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>> We're now going to review multivariate distributions.

Â We'll talk about multivariate CDFs, multivariate PDFs, conditional

Â distributions, and so on. Much of this material is a little tedious,

Â a little dull, but we thought it was worthwhile collecting it all and having it

Â in one place for your review if it crops up later in the course.

Â Okay, so let's get started. Let x be a vector of random variables, x 1

Â up to x n. We say the joint CDF of x is given to us

Â by the following. So, the joint CDF, Fx of little x is equal

Â to the probability that X1 is less than or equal to little x1.

Â X2 is less than or equal to little x2. Up to xn being less than or equal to

Â little x n. And from this joint CDF, we can actually

Â calculate the marginal CDF, so for example, the marginal CDF of Xi is given

Â to us by just plugging infinity into all of the components in the joint CDF except

Â for the ith component which is little xi. Okay, so we can go from the joint CDF to

Â the marginal CDF. It is also straightforward to generalize

Â the previous definition to joint marginal distributions.

Â So for example, if I want the joint CDF of just xi and xj, I can also recover that

Â from the joint CDF of 1 up to xn by placing infinity in all of the arguments,

Â except for the ith argument where I have xi and the jth, where I have xj.

Â We also say that x is a joint PDF or probability density function, f subscript

Â x, if we can write the joint CDF as an integral like this.

Â So, this is just the way we, we, we capture our joint CDF by integrating out

Â the density function by appropriate limits.

Â Okay. We can also talk about conditional CDFs.

Â So what we're going to do is we're going to partition our vector x1 up to xn into

Â two components. The first component is x1, which contains

Â x1 up to xk. And the second component is this boldface

Â x2 which contains xk plus 1 up to xn. And then we can talk about the conditional

Â CDF of x2 given x1, and in fact, it's defined as following, as follows.

Â So the conditional CDF of x2 given x1 is equal to the probability that the random

Â vector x2 is less than or equal to little x2 conditional on x1 being equal to little

Â x1. If x is a PDF, f of x, then the

Â conditional PDF of xx2 is given to us by this quantity here.

Â So it's the joint PDF divided by the marginal PDF of x1, which we can also

Â write like this. Okay.

Â And the conditional CDF, f of x2 given x1, can be determined by integrating the

Â conditional PDF. So this is our conditional PDF, and we can

Â integrate this out with respect to uk1 up to un and that will give us our

Â conditional CDF. Okay, independence.

Â We say the collection x is independent if the joint CDF can be factored into the

Â product of marginal CDFs. So in particular the joint CDF here in the

Â left hand side is equal to the product of the marginal PDFs over here on the right

Â hand side. Similarly, actually, this implies that if

Â x is a PDF fx, then we can also factorize the joint PDF into the product of the

Â marginal PDFs over here. We can also see from one, and one is here

Â on the previous slide. So we can use this, okay, to see that if

Â x1 and x2 are independent, then the conditional PDF of x2 given x1, well by 1,

Â that's equal to this ratio here. So the joint PDF of x divided by the

Â margin PDF of x 1 and by independence here, we can replace the joint PDF by the

Â product. Then these two cancel and we're left with

Â the marginal PDF of x2. So what we're saying here is that if x1

Â and x2 are independent then the conditional PDF of x2 given x1 is simply

Â the marginal PDF, f of x2. In other words, having information about

Â x1 tells you nothing about x2 when x1 and x2 are independent.

Â Okay. Some implications of independence.

Â Well,and I expect we're all familiar with this, but let's, let's go through it

Â anyway. Let X and Y be independent random

Â variables. Then for any events A and B, the

Â probability that X is in A and Y is in B, well, that factorize into the product of

Â the probability of X being an A times the probability of Y being in B.

Â More generally, for any functions, f and g, independence of X and Y implies the

Â expected value of f of X times g of Y is equal to the expected value of f of X

Â times the expected value of g of Y. And in fact, 2 follows from 3, okay?

Â So the implication goes that way and it's easy to see this, because we can write

Â this probability of X being in A and Y being in B as the expected value of the

Â indicator function of X and A times the indicator function of Y and B.

Â Just to remind ourselves what is this indicator function, while it takes on two

Â possible values, it takes on the value 1 if X is in A and it takes on the value 0

Â otherwise. So therefore, the product of these two

Â indicator functions is 1 or 0 and will only be 1 if X is in A and Y is in B.

Â Okay? That occurs with probability X and A, and

Â Y and B. So this statement here is correct.

Â Okay, so we've got this first line. And now we can use the independents X and

Â Y in condition three to break down this expectation down into the product of these

Â two seperate expectations. Okay.

Â But of course, this expectation is the probability that X is in A and this

Â expectation is the probability that Y is in B.

Â So indeed we do see that we can go from three to two.

Â Okay. More generally, if X1 up to Xn are

Â independent random variables, then we can write the expected value of f1 of X1, f2

Â of X2, and so on up to fn of Xn. That factorizes into a product of n

Â separate expectations. The expected value of f1 of X1 times the

Â expected value of f2 of X2 and so on. Random variables can also be conditionally

Â independent. For example, we say that X and Y are

Â conditionally independent given Z, if the expected value of f of X times g of Y

Â given Z is equal to the expected value of f of X given Z times the expected value of

Â g of Y given Z and I should mention this is for all functions f and g.

Â Okay, and in fact, this idea of conditioned independence, we're going to

Â see later in the course, because it's used in the, well, the now infamous Gaussian

Â copula model for pricing CDOs. So just to give you a brief idea of how it

Â might be used in a bond context or a CDO context, let Di be the event that the ith

Â bond in a portfolio defaults, okay? So we'll assume that there is a portfolio

Â of n bonds. Okay.

Â It's not reasonable to assume that the Di's are independent.

Â You might ask, why is that? Well, if you think about it, there will be

Â all sorts of macroeconomic factors or industry specific factors, which will

Â cause defaults to actually be dependent. So for example, maybe some industry

Â crashes that might cause not just one firm to default but multiple firms in that

Â industry to default. And so, it doesn't make sense to assume

Â that these events, these Di's are independent.

Â But, we might be able to say that they're conditionally independent given some other

Â random variable zed. Zed, for example, might reflect some

Â industry factor. Some, some factor that governs how well a

Â particular industry is doing. In that case, if we assume that the

Â default events are conditionally independent given zed, then we can write

Â the probability of D1 up to Dn given Z as being the product of these factors here,

Â probability of D1 given Z up to probability of Dn given Z.

Â And it's actually often easy to compute these quantities.

Â So we'll actually be using this kind of idea later in the course, as I said, when

Â we discuss the Gaussian copula model for pricing CDOs.

Â We'll also see it in a couple of other applications as well.

Â Okay, so very briefly, I also want to mention the mean vector and covariance

Â matrix of a vector round the variables X. I hope we're all familiar with this

Â already, but let's go through it anyway. So the mean vector of X is simply the

Â vector of expected values, expected value of X1 up to expected value of Xn and the

Â covariance matrix of x is. Well, this matrix of covariances.

Â Okay, so, formula is expected value of X minus expected value of X times X minus

Â expected value of X transposed. And just to be clear, this is an n by 1

Â vector, and this is a 1 by n vector, so the product is n by n.

Â And we get an n by n covariance matrix, with the i, jth element of sigma being the

Â covariance of Xi and Xj. The covariance matrix is symmetric that of

Â course is because the covariance of Xi, Xj is equal to the covariance of Xj and Xi.

Â And this diagonal element satisfies sigma i greater or equal to 0, and of course,

Â the diagonal elements are just the variances.

Â So this is equal to the variance of Xi and variances are always nonnegative.

Â It is also positive semi-definite, so this is a, an important well-known property of

Â a covariance matrix, in particular, it means that X transpose sigma X is greater

Â than or equal to 0 for all vectors X and Rn.

Â The correlation matrix row X is similar to the covariance matrix except it has as its

Â i, jth element, the correlation Xi with Xj itself is symmetric, positive

Â semi-definite, and has 1's along the diagonal.

Â And just to remind ourselves, the correlation of Xi and Xj, is equal to the

Â covariance of Xi and Xj, divided by, well, the square root of the variance of Xi

Â times the variance of Xj. Okay.

Â For any matrix A, which is a k by n matrix and a k by 1 vector A, we can take a

Â linear combination of AX plus little a and we can compute the mean of this vector.

Â So the mean is a times expected value of X plus little a and the covariance matrix of

Â this new vector of random variables is a times the covariance of X times A

Â transpose. And of course, five actually implies this

Â result, which you're probably familiar with, that is the variance of aX plus bY

Â equals a squared variance of X plus b squared variance of Y plus 2ab the

Â covariance of X, Y. Note that if X and Y are independent, then

Â the covariance of X, Y equals 0, but the converse is not true in general.

Â And some people tend to forget this, but it is not in general true that if the

Â covariance of two random variables equals zero, then those two random variables are

Â Independent. That is not true.

Â