Pre-U Pre-U 9795/2 2012 June — Question 1 5 marks

Exam BoardPre-U
ModulePre-U 9795/2 (Pre-U Further Mathematics Paper 2)
Year2012
SessionJune
Marks5
TopicMoment generating functions
TypeDerive MGF from PDF
DifficultyStandard +0.8 This is a Further Maths statistics question requiring integration of an exponential function with parameter t to derive an MGF, then differentiation to find moments. While the exponential integral is standard, working with the MGF framework and parameter constraints requires solid technique beyond typical A-level, placing it moderately above average difficulty.
Spec5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03c Calculate mean/variance: by integration

1 The random variable \(X\) has probability density function \(\mathrm { f } ( x )\), 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.
  1. Show that the moment generating function of \(X\) is \(\mathrm { M } _ { X } ( t ) = k ( k - t ) ^ { - 1 } , t < k\).
  2. Use the moment generating function to find \(\mathrm { E } ( X )\) and \(\operatorname { Var } ( X )\).

Question 1(i)
\(M_X(t) = \int_0^{\infty} e^{tx} k e^{-kx} dx\) (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
\(= \frac{-k}{k-t} \left[ e^{-(k-t)x} \right]_0^{\infty} = \frac{k}{k-t}\) (AG) A1 [3]
Question 1(ii)
\(M_X'(t) = \frac{k}{(k-t)^2} \Rightarrow E(X) = M_X'(0) = \frac{1}{k}\) M1A1
\(M_X''(t) = \frac{2k}{(k-t)^3} \Rightarrow E(X^2) = M_X''(0) = \frac{2}{k^2}\) M1A1
(A1 ft if double sign error when differentiating twice, but CA0) A1 [5]
Alternatively:
\(M_X(t) = \left(1 - \frac{t}{k}\right)^{-1} = 1 + \frac{t}{k} + \frac{t^2}{k^2} + \ldots = 1 + \frac{1}{k}t + \frac{2}{k^2}t^2 + \ldots\) M1A1
\(E(X) = \frac{1}{k}\) A1
\(E(X^2) = \frac{2}{k^2} \Rightarrow \text{Var}(X) = \frac{2}{k^2} - \left(\frac{1}{k}\right)^2 = \frac{1}{k^2}\) M1A1 [5] [8]
**Question 1(i)**
$M_X(t) = \int_0^{\infty} e^{tx} k e^{-kx} dx$ (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**

$= \frac{-k}{k-t} \left[ e^{-(k-t)x} \right]_0^{\infty} = \frac{k}{k-t}$ (AG) **A1** [3]

**Question 1(ii)**
$M_X'(t) = \frac{k}{(k-t)^2} \Rightarrow E(X) = M_X'(0) = \frac{1}{k}$ **M1A1**

$M_X''(t) = \frac{2k}{(k-t)^3} \Rightarrow E(X^2) = M_X''(0) = \frac{2}{k^2}$ **M1A1**

(A1 ft if double sign error when differentiating twice, but CA0) **A1** [5]

**Alternatively:**
$M_X(t) = \left(1 - \frac{t}{k}\right)^{-1} = 1 + \frac{t}{k} + \frac{t^2}{k^2} + \ldots = 1 + \frac{1}{k}t + \frac{2}{k^2}t^2 + \ldots$ **M1A1**

$E(X) = \frac{1}{k}$ **A1**

$E(X^2) = \frac{2}{k^2} \Rightarrow \text{Var}(X) = \frac{2}{k^2} - \left(\frac{1}{k}\right)^2 = \frac{1}{k^2}$ **M1A1** [5] **[8]**
1 The random variable $X$ has probability density function $\mathrm { f } ( x )$, 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.\\
(i) Show that the moment generating function of $X$ is $\mathrm { M } _ { X } ( t ) = k ( k - t ) ^ { - 1 } , t < k$.\\
(ii) Use the moment generating function to find $\mathrm { E } ( X )$ and $\operatorname { Var } ( X )$.

\hfill \mbox{\textit{Pre-U Pre-U 9795/2 2012 Q1 [5]}}