Let $g:\mathbb R \to \mathbb R$ be $\mathcal C^\infty$ around $0$, and square-integrable w.r.t the standard normal distribution $N(0,1)$. Let $a_n:=g^{(n)}(0)/n!$ be the $n$ Maclaurin coefficient of $g$ and let $b_n := \mathbb E_{z \sim N(0,1)}[g(z)H_n(z)]$ be the $n$th Hermite coefficient of $g$, where $H_n$ is the probabilist's $n$th Hermite polynomial.
Question. Is there any relationship whatsoever between the $a_n$'s and $b_n$'s ?
Of course, $g(x) = \sum_{n=0}^\infty b_n H_n(x)$ for all $x \in \mathbb R$, and so
- $a_0 = g(0) = \sum_{n=0}^\infty b_n H_n(0) = b_0-b_2+3b_4 + \ldots$
- $a_1 = g'(0) = \sum_{n=0}^\infty b_n(g') H_n(0) = \sum_{n=0}^\infty b_{n+1} H_n(0)$, since $b_n(g') = b_{n+1}(g)$ for all $n$.
- ...
- More generall, $a_k := g^{(k)}(0)/ k! = (1/k!)\sum_{n=0}^\infty b_{n+k} H_n(0)$, for all $k \ge 0$,
I was just wondering if this can be made any simpler / more compact...