1
$\begingroup$

I obviously don't understand multivariate gaussian variables as well as I thought. I have k+1 variables. One which is special, call it X, and I want to find the mean of, given necessary and sufficient information about the k others, call them Y1 through Yk. They are all dependent 0 mean gaussian random variables; suppose they can be completely described by a covariance matrix. Now, if nothing is known about any of them, then they all have 0 mean, and that includes X. But because they are dependent, if something is known about the k Y-variables, that pushes the expected value of X off from 0. For merely the case that k=1, so there are two dependent 0-mean gaussian variables, X and Y, I find that the mean of X given Y is: $$E(X|Y)=\frac{Y*E(X*Y)}{E(Y^2)}=\frac{Y_1*\sigma_{1x}}{\sigma_{11}}$$

OK, that's fine and dandy. But then when you throw in others, first of all, the covariance matrix looks like this: $$[\sigma_{11}, \sigma_{12},...\sigma_{1k},\sigma_{1x}$$$$\sigma_{21},\sigma_{22},...\sigma_{2k},\sigma_{2x}$$$$...$$$$\sigma_{k1},\sigma_{k2},...\sigma_{kk},\sigma_{kx}$$$$\sigma_{1x},\sigma_{2x},...\sigma_{kx},\sigma_{xx}]$$

where the first k variables in the vector are the Y-variables and the last one in the vector is X. Now these gaussian random variables are supposed to be distributed by:$$\frac{1}{(2*\pi)^\frac{k+1}{2}*det(cov)}*e^{\frac{1}{2}u^T*cov^{-1}*u}$$ where u is the vector [Y1, Y2, ... Yk, X], det means determinant and cov^-1 is of course the inverse of the covariance matrix above.

Now, what my goal is, is to find the expected value of X, given that I know how the Y variables are correlated to each other, and given that I know E(X|Yj) for each of the k values of j. So basically, X is 0-mean normally, but I have k variables which are known, and each one pushes the expectation of X off to the side - and these k variables are themselves dependent. (Is that enough information to do this, first of all?)

The two extreme cases, if the k Y-variables were completely correlated, they would all be scalar multiples of each other, and the expected value of X given all of them would equal the expected value of X given just one of them, and they would all give the same answer - or the other extreme, if the k Y-variables were all completely uncorrelated to each other (which is theoretically possible, even though they are all correlated to X, because they could be pairwise independent), then I would add the expected value of X given by each of the Y variables, so if one Y-variable indicates X has a mean of 2 and another said it was 1, given both of them, X's mean would be 3. Presumably with the k Y-variables somewhat correlated, it should be something between these 2 extremes. Not as all the individual Yj variables' resultant expectations on X added together, but not just one of them either.

But in trying to do this, I'm running into a weirdness I can't cope with. The expression above for the distribution seems to imply that the expected value of X given Yj is $$\frac{-Y_j*V_{jx}}{V_{jj}}$$, where V=cov^-1. But it can't be BOTH things, right? It can't be $$\frac{Y_j*\sigma_{jx}}{\sigma_{jj}}$$ as WELL, right? Because they're only the same thing when it's a 2x2 matrix, it doesn't make any sense that it could be both things if there are 3 or more variables involved. Can someone point me in the right direction on this problem?

$\endgroup$
1
$\begingroup$

Never mind, I have the answer I was looking for. The answer is just $$\frac{Y_j*\sigma_{jx}}{\sigma_{jj}}$$. The terms derived from the inverse of the big covariance matrix should only be used when dealing with all the variables together, not the relationship between just two of them. And incidentally, $$\sigma_{jx}=\sigma_{jj}, \sigma_{xx}=T$$ for a large T (the limit as T goes to infinity) actually gets me the answer I was looking for - you invert that matrix and then multiply the result by T, and the negative of last column without the corner are the coefficients to multiply by the Y variables' respective expections on X to get their combined expectation on X what with them being dependent with each other, though I may have not said enough about the particular situation I have here for that to follow; and the other sigma variables are just the variances and covariances between the Y-variables.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.