# Showing that a posterior is Normal given improper prior

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.

• 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 Dec 3, 2020 at 7:37
• Please add the self-study tag. Dec 3, 2020 at 8:30
• The term "likelihood distribution" is terrible and prone to confuse any student exposed to it! Dec 3, 2020 at 8:32

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.