Point null hypothesis in Bayesian statistics

askazy

New Member
#1
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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