(i)     Find an expression for $c$ in terms of $n$ such that $U = c\sum\limits_{i = 1}^n {{X_i}}$ is an unbiased estimator for $a$ .

(ii)     Determine an expression for ${\text{Var}}(U)$ in this case.

(i)     Show that ${\rm{P}}(Y \le y) = {\left( {\frac{y}{a}} \right)^{3n}},0 \le y \le a$ and deduce an expression for the probability density function of $Y$ .

(ii)     Find ${\rm{E}}(Y)$ .

(iii)     Show that ${\rm{Var}}(Y) = \frac{{3n{a^2}}}{{(3n + 2){{(3n + 1)}^2}}}$ .

(iv)     Find an expression for $d$ in terms of $n$ such that $V = dY$ is an unbiased estimator for $a$ .

(v)     Determine an expression for ${\rm{Var}}(V)$ in this case.

Show that $\frac{{{\rm{Var}}(U)}}{{{\rm{Var}}(V)}} = \frac{{3n + 2}}{5}$ and hence state, with a reason, which of $U$ or $V$ is the more efficient estimator for $a$ .

(i)     ${\rm{E}}(U) = c \times n \times \frac{{3a}}{4} = a \Rightarrow c = \frac{4}{{3n}}$     M1A1

(ii)     $Var(U) = \frac{{16}}{{9{n^2}}} \times n \times \frac{{3{a^2}}}{{80}} = \frac{{{a^2}}}{{15n}}$     M1A1

[4 marks]

(i)     ${\rm{P}}(Y \le y) = {\rm{P}}({\text{all }}Xs \le y)$     M1

$= \left[ {\rm{P}} \right.{\left. {(X \le y)} \right]^n}$    (A1)

$= {\left( {{{\left( {\frac{y}{a}} \right)}^3}} \right)^n}$     (A1)

Note: Only one of the two A1 marks above may be implied.

$= {\left( {\frac{y}{a}} \right)^{3n}}$     AG

$g(y) = \frac{{\rm{d}}}{{{\rm{d}}y}}{\left( {\frac{y}{a}} \right)^{3n}} = \frac{{3n{y^{3n - 1}}}}{{{a^{3n}}}},(0 < y < a)$     M1A1

($g(y) = 0$ otherwise)

(ii)     ${\rm{E}}(Y) = \int_0^a {\frac{{3n{y^{3n}}}}{{{a^{3n}}}}} {\rm{d}}y$     M1

$= \left[ {\frac{{3n{y^{3n + 1}}}}{{(3n + 1){a^{3n}}}}} \right]_0^a$     A1

$= \frac{{3na}}{{3n + 1}}$     A1

(iii)     ${\rm{Var}}(Y) = \int_0^a {\frac{{3n{y^{3n + 1}}}}{{{a^{3n}}}}} {\rm{d}}y - {\left( {\frac{{3na}}{{3n + 1}}} \right)^2}$     M1

$= \left[ {\frac{{3n{y^{3n + 2}}}}{{(3n + 2){a^{3n}}}}} \right]_0^a - {\left( {\frac{{3na}}{{3n + 1}}} \right)^2}$     A1

$= \frac{{3n{a^2}}}{{3n + 2}} - \frac{{9{n^2}{a^2}}}{{{{(3n + 1)}^2}}}$     M1

$= \frac{{3n{a^2}(9{n^2} + 6n + 1) - 9{n^2}{a^2}(3n + 2)}}{{(3n + 2)(3n + 1)}}$     A1

$= \frac{{3n{a^2}}}{{(3n + 2){{(3n + 1)}^2}}}$     AG

(iv)     ${\rm{E}}(V) = d \times \frac{{3na}}{{3n + 1}} = a \Rightarrow d = \frac{{3n + 1}}{{3n}}$     M1A1

(v)     $Var(V) = {\left( {\frac{{3n + 1}}{{3n}}} \right)^2} \times \frac{{3n{a^2}}}{{(3n + 2){{(3n + 1)}^2}}}$     M1

$= \frac{{{a^2}}}{{3n(3n + 2)}}$     A1

[16 marks]

$\frac{{{\rm{Var}}(U)}}{{{\rm{Var}}(V)}} = \frac{{\frac{{{a^2}}}{{15n}}}}{{\frac{{{a^2}}}{{3n(3n + 2)}}}}$     A1

$= \frac{{3n + 2}}{5}$     AG

$V$ is the more efficient estimator because $3n + 2 > 5$ (for $n > 1$ )     R1

[2 marks]