# Finding the distribution of a function of $n$ random, normally distributed correlated variables

Given a random vector $X$ of $n$ normally distributed random variables, and an $n \times n$ covariance matrix of those variables with non-zero correlation terms, what is the generalized methodology to find the distribution of a non-linear function $f(X_1,X_2,\dots,X_n)$ of the random variables of $X$?

That's the general formulation of a problem I'm trying to solve. More specifically, given $6$ normally distributed random variables $x_1 \dots x_6$, what is the probability distribution of $$\sqrt{(x_1 - x_4)^2 + (x_2 - x_5)^2 + (x_3 - x_6)^2}$$ where $x_1,x_2,x_3$ are correlated and $x_4,x_5,x_6$ are correlated (i.e. the upper right and lower left $3 \times 3$ correlation terms are zero, but all other correlation terms are not).

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btw if it simplifies things...finding the distribution without the square root would be useful as well – user109078 Sep 6 '11 at 16:27
Does "normally distributed" mean each one separately is normally distributed, or that that they are jointly normally distributed? In the former case, you haven't given enough information; in the latter case, it wouldn't hurt to mention that. – Michael Hardy Sep 6 '11 at 21:11

Assuming the $X_i$ have mean 0, $(X_1 - X_4)^2 + (X_2 - X_5)^2 + (X_3 - X_6)^2$ (or any other quadratic form in the $X_i$) has a generalized chi-square distribution. See http://en.wikipedia.org/wiki/Generalized_chi-square_distribution. If $Y$ is a nonnegative random variable with density $f_Y(y)$, then $S = \sqrt{Y}$ has density $f_S(s) = 2 s f_Y(s^2)$.

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Xi do not have a mean of zero...does that reference still apply? In any case the reference says "is the distribution of a sum of the squares of k independent standard normal random variables"...but I already said they aren't independent, they're correlated. – user109078 Sep 6 '11 at 17:57
In this case, since $X_i$ is not correlated with $X_{i+3}$, yes the chi-square still applies. You have to add $(\bar{X_1} - \bar{X_4})^2 + (\bar{X_2} - \bar{X_5})^2 + (\bar{X_3} - \bar{X_6})^2$ to your chi-square variable $Y$ of course, to make sure that the new number has the correct mean. – Craig Sep 6 '11 at 20:03
There is such a thing as a non-central chi-square distribution. en.wikipedia.org/wiki/Noncentral_chi-square_distribution Things like that and the non-central F-distribution en.wikipedia.org/wiki/Non-central_F-distribution are used for finding statistical power. en.wikipedia.org/wiki/Statistical_power – Michael Hardy Sep 7 '11 at 4:08
@Craig: A random variable with a non-central chi-square distribution is not simply a constant plus a random variable with a central chi-square distribution. In particular, the lower boundary of the support of the probability distribution is still zero. – Michael Hardy Sep 7 '11 at 4:10

If the covariance matrix of the random column vector $X^T=(X_1,\ldots,X_n)^T$ is a nonsingular matrix $V$, and if they are jointly, not merely separately normally distributed, then $V$ has a positive-definite symmetric square root (found by first doing the spectral decomposition). Call that $V^{1/2}$. Then $V^{-1/2}X$ is normally distributed and its entries are not correlated, but in fact are independent, and all the variances are equal to 1. (If the variables are separately, but not jointly, normally distributed, then this transformation will still make the variances 1 and the covariances 0, but then one might not have independence, and there are other complications.) This reduces the problem to that of independent standard normals, provided the expected values are 0.

For quadratic forms, we would then have chi-square distributions or (if the means are not all 0) non-central chi-square distributions.

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