Multivariate normal density function of function of random variable

Question

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?

Answer 1

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 $n(n-1)$.

P.S.: I have confirmed my guess about the eigenvalues.