Determine ${\text{E}}(X)$ and show that ${\text{Var}}(X) = 6\theta - 16{\theta ^2}$.

In order to estimate $\theta$, a random sample of n observations is obtained from the distribution of X .

(i)     Given that ${\bar X}$ denotes the mean of this sample, show that

$${{\hat \theta }_1} = \frac{{3 - \bar X}}{4}$$

is an unbiased estimator for $\theta$ and write down an expression for the variance of ${{\hat \theta }_1}$ in terms of n and $\theta$.

(ii)     Let Y denote the number of observations that are equal to 1 in the sample. Show that Y has the binomial distribution ${\text{B}}(n,{\text{ }}\theta )$ and deduce that ${{\hat \theta }_2} = \frac{Y}{n}$ is another unbiased estimator for $\theta$. Obtain an expression for the variance of ${{\hat \theta }_2}$.

(iii)     Show that ${\text{Var}}({{\hat \theta }_1}) < {\text{Var}}({{\hat \theta }_2})$ and state, with a reason, which is the more efficient estimator, ${{\hat \theta }_1}$ or ${{\hat \theta }_2}$.

${\text{E}}(X) = 1 \times \theta + 2 \times 2\theta + 3(1 - 3\theta ) = 3 - 4\theta$     M1A1

${\text{Var}}(X) = 1 \times \theta + 4 \times 2\theta + 9(1 - 3\theta ) - {(3 - 4\theta )^2}$     M1A1

$= 6\theta - 16{\theta ^2}$     AG

[4 marks]

(i)     ${\text{E}}({\hat \theta _1}) = \frac{{3 - {\text{E}}(\bar X)}}{4} = \frac{{3 - (3 - 4\theta )}}{4} = \theta$     M1A1

so ${\hat \theta _1}$ is an unbiased estimator of $\theta$     AG

${\text{Var}}({{\hat \theta }_1}) = \frac{{6\theta  - 16{\theta ^2}}}{{16n}}$     A1

(ii)     each of the n observed values has a probability $\theta$ of having the value 1     R1

so $Y \sim {\text{B}}(n,{\text{ }}\theta )$     AG

${\text{E}}({{\hat \theta }_2}) = \frac{{{\text{E}}(Y)}}{n} = \frac{{n\theta }}{n} = \theta$     A1

${\text{Var}}({{\hat \theta }_2}) = \frac{{n\theta (1 - \theta )}}{{{n^2}}} = \frac{{\theta (1 - \theta )}}{n}$     M1A1

(iii)     ${\text{Var}}({{\hat \theta }_1}) - {\text{Var}}({{\hat \theta }_2}) = \frac{{6\theta  - 16{\theta ^2} - 16\theta  + 16{\theta ^2}}}{{16n}}$     M1

$= \frac{{ - 10\theta }}{{16n}} < 0$     A1

${{\hat \theta }_1}$ is the more efficient estimator since it has the smaller variance     R1

[10 marks]