I am having difficulty showing the following problem and I suspect it has something to do with my lack of understanding of the question. The question is this:

Suppose we have an improper prior distribution with $f_\Theta(\theta) = 1 $ for all $\theta \in \mathbb{R}$ and the likelihood distribution $X | \Theta = \theta \sim N(\theta, 1)$. Show that the posterior is Normal.

I understand that the posterior distribution is proportional to the product of the prior and likelihood, but since the question did not specify, I am not sure how to interpret the likelihood distribution. In other words, I don't know if I should assume that I have 1 sample coming from that likelihood, or $n$ $iid$ samples. If $n=1$, then the question is easy enough, but otherwise I am having trouble showing that the posterior is normal.

My attempt so far is the following: $$ f_\Theta(\theta|x) \propto f_\Theta(\theta) \cdot L(X=x_1, \ldots, x_n|\Theta = \theta) $$

$$ \propto 1 \cdot \prod_{i=1}^n f(x_i|\Theta = \theta) = \prod_{i=1}^n\frac{1}{\sqrt{2\pi}} \exp \big(-\frac{1}{2}(x_i - \theta)^2\big) $$ $$ \propto \exp \big(-\frac{1}{2} \sum_{i=1}^n \big(x_i-\theta\big)^2\big) $$

And this is where I get stuck trying to show that it is normal. Any suggestion or correction to my understanding of the problem is appreciated.

  • 1
    $\begingroup$ You need to expand the square and the sum and then push some stuff that doesnt depend on $\theta$ into the normalising constant/$\propto$ sign. Then you might have to complete the square on the resulting formula and you will get something that looks like a normal up to proportionality $\endgroup$
    – jcken
    Dec 3, 2020 at 7:37
  • $\begingroup$ Please add the self-study tag. $\endgroup$
    – Xi'an
    Dec 3, 2020 at 8:30
  • $\begingroup$ The term "likelihood distribution" is terrible and prone to confuse any student exposed to it! $\endgroup$
    – Xi'an
    Dec 3, 2020 at 8:32

1 Answer 1


Hint: The number $n$ of observations actually matters very little in the sense that, when observing$$(X_1,\ldots,X_n)\sim\mathcal N(\theta,1)$$it is equivalent to observing a single Normal$$\bar{X}_n\sim\mathcal N(\theta,1/n)$$since the sample average is sufficient. If this is unclear, think factorisation theorem.


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.