| Exam Board | OCR MEI |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2007 |
| Session | June |
| Marks | 24 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Probability Generating Functions |
| Type | Derive standard distribution PGF |
| Difficulty | Challenging +1.2 This is a structured, multi-part question on PGFs and MGFs that guides students through a proof of the normal approximation to the binomial. Parts (i)-(iv) are standard bookwork requiring routine application of definitions and formulas. Part (v) requires careful algebraic manipulation with Taylor series but provides the key limit result. While lengthy (7 parts), each step is scaffolded and uses techniques expected in S4, making it moderately above average difficulty but not requiring novel insight. |
| Spec | 5.02a Discrete probability distributions: general5.02b Expectation and variance: discrete random variables5.05a Sample mean distribution: central limit theorem |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(G(t) = E[t^X] = \sum_{x=0}^{n}\binom{n}{x}(pt)^x(1-p)^{n-x}\) | M1 | |
| \(= [(1-p) + pt]^n\) | 2 | Available as B2 for write-down or as 1+1 for algebra |
| \(= (q + pt)^n\) | 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\mu = G'(1)\), \(G'(t) = np(q+pt)^{n-1}\) | 1 | |
| \(G'(1) = np \times 1 = np\) | 1 | |
| \(\sigma^2 = G''(1) + \mu - \mu^2\) | 1 | |
| \(G''(t) = n(n-1)p^2(q+pt)^{n-2}\) | ||
| \(G''(1) = n(n-1)p^2\) | 1 | |
| \(\therefore \sigma^2 = n^2p^2 - np^2 + np - n^2p^2\) | M1 | |
| \(= -np^2 + np = npq\) | 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(Z = \dfrac{X-\mu}{\sigma}\), Mean 0, Variance 1 | B1 | For BOTH |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(M(\theta) = G(e^\theta) = (q + pe^\theta)^n\) | 1 | |
| \(Z = aX + b\) with: \(a = \dfrac{1}{\sigma} = \dfrac{1}{\sqrt{npq}}\) and \(b = -\dfrac{\mu}{\sigma} = -\sqrt{\dfrac{np}{q}}\) | ||
| \(M_Z(\theta) = e^{b\theta}M_X(a\theta)\) | M1 | |
| \(\therefore M_Z(\theta) = e^{-\sqrt{\frac{np}{q}}\theta}\left(q + pe^{\frac{1}{\sqrt{npq}}\theta}\right)^n\) | 1 | |
| \(= \left(qe^{-\frac{p\theta}{\sqrt{npq}}} + pe^{\frac{1-p}{\sqrt{npq}}\theta}\right)^n\) | 1 | BEWARE PRINTED ANSWER |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(M_Z(\theta) = \left(q - \dfrac{qp\theta}{\sqrt{npq}} + \dfrac{qp^2\theta^2}{2npq} + \cdots\right)^n\) | M1 | For expansion of exponential terms |
| terms in \(n^{-3/2}\), \(n^{-2}\), … | M1 | For indication these can be neglected as \(n \to \infty\). Use of result given in question |
| \(= \left(p + \dfrac{pq\theta}{\sqrt{npq}} + \dfrac{pq^2\theta^2}{2npq} + \cdots\right)^n\) | ||
| \(\left(1 + \dfrac{\theta^2}{2n} + \cdots\right)^n \to\) | 1 | |
| \(e^{\theta^2/2}\) | 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(N(0,1)\) | 1 | |
| Because \(e^{\theta^2/2}\) is the mgf of \(N(0,1)\) | E1 | |
| and the relationship between distributions and their mgfs is unique | E1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| "Unstandardising", \(N(\mu, \sigma^2)\) ie \(N(np, npq)\) | 1 | Parameters need to be given |
# Question 2:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $G(t) = E[t^X] = \sum_{x=0}^{n}\binom{n}{x}(pt)^x(1-p)^{n-x}$ | M1 | |
| $= [(1-p) + pt]^n$ | 2 | Available as B2 for write-down or as 1+1 for algebra |
| $= (q + pt)^n$ | 1 | |
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\mu = G'(1)$, $G'(t) = np(q+pt)^{n-1}$ | 1 | |
| $G'(1) = np \times 1 = np$ | 1 | |
| $\sigma^2 = G''(1) + \mu - \mu^2$ | 1 | |
| $G''(t) = n(n-1)p^2(q+pt)^{n-2}$ | | |
| $G''(1) = n(n-1)p^2$ | 1 | |
| $\therefore \sigma^2 = n^2p^2 - np^2 + np - n^2p^2$ | M1 | |
| $= -np^2 + np = npq$ | 1 | |
## Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $Z = \dfrac{X-\mu}{\sigma}$, Mean 0, Variance 1 | B1 | For BOTH |
## Part (iv)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $M(\theta) = G(e^\theta) = (q + pe^\theta)^n$ | 1 | |
| $Z = aX + b$ with: $a = \dfrac{1}{\sigma} = \dfrac{1}{\sqrt{npq}}$ and $b = -\dfrac{\mu}{\sigma} = -\sqrt{\dfrac{np}{q}}$ | | |
| $M_Z(\theta) = e^{b\theta}M_X(a\theta)$ | M1 | |
| $\therefore M_Z(\theta) = e^{-\sqrt{\frac{np}{q}}\theta}\left(q + pe^{\frac{1}{\sqrt{npq}}\theta}\right)^n$ | 1 | |
| $= \left(qe^{-\frac{p\theta}{\sqrt{npq}}} + pe^{\frac{1-p}{\sqrt{npq}}\theta}\right)^n$ | 1 | BEWARE PRINTED ANSWER |
## Part (v)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $M_Z(\theta) = \left(q - \dfrac{qp\theta}{\sqrt{npq}} + \dfrac{qp^2\theta^2}{2npq} + \cdots\right)^n$ | M1 | For expansion of exponential terms |
| terms in $n^{-3/2}$, $n^{-2}$, … | M1 | For indication these can be neglected as $n \to \infty$. Use of result given in question |
| $= \left(p + \dfrac{pq\theta}{\sqrt{npq}} + \dfrac{pq^2\theta^2}{2npq} + \cdots\right)^n$ | | |
| $\left(1 + \dfrac{\theta^2}{2n} + \cdots\right)^n \to$ | 1 | |
| $e^{\theta^2/2}$ | 1 | |
## Part (vi)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $N(0,1)$ | 1 | |
| Because $e^{\theta^2/2}$ is the mgf of $N(0,1)$ | E1 | |
| and the relationship between distributions and their mgfs is unique | E1 | |
## Part (vii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| "Unstandardising", $N(\mu, \sigma^2)$ ie $N(np, npq)$ | 1 | Parameters need to be given |
---2 The random variable $X$ has the binomial distribution with parameters $n$ and $p$, i.e. $X \sim \mathrm {~B} ( n , p )$.\\
(i) Show that the probability generating function of $X$ is $\mathrm { G } ( t ) = ( q + p t ) ^ { n }$, where $q = 1 - p$.\\
(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 $X$ and use the linear transformation result to show that the moment generating function of $Z$ is
$$\mathrm { M } _ { Z } ( \theta ) = \left( q \mathrm { e } ^ { - \frac { p \theta } { \sqrt { n p q } } } + p \mathrm { e } ^ { \frac { q \theta } { \sqrt { n p q } } } \right) ^ { n } .$$
(v) By expanding the exponential terms in $\mathrm { M } _ { Z } ( \theta )$, show that the limit of $\mathrm { M } _ { Z } ( \theta )$ as $n \rightarrow \infty$ is $\mathrm { e } ^ { \theta ^ { 2 } / 2 }$. You may use the result $\lim _ { n \rightarrow \infty } \left( 1 + \frac { y + \mathrm { f } ( n ) } { n } \right) ^ { n } = \mathrm { e } ^ { y }$ provided $\mathrm { f } ( n ) \rightarrow 0$ as $n \rightarrow \infty$.\\
(vi) What does the result in part (v) imply about the distribution of $Z$ as $n \rightarrow \infty$ ? Explain your reasoning briefly.\\
(vii) What does the result in part (vi) imply about the distribution of $X$ as $n \rightarrow \infty$ ?
\hfill \mbox{\textit{OCR MEI S4 2007 Q2 [24]}}