1 The random variable $X$ has the Normal distribution with mean 0 and variance $\theta$, so that its probability density function is

$$\mathrm { f } ( x ) = \frac { 1 } { \sqrt { 2 \pi \theta } } \mathrm { e } ^ { - x ^ { 2 } / 2 \theta } , \quad - \infty < x < \infty$$

where $\theta ( \theta > 0 )$ is unknown. A random sample of $n$ observations from $X$ is denoted by $X _ { 1 } , X _ { 2 } , \ldots , X _ { n }$.\\
(i) Find $\hat { \theta }$, the maximum likelihood estimator of $\theta$.\\
(ii) Show that $\hat { \theta }$ is an unbiased estimator of $\theta$.\\
(iii) In large samples, the variance of $\hat { \theta }$ may be estimated by $\frac { 2 \hat { \theta } ^ { 2 } } { n }$. Use this and the results of parts (i) and (ii) to find an approximate $95 \%$ confidence interval for $\theta$ in the case when $n = 100$ and $\Sigma X _ { i } ^ { 2 } = 1000$.

\hfill \mbox{\textit{OCR MEI S4 2011 Q1 [24]}}