How to modify Gauss-Hermite quadrature rule when the weight function is slightly generalized hope this is the right forum. Consider a slightly modified version of the Gauss-Hermite quadrature rule, where the weight function is not $\exp(-\frac{x^2}{2})$ as in the standard Gauss-Hermite rule, but it is $\frac{1}{\sqrt{2\pi}}\exp(-\frac{x^2}{2})$. Let's indicate the corresponding nodes and weights respectively by $x_i$ and $w_i$. If I consider a more general weight function of the form  $\frac{1}{\sqrt{2\pi}\sigma}\exp(-\frac{(x-\mu)^2}{2\sigma^2})$, where $\sigma>0$, then how do $x_i$ and $w_i$ change? If I'm not wrong, the weights $w_i$ stay the same, while the nodes $x_i$ become $x'_i=\sigma x_i+\mu$. Do you agree? 
 A: To deal with the constant, just factor it out, do the integration with the more well-known rule, and then multiply it back in. To deal with the case of $\sigma \neq 1$ and/or $\mu \neq 0$, just change variables to $y=\frac{x-\mu}{\sigma}$. One can prove that the Gaussian quadrature rule generated by integration with respect to $y$ is (after rescaling) equivalent to the Gaussian quadrature rule that would have been generated by integrating with respect to $x$. 
Put another way, if we change variables to $u$ where $u$ is a (nonconstant) linear function of $x$, then the nodes in $u$ space are the image of the nodes in $x$ space, and the weights get rescaled by the slope of the transformation (or its reciprocal, I didn't write out the details). This is more or less because the integration functional $I$ and the quadrature functionals $Q_n$ are all linear.
A: The weights will scale with the inner product of the zeroth Hermite polynomial.  Two ways to see this.  First, is more intuitive: if you look at the "tower equations" (nomenclature from Stoer and Bulirsh)
$$
\sum_{i=0}^{N-1}w_iH_i(x_j) = \langle H_0(x),H_0(x)\rangle \delta_{j0}
$$
Of course $H_0(x)=1$ so this is the integral of the weighting function over the entire real line, and the zeroth equation that solves for the weights is
$$
\sum_{i=0}^{N-1}w_i = \int_{-\infty}^\infty \omega(x)dx.  
$$
Since a scaling will not affect the other equations (since the right hand side of all the other equations equals zero), then we can just scale all the weights to ensure this equation is satisfied.
Therefore, changing the weighting function scales the weights by the integral of the weighting function.
$$
\frac{w^{(1)}_i}{w^{(2)}_i}=\frac{\int_{-\infty}^\infty \omega^{(1)}(x)dx}{\int_{-\infty}^\infty \omega^{(2)}(x)dx}.
$$
The second is to look at Equation (2.6) in the paper by Golub and Welsch where the weights are calculated by casting the three-term recurrence relation in terms of a tridiagonal matrix and noting that the eigenvalues are the roots of $H_{N}(x)$ and the squared components of the eigenvector corresponding to the lowest eigenvalue is proportional to the weights
$$
w_i = q_{1,i}^2 \langle H_0(x), H_0(x)\rangle = q_{1,i}^2\int_{-\infty}^\infty \omega(x)dx.
$$
Since the second relies on the Christoffel-Darboux identity, I think the first approach is more intuitive.
