# Point null hypothesis in Bayesian statistics

##### New Member
Let $$X\sim N(\theta,1)$$ and 5 independent observations $$X=(4.9,5.6,5.1,4.6,3.6)$$. The prior probability that $$\theta=4.01$$ is
$$0.5$$. The remain values of $$\theta$$ are given the density of $$g(\theta)$$.

a)Assume $$g(\theta)\sim N(4.01,1)$$ test the hypothesis
$$H_0:\theta=4.01\space vs\space H_1:\theta\neq 4.01$$​

From what I learn to make a hypothesis test I need to find
$$a_0=P(\theta\in\Theta_0|x)$$​
such that
$$a_0+a_1=1$$​
and reject $$H_0$$ if $$a_0<a_1$$
In the cases where the null hypothesis is not a point I can make, but in this case I have a few doubts.

From the notes that I take there is the theorem below

Theorem: For any prior $$\pi(\theta)=\pi_0\space \text{if}\space \theta=\theta_0$$ $$\pi(\theta)=\pi_1 h(\theta)\space\text{if}\space \theta\neq \theta_0$$ such that
$$\int_{\theta\neq \theta_0}h(\theta)d(\theta)=1$$​
then
$$a_0=f(\theta|x)\geq [1+\frac{1-\pi_0}{\pi_0}\frac{r(x)}{f(x|\theta_0)}]^{-1}$$​
where
$$r(x)=sup_{\theta\neq\theta_0}f(x|\theta)$$​
usually
$$r(x)=f(x|\hat{\theta})$$​

In this case $$\hat{\theta}=\overline{X}$$ but the distribution of $$f(x|\overline{X})$$ doesn't make sense to me, in one example that I look they take $$f(\overline{x}|\hat{\theta})$$ but I don't understood the logic.

I need to use the distribution of the likelihood estimator supposing that $$\theta=\hat{\theta}$$?

If someone can give me a explanation with details on how it works I really appreciate, I already read in the textbook but I don't understood.

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