| Exam Board | OCR MEI |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2008 |
| Session | June |
| Marks | 24 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Probability Generating Functions |
| Type | Find PGF from probability distribution |
| Difficulty | Challenging +1.2 This is a structured multi-part question on maximum likelihood estimation and estimator efficiency for the Poisson distribution. Parts (i)-(iii) involve standard techniques (MLE derivation, recognizing binomial structure, computing expectation/variance). Part (iv) requires algebraic manipulation and asymptotic analysis but follows a clear template. While requiring several statistical concepts, each step is guided and uses well-established methods typical of Further Maths Statistics modules. |
| Spec | 5.02i Poisson distribution: random events model5.02k Calculate Poisson probabilities5.05b Unbiased estimates: of population mean and variance |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Relative efficiency of \(\tilde{\lambda}\) wrt ML est \(= \dfrac{\text{Var(ML Est)}}{\text{Var}(\tilde{\lambda})}\) | M1 | Any attempt to compare variances |
| \(= \dfrac{\theta e^{-2\theta}}{n} \cdot \dfrac{n}{e^{-\theta}(1-e^{-\theta})} = \dfrac{\theta}{e^{\theta}-1}\) | M1, A1 | If correct. BEWARE PRINTED ANSWER |
| Expression is \(\dfrac{\theta}{\theta + \dfrac{\theta^2}{2!} + \ldots}\) | M1 | |
| Always \(< 1\) | E1 | |
| \(\approx 1\) if \(\theta\) is small | E1 | |
| \(\approx 0\) if \(\theta\) is large | E1 | Allow statement that \(\dfrac{\theta}{e^{\theta}-1} \to 0\) as \(\theta \to \infty\) |
# Question 1 (Part iv):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Relative efficiency of $\tilde{\lambda}$ wrt ML est $= \dfrac{\text{Var(ML Est)}}{\text{Var}(\tilde{\lambda})}$ | M1 | Any attempt to compare variances |
| $= \dfrac{\theta e^{-2\theta}}{n} \cdot \dfrac{n}{e^{-\theta}(1-e^{-\theta})} = \dfrac{\theta}{e^{\theta}-1}$ | M1, A1 | If correct. BEWARE PRINTED ANSWER |
| Expression is $\dfrac{\theta}{\theta + \dfrac{\theta^2}{2!} + \ldots}$ | M1 | |
| Always $< 1$ | E1 | |
| $\approx 1$ if $\theta$ is small | E1 | |
| $\approx 0$ if $\theta$ is large | E1 | Allow statement that $\dfrac{\theta}{e^{\theta}-1} \to 0$ as $\theta \to \infty$ |
---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 $X _ { 1 } , X _ { 2 } , \ldots , X _ { n }$.\\
(i) Find $\hat { \theta }$, the maximum likelihood estimator of $\theta$.
The value of $\mathrm { P } ( X = 0 )$ is denoted by $\lambda$.\\
(ii) Write down an expression for $\lambda$ in terms of $\theta$.\\
(iii) 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 $\frac { \mathrm { e } ^ { - \theta } \left( 1 - \mathrm { e } ^ { - \theta } \right) } { n }$.\\
(iv) 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.
\hfill \mbox{\textit{OCR MEI S4 2008 Q1 [24]}}