0
$\begingroup$

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?

$\endgroup$
1

1 Answer 1

1
$\begingroup$

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.

$\endgroup$
2
  • 1
    $\begingroup$ Xi'an, perhaps $p(\mu) $ in the second posterior distribution might be a typo (or am I missing something?). $\endgroup$ Commented Feb 2, 2023 at 11:59
  • 1
    $\begingroup$ Thanks for the quick fix. +1. $\endgroup$ Commented Feb 2, 2023 at 17:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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