OCR MEI S4 2013 June — Question 1 24 marks

Exam BoardOCR MEI
ModuleS4 (Statistics 4)
Year2013
SessionJune
Marks24
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicProbability Generating Functions
TypeMoment generating function problems
DifficultyChallenging +1.2 This is a multi-part question on maximum likelihood estimation with Poisson distributions. Parts (i)-(iii) are standard textbook material (basic Poisson MLE and interpretation). Part (iv) requires deriving a truncated Poisson distribution and its MLE, which involves calculus and algebraic manipulation but follows a predictable pattern for S4 students. The question is longer than average and requires careful work, but each step is methodical without requiring novel insight—slightly above average difficulty for Further Maths Statistics.
Spec5.02i Poisson distribution: random events model5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02k Calculate Poisson probabilities

1 Traffic engineers are studying the flow of vehicles along a road. At an initial stage of the investigation, they assume that the average flow remains the same throughout the working day. An automatic counter records the number of vehicles passing a certain point per minute during the working day. A random sample of these records is selected; the sample values are denoted by \(x _ { 1 } , x _ { 2 } , \ldots , x _ { n }\).
  1. The engineers model the underlying random variable \(X\) by a Poisson distribution with unknown parameter \(\theta\). Obtain the likelihood of \(x _ { 1 } , x _ { 2 } , \ldots , x _ { n }\) and hence find the maximum likelihood estimate of \(\theta\).
  2. Write down the maximum likelihood estimate of the probability that no vehicles pass during a minute.
  3. The engineers note that, in a sample of size 1000 with sample mean \(\bar { x } = 5\), there are no observations of zero. Suggest why this might cast some doubt on the investigation.
  4. On checking the automatic counter, the engineers find that, due to a fault, no record at all is made if no vehicle passes in a minute. They therefore model \(X\) as a Poisson random variable, again with an unknown parameter \(\theta\), except that the value \(x = 0\) cannot occur. Show that, under this model, $$\mathrm { P } ( X = x ) = \frac { \theta ^ { x } } { \left( \mathrm { e } ^ { \theta } - 1 \right) x ! } , \quad x = 1,2 , \ldots$$ and hence show that the maximum likelihood estimate of \(\theta\) satisfies the equation $$\frac { \theta \mathrm { e } ^ { \theta } } { \mathrm { e } ^ { \theta } - 1 } = \bar { x }$$

1 Traffic engineers are studying the flow of vehicles along a road. At an initial stage of the investigation, they assume that the average flow remains the same throughout the working day. An automatic counter records the number of vehicles passing a certain point per minute during the working day. A random sample of these records is selected; the sample values are denoted by $x _ { 1 } , x _ { 2 } , \ldots , x _ { n }$.\\
(i) The engineers model the underlying random variable $X$ by a Poisson distribution with unknown parameter $\theta$. Obtain the likelihood of $x _ { 1 } , x _ { 2 } , \ldots , x _ { n }$ and hence find the maximum likelihood estimate of $\theta$.\\
(ii) Write down the maximum likelihood estimate of the probability that no vehicles pass during a minute.\\
(iii) The engineers note that, in a sample of size 1000 with sample mean $\bar { x } = 5$, there are no observations of zero. Suggest why this might cast some doubt on the investigation.\\
(iv) On checking the automatic counter, the engineers find that, due to a fault, no record at all is made if no vehicle passes in a minute. They therefore model $X$ as a Poisson random variable, again with an unknown parameter $\theta$, except that the value $x = 0$ cannot occur. Show that, under this model,

$$\mathrm { P } ( X = x ) = \frac { \theta ^ { x } } { \left( \mathrm { e } ^ { \theta } - 1 \right) x ! } , \quad x = 1,2 , \ldots$$

and hence show that the maximum likelihood estimate of $\theta$ satisfies the equation

$$\frac { \theta \mathrm { e } ^ { \theta } } { \mathrm { e } ^ { \theta } - 1 } = \bar { x }$$

\hfill \mbox{\textit{OCR MEI S4 2013 Q1 [24]}}