Prove moment/cumulant identity for gaussian random variables

Question

I am a physicist so please excuse my lack of formal mathematical language. In my studies I have stumbled across this identity that needs to be true in order for the uniqueness of my physical solution, but I am unable to prove it. Help is very much appreciated.

Assume you have two Gaussian random variables $X_1$ and $X_2$, which are in general correlated, i.e. $\kappa_2({X_1,X_2})=\langle X_1 X_2\rangle -\langle X_1\rangle\langle X_2\rangle\neq 0 $. I would like to prove that \begin{align} \frac{ \langle X_1 f(X_1) \rangle -\langle X_1 \rangle \langle f(X_1) \rangle }{\kappa_2(X_1)} = \frac{ \langle X_2 f(X_1) \rangle -\langle X_2 \rangle \langle f(X_1) \rangle }{\kappa_2(X_1,X_2)}; \end{align} $f$ is a continuous function. If it helps, you can also assume that it is a finite polynomial function.

Answer 1

If we assume that the pair $(X_1,X_2)$ are bivariate Normal then proving this is not too hard. In this case, $X_2=aX_1+Z$ for some constant $a$ and Normal r.v. $Z$ independent of $X_1$. In fact $a=\frac{\kappa_2( X_1,X_2)}{\kappa_2(X_1)}$. You're right to assume $f$ is a finite polynomial, and by linearity you need only prove that the equation is true for $f(x)=x^n$ for all $n$. In this case you get the LHS equal to $\frac{\langle X_1^{n+1}\rangle-\langle X_1\rangle\langle X_1^n\rangle}{\kappa_2(X_1)}$ and the RHS equal to $\frac{a\left(\langle X_1^{n+1}\rangle+\langle ZX_1^n\rangle\right)-(a\langle X_1\rangle+\langle Z\rangle)\langle X_1^n\rangle}{a\kappa_2(X_1)}$. As $\langle ZX_1^n\rangle=\langle Z\rangle\langle X_1^n\rangle$ by independence you're done.

However in the general case ($X_1$, $X_2$ Normal does not imply $(X_1,X_2)$ bivariate Normal) then I'm not sure how to proceed, and in fact I'm not even convinced it's true.