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 $\alpha$
"""
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$\begingroup$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: . 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 $x=y$.
$\endgroup$ 9 $\begingroup$The Hermite expansion of the probability distribution function for $\mathcal{N}(0,\sigma^2)$$$ \omega_\sigma(x) = \frac{1}{\sqrt{2\pi\sigma^2}}e^{-\frac{x^2}{2\sigma^2}} $$is$$ \omega_\sigma(x)=\sum_{m=0}^\infty \frac{(-1)^m}{m!2^m \sqrt{2\pi\left(\sigma^2+1\right)^{2m+1}}}H_{2m}(x). $$
The starting point is the expression for the general coefficient for any function $f(x)=\sum_n d_n H_n(x)$ is given by the expression$$ d_n = \frac{1}{n!}\int_{-\infty}^\infty \frac{\partial^n f(x)}{\partial x^n} \omega(x) dx, $$which, by the definition of the stretched Hermite polynomials $ H_n\left(\frac{x}{\sigma}\right) = \frac{(-\sigma)^n}{\omega_\sigma(x)}\frac{\partial^n}{\partial x^n}\omega_\sigma(x)$, reduces to$$ d_n = \frac{1}{(-\sigma)^nn!\sqrt{2\pi\sigma^2}}\int_{-\infty}^\infty H_n\left(\frac{x}{\sigma}\right) \omega_\sigma(x)\omega(x)dx. $$The product $\omega_\sigma(x)\omega(x)$ can be transformed into $\omega(y)$ by the change of variables $y = \frac{x}{\sigma}\sqrt{\sigma^2+1}$.
The integral of a general stretched Hermite polynomial$$ I_\alpha =\int_{-\infty}^\infty H_\alpha(\gamma x)\omega(x)dx $$can be done via a Taylor expansion$$ I_\alpha = \sum_{\beta=0}^{\alpha} {\alpha \choose\beta} H_{\alpha-\beta}(0)\gamma^\beta \int_{-\infty}^\infty x^\beta\omega(x)dx $$with result$$ I_{\alpha} = (\alpha-1)!!\left(\gamma^2-1\right)^{\frac{\alpha}{2}}. $$Putting this into the formula for $d_n$, we get$$ d_n = \begin{cases} \frac{1}{(\sigma)^nn!\sqrt{2\pi(\sigma^2+1)}}(n-1)!!\left(\frac{-\sigma^2}{\sigma^2+1}\right)^{\frac{n}{2}} & n\textrm{ even}\\ 0 & n \textrm{ odd}, \end{cases} $$which, after a reindex of the summation leads to the expansion given at the start of this answer.
Please read my recently submitted manuscript for all of the details.
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