Numerical Quadrature Rules for Common Distributions

Gauss-Laguerre Quadrature for Exponential Integrals

For integrals of the form ${\int }_0^{\mathrm{\infty }}{e}^{-\zeta }\varphi (\zeta )d\zeta$ for some function $\varphi$, Gauss-Laguerre quadrature of order $n$ provides $n$ abscissæ ${\zeta }_i$ and weights ${\omega }_i$ for constructing the following linear approximation: $${\int }_0^{\mathrm{\infty }}{e}^{-\zeta }\varphi (\zeta )d\zeta \approx \sum _{i=1}^n{\omega }_i\varphi ({\zeta }_i).$$

This form of quadrature is useful for approximating the expectation of a nonlinear function of an exponentially-distributed random variable. Let $X$ be an exponential random variable with rate parameter $\lambda$. Then, with a simple transformation, we can write the expectation of $f(X)$ in the form above. The expectation of $f(X)$ is $$\mathrm{E}[f(X)]={\int }_0^{\mathrm{\infty }}\lambda e^{-\lambda x}f(x)\mathrm{dx}.$$ If we let $\zeta =\lambda x$ and $\varphi (\cdot )=f(\cdot /\lambda )$ (and note that $d\zeta =\lambda \mathrm{dx}$), then, by the change of variables, $$\mathrm{E}[f(X)]={\int }_0^{\mathrm{\infty }}{e}^{-\zeta }\varphi (\zeta )d\zeta .$$ Applying the Gauss-Laguerre quadrature rule above gives the approximation $$\mathrm{E}[f(X)]\approx \sum _{i=1}^n{\omega }_{i}f({\zeta }_i/\lambda ).$$

For weights and abscissæ, see the Digital Library of Mathematical Functions or the calculator at eFunda.

Gauss-Hermite Quadrature for Normal Integrals

Similarly, Gauss-Hermite quadrature provides weights ${\omega }_i$ and abscissæ ${\zeta }_i$ for integral approximations of the form: $${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-{\zeta }^2}\varphi (\zeta )d\zeta \approx \sum _{i=1}^n{\omega }_i\varphi ({\zeta }_i).$$ If $X$ is a normally distributed random variable with mean $\mu$ and variance ${\sigma }^2$, then we can approximate the expectation of $f(X)$ using quadrature by applying the transformation $\zeta =(x-\mu )/(\sqrt{2}\sigma )$, $\varphi (\zeta )=f(\mu +\sqrt{2}\sigma \zeta )/\sqrt{\pi }$, and thus, $d\zeta =\mathrm{dx}/\sqrt{2{\sigma }^2}$. Then, $$E[f(X)]={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{1}{\sqrt{2\pi {\sigma }^2}}{e}^{-\frac{(x-\mu {)}^2}{2{\sigma }^2}}f(x)\mathrm{dx}\approx \sum _{i=1}^n\frac{{\omega }_i}{\sqrt{\pi }}f(\mu +\sqrt{2}\sigma {\zeta }_i).$$

For weights and abscissæ, see the Digital Library of Mathematical Functions or the calculator at eFunda.