(i)     Find \({\text{E}}({X_1})\).

(ii)     Hence obtain an unbiased estimator for \(p\).

The die is rolled a second time, and the score \({X_2}\) is noted.

(i)     Show that \(k({X_1} - 3) + \left( {\frac{1}{3} - k} \right)({X_2} - 3)\) is also an unbiased estimator for \(p\) for all values of \(k \in \mathbb{R}\).

(ii)     Find the value for \(k\), which maximizes the efficiency of this estimator.

let \(X\) denote the score on the die

(i)     \({\text{P}}(X = x) = \left\{ {\begin{array}{*{20}{c}} {\frac{{1 - p}}{5},}&{x = 1,{\text{ 2}},{\text{ 3}},{\text{ 4}},{\text{ 5}}} \\ {p,}&{x = 6} \end{array}} \right.\)     (M1)

\(E({X_1}) = (1 + 2 + 3 + 4 + 5)\frac{{1 - p}}{5} + 6p\)     M1

\( = 3 + 3p\)     A1

(ii)     so an unbiased estimator for \(p\) would be \(\frac{{{X_1} - 3}}{3}\)     A1

[4 marks]

(i)     \(E\left( {k({X_1} - 3) + \left( {\frac{1}{3} - k} \right)({X_2} - 3)} \right)\)     M1

\( = kE({X_1} - 3) + \left( {\frac{1}{3} - k} \right)E({X_2} - 3)\)     M1

\( = k(3p) + \left( {\frac{1}{3} - k} \right)(3p)\)     A1

any correct expression involving just \(k\) and \(p\)

\( = p\)     AG

hence \(k({X_1} - 3) + \left( {\frac{1}{3} - k} \right)({X_2} - 3)\) is an unbiased estimator of \(p\)

(ii)     \({\text{Var}}\left( {k({X_1} - 3) + \left( {\frac{1}{3} - k} \right)({X_2} - 3)} \right)\)     M1

\( = {k^2}{\text{Var}}({X_1} - 3) + {\left( {\frac{1}{3} - k} \right)^2}{\text{Var}}({X_2} - 3)\)     A1

\( = \left( {{k^2} + {{\left( {\frac{1}{3} - k} \right)}^2}} \right){\sigma ^2}\) (where \({\sigma ^2}\) denotes \({\text{Var}}(X)\))

valid attempt to minimise the variance     M1

\(k = \frac{1}{6}\)      A1

Note:     Accept an argument which states that the most efficient estimator is the one having equal coefficients of \({X_1}\) and \({X_2}\).

[7 marks]

Total [11 marks]