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Thread: Point null hypothesis in Bayesian statistics

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    Point null hypothesis in Bayesian statistics

    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
    such that
    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
    a_0=f(\theta|x)\geq [1+\frac{1-\pi_0}{\pi_0}\frac{r(x)}{f(x|\theta_0)}]^{-1}

    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.
    Last edited by askazy; 04-28-2016 at 05:46 PM.
    The difference between stupidity and genius is that genius has its limits.
    "Albert Einstein"

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