How to calculate the posterior distribution with a normal likelihood function and a prior that involves sigma

Question

In the problem, the data X follows a normal distribution, or $f(x|\mu,\sigma^2) = \frac{1}{\sqrt{2\pi\sigma^2}}\exp(-\frac{1}{2}(\frac{x-\mu}{\sigma})^2)$. Let's say I know the value of $\sigma^2$ and want to find the posterior distribution of $\mu$, and I choose a prior distribution of $p(\mu, \sigma^2) = \frac{1}{\sigma^2}$. My understanding here is that the posterior distribution is calculated as $p(\mu|x, \sigma^2) = f(x|\mu,\sigma^2)p(\mu, \sigma^2)$, or basically $\frac{1}{\sqrt{2\pi\sigma^2}}\exp(-\frac{1}{2}(\frac{x-\mu}{\sigma})^2)$ times $\frac{1}{\sigma^2}$. And I understand that the goal here would be to arrive at something that still looks like a normal distribution for my posterior distribution, so that I can just draw from this normal distribution with some modified parameters to do what I want with the posterior.

What I'm stuck on is how I multiply those two equations together in a way that achieves that. How do I work in that $\sigma^2$ quantity and arrive at a posterior distribution that utilizes it somehow?

Answer 1

The posterior distribution on $(\mu,\sigma)$ is $$p(\mu,\sigma|x)\propto p(\mu,\sigma)f(x|p(\mu,\sigma))=\frac{1}{\sqrt{2\pi\sigma^2}\sigma^2}\exp\left\{-\frac{1}{2}\left(\frac{x-\mu}{\sigma}\right)^2\right\}$$ However, the rhs integrates to infinity, hence one cannot use this improper prior with a single observation $x$.

If the only parameter is $\mu$, meaning that the value of $\sigma^2$ is known, then, using a flat prior on $\mu$, $$p(\mu|x)\propto p(\mu) f(x|\mu)=\frac{1}{\sqrt{2\pi\sigma^2}}\exp\left\{-\frac{1}{2}\left(\frac{x-\mu}{\sigma}\right)^2\right\}$$ is a well-defined probability distribution, meaning that the posterior on $\mu$ is a Normal $\mathcal N(x,\sigma^2)$ distribution.