0
$\begingroup$

I want to compute the posterior distribution for a Jeffreys prior of a normal with unknown mean (and known variance $\sigma^2$)

My thoughts

Given a normal distribution $N(\mu,\sigma^2)$ with unknown $\sigma^2$, we have that the Fisher information is $I(\mu) = 1/\sigma^2$. Therefore, the Jeffreys prior should be $\pi(\mu) \propto 1/\sigma$.

It is clear that this is an improper distribution since $\int_{\mathbb{R}} 1/\sigma d\mu = \infty$ and as usually this is marked adding a positive constant $c$. I write $\pi(\mu) = c/\sigma$ with $c > 0$.

For the posterior distribution I compute $\pi(\mu|x) = \frac{f(x|\mu)\pi(\mu)}{\int_{\mathbb{R}} f(x|\mu)\pi(\mu) d\mu} = \frac{f(x|\mu)}{\int_{\mathbb{R}} f(x|\mu) d\mu}= f(x|\mu)$ since the denominator integrates one as it is the density function of a probability distribution.

However, I notice that this is not a normal distribution since it represents a conjoint distribution of a sample (here x is a vector). I read in a text that the posterior is a normal distribution with mean $\overline{x}$ and variance $\sigma^2/n$. Why is it the case?

Partial solution

I have encountered this situation previously and I think the trick will work here:

$f(x|\mu)\pi(\mu) = B(\sigma,n,c) \frac{1}{\sqrt{2\pi}\sigma^2/n} e^\frac{-(\overline{x}-\mu)}{2 \sigma^2/n}$

if i integrate with respect to $\mu$ then I get that the marginal is equal to the function $B(\sigma,n,c)$ which since $\sigma^2$ is known can be considered as a constant function.

Do you think this is the right answer?

$\endgroup$
5
  • $\begingroup$ What text are you following? Also can you clarify why you are stating the Jeffrey's prior as $\pi(\mu)$ when $\mu$ is known, and the unknown variable is $\sigma^2$. Similarly when you are referring to $\pi(\mu|x)$ should this not be $\pi(\sigma^2|x)$? $\endgroup$ – owen88 Jan 2 '18 at 16:29
  • $\begingroup$ @owen88 the text is in spanish and difficult to find "lecciones de inferencia estadística" josé antonio cristóbal but in order to establish this point just gives a table with the models, priors, posteriors...also, i say unknown mean ($\mu$) and i assume known variance $\sigma^2$ but ill rephrase it so that it is clearer $\endgroup$ – Rodrigo Jan 2 '18 at 16:36
  • 1
    $\begingroup$ Aha ok; I realised I had mis-understood your question a bit. From re-reading I think you are interested in the case of uknown mean, $mu$. But in that case why, are you referring to the Fisher Information for unknown $\sigma^2$? $\endgroup$ – owen88 Jan 2 '18 at 16:37
  • $\begingroup$ @owen88 unless i've made a mistake what i write is the fisher information for known $\sigma^2$. i've rephrased it to make it clearer but you think this is not the fisher information of known $\sigma^2$? $\endgroup$ – Rodrigo Jan 2 '18 at 16:40
  • $\begingroup$ Yes, the important point that you are missing in your original post is that the Jeffrey's prior $\mathcal{J}(\mu) = 1/\sigma^2 \propto 1$, as a function of $\theta$. Hence it can be ignored from the computations. $\endgroup$ – owen88 Jan 2 '18 at 17:02
1
$\begingroup$

According to page 3 here the Jeffrey's prior is $P[\mu] = c\sqrt{\frac{n}{\sigma^2}}$, which is still a constant so it doesn't matter.

Noting that $P[\bar{X}|\mu] = \frac{1}{\sqrt{2\pi\frac{\sigma^2}{n}}}\textrm{Exp}[\frac{(\bar{X}-\mu)^2}{2\frac{\sigma^2}{n}}]$.

Multiplying these two expressions together gives us:

\begin{equation} P[\bar{X},\mu] = \frac{c\sqrt{\frac{n}{\sigma^2}}}{\sqrt{2\pi\frac{\sigma^2}{n}}}\textrm{Exp}[\frac{(\bar{X}-\mu)^2}{2\frac{\sigma^2}{n}}] \end{equation}

Note that all of the stuff outside of the exponential is just a constant, so examining only the exponential kernel, we can see that the expression is a normal kernel for $\mu | \bar{X} \sim \textrm{N}(\bar{X},\frac{\sigma^2}{n})$.

So this is our posterior distribution.

$\endgroup$

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.