Given $n$ normally distributed observations $f(X|\sigma_X)=\mathcal{N}(\mu_X, \sigma_X^2)$ and assuming a uniform prior on $\log(\sigma_X)$ and known $\mu_x$, I'm trying to find marginal posterior distribution for $\sigma^2_X$. I know how to approach for finding marginal posterior distribution for $\sigma_X$, but not $\sigma^2_X$ and even in that case, I don't find any nice form for the posterior, so I was not sure if my approach is correct.

Here is my try:


$$f(log(\sigma_X))\propto 1 \qquad \text{therefore, }\qquad f(\sigma_X)\propto\frac{1}{\sigma_X}$$

$$P(\sigma_X|X)\propto \frac{1}{\sigma_X}\prod_i^n\frac{1}{\sqrt{2\pi\sigma^2_X}}\exp(-\frac{(x_i-\mu_X)^2}{2\sigma_X^2})$$

Which seems to me that is no particular distribution form. My questions are:

1) Is my interpretation of the uniform prior on log scale correct?

2) Does this final form represent any particular distribution function?

3) Is there a way to find marginal posterior distribution for $\sigma^2_X$ or should I assume that this was a typo and we only can find marginal posterior distribution for $\sigma_X$?

  • $\begingroup$ You need to integrate out $\mu_X$ $\endgroup$ – Xi'an Sep 30 '18 at 15:17
  • $\begingroup$ In general, yes. But $\mu_X$ is a known value here. $\endgroup$ – Blade Sep 30 '18 at 15:29

In the case where $\mu$ is known, there is no "marginal" posterior distribution, only a posterior distribution, and you already have it. To see this, start out by changing your prior to be on $\sigma^2$; as it happens, making it uniform on $\log \sigma^2$ gives you $f(\sigma^2) \propto 1/\sigma$, as before. (The answer to your first question is yes, you are interpreting the uniform prior on the log scale correctly.) Your posterior is then $P(\sigma^2|X)$ with exactly the same functional form as you have already derived; perform the product multiplication from $1$ to $n$ to get:

$$P(\sigma^2|X,\mu) \propto (\sigma^2)^{-(n+1)/2}\exp\left(-{\sum(x_i-\mu)^2 \over 2\sigma^2}\right)$$

which is simply a rewritten version of what you already have.

As for what distribution this is, if you look at the functional form, it looks a lot like that of a Gamma distribution, just with the variable of interest in the denominator everywhere instead of the numerator. This leads us to the inverse gamma distribution (Wikipedia link.)


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