# Bayesian Homework: Uniform Prior

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?

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

## 1 Answer

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.