Explain briefly the meaning of

(i)     an estimator of $\mu$;

(ii)     an unbiased estimator of $\mu$.

A random sample ${X_1},{\text{ }}{X_2},{\text{ }}{X_3}$ of three independent observations is taken from the distribution of $X$.

An unbiased estimator of $\mu ,{\text{ }}\mu  \ne 0$, is given by $U = \alpha {X_1} + \beta {X_2} + (\alpha  - \beta ){X_3}$,

where $\alpha ,{\text{ }}\beta  \in \mathbb{R}$.

(i)     Find the value of $\alpha$.

(ii)     Show that ${\text{Var}}(U) = {\sigma ^2}\left( {2{\beta ^2} - \beta  + \frac{1}{2}} \right)$ where ${\sigma ^2} = {\text{Var}}(X)$.

(iii)     Find the value of $\beta$ which gives the most efficient estimator of $\mu$ of this form.

(iv)     Write down an expression for this estimator and determine its variance.

(v)     Write down a more efficient estimator of $\mu$ than the one found in (iv), justifying your answer.

(i)     an estimator $T$ is a formula (or statistic) that can be applied to the values in any sample, taken from $X$     A1

to estimate the value of $\mu$     A1

(ii)     an estimator is unbiased if ${\text{E}}(T) = \mu$     A1

[3 marks]

(i)     using linearity and the definition of an unbiased estimator     M1

$\mu  = \alpha \mu  + \beta \mu  + (\alpha  - \beta )\mu$     A1

obtain $\alpha  = \frac{1}{2}$     A1

(ii)     attempt to compute ${\text{Var}}(U)$ using correct formula     M1

${\text{Var}}(U) = \frac{1}{4}{\sigma ^2} + {\beta ^2}{\sigma ^2} + {\left( {\frac{1}{2} - \beta } \right)^2}{\sigma ^2}$     A1

${\text{Var}}(U) = {\sigma ^2}\left( {2{\beta ^2} - \beta  + \frac{1}{2}} \right)$     AG

(iii)     attempt to minimise quadratic in $\beta$ (or equivalent)     (M1)

$\beta  = \frac{1}{4}$     A1

(iv)     $(U) = \frac{1}{2}{X_1} + \frac{1}{4}{X_2} + \frac{1}{4}{X_3}$     A1

${\text{Var}}(U) = \frac{3}{8}{\sigma ^2}$     A1

(v)     $\frac{1}{3}{X_1} + \frac{1}{3}{X_2} + \frac{1}{3}{X_3}$     A1

${\text{Var}}\left( {\frac{1}{3}{X_1} + \frac{1}{3}{X_2} + \frac{1}{3}{X_3}} \right) = \frac{3}{9}{\sigma ^2}$     A1

$< {\text{Var}}(U)$     R1

Note:     Accept $\sum\limits_{i = 1}^3 {{\lambda _i}{X_i}}$ if $\sum\limits_{i = 1}^3 {{\lambda _i} = 1}$ and $\sum\limits_{i = 1}^3 {\lambda _i^2 < \frac{3}{8}}$ and follow through to the variance if this is the case.

[12 marks]

Total [15 marks]

In general, solutions to (a) were extremely disappointing with the vast majority unable to give correct explanations of estimators and unbiased estimators. Solutions to (b) were reasonably good in general, indicating perhaps that the poor explanations in (a) were due to an inability to explain what they know rather than a lack of understanding.

Solutions to (b) were reasonably good in general, indicating perhaps that the poor explanations in (a) were due to an inability to explain what they know rather than a lack of understanding.