marginal posterior distribution

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:

$$P(\sigma_X|X)=P(X|\sigma_X)P(\sigma_X)$$

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

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

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 $$\sigma_X$$?

• You need to integrate out $\mu_X$ Sep 30 '18 at 15:17
• In general, yes. But $\mu_X$ is a known value here. 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:
$$P(\sigma^2|X,\mu) \propto (\sigma^2)^{-(n+1)/2}\exp\left(-{\sum(x_i-\mu)^2 \over 2\sigma^2}\right)$$