Question

Consider the Bayes model \(X_{i} \mid \theta, i=1,2, \ldots, n \quad \sim\) iid with distribution \(b(1, \theta), 0<\theta<1\) $$ \Theta \sim h(\theta)=1 \text { . } $$ (a) Obtain the posterior pdf. (b) Assume squared-error loss and obtain the Bayes estimate of \(\theta\).

Short answer

The posterior density function is a Beta distribution with parameters \(\sum x_i+1, n-\sum x_i+1\), and the Bayes estimate of \(\theta\) under squared error loss is \(\frac{\sum x_i +1}{n+2}\).

Step-by-step solution

Write out the prior and likelihood

First, write out the prior distribution for \(\Theta\) and the likelihood \(L(\theta | X)\) based on the binomial distribution. In this case, the prior is a constant, \(h(\theta) = 1\), and the likelihood will be given by the product of the binomial probability mass functions for each \(X_i\), which is \((\theta)^x (1-\theta)^{1-x}\).

Obtain the Posterior distribution

Using Bayes’ theorem, combine the prior and likelihood to obtain the posterior distribution of \(\Theta | X\). The posterior density function is proportional to the product of the likelihood and the prior, that is, \(f(\theta | X) \propto L(\theta | X) \cdot h(\theta)\). Simplifying, we get a Beta posterior distribution as Beta (\( \sum x_i+1, n-\sum x_i+1\)).

Obtain the Bayes estimate

The Bayes estimate of \(\theta\) under squared error loss is given by the posterior mean. For a Beta (\(a, b\)) distribution, the mean is given by \(\frac{a}{a + b}\). Substitute the parameters of the posterior distribution into this formula to obtain the Bayes estimate of \(\theta\) which would be \(\frac{\sum x_i +1}{n+2}\).