I am reading a paper that makes the following claim (without proof):
Claim: Suppose that $f$ is a degree-$p$ polynomial, i.e., $f = \sum_{i = 0}^p c_i x^i$. Then its Hermite expansion can be written as: $$ f(x) = \sum_{i = 0}^p \frac{1}{i!} C_i He_i(x), $$ where $C_i = \mathbb{E}_{X \sim \mathcal{N}(0, 1)}[f^{(i)}(X)]$ is the expected $i$-th derivative of $f$ under the Gaussian measure and $He_i$ is the $i$-th Chebyshev-Hermite polynomial.
I am struggling to see why this is the case. My first thought was to write down the monomials $x^i$ in terms of their Hermite expansion:
$$ x^i = i! \sum_{j = 0}^{\lfloor i/2 \rfloor} \frac{1}{2^j j!(i - 2j)!} He_{i-2j}(x) $$
and use this to calculate the Hermite coefficients $a_n = \langle f, He_n \rangle$. However, the resulting sum seems nontrivial to simplify. I was wondering if there is an "obvious" proof that I am missing (or a reference that includes a rigorous proof).
Define an inner product on polynomials by $$\langle f, g \rangle = \mathbb E(f(X)g(X)).$$ Then $$\langle \mathrm{He}_j, \mathrm{He}_k \rangle = k! \,\delta_{jk}$$ for all $j,k\in \mathbb N$. So a polynomial $f$ of degree $n$ can be expanded as $$f = \sum_{k=0}^n \frac1{k!}\langle f, \mathrm{He}_k\rangle \mathrm{He}_k.$$ Finally, repeated partial integration shows that $$\langle f, \mathrm{He}_k \rangle = \langle f^{(k)}, 1\rangle = \mathbb E(f^{(k)}(X)).$$