I am trying to find an expansion of centered Gaussian – $\frac{1}{\sqrt{2\pi}\sigma}\exp({-\frac{x^2}{2\sigma^2})}$ in terms of Hermite polynomials.
Namely to calculate $a_n=\frac{1}{\sqrt{2\pi}\sigma}\int_{-\infty}^{\infty}{\exp({-\frac{x^2}{2\sigma^2}})}H_{n}(x)\exp({-\frac{x^2}{2})}dx$
Any comments are welcome.
Edited later:
"""
Equivalently, I am looking for the value of –
$a_n=\int_{-\infty}^{\infty}{\exp({-\frac{x^2}{\alpha}})}H_{n}(x)dx$
for some arbitrary $\alpha$
"""
Best Answer
If $\hat{H}_{n}$ are the normalized Hermite functions, and $-1 < r < 1$, then $$ \begin{align} \sum_{n=0}^{\infty}r^{n}\hat{H}_{n}(x)^{2} & =\frac{1}{\sqrt{\pi(1-r^{2})}}\exp\left(-\frac{1-r}{1+r}x^{2}+x^{2}\right) \\ & = \frac{1}{\sqrt{\pi(1-r^{2})}}\exp\left(\frac{2r}{1+r}x^{2}\right). \end{align} $$ The normalized $\hat{H}_{n}$ are chosen so that $\int_{-\infty}^{\infty}\hat{H}_{n}(x)e^{-x^{2}}dx = 1$. By choosing $r$ appropriately, you can get what you want. Look for Mehler's kernel on this Wikipedia page: http://en.wikipedia.org/wiki/Hermite_polynomials#Hermite_functions . You'll find more information about how one derives the above special case of Mehler's kernel $\sum_{n=0}^{\infty}r^{n}H_{n}(x)H_{n}(y)$ where $x=y$.