| Exam Board | OCR |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2012 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moment generating functions |
| Type | Derive MGF from PDF |
| Difficulty | Standard +0.3 This is a straightforward application of MGF theory requiring integration by parts for part (i), then simple substitution and multiplication properties of MGFs for parts (ii) and (iii). While it involves Further Maths Statistics content, the techniques are standard and well-practiced, making it slightly easier than average overall. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03c Calculate mean/variance: by integration |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(\int_0^e 4xe^{(2-t)x}dx\) oe | M1 | |
| \(= \left[\frac{-4}{2-t}xe^{(2-t)x}\right]_0^e + \frac{4}{2-t}\int_0^e e^{(2-t)x}dx\) oe | M1 | Using integration by parts (Allow omission of limits for M1M1) |
| \(= \frac{-4}{(2-t)^2}\left[e^{(2-t)x}\right]_0^e\) oe | A1 | |
| \(= \frac{4}{(2-t)^2}\) AG | A1 | Allow \(\frac{4}{(t-2)^2}\) |
| [4] | ||
| (ii) Requires \(E(e^{-tM}) = \) which is \(E(e^{t})\) with \(-t\) for \(t\) | B1 | Or from mgfs |
| [1] | ||
| (iii) \(16/(4-t)^2\) | B1 | AEF, ISW |
| [1] |
**(i)** $\int_0^e 4xe^{(2-t)x}dx$ oe | M1 |
$= \left[\frac{-4}{2-t}xe^{(2-t)x}\right]_0^e + \frac{4}{2-t}\int_0^e e^{(2-t)x}dx$ oe | M1 | Using integration by parts (Allow omission of limits for M1M1)
$= \frac{-4}{(2-t)^2}\left[e^{(2-t)x}\right]_0^e$ oe | A1 |
$= \frac{4}{(2-t)^2}$ AG | A1 | Allow $\frac{4}{(t-2)^2}$
| [4] |
**(ii)** Requires $E(e^{-tM}) = $ which is $E(e^{t})$ with $-t$ for $t$ | B1 | Or from mgfs | or from $\int_0^e 4xe^{-t(1+2)}dx$ | Must be $-x(t+2)$, not $-xt-2x$
| [1] |
**(iii)** $16/(4-t)^2$ | B1 | AEF, ISW
| [1] |2 The continuous random variable $X$ has probability density function given by
$$f ( x ) = \begin{cases} 4 x e ^ { - 2 x } & x \geqslant 0 \\ 0 & \text { otherwise } \end{cases}$$
(i) Show that the moment generating function ( mgf ) of $X$ is
$$\frac { 4 } { ( 2 - t ) ^ { 2 } } , \text { where } | t | < 2$$
(ii) Explain why the $\operatorname { mgf }$ of $- X$ is $\frac { 4 } { ( 2 + t ) ^ { 2 } }$.\\
(iii) Two random observations of $X$ are denoted by $X _ { 1 }$ and $X _ { 2 }$. What is the $\operatorname { mgf }$ of $X _ { 1 } - X _ { 2 }$ ?
\hfill \mbox{\textit{OCR S4 2012 Q2 [6]}}