A question about posterior distribution $p(\mu|\mathcal{D}) = \mu|x_1, \ldots, x_n$

On these course notes, we are given the distribution of the posterior distribution https://people.eecs.berkeley.edu/~jordan/courses/260-spring10/lectures/lecture5.pdf

This famous result can be found in many other places, such as https://www.cs.ubc.ca/~murphyk/Papers/bayesGauss.pdf. However, I am confused about the coefficient (i.e., the $\dfrac{1}{\sqrt{2\pi}\sigma}$) term associated with the Gaussian for the posterior distribution. Here is what I am having trouble with.

We know that given $\mathcal{D} = (x_1, \ldots, x_n), x_i$ iid,

$$p(\mu|\mathcal{D}) \propto p(D|\mu)p(\mu)$$

Suppose that $$p(\mu) = \dfrac{1}{\sqrt{2\pi}\sigma_o} \exp{\dfrac{(x - \mu_o)^2}{2\sigma_o^2}}.$$ and $$p(D|\mu)= \dfrac{1}{(2\pi\sigma^2)^\frac{N}{2}} \exp(-\dfrac{1}{2\sigma^2}\sum\limits_{n = 1}^N (x_n - \mu))$$

then multiplying the expressions together, we obtain:

$$\dfrac{1}{(2\pi\sigma^2)^\frac{n}{2}} \exp{\left[-\dfrac{1}{2\sigma^2}\sum\limits_{n = 1}^N (x_n - \mu)\right]} \dfrac{1}{\sqrt{2\pi}\sigma_o} \exp{\left[\dfrac{(x - \mu_o)^2}{2\sigma_o^2}\right]}$$

While we can perform a complete the square inside of the exponential, what about the constants $\dfrac{1}{(2\pi\sigma^2)^\frac{n}{2}}$ and $\dfrac{1}{\sqrt{2\pi}\sigma_o}$?

I don't see how $\dfrac{1}{(2\pi\sigma^2)^\frac{n}{2}} *\dfrac{1}{\sqrt{2\pi}\sigma_o} = \dfrac{1}{\sqrt{2\pi}\sigma_n^2}$

where $\sigma_n^2 = (1/\sigma_o^2 + n/\sigma^2)^{-1}$ as shown in Lemma 6.

I tried playing around with the terms but I couldn't make them equal, even when $n = 1$. Have I made a mistake or is this because $p(\mu|\mathcal{D})$ is not a "true" probability distribution (i.e., doesn't integrate to 1)?

If so, how would people deal with this leading coefficient during simulation?

Bishop Pattern Reconigition and Machine Learning (2006) Pg. 98

• It's just a constant of integration. Note that symbol "$\propto$" in your initial expression; it means "proportional to", as in "don't worry about the constant of integration until you're done with everything else, then it's whatever is needed to make the posterior integrate to one". – jbowman Jan 26 '18 at 4:25
• It is $1/\sqrt{2\pi}\sigma_n$. – jbowman Jan 26 '18 at 4:28
• To expand on that... if the constant is not $1/\sqrt{2\pi}\sigma_n$, what it really means is that $p(\mu|D)$ doesn't integrate to 1, which tells you that you have the wrong constant. The kernel / functional form is still that of a Gaussian distribution, though. – jbowman Jan 26 '18 at 4:36
• @StackexchangeHouseNinja Did you read jbowman's comments, especially the first one? What do you think "$\propto$" means in $p(\mu \mid D) \propto p(\mu) p(D \mid \mu)$? – Juho Kokkala Jan 26 '18 at 20:09
• I suspect I see your problem. It appears to me that you think $p(D|\mu)p(\mu)$ is a Normal distribution. It isn't. It's proportional to a Normal distribution. That is why we use the $\propto$ symbol instead of the $=$ symbol in the expression $p(\mu|D) \propto p(D|\mu)p(\mu)$. More specifically, $p(D|\mu)p(\mu) \propto N(\mu_n, \sigma^2_n) = p(\mu|D)$. – jbowman Jan 27 '18 at 2:25

So, just to note, $\Pr(\mu|D)$ is a real probability distribution or at least a distribution function. Read lines 8-12 in the commentary of