Pre-U Pre-U 9795/2 2016 Specimen — Question 1 9 marks

Exam BoardPre-U
ModulePre-U 9795/2 (Pre-U Further Mathematics Paper 2)
Year2016
SessionSpecimen
Marks9
TopicMoment generating functions
TypeDerive MGF from PDF
DifficultyStandard +0.3 This is a straightforward application of MGF definition to the exponential distribution. Part (i) requires evaluating a standard integral ∫₀^∞ e^{tx}·ke^{-kx}dx = k∫₀^∞ e^{(t-k)x}dx, which is routine for Further Maths students. Part (ii) involves differentiating the MGF and substituting t=0, which is direct recall of the MGF method. While MGFs are a Further Maths topic, this question involves no novel insight or complex manipulation—just applying standard techniques to a textbook example.
Spec5.03a Continuous random variables: pdf and cdf5.03c 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 \(\operatorname { Var } ( X )\).

(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
(ii) \(M_X'(t) = \frac{k}{(k-t)^2} \Rightarrow E(X) = M_X'(0) = \frac{1}{k}\) M1 A1
\(M_X''(t) = \frac{2k}{(k-t)^3} \Rightarrow E(X^2) = M_X''(0) = \frac{2}{k^2}\) M1 A1
(A1 ft if double sign error when differentiating twice, but CAO) A1
Alternatively:
\(M_X(t) = \left(1 - \frac{t}{k}\right)^{-1} = 1 + \frac{t}{k} + \frac{t^2}{k^2} + \ldots = 1 + \frac{1}{k}t + \frac{2}{k^2}t^2 + \ldots\) M1 A1
\(E(X) = \frac{1}{k}\) A1
\(E(X^2) = \frac{2}{k^2} \Rightarrow \text{Var}(X) = \frac{2}{k^2} - \left(\frac{1}{k}\right)^2 = \frac{1}{k^2}\) M1 A1
**(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**

**(ii)** $M_X'(t) = \frac{k}{(k-t)^2} \Rightarrow E(X) = M_X'(0) = \frac{1}{k}$ **M1 A1**

$M_X''(t) = \frac{2k}{(k-t)^3} \Rightarrow E(X^2) = M_X''(0) = \frac{2}{k^2}$ **M1 A1**

(A1 ft if double sign error when differentiating twice, but CAO) **A1**

**Alternatively:**

$M_X(t) = \left(1 - \frac{t}{k}\right)^{-1} = 1 + \frac{t}{k} + \frac{t^2}{k^2} + \ldots = 1 + \frac{1}{k}t + \frac{2}{k^2}t^2 + \ldots$ **M1 A1**

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

$E(X^2) = \frac{2}{k^2} \Rightarrow \text{Var}(X) = \frac{2}{k^2} - \left(\frac{1}{k}\right)^2 = \frac{1}{k^2}$ **M1 A1**
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 2016 Q1 [9]}}