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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.

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    $\begingroup$ 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

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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.

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