| Exam Board | OCR |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2014 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moment generating functions |
| Type | Derive MGF from PDF |
| Difficulty | Challenging +1.2 This is a multi-part Further Maths Statistics question requiring MGF derivation via integration by parts, MGF properties for sums, and series expansion. While technically demanding, it follows standard S4 procedures with clear signposting through the parts, making it moderately above average difficulty but not requiring novel insight. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03c Calculate mean/variance: by integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\int_0^1 xe^{tx}\,dx + \int_1^2 (2-x)e^{tx}\,dx\) | M1 | |
| \(\left[\frac{xe^{tx}}{t}\right] + \left[\frac{-e^{tx}}{t^2}\right]\), \(\left[\frac{(2-x)e^{tx}}{t}\right] + \left[\frac{e^{tx}}{t^2}\right]\) | M1 | Integration by parts, either integral. No need for limits for this mark |
| \(\dfrac{e^t}{t} - \dfrac{e^t}{t^2} + \dfrac{1}{t^2}\) | A1 | |
| \(+\dfrac{e^{2t}}{t^2} - \dfrac{e^t}{t^2} - \dfrac{e^t}{t}\) | A2 | A1 for one error in 2nd integral |
| \(\dfrac{(e^t-1)^2}{t^2}\) AG | A1 | cwo |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\dfrac{e^t - 1}{t}\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\dfrac{1}{t}\!\left(t + \dfrac{t^2}{2} + \dfrac{t^3}{6} + \dfrac{t^4}{24} + \ldots\right)\) | M1 | Need attempt at first 3 terms in brackets |
| \(1 + \dfrac{t}{2} + \dfrac{t^2}{6}\) | A1 | Allow use of ! |
| \(E(Y) = \tfrac{1}{2}\), \(E(Y^2) = \tfrac{1}{3}\) | B1ft | For both, ft coeff of \(t\), \(2\times\) coeff of \(t^2\) |
| \(\text{Var}(Y) = \tfrac{1}{3} - \tfrac{1}{4}\) | M1 | |
| \(= \tfrac{1}{12}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\tfrac{1}{6}\) | B1ft | ft \(2\text{Var}(Y)\) |
# Question 4:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\int_0^1 xe^{tx}\,dx + \int_1^2 (2-x)e^{tx}\,dx$ | M1 | |
| $\left[\frac{xe^{tx}}{t}\right] + \left[\frac{-e^{tx}}{t^2}\right]$, $\left[\frac{(2-x)e^{tx}}{t}\right] + \left[\frac{e^{tx}}{t^2}\right]$ | M1 | Integration by parts, either integral. No need for limits for this mark |
| $\dfrac{e^t}{t} - \dfrac{e^t}{t^2} + \dfrac{1}{t^2}$ | A1 | |
| $+\dfrac{e^{2t}}{t^2} - \dfrac{e^t}{t^2} - \dfrac{e^t}{t}$ | A2 | A1 for one error in 2nd integral |
| $\dfrac{(e^t-1)^2}{t^2}$ AG | A1 | cwo |
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\dfrac{e^t - 1}{t}$ | B1 | |
## Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\dfrac{1}{t}\!\left(t + \dfrac{t^2}{2} + \dfrac{t^3}{6} + \dfrac{t^4}{24} + \ldots\right)$ | M1 | Need attempt at first 3 terms in brackets |
| $1 + \dfrac{t}{2} + \dfrac{t^2}{6}$ | A1 | Allow use of ! |
| $E(Y) = \tfrac{1}{2}$, $E(Y^2) = \tfrac{1}{3}$ | B1ft | For both, ft coeff of $t$, $2\times$ coeff of $t^2$ |
| $\text{Var}(Y) = \tfrac{1}{3} - \tfrac{1}{4}$ | M1 | |
| $= \tfrac{1}{12}$ | A1 | |
## Part (iv)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\tfrac{1}{6}$ | B1ft | ft $2\text{Var}(Y)$ |
---4 The continuous random variable $X$ has probability density function
$$f ( x ) = \left\{ \begin{array} { c c }
x & 0 \leqslant x \leqslant 1 \\
2 - x & 1 \leqslant x \leqslant 2 \\
0 & \text { otherwise }
\end{array} \right.$$
(i) Show that the moment generating function of $X$ is $\frac { \left( \mathrm { e } ^ { t } - 1 \right) ^ { 2 } } { t ^ { 2 } }$.\\
$Y _ { 1 }$ and $Y _ { 2 }$ are independent observations of a random variable $Y$. The moment generating function of $Y _ { 1 } + Y _ { 2 }$ is $\frac { \left( \mathrm { e } ^ { t } - 1 \right) ^ { 2 } } { t ^ { 2 } }$.\\
(ii) Write down the moment generating function of $Y$.\\
(iii) Use the expansion of $\mathrm { e } ^ { t }$ to find $\operatorname { Var } ( Y )$.\\
(iv) Deduce the value of $\operatorname { Var } ( X )$.
\hfill \mbox{\textit{OCR S4 2014 Q4 [13]}}