Consider the following:$$$$ If $S^2$ is the sample variance based on a sample of size n from a normal population, we know that $\displaystyle\frac{(n-1)S^2}{\sigma ^2} \sim \chi ^2_{n-1}$. The conjugate prior for $\sigma ^2$ is the inverted gamma pdf $IG(\alpha,\beta)$, that is: $$ \displaystyle \pi(\sigma^2)=\frac{1}{\Gamma(\alpha)\beta^\alpha(\sigma^2)^{\alpha+1}}e^{-\frac{1}{\beta \sigma^2}} $$ I would like to show that the posterior distribution of $\sigma ^2$ is $\displaystyle IG(\alpha+\frac{n-1}{2},\Big[\frac{(n-1)S^2}{2}+\frac{1}{\beta}\Big]^{-1})$


What exactly is a prior distribution, posterior distribution, and conjugate prior? If you can, please state both a formal and informal definition/description.


1 Answer 1


If you have a likelihood with a specified density $f(x|\theta)$, and you put a distribution on $\theta$ with density $\pi(\theta)$ (the prior), you can get a distribution on $\theta$ conditional on the observed data using Bayes rule.

\begin{equation} \pi(\theta|x) = \frac{f(x|\theta)\pi(\theta)}{f(x)} \end{equation}

This is the posterior distribution, the distribution of the parameters conditional on the data. Note that the prior represents what you think the distribution of the parameter should be a priori, and the posterior represents what you think the parameter should be once you observe data (taking into account your prior beliefs).

A conjugate prior is one in which the form of the posterior is in the same family of distributions as the prior; with adjusted parameters. This makes computing the posterior distribution much easier.

We have that if $\frac{(n-1)S^2}{\sigma^2}\sim\chi^2_{n-1}$ then $\frac{(n-1)S^2}{2} \sim \textrm{Gamma}(\frac{n-1}{2},\sigma^2)$.

Using the conjugacy of the Gamma to the second parameter of a Gamma distribution, we place a $\sigma^2 \sim \textrm{Gamma}(\alpha,\beta)$ prior on $\sigma^2$.

Which gives us $\sigma^2 | S^2 \sim \textrm{Gamma}(\alpha + \frac{n-1}{2},\beta + \frac{(n-1)S^2}{2})$

You can see the formula for how I got this here.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.