Question

Let \(Y\) have a binomial distribution in which \(n=20\) and \(p=\theta\). The prior probabilities on \(\theta\) are \(P(\theta=0.3)=2 / 3\) and \(P(\theta=0.5)=1 / 3\). If \(y=9\), what are the posterior probabilities for \(\theta=0.3\) and \(\theta=0.5\) ?

Short answer

The posterior probabilities for \(\theta=0.3\) and \(\theta=0.5\) are calculated by applying Bayes' theorem and normalizing them. Detailed calculations depend on the exact values of binomial coefficients and likelihood expressions.

Step-by-step solution

Calculate the likelihoods

Given that \(Y\sim Bin(n=20,\theta)\), the likelihoods for \(\theta=0.3\) and \(\theta=0.5\) when \(y=9\) are \(L(0.3 | y=9) = \binom{20}{9}*(0.3)^9*(1-0.3)^{20-9} \) and \(L(0.5 | y=9) = \binom{20}{9}*(0.5)^9*(1-0.5)^{20-9} \), respectively.

Calculate the priors

The prior probabilities are given as \(P(\theta=0.3)=2/3\) and \(P(\theta=0.5)=1/3\). These probabilities represent the belief about \(\theta\) before seeing the data.

Apply the Bayes' theorem

The posterior probability is calculated using Bayes' theorem: \(P(\theta | y) = \frac{L(\theta | y)P(\theta)}{P(y)}\). However as \(P(y)\) remains constant for each \(\theta\) (and might be hard to calculate), we can focus on the numerator \(L(\theta | y)P(\theta)\). For \(\theta=0.3\), this yields \(\frac{2}{3}*L(0.3|y=9)\), and for \(\theta=0.5\), it yields \(\frac{1}{3}*L(0.5|y=9)\)

Normalize posterior probabilities

Finally normalize the probabilities by dividing them by the sum of probabilities, to ensure they sum to 1