Suppose that $p:\mathbb{R} \rightarrow \mathbb{R}$ is a polynomial and let $z$ be a standard Gaussian variable $(z \sim \mathcal{N}(0,1))$. I am interested in the behavior of the expectation $\mathbb{E}[p(z)^kz]$ for some natural number $k \in \mathbb{N}$. It is clear that when $p$ is an even polynomial, this expectation is always zero due to the symmetry of the Gaussian measure.
However, for any non-even polynomial $p$, I hypothesize that there exists at least one $k \in \mathbb{N}$ such that: $$\mathbb{E}[p(z)^kz]\neq 0.$$
Could someone provide insights or a proof to show whether this is the case for all non-even polynomials or a counter-example if it’s not? Any suggestions or references to relevant theorems or techniques would also be quite helpful.
Observation 1: The expectation being zero is equivalent to $\int_0^\infty (p^k(z)-p^{k}(-z))\phi(z) dz=0$.
Observation 2: it suffices to prove that the limit as $k\to\infty$ cannot be zero.
Observation 3: the limit of $p^k(z)-p^{k}(-z)$ will be infinite as $z\to\infty$
Observation 4: for odd $k$ and fixed $z$, the sign remains the same
Observation 5: as $k$ becomes arbitrarily large, only the biggest values are relevant for the value of the integral
Combining these observations should give you a proof.