Deducing the posterior distribution

Recall that we found out that the posterior distribution in the basketball example is given by:

\[p(\pi|X)\propto \pi^{a_1-1}(1-\pi)^{b_1-1}\]

Let’s assume that this proportionality constant is \(K_3\). Therefore, we can replace the “\(\propto\)” sign by an equal sign:

\[p(\pi|X)=K_3 \pi^{a_1-1}(1-\pi)^{b_1-1}\]

We know that the posterior has to integrate to one \[\int p(\pi|X)d\pi=1\]

Therefore: \[\int \pi^{a_1-1}(1-\pi)^{\beta_1-1}d\pi=1/K_3\]

We also know that for a generic beta distribution: \[\int \frac{\pi^{c-1}(1-\pi)^{d-1}}{B(c,d)}d\pi=1\] \[\int \pi^{c-1}(1-\pi)^{d-1}d\pi=B(c,d)\]

This implies that

\[\int \pi^{a_1-1}(1-\pi)^{\beta_1-1}d\pi=B(\alpha_1,\beta_1)\]

and therefore,

\[K_3=\frac{1}{B(\alpha_1,\beta_1)}\]

Taken together, this implies that the posterior distribution is given by a beta distribution:

\[p(\pi|X)=\frac{1}{B(\alpha_1,\beta_1)} \pi^{a_1-1} (1-\pi)^{\beta_1-1}\]

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