When is a function on symmetric positive definite matrices an expectation of Gaussian?

Question

Is there some characterization of real-valued functions of the form $\Phi(C)=\mathbb{E}F(X)$, where $X$ has the Gaussian $N(0,C)$ distribution, on the space of symmetric positive semidefinite $n\times n$ matrices $C$? In other words, given $\Phi\colon \mathrm{SPD}\to \mathbb{R}$, is there a way to tell whether such $F$ exists?

Edit: For example, is it clear that $\det(C)$ is not of this form?

Edit 2: Answer below shows that $\det(C)$ is not of this form even for $n=2$.

Answer 1

Here's an approach to determine if $\det(C)$ is of this form with $F$ analytic and $n=2$.

By Sylvester's criterion a matrix $\begin{pmatrix} a& b \\ b & c\end{pmatrix}$ is symmetric positive semidefinite iff $a\ge 0$ and $c\ge 0$, and $ac-b^2\ge 0$.

Let $X=(X_1,X_2)$ be normal with mean 0 and $E(X_1X_2)=b$, $E(X_1^2)=a$, $E(X_2^2)=c$. The Pearson correlation coefficient is $\rho=b/\sqrt{ac}$. Suppose $F(x,y)=\sum c_{mn}x^my^n$.

Is there an $F$ with $E(F(X))=ac-b^2$? We have $X=\sqrt{a}Z_1$, $Y=\sqrt{c}(\rho Z_1+\sqrt{1-\rho^2}Z_2)$ where $Z_i$ are independent standard normal, so using $$E\left[(\rho Z_1+\sqrt{1-\rho^2}Z_2)^{2k}\right]=E\sum_{t=0}^{2k}\binom{2k}{t}\rho^t(1-\rho^2)^{(2k-t)/2}Z_1^t Z_2^{2k-t}$$ $$=\sum_{t=0}^{2k}\binom{2k}{t}\rho^t(1-\rho^2)^{(2k-t)/2}E[Z_1^t Z_2^{2k-t}] =\sum_{u=0}^{k}\binom{2k}{2u}\rho^{2u}(1-\rho^2)^{(2k-2u)/2}E[Z_1^{2u} Z_2^{2k-2u}]$$ $$=\sum_{u=0}^{k}\binom{2k}{2u}\rho^{2u}(1-\rho^2)^{(2k-2u)/2}(2u-1)!!(2(k-u)-1)!!$$ (where $(-1)!!=1$) we have $$E(X^{2\ell}Y^{2k})=a^\ell c^k\sum_{u=0}^{k}\binom{2k}{2u}\rho^{2u}(1-\rho^2)^{k-u}(2(\ell+u)-1)!!(2(k-u)-1)!!$$

• When $k=\ell=1$, it is $ac((1-\rho^2)+3\rho^2)=ac+2b^2$ since $\rho^2=b^2/ac$.
• When $k=0$ and $\ell=2$, it is $3a^2$.
• When $k=2$ and $\ell=0$, it is $3c^2-4b^4/a^2$.
• I guess we should also do the case $E(X^3Y^1)$ and a couple of others.

A Taylor series in variables $a,b,c$ defines the zero function only if all coefficients are zero. So now by calculating these expectations $E(X^pY^q)$ we can determine whether $ac-b^2$ is obtainable.