| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/2 (Pre-U Further Mathematics Paper 2) |
| Year | 2019 |
| Session | Specimen |
| Marks | 3 |
| Topic | Moment generating functions |
| Type | MGF of transformed variable |
| Difficulty | Standard +0.3 This is a structured MGF question with standard exponential distribution. Parts (a)-(c) involve routine integration and MGF manipulation. Part (d) requires recognizing that M_{X₁-X₂}(t) = M_{X₁}(t)·M_{-X₂}(t) and differentiating twice, which is methodical rather than insightful. Slightly above average due to the multi-part nature and MGF transformations, but all techniques are standard for Further Maths statistics. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03c Calculate mean/variance: by integration5.03d E(g(X)): general expectation formula |
**Question 5(a)**
$M_X(t) = \int_0^{\infty} e^{tx} k e^{-kx} dx$ (Limits required) [M1]
$= k\int_0^{\infty} e^{(t-k)x} dx = k\int_0^{\infty} e^{-(k-t)x} dx$ (Limits not required) [M1]
$= \dfrac{-k}{k-t}\left[e^{-(k-t)x}\right]_0^{\infty} = \dfrac{k}{k-t}$ **AG** [A1]
**Total: 3 marks**
**Question 5(b)**
$M_X'(t) = \dfrac{k}{(k-t)^2} \Rightarrow E(X) = M_X'(0) = \dfrac{1}{k}$ [M1A1]
$M_X''(t) = \dfrac{2k}{(k-t)^3} \Rightarrow E(X^2) = M_X''(0) = \dfrac{2}{k^2}$ [M1A1]
$\Rightarrow \text{Var}(X) = \dfrac{2}{k^2} - \left(\dfrac{1}{k}\right)^2 = \dfrac{1}{k^2}$ (A1 **ft** if double sign error when differentiating twice, but CAO) [A1]
**OR Alternatively:**
$M_X(t) = \left(1 - \dfrac{t}{k}\right)^{-1} = 1 + \dfrac{t}{k} + \dfrac{t^2}{k^2} + \ldots$ [M1A1]
$E(X) = \dfrac{1}{k}$ [A1]
$E(X^2) = \dfrac{2}{k^2} \Rightarrow \text{Var}(X) = \dfrac{2}{k^2} - \left(\dfrac{1}{k}\right)^2 = \dfrac{1}{k^2}$ [M1A1]
**Total: 5 marks**
**Question 5(c)**
$E(e^{t(-X)}) = E(e^{-tX}) = M_X(-t) = k(k+t)^{-1}$. Or equivalent [B1]
**Total: 1 mark**
**Question 5(d)**
$M_{X_1 - X_2}(t) = M_X(t) \times M_{-X}(t) = k^2(k^2 - t^2)^{-1}$ [M1]
$= 1 + \dfrac{t^2}{k^2} + \dfrac{t^4}{k^4} + \ldots$ **OR** find $M''(0)$ [M1]
$\Rightarrow E(X_1 - X_2)^2 = 2! \times \text{coefficient of } t^2 = \dfrac{2}{k^2}$ [A1]
**Total: 3 marks**5 The random variable $X$ has probability density function $\mathrm { f } ( \mathrm { x } )$, where
$$\mathrm { f } ( x ) = \begin{cases} k \mathrm { e } ^ { - k x } & x \geqslant 0 \\ 0 & x < 0 \end{cases}$$
and $k$ is a positive constant.
\begin{enumerate}[label=(\alph*)]
\item Show that the moment generating function of $X$ is $\mathrm { M } _ { X } ( t ) = k ( k - t ) ^ { - 1 } , t < k$.
\item Use the moment generating function to find $\mathrm { E } ( X )$ and $\operatorname { Var } ( X )$.
\item Show that the moment generating function of $- X$ is $k ( k + t ) ^ { - 1 }$.
\item $X _ { 1 }$ and $X _ { 2 }$ are two independent observations of $X$. Use the moment generating function of $X _ { 1 } - X _ { 2 }$ to find the value of $\mathrm { E } \left[ \left( X _ { 1 } - X _ { 2 } \right) ^ { 2 } \right]$.
\end{enumerate}
\hfill \mbox{\textit{Pre-U Pre-U 9795/2 2019 Q5 [3]}}