| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/2 (Pre-U Further Mathematics Paper 2) |
| Year | 2014 |
| Session | June |
| Marks | 10 |
| Topic | Moment generating functions |
| Type | MGF series expansion |
| Difficulty | Challenging +1.2 This is a structured Further Maths statistics question requiring integration to verify the pdf normalizes, standard MGF calculation, series expansion using given formulas, and reading off the mean from coefficients. While it involves multiple techniques (integration, MGF, series expansion), each step is clearly signposted and uses standard methods without requiring novel insight. The algebraic manipulation is moderately involved but routine for Further Maths students. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03c Calculate mean/variance: by integration |
The continuous random variable $X$ has probability density function given by
$$f(x) = \begin{cases} 3e^{-x} & 0 \leq x \leq k, \\ 0 & \text{otherwise,} \end{cases}$$
where $k$ is a constant.
\begin{enumerate}[label=(\roman*)]
\item Show that $e^{-k} = \frac{2}{3}$. [2]
\item Show that the moment generating function of $X$ is given by $M_X(t) = \frac{3}{1-t}\left(1 - \frac{2}{3}e^{kt}\right)$. [4]
\item By expanding $M_X(t)$ as a power series in $t$, up to and including the term in $t^2$, show that
$$M_X(t) = 1 + (1 - 2k)t + (1 - 2k - k^2)t^2 + \ldots.$$
[3]
[You may use the standard series for $(1-t)^{-1}$ and $e^{kt}$ without proof.]
\item Deduce that the exact value of E$(X)$ is $1 - 2\ln\left(\frac{2}{3}\right)$. [1]
\end{enumerate}
\hfill \mbox{\textit{Pre-U Pre-U 9795/2 2014 Q4 [10]}}