How to calculate the mean of standard deviation when data are drawn from a Gaussian population?

Question

This is a pure algebraic question. Drawing a sample of size $n$ from a Gaussian population $N(0, \sigma^2)$, the posterior probability of $\sigma$ is proportional to $\frac{1}{\sigma^{n+1}} e^{-(s/\sigma)^2}$, where s is the standard deviation of the sample. To get the mean standard deviation,

\begin{equation} \begin{split} \langle\sigma^2\rangle & = C\int_0^\infty \sigma^2\sigma^{-n-1}e^{-n(s/\sigma)^2} \, d\sigma \\[8pt] & = \frac{ns^2}{n-2} \\[8pt] & = \frac{1}{n-2}\sum_i (x-\mu)^2 \end{split} \end{equation}

Does anyone have idea how the integral is made? It is not a regular integral that can be found in a common Gaussian integral table.

Answer 1

As Xi'an points out in comments, the substitution $\xi=\sigma^{-2}$ is called for. \begin{align} & \int_0^\infty \sigma^2\sigma^{-n-1}e^{-n(s/\sigma)^2} \, d\sigma \\[10pt] = {} & \int^0_\infty \xi^{-1} \xi^{(n+1)/2} e^{-ns^2\xi} \left( \frac{-\xi^{-3/2}\, d\xi} 2 \right) \\ & \qquad \text{where } \xi = \sigma^{-2} \\[10pt] = {} & \frac 1 2 \int_0^\infty \xi^{(n-4)/2} e^{-ns^2\xi} \, d\xi \\[8pt] = {} & \frac 1 2 \int_0^\infty \left( \frac \tau {ns^2} \right)^{(n-4)/2} e^{-\tau} \left( \frac{d\tau}{ns^2} \right) \\ & \qquad \text{where } \tau = ns^2 \xi \\[10pt] = {} & \frac 1 {2(ns^2)^{(n-2)/2}} \int_0^\infty \tau^{(n-2)/2 - 1} e^{-\tau} \, d\tau \\[8pt] = {} & \frac 1 {2(ns^2)^{(n-2)/2}} \Gamma\left( \frac{n-2} 2 \right). \end{align} One could of course do this as just one substitution: $\tau = n(s/\sigma)^2.$

How to use ordinary substitutions to reduce and integral to standard form can be useful.