Formula for Gaussian random variable

Question

Assume that $\xi=(\xi_1,\xi_2,\ldots,\xi_n)$ is a random Gaussian vector. Prove that $$ E\{f(\xi)\xi\}=E\{\xi\xi^T\}E\{\nabla f(\xi)\}$$ for every differentiable function $f(x)$ and $E\{\xi\}=0$. I will be grateful for any help, at least for $n=1$.

Answer 1

As I mentioned, this is only true if $E \xi=0$, which I will assume.

The $n=1$ case is done using integration by parts. Suppose $\xi\sim N (0,\sigma^2)$. I will also assume that $f$ is compactly supported. The result can be extended to general $f$ by approximating $f$ by its product with bump functions.

Let $\phi(x)=\frac1{\sqrt{2\pi\sigma^2}}\exp\left(-\frac{x^2}{2\sigma^2}\right)$ be the pdf of $\xi$. You can verify that $$ \phi'(x)=-\frac{x}{\sigma^2}\phi(x), $$ so $$ \begin{align} E[\xi f(\xi)] &=\sigma^2\int_{-\infty}^\infty \frac{x}{\sigma^2}f(x)\,\phi(x)\,dx \\&=\require{cancel}-\sigma^2\int f(x)\phi'(x)\,dx \\&=\cancelto{}{-{\sigma^2f(x)\phi(x)\bigg|^\infty_{-\infty}}}+\sigma^2\int f'(x)\phi(x)\,dx \\&=\sigma^2E[f'(X)]. \end{align} $$ as desired.

Now, for general $n$, the trick is to apply the $n=1$ case to each component of $\xi f(\xi)$ while conditioning all but one of the components of $\xi$.

First assume that $\xi\sim N(0,I_n)$, so that its components $\xi_1,\dots,\xi_n$ are independent $N(0,1)$. Let $E_i$ denote the expectation conditional on all of these components except for $X_i$. Then the $i^th$ component of $E[\xi f(\xi)]$ is $$ E[\xi_i f(\xi)] = E[E_i[\xi_i f(\xi)]] $$ Now, let $f_i(x)$ be the function $$f_i(x) = f(\xi_1,\xi_2,\dots,\xi_{i-1},x,\xi_{i+1}\dots,\xi_n)$$ Then by applying the $n=1$ case to $E_i[\xi_i f(\xi)]=E_i[\xi_i f_i(\xi_i)]$, we have $$ E[E_i[\xi_i f(\xi)]] = E\left[E_i\left[f_i'(\xi_i)\right]\right]=E[\left[f_i'(\xi_i)\right]=E\frac{\partial f}{\partial x_i}(\xi) $$ This is true for each components, so $E[\xi f(\xi)]=E[\nabla f(\xi)]$.

I leave it to you to work out what to do in the general case when $\xi\sim N(0,\Sigma)$ with $\Sigma$ invertible. Hint: $f(\xi)=f(\Sigma^{1/2}\Sigma^{-1/2}\xi)$, where $\Sigma=E[\xi\xi^T]$, so that $ \Sigma^{-1/2}\xi\sim N(0,1)$.