Hermite polynomials, expected values, and integration

In a previous post, I alluded to using Hermite polynomials in conjunction with higher-order Laplace approximation. In this post I’ll expand on what that means.

Hermite polynomials are orthogonal polynomials over the real line with respect to the weight given by the standard normal distribution. (There are two conventions for defining Hermite polynomials, what Wikipedia calls the physicist convention and the probabilist convention. We’re using the latter here.)

The first few Hermite polynomials are 1, x, x2 – 1, x3 – 3x, … . You can find the rest of the Hermite polynomials via the recurrence relation

Hn+1(x) = x Hnn Hn-1(x).

You could think of the Hermite polynomials as the right basis to use when working with normal probability distributions. Writing a polynomial as a linear combination of Hermite polyn0mials is a change of basis that makes integration easy:

\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty \left(\sum_{k=0}^N a_k\,H_k(x)\right) \exp(-x^2/2) \,dx = \sum_{k=0}^N a_k \, [k\, \mathrm{ even} ] \,k!!

Here [k even] is the function that returns 1 if k is even and 0 otherwise. This notation was introduced by Kenneth Iverson in the APL programming language and has become moderately common in mathematics.


One thought on “Hermite polynomials, expected values, and integration

  1. Arjun Viswanathan

    John, i just discovered your _amazing_ blog in 2018 – it’s a real mine of mathematical gems, like a daily shot of mathematical espresso.

    I think there is an error in the post above, or i am reading it wrong …

    The post expresses a polynomial as a sum of probabilist Hermite polynomials, ie P(x) = sum(k = 1 to n) a_k*H_k , and asserts:

    1/sqrt(2pi) * integral (-infinity to infinity) of a_k * H_k * exp(-x^2/2) dx = sum (k = 0 to n, k even) a_k * (k-1) !!

    However, as

    a) the family of H_ks are orthogonal w.r.t the weight function exp(-x^2/2), and

    b) the constant polynomial 1 = H_0,

    then for any k > 0,
    c) integral (-infinity to infinity) of a_k * H_k * 1 * exp(-x^2/2) dx = 0

    in fact the integral should neatly reduce to a_0

    I think (source: 5 min of doodling, so not entirely sure) – the double factorial sum above applies when a polynomial is expanded the standard way as P(x) = sum (k = 1 to n) a_k x^k

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