# Getting a Hermite polynomial expansion of Gaussian with given variance.

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 $a_n=\frac{1}{\sqrt{2\pi}\sigma}\int_{-\infty}^{\infty}{\exp({-\frac{x^2}{2\sigma^2}})}H_{n}(x)\exp({-\frac{x^2}{2})}dx$

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 $\alpha$

"""

• The integrands of two parts of the question are not the same. In the first part, the integrand is: $exp(.)H[x] exp(.)$. In the second part, the integrand is $exp(.) H[x]$. – wolfies Jan 17 '15 at 5:00
• If you add exponents you get the general form as in the second part. – them May 15 '16 at 5:05

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 $x=y$.