Suppose posterior density of parameter $\theta$ is

$$\pi(\theta|\mathbf x)=\frac{\Gamma(5)}{\Gamma(3)\Gamma(2)}\theta^{3-1}(1-\theta)^{2-1}.$$

Now I have to find which of the two hypotheses $H_1:\theta\le0.5$ and $H_2:\theta>0.5,$ has greater posterior probability under the uniform prior?

I have the solution of the question too, but I didn't understand. It said:

$P(H_1 \text{is true}|\mathbf x)=\int_{0}^{0.5}\frac{\Gamma(5)}{\Gamma(3)\Gamma(2)}\theta^{3-1}(1-\theta)^{2-1}d\theta$

What is uniform prior? Why didn't they incorporate the information of uniform prior in the above integration?

  • 1
    $\begingroup$ Sounds like a trick question to me: since you are given the posterior density, you don't need to know what prior was used. $\endgroup$
    – whuber
    Jan 18, 2020 at 19:40

1 Answer 1


As also pointed out in the comments, you don't need prior since all you need is the posterior: $$P(\text{$H_1$ is true}|\mathbf{x})=P(\theta\leq0.5|\mathbf{x})=\int_0^{0.5} \pi(\theta|\mathbf{x})d\theta$$

Since this is Beta distribution, $0\leq\theta\leq1$, a uniform prior on $\theta$ would be $\pi(\theta)=1$ and you wouldn't notice it in the integration even if it was used mistakenly.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.