Discussion
Let \(X_{1}, X_{2}, \ldots, X_{n}\) be i.i.d. Bernoulli \((\theta)\). Let the loss function for the problem of estimation of \(\theta\) be given by
\[ L(\theta, a) = \frac{(\theta - a)^{2}}{\theta(1 - \theta)}. \]
Let the pdf of the prior for \(\theta\) (\(\pi(\theta)\)) be given by \(\pi(\theta) = 1\), \(0 \le \theta \le 1\).
Let \(\overline{X}\) be the sample mean. Then :
Answer: C
\(\bar{X}\) is both Bayes and minimax.
Under the weighted loss \(L = (\theta - a)^{2}/[\theta(1 - \theta)]\), the Bayes estimator is \(E[\theta w(\theta) \mid x]/E[w(\theta) \mid x]\) with \(w = 1/[\theta(1 - \theta)]\). With the uniform prior the posterior is Beta(s + 1, n - s + 1), where \(s = \sum x_{i}\); multiplying by \(w\) gives a Beta(s, n - s) kernel, whose mean is \(s/n = \bar{X}\). So \(\bar{X}\) is Bayes.
Its risk is constant:
\[ \begin{aligned} \frac{E(\bar{X} - \theta)^{2}}{\theta(1 - \theta)} &= \frac{\theta(1 - \theta)/n}{\theta(1 - \theta)} \\ &= \frac{1}{n} \end{aligned} \]
A Bayes estimator with constant risk is minimax, and a unique Bayes estimator is admissible, so A, B and D are wrong.
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