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I try to develop Bayesian estimation for one dimensional Gaussian with unknown $\mu$ and known $\sigma$. I got \begin{align} p(x|D) &= \int p(x|\mu)p(\mu|D) d\mu \\ &=\int \frac{1}{\sigma \sqrt{2\pi}}\exp\left(-\frac{1}{2}\left(\frac{x-\mu}{\sigma}\right)^2\right)\frac{1}{\sigma_n\sqrt{2\pi}}\exp\left(-\frac{1}{2}\left(\frac{\mu-\mu_n}{\sigma_n}\right)^2\right)d\mu \\ &= \frac{1}{2\pi\sigma\sigma_n}\exp\left(-\frac{1}{2}\frac{\left(x-\mu_n\right)^2}{\sigma^2+\sigma_n^2}\right)f(\sigma,\sigma_n) \\ &\text{ where } \\ f(\sigma,\sigma_n) &= \int \exp\left(-\frac{1}{2}\frac{\sigma^2+\sigma_n^2}{\sigma^2\sigma_n^2}\left(\mu-\frac{\sigma_n^2x+\sigma^2\mu_n}{\sigma^2+\sigma_n^2}\right)^2\right)d\mu \end{align}

Now, It says that $p(x|D)$ is proportional to $$\exp\left(-\frac{1}{2}\frac{\left(x-\mu_n\right)^2}{\sigma^2+\sigma_n^2}\right)$$ as a function of x and hence $p(x|D)$ is normally distributed with mean $\mu_n$ and variance $\sigma^2+\sigma_n^2$.

I am asking for an explanation and/or proof for why $p(x|D)$ is proportional to the $\exp$ term $\Rightarrow p(x|D)$ distibuted normally.

Edit: this answer I got is general. I am asking for an explanation for this case.

gung - Reinstate Monica
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George
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  • Hint: given that all densities must represent a total probability of $1$, what are the possible constants of proportionality? – whuber Dec 11 '14 at 23:38
  • can you elaborate? I have x in $f(\sigma,\sigma_n)$, so it is not clear because this is not a constant. – George Dec 11 '14 at 23:46
  • Better: I found another thread with an expansive answer. – whuber Dec 11 '14 at 23:53
  • can you explaing in more details? this is exp times something that is not constant because there is x in $f(\sigma,\sigma_n)$. Thus, I don't understand. – George Dec 12 '14 at 06:08
  • Even though $x$ appears in the formula for $f$, $f$ does not depend on $x$. If you solve the integral in $f$ this will be clear. – Tom Minka Dec 14 '14 at 11:46
  • Can you show how to integrate this and get that $f$ does not depend on $x$? thank you. – George Dec 14 '14 at 18:24
  • Also, Can someon remove the duplication mark? this is not the same question and the other question is too general and don't address this case. – George Dec 14 '14 at 18:26

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