Weights versus xi for four choices of n

In numerical analysis, Gauss–Hermite quadrature is a form of Gaussian quadrature for approximating the value of integrals of the following kind:

${\displaystyle \int _{-\infty }^{+\infty }e^{-x^{2}}f(x)\,dx.}$

In this case

${\displaystyle \int _{-\infty }^{+\infty }e^{-x^{2}}f(x)\,dx\approx \sum _{i=1}^{n}w_{i}f(x_{i})}$

where n is the number of sample points used. The xi are the roots of the physicists' version of the Hermite polynomial Hn(x) (i = 1,2,...,n), and the associated weights wi are given by [1]

${\displaystyle w_{i}={\frac {2^{n-1}n!{\sqrt {\pi }}}{n^{2}[H_{n-1}(x_{i})]^{2}}}.}$

## Example with change of variable

Consider a function h(y), where the variable y is Normally distributed: ${\displaystyle y\sim {\mathcal {N}}(\mu ,\sigma ^{2})}$. The expectation of h corresponds to the following integral:

${\displaystyle E[h(y)]=\int _{-\infty }^{+\infty }{\frac {1}{\sigma {\sqrt {2\pi }}}}\exp \left(-{\frac {(y-\mu )^{2}}{2\sigma ^{2}}}\right)h(y)dy}$

As this doesn't exactly correspond to the Hermite polynomial, we need to change variables:

${\displaystyle x={\frac {y-\mu }{{\sqrt {2}}\sigma }}\Leftrightarrow y={\sqrt {2}}\sigma x+\mu }$

Coupled with the integration by substitution, we obtain:

${\displaystyle E[h(y)]=\int _{-\infty }^{+\infty }{\frac {1}{\sqrt {\pi }}}\exp(-x^{2})h({\sqrt {2}}\sigma x+\mu )dx}$

${\displaystyle E[h(y)]\approx {\frac {1}{\sqrt {\pi }}}\sum _{i=1}^{n}w_{i}h({\sqrt {2}}\sigma x_{i}+\mu )}$