiotto: So, literally, a function which is defined through an integral. And in fact, well, let me refresh you something about this. In general.
14:57:320Paolo Guiotto: an integral function.
15:02:820Paolo Guiotto: is… Function.
15:08:620Paolo Guiotto: defined in this way. Well, let's use,
15:14:670Paolo Guiotto: Let's use not the letter F.
15:18:100Paolo Guiotto: but say, capital I,
15:21:430Paolo Guiotto: Capital I… well, let's use the letter Y. Well, let's use… not the letter Y, because it would be used. So, let's say I of Z…
15:32:430Paolo Guiotto: is the integral from a fixed value C to Z of a certain function, let's say we can say…
15:43:220Paolo Guiotto: F of, X, DX.
15:47:960Paolo Guiotto: An integral function is an object like this. So, we call this integral function, because integral function, we move… the value here is one of the n components of the integral.
15:59:510Paolo Guiotto: Okay? So, yeah, I mean, we buy one domain, you know?
16:06:510Paolo Guiotto: First year calculus, you learned that if the function F.
16:12:630Paolo Guiotto: is a continuous function, then this is the fundamental.
16:18:710Paolo Guiotto: Theorem.
16:20:630Paolo Guiotto: off.
16:21:540Paolo Guiotto: Integrate.
16:23:330Paolo Guiotto: calculus.
16:25:360Paolo Guiotto: The integral function is differentiable.
16:29:130Paolo Guiotto: So there exists a derivative, I prime of Z, and that's exactly the integrand function, the function little f.
16:38:720Paolo Guiotto: Okay?
16:40:390Paolo Guiotto: Now, here we have something that looks like, at least in flavor, let's say, this situation, because you see, you have the F…
16:50:350Paolo Guiotto: Y is an integral, where actually, the two endpoints depend both on Y, so they are both viable. Other function, which is the F, so the ingredients are F is that function, the density, the density of the Bible that I said.
17:08:800Paolo Guiotto: And this is not exactly this one, but it can be transformed easily, because in our case, so let's close this parentheses.
17:20:430Paolo Guiotto: In our case, So, our FY of Y…
17:28:480Paolo Guiotto: is the integral from minus y squared plus y squared of, let's just keep FXX DX.
17:36:210Paolo Guiotto: Now, first of all, we can always split this integral in two integrals to make one endpoint fixed. So, for example, I can say I integrated from minus y squared to 0 plus integral from 0 to y squared. And this way, I made
17:52:160Paolo Guiotto: One of the two endpoints only moving.
17:56:280Paolo Guiotto: If you want to see that the moving point is always the upper end point, you can just flip the two endpoints here. You know that when you flip the order, you have to change the sign to the integral. So this is minus the integral from 0 to minus Y squared.
18:12:210Paolo Guiotto: So at the end, I have integral from 0 to y squared of the density minus
18:20:620Paolo Guiotto: The integral from 0 to minus y square of the density.
18:25:30Paolo Guiotto: And now, this looks more similar to this one, no? You see, one of the endpoints is fixed, it's zero, the other endpoint is variable, but it is not the variable.
18:36:550Paolo Guiotto: Yeah, okay, but if you use the notation we used here, it's like this function, where instead of having Z, I have Z squared. So it's like I of Z squared
18:51:200Paolo Guiotto: And this is minus i of minus z squared, where now I is exactly what I have there. I of, of, sorry, Y squared.
19:04:980Paolo Guiotto: One sec.
19:06:50Paolo Guiotto: Y of Z, I of Z is just now the integral from 0 to Z of the density FX.
19:13:650Paolo Guiotto: Okay, so at the end, my FY…
19:18:740Paolo Guiotto: of Y can be represented as this function i, evaluated at Y squared minus the same function i, but now evaluated at minus
19:30:420Paolo Guiotto: Y squared with i of Z equal integer from 0 to Z of little f.
19:39:920Paolo Guiotto: So now I can compute the derivative of capital F, because I can compute the derivative of the capital I.
19:46:990Paolo Guiotto: So the derivative of capital F, Why? Exists?
19:54:890Paolo Guiotto: Because, well, when you do the derivative of this, you use the chain rule, no? That's a composition. It's i of y squared. So you do the derivative of I, you evaluate at Y squared, then you multiply by the derivative of the argument, which is 2Y.
20:13:230Paolo Guiotto: And here, minus… again, you do the derivative of i, you evaluate at the value of the argument, which is minus Y squared, times the derivative of the argument, which is minus 2Y.
20:27:250Paolo Guiotto: So, at the end, I have 2Y that multiplies I prime of Y squared minus minus plus I prime of minus Y squared.
20:41:920Paolo Guiotto: And we know what is I prime, because I prime…
20:46:530Paolo Guiotto: of Z in general, by the fundamental theorem of integral calculus, that's exactly the integrand function at point Z. So this is now 2Y
20:58:140Paolo Guiotto: FX evaluated at Y squared plus FX evaluated at minus Y squared.
21:08:280Paolo Guiotto: And this formula holds for every Y.
21:13:440Paolo Guiotto: So this says that there exists a derivative, if there exists a derivative, we check off continuity is automatic, and the derivative is equal to this quantity here, if you want to care also right, because you have
21:29:840Paolo Guiotto: is the explicit formula. 40?
21:33:60Paolo Guiotto: So this means that this guy here is now the density of Y.
21:40:60Paolo Guiotto: So…
21:41:740Paolo Guiotto: We can say that it exists, we don't need to verify that the integral is 1, because it's automatically equal to 1. That's the integral, it's the derivative of the CDF.
21:52:840Paolo Guiotto: If you integrate, it comes the value of CDF at plus infinity minus the value at minus infinity, which is 1.
21:59:640Paolo Guiotto: So there exists the density of Y, and it is this function, 2Y,
22:07:240Paolo Guiotto: FX of Y squared plus FX minus Y squared, that if you want, we can explicitly write this 2Y
22:20:380Paolo Guiotto: So, we have a common factor, scaling factor, 2 pi sigma squared, and we have here E minus, so FX is X minus M, so this will be Y squared minus m squared divided 2 sigma squared.
22:39:400Paolo Guiotto: And this is plus E minus minus X squared, so if you want, factorizing a minus becomes like this.
22:49:140Paolo Guiotto: Y squared plus M squared divided to sigma squared.
22:53:180Paolo Guiotto: And that's the density of this variable.
23:01:380Paolo Guiotto: Okay.
23:03:570Paolo Guiotto: So, in other words, you have to…
23:06:10Paolo Guiotto: adapt ideas. If you are not exactly in the case, you can apply a straight,
23:15:320Paolo Guiotto: in a straightforward way, the… the…
23:24:20Paolo Guiotto: The… the change of variable formula.
23:26:910Paolo Guiotto: Okay, so do the other one.
23:29:990Paolo Guiotto: Duel.
23:31:120Paolo Guiotto: The 346, the number 2.
23:36:270Paolo Guiotto: And then do also 347.
23:41:750Paolo Guiotto: Et. Et.
23:44:480Paolo Guiotto: If you want, you can do the 9-digit theoretical exercise, and do the remaining… all the remaining exercises we left before.
23:55:560Paolo Guiotto: Okay, so, let's see… we started at 10.30, so we have about 7 minutes.
24:06:120Paolo Guiotto: So, it's not sufficient,
24:13:660Paolo Guiotto: I added a section, and you will see there is a section 3-3 at the end of the chapter.
24:20:270Paolo Guiotto: Maybe I will, I will tell of this another time.
24:24:820Paolo Guiotto: It's just an application of the CDF function to this problem, which is a very important problem in applied world, which is the classical news vendor model.
24:37:290Paolo Guiotto: Maybe I will tell you another time, because we have no time now to start this.
24:43:330Paolo Guiotto: Okay, well, let's just see the first definitions of the next part, which is just a little extension. So we have seen what is a random variable. A random variable, from the mathematical point of view, is a measurable function.
25:01:10Paolo Guiotto: And,
25:04:770Paolo Guiotto: It represents the… we may think, a random variable as the outcome of an experiment that is a numerical value, okay?
25:16:50Paolo Guiotto: Now, imagine that you have n experiments, each yielding a numerical value, and together you have an array of numbers that we still call random variable, even if the values are vector.
25:31:380Paolo Guiotto: In this case, actually, we use this, or in probability, they use this term of multivariate random variables. So, what is a multivariate?
25:40:670Paolo Guiotto: a random variable. We have just a trivial definition. If we have omega F… P… be a probability… space.
25:57:920Paolo Guiotto: We have a X, which is an array, X1, say, to Xena.
26:06:380Paolo Guiotto: Defined on omega with values in Rn.
26:11:920Paolo Guiotto: We say, well, we can give equivalently… equivalent definitions, basically. We say that,
26:23:620Paolo Guiotto: X is L.
26:27:500Paolo Guiotto: multivariate.
26:34:590Paolo Guiotto: random variable.
26:37:200Paolo Guiotto: If a… Well, let's say we mention two equivalent conditions if equivalently.
26:50:730Paolo Guiotto: One condition is that each of the components is a random variable.
26:56:650Paolo Guiotto: Each… XJ.
27:02:430Paolo Guiotto: Is a random variable, so it's just an array of random variables.
27:07:640Paolo Guiotto: or you can give a direct condition here. Now, you remind that for a random variable, we may say that this happens, that the set where X belongs to E
27:25:620Paolo Guiotto: is a boreal set.
27:28:670Paolo Guiotto: of R for every E that is a Borel set. This is a slight extension of the property that says the set where X belongs to an interval is a Boren measurable set.
27:42:780Paolo Guiotto: Now, we can, because for certain cases it is convenient to have this already written in this way, we can say the same.
27:53:710Paolo Guiotto: by thinking in terms of array, so that is, for every set E, which is now, since the values of X are vectors, this set E will be a set of vectors, so it will be a subset of Rn.
28:10:250Paolo Guiotto: So, we… we can say that the set of omegas, no, this is the set of omegas in the sample space capital omega, such that X of omega belongs to the set E, which is a subset of R, capital N,
28:29:340Paolo Guiotto: This set is, measurable, is an event. Sorry, here, I did a mistake, would be in F, is an event.
28:39:820Paolo Guiotto: for every set E, which is a Boreal set of Rn. Boreal set of Rn means what? Means the sigma algebra generated by
28:51:770Paolo Guiotto: Well, in R, we may say intervals, or equivalently also by open intervals, or equivalently by open sets. And in fact, this is the definition we have here. Open sets…
29:07:390Paolo Guiotto: So, whatever is the open set.
29:11:60Paolo Guiotto: a set of omega, sorry, whatever is the SETI Borel set, so in the sigma generated by open sets.
29:19:960Paolo Guiotto: This is an event, the set where X belongs to E is an event. So, in particular, this will allow to define the probability of this event that will be the law of X on the set E, but this will… we will return later.
29:38:160Paolo Guiotto: So, let's say that.
29:44:480Paolo Guiotto: We say also that.
29:47:210Paolo Guiotto: We say, that X is in L1.
29:56:300Paolo Guiotto: Now, X is an array, we will say that it is in L1 if and only if all the components XJ are in L1 for every J equal 1 to N.
30:06:40Paolo Guiotto: And we define the expected value of X,
30:10:620Paolo Guiotto: Now, this is a function which is vector-valued, but as you may imagine, this will be an array made of the expectations of the components of X.
30:22:590Paolo Guiotto: So, without, no particular fantasy.
30:29:370Paolo Guiotto: So the expectation… the unique thing to keep in mind is that the expected value is not a number now, for a vector value will be a vector of the same nature of the values of X as it should be.
30:47:750Paolo Guiotto: Mmm…
30:55:350Paolo Guiotto: Okay, an important object is the so-called covariance metrics. We defined the covariance between two random variables. So now,
31:08:720Paolo Guiotto: we define
31:13:740Paolo Guiotto: covariance.
31:18:980Paolo Guiotto: metrics.
31:23:140Paolo Guiotto: So it would be a matrix of numbers. It is an n by n matrix.
31:33:810Paolo Guiotto: the matrix C with the entries, say, C, I, J,
31:39:770Paolo Guiotto: With the two indexes, ranging from one to N.
31:45:190Paolo Guiotto: Which are defined in this way. CIJ is the covariance between Xi and XJ.
31:54:310Paolo Guiotto: Xi, XJ are two components of the array, so they are two numerical values. The covariance is well-defined. It is, I remind you, we have not yet used
32:05:910Paolo Guiotto: Particularly, this is the value, the expected value of Xi minus its expectation.
32:16:290Paolo Guiotto: times XJ minus its expectation.
32:23:200Paolo Guiotto: In order these quantities be defined, I remind you that the technical condition is that the XJ must be L2, so the expected value of the square is defined. So, which is…
32:41:660Paolo Guiotto: Well, they find… provided if… the XJ are in L2 omega.
32:53:300Paolo Guiotto: J equal 1 to capital N.
33:00:330Paolo Guiotto: Now, let's do just one thing, one remark, an important factor.
33:08:60Paolo Guiotto: is that this matrix, D, covariance… metric, sir?
33:17:300Paolo Guiotto: is always… symmetric.
33:29:550Paolo Guiotto: And this is evident by the complete the order of the two indexes, you see that nothing changed, because the problem is commutative, and positive definite.
33:43:120Paolo Guiotto: This is not evident.
33:45:460Paolo Guiotto: and positive.
33:48:210Paolo Guiotto: definite.
33:49:790Paolo Guiotto: What does it mean, this? It means that if you take your matrix C, you apply on a vector V of Rn.
33:58:760Paolo Guiotto: you get a vector of Rn. You do this color product with the vector V itself, you get now a number.
34:06:790Paolo Guiotto: Now, this quantity is always greater or equal than zero for every vector V in Rn.
34:14:990Paolo Guiotto: This can be easily checked, because if you do CV.V, it means that you are doing, let's say, sum…
34:26:920Paolo Guiotto: D, some of the products of, let's say, when i goes from 1 to N, of the ith component of CV times the ith component of V. This is the stellar product.
34:43:720Paolo Guiotto: But what is the height component of CV?
34:47:840Paolo Guiotto: The height component is the product between a line, and precisely the it line, and the vector V. So this is, since you are doing, no, the matrix is C11, etc, C1N, etc, CN1, CNN,
35:07:600Paolo Guiotto: you multiply by V, which is a vector V1VN,
35:12:910Paolo Guiotto: If you want to get the height component of the product, you know that you do the product lines by column, right?
35:20:510Paolo Guiotto: Well, the height component is obtained by multiplying the height line of the matrix by the column vector V. So you are doing a sum, i going from 1 to N,
35:34:90Paolo Guiotto: Now, the it component will have a component C, I, 1, CI2, etc, CIN. So you have something like sum, J going from 1 to N, C, I, J, VJ, this is the product, line by column.
35:55:690Paolo Guiotto: that yields the height component of this vector, times VI. So at the end, you have this sum, double sum, sum over i and j, C, I, J, VI, VJ.
36:11:130Paolo Guiotto: If you now write down what are these coefficients, CIJ, these are expectations of
36:18:770Paolo Guiotto: Well, to… to write shortly, instead of writing Xi minus its average, it's expected, let's write XI bar for this, IXJ bar for that, okay? So, in such a way, we write shortly, Xi
36:37:150Paolo Guiotto: By XJ. Ba… And this is VIVJ outside.
36:44:60Paolo Guiotto: Now, these are colors we carry inside the expectation. Also, we carry inside the sum into the expectation. We have a unique big expectation of a sum of IJ, Xi, say, bar, VIXJ bar VJ.
37:06:150Paolo Guiotto: But this product comes, when you do this, you take X1 bar V1 plus X2, V2, etc, plus X3VT, etc. plus XNVN,
37:21:840Paolo Guiotto: And you multiply by itself.
37:25:370Paolo Guiotto: Because if you multiply by itself, huh?
37:33:960Paolo Guiotto: What is this predator?
37:35:900Paolo Guiotto: It's exactly the sum of all possible products of one factor taken here, and another factor taken here. So, here you pick an Xi Bi, and here we pick an XJ, and you multiply the sum of all possible pairsi, and we get X. But this is that that expression is necessary enough to multiply by itself, no? So it's a square.
37:59:180Paolo Guiotto: So, in fact, this is the expectation of the square of the sum of these XJ bar VJ, so whatever it is, is greater than or equal than zero. That's why it's paused.
38:16:90Paolo Guiotto: So the covariance matrix is a linear matrix, symmetric and positive definite.
38:24:310Paolo Guiotto: Okay, we stop. Yeah.
38:28:840Paolo Guiotto: And we'll see you on Fridays, right? I'll play the West Nation.
38:34:10Paolo Guiotto: Okay, have an answer.
38:38:170Paolo Guiotto: Can I sleep with you?
38:42:770Paolo Guiotto: Sort of.