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?

  • $\begingroup$ 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, 2018 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, 2018 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, 2018 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, 2018 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, 2018 at 17:02

1 Answer 1


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.


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