Discussion

Suppose

\[ P(Y=y/\theta)=\binom{n}{y}\theta^{y}(1-\theta)^{n-y}, \]

\(y \in \{0, 1, 2, \ldots, n\}\). If \(\theta\) ~ uniform (0, 1), then the posterior distribution will be :

Answer: C

The posterior is Beta(y + 1, n − y + 1).

The uniform prior on (0, 1) has density 1, so the posterior is proportional to the likelihood:

\[ \pi(\theta \mid y) \propto \theta^{y}(1 - \theta)^{n-y} \]

This is the kernel of a Beta distribution with parameters \(y + 1\) and \(n - y + 1\) (the Beta(a, b) kernel is \(\theta^{a-1}(1 - \theta)^{b-1}\)).

Beta(y, n − y) drops the 1 contributed by the uniform prior, which is Beta(1, 1). The uniform distributions ignore the data.

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