| Exam Board | OCR MEI |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2006 |
| Session | June |
| Marks | 24 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | The Gamma Distribution |
| Type | Deriving moment generating function |
| Difficulty | Standard +0.8 This is a Further Maths Statistics question requiring derivation of MGF through integration by substitution, application of MGF properties for sums, and differentiation to find moments. While the integral result is given, students must handle the algebraic manipulation of the gamma distribution MGF and apply standard MGF theory. The multi-part structure and requirement to work with abstract parameters (λ, k, n) elevates this above routine A-level questions, but it follows a standard MGF derivation template taught in S4. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation5.03a Continuous random variables: pdf and cdf5.03c Calculate mean/variance: by integration5.04a Linear combinations: E(aX+bY), Var(aX+bY)5.05a Sample mean distribution: central limit theorem |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(M_X(\theta) = E[e^{\theta x}]\) | M1 | |
| \(= \int_0^\infty \frac{\lambda^{k+1}}{k!} x^k e^{-(\lambda-\theta)x} dx\) | M1 | |
| Put \((\lambda - \theta)x = u\) | M1 | |
| \(= \frac{\lambda^{k+1}}{k!(\lambda-\theta)^{k+1}} \int_0^\infty u^k e^{-u} du\) | A1 | For obtaining this expression after substitution |
| Take out constants | A1 | Dep on subst. |
| \(= \left(\frac{\lambda}{\lambda-\theta}\right)^{k+1}\) | A1, A1 | Apply "given": integral \(= k!\) (Dep on subst.) BEWARE PRINTED ANSWER |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(Y = X_1 + X_2 + \ldots + X_n\); by convolution theorem mgf of \(Y\) is \(\{M_X(\theta)\}^n\), i.e. \(\left(\frac{\lambda}{\lambda-\theta}\right)^{nk+n}\) | B1 | |
| \(M'(\theta) = \lambda^{nk+n}(-nk-n)(\lambda-\theta)^{-nk-n-1}(-1)\) | M1, A1 | |
| \(\therefore \mu = \frac{nk+n}{\lambda}\) | A1 | |
| \(M''(\theta) = (nk+n)\lambda^{nk+n}(-nk-n-1)(\lambda-\theta)^{-nk-n-2}(-1)\) | M1 | |
| \(\therefore M''(0) = (nk+n)(nk+n+1)/\lambda^2\) | A1 | |
| \(\therefore \sigma^2 = \frac{(nk+n)(nk+n+1)}{\lambda^2} - \frac{(nk+n)^2}{\lambda^2}\) | M1 | |
| \(= \frac{nk+n}{\lambda^2}\) | A1 | [8] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Note that \(M_Y(t)\) is of the same functional form as \(M_X(t)\) with \(k+1\) replaced by \(nk+n\), i.e. \(k\) replaced by \(nk+n-1\) | ||
| Pdf of \(Y\) is \(\frac{\lambda^{nk+n}}{(nk+n-1)!} \times y^{nk+n-1} \times e^{-\lambda y}\) for \(y > 0\) | B1, B1, B1 | One mark for each factor. Third factor mark depends on at least one of the other two [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\lambda=1,\ k=2,\ n=5\); Exact \(P(Y>10) = 0.9165\) | ||
| Use of \(N(15, 15)\) | M1, M1 | Mean ft (ii); Variance ft (ii) |
| \(P\left(N(0,1) > \frac{10-15}{\sqrt{15}} = -1.291\right)\) | A1 | c.a.o. |
| \(= 0.9017\) | A1 | c.a.o. |
| Reasonably good agreement – CLT working for only small \(n\) | E2 | (E1, E1) or other sensible comments [6] |
# Question 2:
## Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $M_X(\theta) = E[e^{\theta x}]$ | M1 | |
| $= \int_0^\infty \frac{\lambda^{k+1}}{k!} x^k e^{-(\lambda-\theta)x} dx$ | M1 | |
| Put $(\lambda - \theta)x = u$ | M1 | |
| $= \frac{\lambda^{k+1}}{k!(\lambda-\theta)^{k+1}} \int_0^\infty u^k e^{-u} du$ | A1 | For obtaining this expression after substitution |
| Take out constants | A1 | Dep on subst. |
| $= \left(\frac{\lambda}{\lambda-\theta}\right)^{k+1}$ | A1, A1 | Apply "given": integral $= k!$ (Dep on subst.) BEWARE PRINTED ANSWER | **[7]** |
## Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $Y = X_1 + X_2 + \ldots + X_n$; by convolution theorem mgf of $Y$ is $\{M_X(\theta)\}^n$, i.e. $\left(\frac{\lambda}{\lambda-\theta}\right)^{nk+n}$ | B1 | |
| $M'(\theta) = \lambda^{nk+n}(-nk-n)(\lambda-\theta)^{-nk-n-1}(-1)$ | M1, A1 | |
| $\therefore \mu = \frac{nk+n}{\lambda}$ | A1 | |
| $M''(\theta) = (nk+n)\lambda^{nk+n}(-nk-n-1)(\lambda-\theta)^{-nk-n-2}(-1)$ | M1 | |
| $\therefore M''(0) = (nk+n)(nk+n+1)/\lambda^2$ | A1 | |
| $\therefore \sigma^2 = \frac{(nk+n)(nk+n+1)}{\lambda^2} - \frac{(nk+n)^2}{\lambda^2}$ | M1 | |
| $= \frac{nk+n}{\lambda^2}$ | A1 | **[8]** |
## Part (iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Note that $M_Y(t)$ is of the same functional form as $M_X(t)$ with $k+1$ replaced by $nk+n$, i.e. $k$ replaced by $nk+n-1$ | | |
| Pdf of $Y$ is $\frac{\lambda^{nk+n}}{(nk+n-1)!} \times y^{nk+n-1} \times e^{-\lambda y}$ for $y > 0$ | B1, B1, B1 | One mark for each factor. Third factor mark depends on at least one of the other two **[3]** |
## Part (iv):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\lambda=1,\ k=2,\ n=5$; Exact $P(Y>10) = 0.9165$ | | |
| Use of $N(15, 15)$ | M1, M1 | Mean ft (ii); Variance ft (ii) |
| $P\left(N(0,1) > \frac{10-15}{\sqrt{15}} = -1.291\right)$ | A1 | c.a.o. |
| $= 0.9017$ | A1 | c.a.o. |
| Reasonably good agreement – CLT working for only small $n$ | E2 | (E1, E1) or other sensible comments **[6]** |
---2 [In this question, you may use the result $\int _ { 0 } ^ { \infty } u ^ { m } \mathrm { e } ^ { - u } \mathrm {~d} u = m$ ! for any non-negative integer $m$.]\\
The random variable $X$ has probability density function
$$\mathrm { f } ( x ) = \begin{cases} \frac { \lambda ^ { k + 1 } x ^ { k } \mathrm { e } ^ { - \lambda x } } { k ! } , & x > 0 \\ 0 , & \text { elsewhere } \end{cases}$$
where $\lambda > 0$ and $k$ is a non-negative integer.\\
(i) Show that the moment generating function of $X$ is $\left( \frac { \lambda } { \lambda - \theta } \right) ^ { k + 1 }$.\\
(ii) The random variable $Y$ is the sum of $n$ independent random variables each distributed as $X$. Find the moment generating function of $Y$ and hence obtain the mean and variance of $Y$. [8]\\
(iii) State the probability density function of $Y$.\\
(iv) For the case $\lambda = 1 , k = 2$ and $n = 5$, it may be shown that the definite integral of the probability density function of $Y$ between limits 10 and $\infty$ is 0.9165 . Calculate the corresponding probability that would be given by a Normal approximation and comment briefly.
\hfill \mbox{\textit{OCR MEI S4 2006 Q2 [24]}}