Discussion
Let \(X_{1}\), \(X_{2}\), ........, \(X_{n}\) be a random sample of size \(n\) from Bernoulli distribution with parameter \(\theta\). Then variance of conditional expectation of \(\left(\dfrac{X_{1} + X_{2}}{2}\right)\) given \(\sum_{i=1}^{n} X_{i}\) is :
Answer: D
The variance is \(\theta(1 - \theta)/n\).
Given \(T = \sum X_{i} = t\), the observations are exchangeable, so \(E(X_{1}|T) = E(X_{2}|T) = T/n\). Hence
\[ \begin{aligned} E\left(\frac{X_{1} + X_{2}}{2}\,\Big|\,T\right) \\ &= \frac{T}{n} = \bar{X} \end{aligned} \]
and \(Var(\bar{X}) = \theta(1 - \theta)/n\). This is Rao-Blackwellisation: conditioning the unbiased estimator \((X_{1} + X_{2})/2\), with variance \(\theta(1 - \theta)/2\), on the sufficient statistic reduces its variance to that of \(\bar{X}\).
\(\theta(1 - \theta)\) is the variance of a single observation, 0 would need a constant, and \(\theta\) is the mean, not a variance.
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