Question

Let \(X_{1}, X_{2}, \ldots, X_{n}\) denote a random sample from a Poisson distribution with mean \(\theta, 0<\theta<\infty\). Let \(Y=\sum_{1}^{n} X_{i} .\) Use the loss function \(\mathcal{L}[\theta, \delta(y)]=\) \([\theta-\delta(y)]^{2}\). Let \(\theta\) be an observed value of the random variable \(\Theta\). If \(\Theta\) has the prior \(\operatorname{pdf} h(\theta)=\theta^{\alpha-1} e^{-\theta / \beta} / \Gamma(\alpha) \beta^{\alpha}\), for \(0<\theta<\infty\), zero elsewhere, where \(\alpha>0, \beta>0\) are known numbers, find the Bayes solution \(\delta(y)\) for a point estimate for \(\theta\).

Short answer

The Bayes solution for a point estimate for \(\theta\) is \(\delta(y) = \frac{(\sum xi +\alpha)}{(1/\beta +n)}\).

Step-by-step solution

Identify Prior and likelihood

We observe that the prior \(h(\theta)\) follows the gamma distribution of parameters \(\alpha, \beta\).The likelihood is given by \[L(\theta:y)=\frac{e^{-n\theta}\theta^{\sum xi}}{(\sum xi)!}\] which is the Poisson distribution.

Compute the Posterior distribution

Multiplying prior and likelihood, we drop some multiplicative constants not involving \(\theta\) and compute the posterior distribution:\[\pi(\theta:y) \propto L(\theta:y)h(\theta)= \theta^{(\sum xi)} e^{-n\theta}\theta^{(\alpha-1)}e^{-\theta/\beta}=\theta^{(\sum xi + \alpha -1)}e^{-\theta(1/\beta + n )}\] This is a gamma distribution with parameters \((\sum xi + \alpha)\) and \((1/\beta + n)\)

Calculate Bayes Estimator

The Bayes estimator minimizes the posterior expected loss. Here, we have a square error loss, so the Bayes estimator of \(\theta\) is the posterior expected value of \(\theta\), which in the case of a gamma distribution is:\[\delta(y) = E[\theta|Y=y]=\frac{(\sum xi +\alpha)}{(1/\beta +n)}\]