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 \(X _ { 1 } , X _ { 2 } , \ldots , X _ { n }\).
- Find \(\hat { \theta }\), the maximum likelihood estimator of \(\theta\).
The value of \(\mathrm { P } ( X = 0 )\) is denoted by \(\lambda\).
- Write down an expression for \(\lambda\) in terms of \(\theta\).
- 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 \(\frac { \mathrm { e } ^ { - \theta } \left( 1 - \mathrm { e } ^ { - \theta } \right) } { n }\).
- 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.