(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]