Let's talk about the conditional distribution associated with the normal distribution. So let me let X, my vector X be equal to X1 X2. And so without loss of generality, we're just going to assume that we want to know what the conditional distribution of X1 given X2 is. So let me write out the answer first. First of all, I'm working under the assumption that x as we're talking about in almost all of the cases from this section of the class. X is multi variant normal distributed. So I'm going to write out the variance covariance matrix this way. Sigma 1 1, sigma 1 2, sigma 1 2 transposed, and sigma 2 2. So what I'd like to know is what's the distribution of X1 given X2. And we know that it's normally distributed, because we've mentioned earlier that all the conditional distributions are normal. But what's the variance and covariance? So, Let me just write out the answers first. And then I'll show you a nifty derivation of this. I'll do the derivation just because I like it. So the expected value of X1 given X2 took the particular value, little x2 is exactly equal to mu 1, okay? It will be weird if that wasn't there plus sigma 1 2 sigma 2 2 inverse X2 minus mu 2. And then the variance of X1 given that X2 equals little x2 is then equal to sigma 1 1. So what we would hope it would be, because that would be easy, just the same as the marginal variance. But of course, there's got to be other stuff minus, sorry, minus sigma 1 2 sigma 2 2 inverse. You could either write it as sigma 2 1 or sigma 1 2 transpose. I'd like to write it as sigma 1 2 transpose, okay? Let's come up with a derivation of this. And there's a really clever one, and I'm not sure how someone figured this out but it's really, I find it really nice. And in fact it was a student in my class, who showed me this, because I was deriving it using inverses of partition matrices which is a lengthy bookkeeping procedure though not hard. And they show me there's a much easier way to do it. And what she said was, define X as Z equal to X1 plus AX2. Where A is equal to negative sigma 1 2, the sigma 2 2 inverse. And then what I would contend is that the covariance between X2 and Z. Let's say, Z and X2 is equal to 0, and let's work that out really quick. So the covariance of Z and X2 is nothing other than the covariance of X1 plus AX2 and X2. Which is equal to the covariance of X1 and X2 plus A covariance of X2 and X2, okay? So that's equal to covariance of X1, X2 is sigma 1 2 plus A, which I have defined as negative sigma 1 2 sigma 2 2 inverse. And covariance X2 and X2, itself, is sigma 2 2, okay? So I think you can pretty much see at this point that this equals 0. So Z is independent of X2. So the distribution of Z given X2 is just the distribution of Z disregarding X2, because Z is independent of X2. So let's calculate, and we know that Z is normal, because it's a linear combination of normals. So let's calculate the expected value of Z. Which is equal to mu 1 plus A mu 2, right? And the variance of Z, Is equal to, A the variance of X2 times A transpose with variance of X2 is sigma 2 2 times A transpose. Now let's fill some of those in that's mu 1 minus sigma 1 2 sigma 2 2 inverse mu 2. And this is equal to the negative sign will cancel out sigma 1 2. And then I get a sigma 2 2 inverse sigma 2 2 sigma 2 2 inverse, so it's just going to be sigma 2 2 inverse sigma 1 2 transpose, okay? Okay, so let's think about this. So the expected value of Z given X2 is equal to little x2 is equal to the expected value of Z, but it's also equal to, Mu 1 minus sigma 1 2 sigma 2 2 inverse mu 2, okay? So we have that much so far, but we can also write out the expected value of Z as the expected value. Remember, the definition of Z which we've written up here. And that works out to be then that we can write this as expected value of X1 plus AX2 given X2 equals little x2, okay? So we then can write, given X2 plus A expected value of X2 given X2. So this is the expected value of X1 given X2 plus A expected value of X2 given X2 is just X2. Okay, so now take this equation and this equation, and we can solve for the expected value of X1 given X2. Right? And it works out to be exactly the formula that we had above. And then you can use the same technique to derive the variance, so the variance of Z given X2 equals little x2. Well that is just exactly equal to the variance of Z, since Z is independent of X2, since Z is independent of X2. And we see right up here the variance of Z is sigma 1 2 sigma 2 inverse, sigma 1 2 sigma 1 2 sigma 2 2 inverse sigma 1 2 transpose. But then we also know that it's the variance of X1 plus AX2 given X2. Because we're just simply rewriting that statement out with the definition of Z. And by the rules of variants, we can then apply the variance formula and apply the rules of variants, and solve and solve again. And I'm going to ask you to complete the steps for homework.
