| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/2 (Pre-U Further Mathematics Paper 2) |
| Year | 2020 |
| Session | Specimen |
| Marks | 3 |
| Topic | Moment generating functions |
| Type | MGF of transformed variable |
| Difficulty | Standard +0.8 This is a structured MGF question requiring integration, differentiation of MGFs, and understanding of MGF properties for linear transformations and sums of independent variables. While the exponential distribution is standard, parts (c) and (d) require non-trivial manipulation of MGFs and understanding that E[(X₁-X₂)²] = Var(X₁-X₂) + [E(X₁-X₂)]². This is moderately challenging for Further Maths students but follows a guided structure. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03c Calculate mean/variance: by integration |
**Question 5(a):**
$M_X(t) = \int_0^\infty e^{tx} k e^{-kx}\,\text{d}x$ (Limits required) [M1]
$= k\int_0^\infty e^{(t-k)x}\,\text{d}x = k\int_0^\infty e^{-(k-t)x}\,\text{d}x$ (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 $\mathrm { f } ( x )$, where
$$\mathrm { f } ( x ) = \begin{cases} k e ^ { - k x } & x \geqslant 0 \\ 0 & x < 0 \end{cases}$$
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 2020 Q5 [3]}}