Question

Let \(X_{1}, X_{2}, \ldots, X_{n}, n>2\), be a random sample from the binomial distribution \(b(1, \theta)\). (a) Show that \(Y_{1}=X_{1}+X_{2}+\cdots+X_{n}\) is a complete sufficient statistic for \(\theta\). (b) Find the function \(\varphi\left(Y_{1}\right)\) that is the MVUE of \(\theta\). (c) Let \(Y_{2}=\left(X_{1}+X_{2}\right) / 2\) and compute \(E\left(Y_{2}\right)\). (d) Determine \(E\left(Y_{2} \mid Y_{1}=y_{1}\right)\).

Short answer

\((a)\) \(Y_{1}\) is a complete sufficient statistic for \(\theta\). \((b)\) The MVUE of \(\theta\) is \(Y_{1}/n\). \((c)\) \(E\left(Y_{2}\right) = \theta\). \((d)\) \(E\left(Y_{2} \mid Y_{1}=y_{1}\right) = \theta\)

Step-by-step solution

Show that \(Y_{1}\) is a complete sufficient statistic for \(\theta\)

We know that \(Y_{1}=X_{1}+X_{2}+\cdots+X_{n}\) has a Binomial distribution being the sum of \(n\) binomial random variables, which is known to be sufficient for \(\theta\). A statistic is complete if every unbiased estimator depends on it. This means that if \(E(g(Y_{1})) = 0 \) for all \(\theta\), then we must have \(g(Y_{1}) = 0\) almost surely. Since any estimator must depend only on \(Y_{1}\), we can therefore say \(Y_{1}\) is a complete sufficient statistic for \(\theta\).

Find the MVUE of \(\theta\)

The Minimum Variance Unbiased Estimator (MVUE) is the estimator that has the smallest variance among all unbiased estimators of \(\theta\). The estimator \(Y_{1}/n\) is unbiased for \(\theta\) as \(E(Y_{1}/n)=(n\theta)/n=\theta\) and depends upon the complete sufficient statistic \(Y_{1}\). Hence, by the Lehmann-Scheffe Theorem, \(Y_{1}/n\) is the MVUE of \(\theta\). Thus, \(\varphi\left(Y_{1}\right) = Y_{1}/n\)

Compute \(E\left(Y_{2}\right)\)

Recall that \(Y_{2}=\left(X_{1}+X_{2}\right) / 2\). The expectation \(E(Y_{2})\) is then found as \(E(Y_{2})=E[\frac{1}{2}(X_{1}+X_{2})] = \frac{1}{2}(E(X_{1})+E(X_{2})). As X's are Binomial random variables with parameters (1, \(\theta\)), \(E(X_{i})=\theta\). Thus, \(E(Y_{2})=(\theta + \theta)/2 = \theta\)

Determine \(E\left(Y_{2} \mid Y_{1}=y_{1}\right)\)

Note \(Y_{1}\) and \(Y_{2}\) are dependent since \(Y_{1}\) includes \(X_1+X_2\). The conditional expectation \(E\left(Y_{2} \mid Y_{1} = y_{1}\right) \) can be computed using the definition of conditional expectation and the distribution of \(Y_{1}\) and \(Y_{2}\). However, as \(Y_{1}\) is a complete sufficient statistic for \(\theta\), \(\theta\) is independent of \(Y_{1}\). Hence, \(E\left(Y_{2} \mid Y_{1} = y_{1}\right) = E(Y_2) = \theta\)