1 The random variable \(X\) has the Poisson distribution with parameter \(\theta\) so that its probability function is $$\mathrm { P } ( X = x ) = \frac { \mathrm { e } ^ { - \theta } \theta ^ { x } } { x ! } , \quad x = 0,1,2 , \ldots$$ where \(\theta ( \theta > 0 )\) is unknown. A random sample of \(n\) observations from \(X\) is denoted by \(X _ { 1 } , X _ { 2 } , \ldots , X _ { n }\).

1. Find \(\hat { \theta }\), the maximum likelihood estimator of \(\theta\). The value of \(\mathrm { P } ( X = 0 )\) is denoted by \(\lambda\).
2. Write down an expression for \(\lambda\) in terms of \(\theta\).
3. Let \(R\) denote the number of observations in the sample with value zero. By considering the binomial distribution with parameters \(n\) and \(\mathrm { e } ^ { - \theta }\), write down \(\mathrm { E } ( R )\) and \(\operatorname { Var } ( R )\). Deduce that the observed proportion of observations in the sample with value zero, denoted by \(\tilde { \lambda }\), is an unbiased estimator of \(\lambda\) with variance \(\frac { \mathrm { e } ^ { - \theta } \left( 1 - \mathrm { e } ^ { - \theta } \right) } { n }\).
4. In large samples, the variance of the maximum likelihood estimator of \(\lambda\) may be taken as \(\frac { \theta \mathrm { e } ^ { - 2 \theta } } { n }\). Use this and the appropriate result from part (iii) to show that the relative efficiency of \(\tilde { \lambda }\) with respect to the maximum likelihood estimator is \(\frac { \theta } { \mathrm { e } ^ { \theta } - 1 }\). Show that this expression is always less than 1 . Show also that it is near 1 if \(\theta\) is small and near 0 if \(\theta\) is large.