Show that $U = a{\bar X_1} + (1 - a){\bar X_2},{\text{ }}a \in \mathbb{R}$, is an unbiased estimator of $\mu$.

Show that ${\text{Var}}(U) = {a^2}\frac{{{\sigma ^2}}}{{{n_1}}} + {(1 - a)^2}\frac{{{\sigma ^2}}}{{{n_2}}}$.

Find, in terms of ${n_1}$ and ${n_2}$, an expression for $a$ which gives the most efficient estimator of this form.

Hence find an expression for the most efficient estimator and interpret the result.

${\text{E}}(U) = E(a{\bar X_1} + (1 - a){\bar X_2}) = a{\text{E}}({\bar X_1}) + (1 - a){\text{E}}({\bar X_2})$     (M1)

${\text{E}}({\bar X_1}) = \mu$ and ${\text{E}}({\bar X_2}) = \mu$

${\text{E}}(U) = a\mu  + (1 - a)\mu$ (or equivalent)      A1

$= \mu$     A1

hence $U$ is an unbiased estimator of $\mu$     AG

[3 marks]

${\text{Var}}(U) = {\text{Var}}(a{\bar X_1} + (1 - a){\bar X_2})$

$= {a^2}{\text{Var}}({\bar X_1}) + {(1 - a)^2}{\text{Var}}({\bar X_2})$     M1

stating that ${\text{Var}}({\bar X_1}) = \frac{{{\sigma ^2}}}{{{n_1}}}$ and ${\text{Var}}({\bar X_2}) = \frac{{{\sigma ^2}}}{{{n_2}}}$     A1

$\Rightarrow {\text{Var}}(U) = {a^2}\frac{{{\sigma ^2}}}{{{n_1}}} + {(1 - a)^2}\frac{{{\sigma ^2}}}{{{n_2}}}$     AG

Note:     Line 3 or equivalent must be seen somewhere.

[2 marks]

let ${\text{Var}}(U) = V$

EITHER

$\frac{{{\text{d}}V}}{{{\text{d}}a}} = 2a\frac{{{\sigma ^2}}}{{{n_1}}} - 2(1 - a)\frac{{{\sigma ^2}}}{{{n_2}}}$     M1

attempting to solve $\frac{{{\text{d}}V}}{{{\text{d}}a}} = 0$ for $a$     R1

Note:     Award M1 for obtaining $a$ in terms of ${n_1},{\text{ }}{n_2}$ and $\sigma$.

OR

forming a quadratic in $a$

$V = \left( {\frac{{{\sigma ^2}}}{{{n_1}}} + \frac{{{\sigma ^2}}}{{{n_2}}}} \right){a^2} - 2\frac{{{\sigma ^2}}}{{{n_2}}}a + \frac{{{\sigma ^2}}}{{{n_2}}}$     M1

attempting to find the axis of symmetry of V     R1

THEN

$a = \frac{{\frac{{2{\sigma ^2}}}{{{n_2}}}}}{{2{\sigma ^2}\left( {\frac{1}{{{n_1}}} + \frac{1}{{{n_2}}}} \right)}}$     (A1)

$a = \frac{{{n_1}}}{{{n_1} + {n_2}}}$     A1 

[4 marks]

substituting $a$ into $U$     (M1)

$U = \frac{{{n_1}{{\bar X}_1} + {n_2}{{\bar X}_2}}}{{{n_1} + {n_2}}}$     A1

Note:     Do not FT an incorrect $a$ for A1, the M1 may however be awarded.

this is an expression for the mean of the combined samples

OR this is a weighted mean of the two sample means     R1

[3 marks]