6
$\begingroup$

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?

|cite|improve this question
$\endgroup$
  • $\begingroup$ 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/…). $\endgroup$ – Budenn Jul 31 '15 at 13:01
  • $\begingroup$ 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. ${}\qquad{}$ $\endgroup$ – Michael Hardy Jul 31 '15 at 13:28
  • $\begingroup$ 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? ${}\qquad{}$ $\endgroup$ – Michael Hardy Jul 31 '15 at 13:32
  • $\begingroup$ 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? ${}\qquad{}$ $\endgroup$ – Michael Hardy Jul 31 '15 at 13:41
  • $\begingroup$ If $n\ge 4$ then $\det C=0$ but otherwise $\det C>1$. ${}\qquad{}$ $\endgroup$ – Michael Hardy Jul 31 '15 at 13:52
2
$\begingroup$

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.

|cite|improve this answer
$\endgroup$
  • $\begingroup$ Thanks for your help. I think, in this special case, we can obtain the density function by formula as gien by @Budenn. However, how about my above question? $\endgroup$ – Jlamprong Aug 3 '15 at 0:34

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.