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Thread: Math Stat question

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    Math Stat question




    im stuck with this question:

    Consider the prior distribution for the Bernouli parameter p with p.d.f g(p) proportional to 1/sqrt(p(1-p) for 0<p<1. Suppose you now observe a success. Give the predictive distribution of p for the next trial.

    Hope someone can help. Thanks!

  2. #2
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    Re: Math Stat question


    Let X|P = p \sim \mathrm{Bernoulli}(p), 
g_P(p) = \frac {c} {\sqrt{p(1-p)}}, p \in (0, 1)
    for some constant c > 0
    If you are familiar with the Beta distribution, you can see actually the prior
    distribution of P is \mathrm{Beta}
\left(\frac {1} {2}, \frac {1} {2} \right)

    It is because the p.d.f. of \mathrm{Beta}(\alpha, \beta) is

    \frac {\Gamma(\alpha + \beta)} {\Gamma(\alpha)\Gamma(\beta)}
p^{\alpha - 1} (1 - p)^{\beta - 1}, p \in (0, 1)


    To determine the required predictive (posterior) distribution, we just need to
    compute the conditional pdf of P|X = 1: \forall p \in (0, 1),

    f_{P|X = 1}(p) = \frac {\Pr\{X = 1|P = p\}g_P(p)} {\Pr\{X = 1\}}
= c' \times p \times \frac {1} {\sqrt{p(1-p)}} 
= c' \times p^{\frac {3} {2} - 1}(1 - p)^{\frac {1} {2} - 1}

    where c' = \frac {c} {\Pr\{X = 1\}} is just another constant
    that is independent from p.

    So we can see the posterior distribution is
    \mathrm{Beta}
\left(\frac {3} {2}, \frac {1} {2} \right)

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