| Exam Board | OCR MEI |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2009 |
| Session | June |
| Marks | 24 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moment generating functions |
| Type | MGF of transformed variable |
| Difficulty | Standard +0.3 This is a structured, multi-part question that guides students through standard MGF derivations. Part (i) is a bookwork result, part (ii) is a straightforward proof using MGF definition, part (iii) applies the previous results mechanically, and part (iv) requires recognizing E[W^k] = E[e^{kX}] = M_X(k) and computing mean/variance. While it requires careful algebra and understanding of MGFs, the question provides a clear path with no novel insights needed, making it slightly easier than average for Further Maths S4. |
| Spec | 5.02a Discrete probability distributions: general5.02b Expectation and variance: discrete random variables |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Mgf of \(Z = E(e^{tZ}) = \int_{-\infty}^{\infty} \frac{1}{\sqrt{2\pi}} e^{tz - \frac{z^2}{2}} dz\) | M1 | |
| Complete the square: \(tz - \frac{z^2}{2} = -\frac{1}{2}(z-t)^2 + \frac{1}{2}t^2\) | M1, A1, A1 | |
| \(= e^{\frac{t^2}{2}} \int_{-\infty}^{\infty} \frac{1}{\sqrt{2\pi}} e^{-\frac{(z-t)^2}{2}} dt = e^{\frac{t^2}{2}}\) | M1, M1, M1 | For taking out factor \(e^{\frac{t^2}{2}}\); For use of pdf of \(N(t,1)\); For \(\int \text{pdf} = 1\) |
| \(\therefore \int = 1\), final answer \(e^{\frac{t^2}{2}}\) | A1 | For final answer \(e^{\frac{t^2}{2}}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(Y\) has mgf \(M_Y(t)\); Mgf of \(aY + b\) is \(E[e^{t(aY+b)}]\) | M1 | |
| \(= e^{bt} E[e^{(at)Y}] = e^{bt} M_Y(at)\) | 1, 1, 1 | For factor \(e^{bt}\); For factor \(E[e^{(at)Y}]\); For final answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(Z = \frac{X - \mu}{\sigma}\), so \(X = \sigma Z + \mu\) | M1, 1 | For factor \(e^{\mu t}\) |
| \(\therefore M_X(t) = e^{\mu t} \cdot e^{\frac{(\sigma t)^2}{2}} = e^{\mu t + \frac{\sigma^2 t^2}{2}}\) | 1, 1 | For factor \(e^{\frac{(\sigma t)^2}{2}}\); For final answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(W = e^X\); \(E(W^k) = E[(e^X)^k] = E(e^{kX}) = M_X(k)\) | M1, A1 | For \(E[(e^X)^k]\); For \(E(e^{kX})\) |
| A1 | For \(M_X(k)\) | |
| \(\therefore E(W) = M_X(1) = e^{\mu + \frac{\sigma^2}{2}}\) | M1 A1 | |
| \(E(W^2) = M_X(2) = e^{2\mu + 2\sigma^2}\) | M1 A1, A1 | |
| \(\therefore \text{Var}(W) = e^{2\mu + 2\sigma^2} - e^{2\mu + \sigma^2} \left[= e^{2\mu + \sigma^2}(e^{\sigma^2} - 1)\right]\) |
# Question 2:
## Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Mgf of $Z = E(e^{tZ}) = \int_{-\infty}^{\infty} \frac{1}{\sqrt{2\pi}} e^{tz - \frac{z^2}{2}} dz$ | M1 | |
| Complete the square: $tz - \frac{z^2}{2} = -\frac{1}{2}(z-t)^2 + \frac{1}{2}t^2$ | M1, A1, A1 | |
| $= e^{\frac{t^2}{2}} \int_{-\infty}^{\infty} \frac{1}{\sqrt{2\pi}} e^{-\frac{(z-t)^2}{2}} dt = e^{\frac{t^2}{2}}$ | M1, M1, M1 | For taking out factor $e^{\frac{t^2}{2}}$; For use of pdf of $N(t,1)$; For $\int \text{pdf} = 1$ |
| $\therefore \int = 1$, final answer $e^{\frac{t^2}{2}}$ | A1 | For final answer $e^{\frac{t^2}{2}}$ | **[8 marks]**
## Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $Y$ has mgf $M_Y(t)$; Mgf of $aY + b$ is $E[e^{t(aY+b)}]$ | M1 | |
| $= e^{bt} E[e^{(at)Y}] = e^{bt} M_Y(at)$ | 1, 1, 1 | For factor $e^{bt}$; For factor $E[e^{(at)Y}]$; For final answer | **[4 marks]**
## Part (iii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $Z = \frac{X - \mu}{\sigma}$, so $X = \sigma Z + \mu$ | M1, 1 | For factor $e^{\mu t}$ |
| $\therefore M_X(t) = e^{\mu t} \cdot e^{\frac{(\sigma t)^2}{2}} = e^{\mu t + \frac{\sigma^2 t^2}{2}}$ | 1, 1 | For factor $e^{\frac{(\sigma t)^2}{2}}$; For final answer | **[4 marks]**
## Part (iv):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $W = e^X$; $E(W^k) = E[(e^X)^k] = E(e^{kX}) = M_X(k)$ | M1, A1 | For $E[(e^X)^k]$; For $E(e^{kX})$ |
| | A1 | For $M_X(k)$ |
| $\therefore E(W) = M_X(1) = e^{\mu + \frac{\sigma^2}{2}}$ | M1 A1 | |
| $E(W^2) = M_X(2) = e^{2\mu + 2\sigma^2}$ | M1 A1, A1 | |
| $\therefore \text{Var}(W) = e^{2\mu + 2\sigma^2} - e^{2\mu + \sigma^2} \left[= e^{2\mu + \sigma^2}(e^{\sigma^2} - 1)\right]$ | | | **[8 marks]**
---2 (i) The random variable $Z$ has the standard Normal distribution with probability density function
$$\mathrm { f } ( z ) = \frac { 1 } { \sqrt { 2 \pi } } \mathrm { e } ^ { - z ^ { 2 } / 2 } , \quad - \infty < z < \infty$$
Obtain the moment generating function of $Z$.\\
(ii) Let $\mathrm { M } _ { Y } ( t )$ denote the moment generating function of the random variable $Y$. Show that the moment generating function of the random variable $a Y + b$, where $a$ and $b$ are constants, is $\mathrm { e } ^ { b t } \mathrm { M } _ { Y } ( a t )$.\\
(iii) Use the results in parts (i) and (ii) to obtain the moment generating function $\mathrm { M } _ { X } ( t )$ of the random variable $X$ having the Normal distribution with parameters $\mu$ and $\sigma ^ { 2 }$.\\
(iv) If $W = \mathrm { e } ^ { X }$ where $X$ is as in part (iii), $W$ is said to have a lognormal distribution. Show that, for any positive integer $k$, the expected value of $W ^ { k }$ is $\mathrm { M } _ { X } ( k )$. Use this result to find the expected value and variance of the lognormal distribution.
\hfill \mbox{\textit{OCR MEI S4 2009 Q2 [24]}}