Question 2 - A-Level Maths

OCR MEI S4 2010 June — Question 2 24 marks

Exam Board	OCR MEI
Module	S4 (Statistics 4)
Year	2010
Session	June
Marks	24
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Probability Generating Functions
Type	Derive standard distribution PGF
Difficulty	Standard +0.8 This is a substantial Further Maths S4 question requiring derivation of PGF from first principles, conversion to MGF, application of transformation results, and a limit proof involving Taylor expansion to show convergence to normal distribution. While the individual techniques are standard for this module, the multi-part structure, the algebraic manipulation in part (v), and the conceptual understanding required in part (vi) make this moderately challenging even for Further Maths students.
Spec	5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02m Poisson: mean = variance = lambda 5.05a Sample mean distribution: central limit theorem

2 The random variable $X$ has the Poisson distribution with parameter $\lambda$.

Show that the probability generating function of $X$ is $\mathrm { G } ( t ) = \mathrm { e } ^ { \lambda ( t - 1 ) }$.
Hence obtain the mean $\mu$ and variance $\sigma ^ { 2 }$ of $X$.
Write down the mean and variance of the random variable $Z = \frac { X - \mu } { \sigma }$.
Write down the moment generating function of $X$. State the linear transformation result for moment generating functions and use it to show that the moment generating function of $Z$ is $$\mathrm { M } _ { Z } ( \theta ) = \mathrm { e } ^ { \mathrm { f } ( \theta ) } \quad \text { where } \mathrm { f } ( \theta ) = \lambda \left( \mathrm { e } ^ { \theta / \sqrt { \lambda } } - \frac { \theta } { \sqrt { \lambda } } - 1 \right)$$
Show that the limit of $\mathrm { M } _ { Z } ( \theta )$ as $\lambda \rightarrow \infty$ is $\mathrm { e } ^ { \theta ^ { 2 } / 2 }$.
Explain briefly why this implies that the distribution of $Z$ tends to $\mathrm { N } ( 0,1 )$ as $\lambda \rightarrow \infty$. What does this imply about the distribution of $X$ as $\lambda \rightarrow \infty$ ?

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Question 2:

Part (i)

Answer	Marks	Guidance
$G(t) = E(t^X) = \sum_{x=0}^{\infty} \frac{e^{-\lambda}(\lambda t)^x}{x!}$	M1
$= e^{-\lambda}\left(1 + \lambda t + \frac{\lambda^2 t^2}{2!} + ...\right)$	A1
$= e^{-\lambda}e^{\lambda t} = e^{\lambda(t-1)}$	A1	Allow omission of previous A1 step and write-down of this for A2 provided opening M1 earned

Part (ii)

Answer	Marks
Mean $= G'(1)$; $G'(t) = \lambda e^{\lambda(t-1)}$	M1
$G'(1) = \lambda$	A1
Variance $= G''(1) +$ mean $-$ mean$^2$; $G''(t) = \lambda^2 e^{\lambda(t-1)}$	M1
$G''(1) = \lambda^2$	A1
$\therefore$ variance $= \lambda^2 + \lambda - \lambda^2 = \lambda$	A1

Part (iii)

Answer	Marks
$Z = \frac{X - \mu}{\sigma}$: mean $0$	B1
variance $1$	B1

Part (iv)

Answer	Marks	Guidance
Mgf of $X$ is $M(\theta) = G(e^\theta) = e^{\lambda(e^\theta - 1)}$	B1
Linear transformation result: $M_{aX+b}(\theta) = e^{b\theta}M_X(a\theta)$	B2	B1 if either factor correct
Use with $a = \frac{1}{\sigma} = \frac{1}{\sqrt{\lambda}}$ and $b = -\frac{\mu}{\sigma} = -\sqrt{\lambda}$	M1
$M_Z(\theta) = e^{-\sqrt{\lambda}\theta}e^{\lambda(e^{\theta/\sqrt{\lambda}}-1)} = e^{\lambda\left(e^{\theta/\sqrt{\lambda}} - \frac{\theta}{\sqrt{\lambda}} - 1\right)}$	A1, A1, A1	NB answer is given

Part (v)

Answer	Marks	Guidance
Consider $\lambda\left(e^{\theta/\sqrt{\lambda}} - \frac{\theta}{\sqrt{\lambda}} - 1\right) = \lambda\left(1 + \frac{\theta}{\sqrt{\lambda}} + \frac{\theta^2}{2!\lambda} + \frac{\theta^3}{3!\lambda^{3/2}} + ... - \frac{\theta}{\sqrt{\lambda}} - 1\right)$	M1
$= \frac{\theta^2}{2} +$ terms in $\lambda^{-1/2}, \lambda^{-1}, \lambda^{-3/2}, ...$	A1
$\rightarrow \frac{\theta^2}{2}$ as $\lambda \rightarrow \infty$	M1	Some explanation required
$\therefore M_Z(\theta) \rightarrow e^{\theta^2/2}$ as $\lambda \rightarrow \infty$	A1	Answer given

Part (vi)

Answer	Marks	Guidance
$e^{\theta^2/2}$ is the mgf of $N(0,1)$	E1
The relationship between distributions and their mgfs is unique	E1
"Unstandardising", $X$ tends to $N(\mu, \sigma^2)$ i.e. $N(\lambda, \lambda)$	B1	Parameters must be given

# Question 2:

## Part (i)
| $G(t) = E(t^X) = \sum_{x=0}^{\infty} \frac{e^{-\lambda}(\lambda t)^x}{x!}$ | M1 | |
|---|---|---|
| $= e^{-\lambda}\left(1 + \lambda t + \frac{\lambda^2 t^2}{2!} + ...\right)$ | A1 | |
| $= e^{-\lambda}e^{\lambda t} = e^{\lambda(t-1)}$ | A1 | Allow omission of previous A1 step and write-down of this for A2 provided opening M1 earned |

## Part (ii)
| Mean $= G'(1)$; $G'(t) = \lambda e^{\lambda(t-1)}$ | M1 | |
|---|---|---|
| $G'(1) = \lambda$ | A1 | |
| Variance $= G''(1) +$ mean $-$ mean$^2$; $G''(t) = \lambda^2 e^{\lambda(t-1)}$ | M1 | |
| $G''(1) = \lambda^2$ | A1 | |
| $\therefore$ variance $= \lambda^2 + \lambda - \lambda^2 = \lambda$ | A1 | |

## Part (iii)
| $Z = \frac{X - \mu}{\sigma}$: mean $0$ | B1 | |
|---|---|---|
| variance $1$ | B1 | |

## Part (iv)
| Mgf of $X$ is $M(\theta) = G(e^\theta) = e^{\lambda(e^\theta - 1)}$ | B1 | |
|---|---|---|
| Linear transformation result: $M_{aX+b}(\theta) = e^{b\theta}M_X(a\theta)$ | B2 | B1 if either factor correct |
| Use with $a = \frac{1}{\sigma} = \frac{1}{\sqrt{\lambda}}$ and $b = -\frac{\mu}{\sigma} = -\sqrt{\lambda}$ | M1 | |
| $M_Z(\theta) = e^{-\sqrt{\lambda}\theta}e^{\lambda(e^{\theta/\sqrt{\lambda}}-1)} = e^{\lambda\left(e^{\theta/\sqrt{\lambda}} - \frac{\theta}{\sqrt{\lambda}} - 1\right)}$ | A1, A1, A1 | NB answer is given |

## Part (v)
| Consider $\lambda\left(e^{\theta/\sqrt{\lambda}} - \frac{\theta}{\sqrt{\lambda}} - 1\right) = \lambda\left(1 + \frac{\theta}{\sqrt{\lambda}} + \frac{\theta^2}{2!\lambda} + \frac{\theta^3}{3!\lambda^{3/2}} + ... - \frac{\theta}{\sqrt{\lambda}} - 1\right)$ | M1 | |
|---|---|---|
| $= \frac{\theta^2}{2} +$ terms in $\lambda^{-1/2}, \lambda^{-1}, \lambda^{-3/2}, ...$ | A1 | |
| $\rightarrow \frac{\theta^2}{2}$ as $\lambda \rightarrow \infty$ | M1 | Some explanation required |
| $\therefore M_Z(\theta) \rightarrow e^{\theta^2/2}$ as $\lambda \rightarrow \infty$ | A1 | Answer given |

## Part (vi)
| $e^{\theta^2/2}$ is the mgf of $N(0,1)$ | E1 | |
|---|---|---|
| The relationship between distributions and their mgfs is unique | E1 | |
| "Unstandardising", $X$ tends to $N(\mu, \sigma^2)$ i.e. $N(\lambda, \lambda)$ | B1 | Parameters must be given |

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Show LaTeX source

2 The random variable $X$ has the Poisson distribution with parameter $\lambda$.\\
(i) Show that the probability generating function of $X$ is $\mathrm { G } ( t ) = \mathrm { e } ^ { \lambda ( t - 1 ) }$.\\
(ii) Hence obtain the mean $\mu$ and variance $\sigma ^ { 2 }$ of $X$.\\
(iii) Write down the mean and variance of the random variable $Z = \frac { X - \mu } { \sigma }$.\\
(iv) Write down the moment generating function of $X$. State the linear transformation result for moment generating functions and use it to show that the moment generating function of $Z$ is

$$\mathrm { M } _ { Z } ( \theta ) = \mathrm { e } ^ { \mathrm { f } ( \theta ) } \quad \text { where } \mathrm { f } ( \theta ) = \lambda \left( \mathrm { e } ^ { \theta / \sqrt { \lambda } } - \frac { \theta } { \sqrt { \lambda } } - 1 \right)$$

(v) Show that the limit of $\mathrm { M } _ { Z } ( \theta )$ as $\lambda \rightarrow \infty$ is $\mathrm { e } ^ { \theta ^ { 2 } / 2 }$.\\
(vi) Explain briefly why this implies that the distribution of $Z$ tends to $\mathrm { N } ( 0,1 )$ as $\lambda \rightarrow \infty$. What does this imply about the distribution of $X$ as $\lambda \rightarrow \infty$ ?

\hfill \mbox{\textit{OCR MEI S4 2010 Q2 [24]}}

This paper (4 questions)

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Q1 24 Q2 24 Q3 24 Q4 24

Answer	Marks	Guidance
\(G(t) = E(t^X) = \sum_{x=0}^{\infty} \frac{e^{-\lambda}(\lambda t)^x}{x!}\)	M1
\(= e^{-\lambda}\left(1 + \lambda t + \frac{\lambda^2 t^2}{2!} + ...\right)\)	A1
\(= e^{-\lambda}e^{\lambda t} = e^{\lambda(t-1)}\)	A1	Allow omission of previous A1 step and write-down of this for A2 provided opening M1 earned

Answer	Marks
Mean \(= G'(1)\); \(G'(t) = \lambda e^{\lambda(t-1)}\)	M1
\(G'(1) = \lambda\)	A1
Variance \(= G''(1) +\) mean \(-\) mean\(^2\); \(G''(t) = \lambda^2 e^{\lambda(t-1)}\)	M1
\(G''(1) = \lambda^2\)	A1
\(\therefore\) variance \(= \lambda^2 + \lambda - \lambda^2 = \lambda\)	A1

Answer	Marks
\(Z = \frac{X - \mu}{\sigma}\): mean \(0\)	B1
variance \(1\)	B1

Answer	Marks	Guidance
Mgf of \(X\) is \(M(\theta) = G(e^\theta) = e^{\lambda(e^\theta - 1)}\)	B1
Linear transformation result: \(M_{aX+b}(\theta) = e^{b\theta}M_X(a\theta)\)	B2	B1 if either factor correct
Use with \(a = \frac{1}{\sigma} = \frac{1}{\sqrt{\lambda}}\) and \(b = -\frac{\mu}{\sigma} = -\sqrt{\lambda}\)	M1
\(M_Z(\theta) = e^{-\sqrt{\lambda}\theta}e^{\lambda(e^{\theta/\sqrt{\lambda}}-1)} = e^{\lambda\left(e^{\theta/\sqrt{\lambda}} - \frac{\theta}{\sqrt{\lambda}} - 1\right)}\)	A1, A1, A1	NB answer is given

Answer	Marks	Guidance
Consider \(\lambda\left(e^{\theta/\sqrt{\lambda}} - \frac{\theta}{\sqrt{\lambda}} - 1\right) = \lambda\left(1 + \frac{\theta}{\sqrt{\lambda}} + \frac{\theta^2}{2!\lambda} + \frac{\theta^3}{3!\lambda^{3/2}} + ... - \frac{\theta}{\sqrt{\lambda}} - 1\right)\)	M1
\(= \frac{\theta^2}{2} +\) terms in \(\lambda^{-1/2}, \lambda^{-1}, \lambda^{-3/2}, ...\)	A1
\(\rightarrow \frac{\theta^2}{2}\) as \(\lambda \rightarrow \infty\)	M1	Some explanation required
\(\therefore M_Z(\theta) \rightarrow e^{\theta^2/2}\) as \(\lambda \rightarrow \infty\)	A1	Answer given

Answer	Marks	Guidance
\(e^{\theta^2/2}\) is the mgf of \(N(0,1)\)	E1
The relationship between distributions and their mgfs is unique	E1
"Unstandardising", \(X\) tends to \(N(\mu, \sigma^2)\) i.e. \(N(\lambda, \lambda)\)	B1	Parameters must be given