Given that $U$ is unbiased, show that $2a + 6b = 1$.

Show that ${\text{Var}}(U) = (39{b^2} - 12b + 1){\sigma ^2}$.

Hence find the value of $a$ and the value of $b$ which give the best unbiased estimator of this form, giving your answers as fractions.

Hence find the variance of this best unbiased estimator.

${\text{E}}(U) = a\left( {{\text{E}}({X_1}) + {\text{E}}({X_2})} \right) + b\left( {{\text{E}}({Y_1}) + {\text{E}}({Y_2}) + {\text{E}}({Y_3})} \right)$     (M1)

$= 2a\mu  + 6b\mu$     A1

(for an unbiased estimator,) ${\text{E}}(U) = \mu$     R1

giving $2a + 6b = 1$     AG

Note:     Condone omission of E on LHS.

[3 marks]

${\text{Var}}(U) = {a^2}\left( {{\text{Var}}({X_1}) + {\text{Var}}({X_2})} \right) + {b^2}\left( {{\text{Var}}({Y_1}) + {\text{Var}}({Y_2}) + {\text{Var}}({Y_3})} \right)$     (M1)

$= 4{a^2}{\sigma ^2} + 3{b^2}{\sigma ^2}$     A1

$= 4{\left( {\frac{{1 - 6b}}{2}} \right)^2}{\sigma ^2} + 3{b^2}{\sigma ^2}$     A1

$= (39{b^2} - 12b + 1){\sigma ^2}$     AG

[3 marks]

the best unbiased estimator (of this form) will be found by minimising ${\text{Var}}(U)$     (R1)

For example, $\frac{{\text{d}}}{{{\text{d}}b}}\left( {{\text{Var}}(U)} \right) = (78b - 12){\sigma ^2}$     (A1)

for a minimum, $b = \frac{{12}}{{78}}\,\,\,\left( { = \frac{2}{{13}}} \right)$ so that $a = \frac{3}{{78}}\,\,\,\left( { = \frac{1}{{26}}} \right)$     A1

[3 marks]

${\text{Var}}U = \left( {39{{\left( {\frac{2}{{13}}} \right)}^2} - 12\left( {\frac{2}{{13}}} \right) + 1} \right){\sigma ^2}$

$= \frac{{{\sigma ^2}}}{{13}}\,\,\,(0.0769{\sigma ^2})$     A1

[1 mark]