Determine \({\text{E}}(X)\) and show that \({\text{Var}}(X) = 6\theta - 16{\theta ^2}\).

In order to estimate \(\theta \), a random sample of n observations is obtained from the distribution of X .

(i)     Given that \({\bar X}\) denotes the mean of this sample, show that

\[{{\hat \theta }_1} = \frac{{3 - \bar X}}{4}\]

is an unbiased estimator for \(\theta \) and write down an expression for the variance of \({{\hat \theta }_1}\) in terms of n and \(\theta \).

(ii)     Let Y denote the number of observations that are equal to 1 in the sample. Show that Y has the binomial distribution \({\text{B}}(n,{\text{ }}\theta )\) and deduce that \({{\hat \theta }_2} = \frac{Y}{n}\) is another unbiased estimator for \(\theta \). Obtain an expression for the variance of \({{\hat \theta }_2}\).

(iii)     Show that \({\text{Var}}({{\hat \theta }_1}) < {\text{Var}}({{\hat \theta }_2})\) and state, with a reason, which is the more efficient estimator, \({{\hat \theta }_1}\) or \({{\hat \theta }_2}\).

\({\text{E}}(X) = 1 \times \theta + 2 \times 2\theta + 3(1 - 3\theta ) = 3 - 4\theta \)     M1A1

\({\text{Var}}(X) = 1 \times \theta + 4 \times 2\theta + 9(1 - 3\theta ) - {(3 - 4\theta )^2}\)     M1A1

\( = 6\theta - 16{\theta ^2}\)     AG

[4 marks]

(i)     \({\text{E}}({\hat \theta _1}) = \frac{{3 - {\text{E}}(\bar X)}}{4} = \frac{{3 - (3 - 4\theta )}}{4} = \theta \)     M1A1

so \({\hat \theta _1}\) is an unbiased estimator of \(\theta \)     AG

\({\text{Var}}({{\hat \theta }_1}) = \frac{{6\theta  - 16{\theta ^2}}}{{16n}}\)     A1

(ii)     each of the n observed values has a probability \(\theta \) of having the value 1     R1

so \(Y \sim {\text{B}}(n,{\text{ }}\theta )\)     AG

\({\text{E}}({{\hat \theta }_2}) = \frac{{{\text{E}}(Y)}}{n} = \frac{{n\theta }}{n} = \theta \)     A1

\({\text{Var}}({{\hat \theta }_2}) = \frac{{n\theta (1 - \theta )}}{{{n^2}}} = \frac{{\theta (1 - \theta )}}{n}\)     M1A1

(iii)     \({\text{Var}}({{\hat \theta }_1}) - {\text{Var}}({{\hat \theta }_2}) = \frac{{6\theta  - 16{\theta ^2} - 16\theta  + 16{\theta ^2}}}{{16n}}\)     M1

\( = \frac{{ - 10\theta }}{{16n}} < 0\)     A1

\({{\hat \theta }_1}\) is the more efficient estimator since it has the smaller variance     R1

[10 marks]