# Multivariate normal density function of function of random variable

Let $X_1,\dots,X_n$ be i.i.d random variables and $g$ be a symmetric function such that $$g(X_i,X_j)\sim N(\mu,\sigma^2)$$ for all $1\le i<j\le n$. I wish to know the density function of the joint random variable $$Z=\left(g(X_1,X_2),g(X_1,X_3),\dots,g(X_{n-1},X_{n})\right)$$ which lies in ${n\choose 2}$-dimensional space. To do this, I assume that $Z$ follows multivariate normal distribution. However, the problem is that the covariance matrix $C$ of $Z$ is singular. Could anyone help me? Any advice or suggestion?

• If the covariance matrix is singular, then the multivariate normal distribution is degenerate and does not have a density w.r.t. the corresponding Lebesgue measure. You could define its density w.r.t. the Lebesgue measure on the subspace containing the distribution's support (see en.wikipedia.org/wiki/…). – Budenn Jul 31 '15 at 13:01
• If I answer this, I'll probably say how to find $\operatorname{E}(Z) \in \mathbb R^{\binom n 2}$ and $C = \operatorname{var}(Z) \in \mathbb R^{\binom n 2 \times\binom n 2}$ and then let others worry about the actual density if they want to. ${}\qquad{}$ – Michael Hardy Jul 31 '15 at 13:28
• So part of the question would be this: For which symmetric functions and which distributions of $X_1$ would $Z$ have a multivariate normal distribution? ${}\qquad{}$ – Michael Hardy Jul 31 '15 at 13:32
• I suspect one can show that no generality would be lost by assuming $X_1$ is normally distributed. If so, then for which symmetric functions $g$ would it be the case that all of $g(X_i,X_j)$ are normally distributed? Clearly it's true if $g(u,v) = a(u+v)$. I wonder if one can eliminate all other possibilities? ${}\qquad{}$ – Michael Hardy Jul 31 '15 at 13:41
• If $n\ge 4$ then $\det C=0$ but otherwise $\det C>1$. ${}\qquad{}$ – Michael Hardy Jul 31 '15 at 13:52

See my comments under the question. What I write below will assume my guesses there are right and that $a=1$.
Your symmetric matrix $C$ will have one row corresponding to each unordered pair $\{i,j\}$.
The entry in row $\{i,j\}$ and column $\{k,\ell\}$ will be $2$ if $\{i,j\}=\{k,\ell\}$. Those are just the diagonal entries in the matrix. In row $\{i,j\}$, there will be a $1$ in each column $\{k,\ell\}$ for which $|\{i,j\}\cap\{k,\ell\}| = 1$, and there are $2n-3$ such columns. The remaining $\dbinom{n-2} 2$ such columns in that row will contain a $0$. This is a $\dbinom n 2 \times \dbinom n 2$ matrix of rank $n$. It can be diagonalized by an orthogonal matrix and then you have the $n$ nonzero eigenvalues on the diagonal.
Software is telling me that when $n=5$ then the eigenvalues are $8,3,3,3,3,0,0,0,0,0$ and when $n=6$ they are $10,4,4,4,4,4,0,0,0,0,0,0,0,0,0$. When $n=7$ they are $12,5,5,5,5,5,5,0,\ldots,0$ ($21$ of them). Here one can guess that the largest eigenvalue is $2(n-1)$ and the next $n-1$ of them are $n-2$, and clearly the rest have to be $0$. The sum of the eigenvalues would then be $n(n-1)$.