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 \(\varphi\left(Y_{1}\right)\) which 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

The statistic \(Y_{1}\) is a complete and sufficient statistic for \(\theta\). The MVUE of \(\theta\) is \(\frac{Y_{1}}{n}\). The expected value of \(Y_{2}\) and the conditional expectation of \(Y_{2}\) given \(Y_{1}=y_{1}\) are both \(\theta\).

Step-by-step solution

Showing \(Y_{1}\) is a complete sufficient statistic.

Given a random sample from the binomial distribution, \(Y_{1}=X_{1}+X_{2}+\cdots+X_{n}\) follows a binomial distribution with parameters \(n\) and \(\theta\). Thus, we can write the likelihood function \(f(x|\theta) = \binom{n}{x}\theta^{x}(1-\theta)^{n-x}\) where \(x = Y_{1}\). The statistic \(Y_{1}\) includes all parameters \(\theta\) which makes it a sufficient statistic. A sufficient statistic is complete when its conditional expectation is a function of the statistic itself, or any unbiased estimator based on it is unique. In this case, for any function \(g\), if \(E[g(Y_{1})] = 0\) for all \(\theta\), then \(g(Y_{1}) = 0\), hence \(Y_{1}\) is a complete sufficient statistic.

Detemining the MVUE of \(\theta\)

To find the minimum variance unbiased estimator (MVUE) of \(\theta\), it is always based on a complete sufficient statistic by Lehmann–Scheffé theorem. An unbiased estimator of \(\theta\) based on the statistic \(Y_{1}\) could be \(T = \frac{Y_{1}}{n}\). As per the expectation operation, \(E[T] = \theta\), thus \(T\) is an unbiased estimator of \(\theta\). Since it is based on the complete sufficient statistic, it is the MVUE of \(\theta\). Hence \(\varphi(Y_{1}) = \frac{Y_{1}}{n}\).

Computing \(E[Y_{2}]\)

Given \(Y_{2}=(X_{1}+X_{2}) / 2\), we know that each \(X_{i}\) follows a binomial distribution with parameters \(1\) and \(\theta\). Therefore, \(E[X_{i}] = \theta\), and \(E[Y_{2}] = E[(X_{1}+X_{2})/2] = (E[X_{1}] + E[X_{2}])/2 = (\theta + \theta)/2 = \theta\).

Calculating \(E[Y_{2} | Y_{1}=y_{1}]\)

When we compute the expectation \(E[Y_{2} | Y_{1}=y_{1}]\), we know that \(Y_{1}\) is the sum of all \(X_{i}\)'s and \(Y_{2}\) is the average of \(X_{1}\) and \(X_{2}\) which are a subset of the variables constituting \(Y_{1}\). Given this, the expectation \(E[Y_{2} | Y_{1}=y_{1}]\) doesn't actually depend on the value of \(Y_{1}\) because \(X_{1}\) and \(X_{2}\) are independent of the rest. The expected value of \(Y_{2}\) remains the same as \(\theta\), even conditionally, hence \(E[Y_{2} | Y_{1}=y_{1}] = \theta\).