Given data, $p(y_1,...,y_n|\theta)$ normally distributed with mean, $\mu=1$ and unknown variance, $\sigma^2$ and the prior$$\theta \sim Beta \ (1,1) \implies p(\theta)=1$$ Find the posterior distribution, $p(\theta|y_1,...,y_n)$.

I tried and got the posterior $$p(\theta|y_1,...,y_n)\sim N(1,\frac{\sigma^2}{n})$$

Don't know if it's correct.

  • $\begingroup$ What is the relation between $\theta$ and $\sigma^2$? Do you intend to say $\sigma^2=\theta$? It seems unlikely that the posterior distribution might be independent of the values of the observations but dependent on how many there are $\endgroup$ – Henry Dec 8 '16 at 21:30
  • $\begingroup$ @Henry. My bad. I meant to put $\sigma^2$. I edited it just now. $\endgroup$ – sucksatmath Dec 8 '16 at 21:41

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