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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?

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    $\begingroup$ 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

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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.

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