| Exam Board | OCR |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2013 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moment generating functions |
| Type | Derive MGF from PDF |
| Difficulty | Standard +0.8 This is a Further Maths S4 question requiring integration by parts to derive an MGF, recognition of convergence conditions, and differentiation of the MGF to find variance. While the integration is moderately technical, it follows a standard template for MGF derivation that S4 students practice extensively. The difficulty is elevated above average due to the Further Maths context and multi-step calculus, but it's a routine exercise within this specification. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03c Calculate mean/variance: by integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(M(t) = \int_0^{\infty} \frac{1}{4}x e^{-\frac{1}{2}x(1-2t)}\,dx\) | M1* | From \(E(e^{tX})\). Need single exponential term, not nec. correct |
| \(= \left[\frac{-xe^{-\frac{1}{2}x(1-2t)}}{2(1-2t)}\right]_0^{\infty} + \int_0^{\infty} \frac{e^{-\frac{1}{2}x(1-2t)}}{2(1-2t)}\,dx\) | *M1A1 | Integration by parts |
| \(= \left[\frac{-e^{-\frac{1}{2}x(1-2t)}}{(1-2t)^2}\right]_0^{\infty}\) | A1 | \(= \left[\frac{-e^{-\frac{1}{2}x(1-2t)}}{4(t-\frac{1}{2})^2}\right]_0^{\infty}\). Allow without limits |
| \(= \text{AG}\ (1-2t)^{-2}\) | A1 | With evidence, cwo |
| Requires \(1-2t > 0\) for correct limits | B1 | Or for convergence of the integral |
| [6] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(M'(t) = 4(1-2t)^{-3}\), \(E(X) = 4\) cwo | B1 | or from \(1+4t\) |
| \(M''(t) = 24(1-2t)^{-4}\), \(E(X^2) = 24\) cwo | B1 | \(+12t^2\) |
| \(\text{Var} = 24 - 16 = 8\) | B1FT | provided \(\text{Var} > 0\) |
| [3] |
# Question 3:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $M(t) = \int_0^{\infty} \frac{1}{4}x e^{-\frac{1}{2}x(1-2t)}\,dx$ | M1* | From $E(e^{tX})$. Need single exponential term, not nec. correct |
| $= \left[\frac{-xe^{-\frac{1}{2}x(1-2t)}}{2(1-2t)}\right]_0^{\infty} + \int_0^{\infty} \frac{e^{-\frac{1}{2}x(1-2t)}}{2(1-2t)}\,dx$ | *M1A1 | Integration by parts |
| $= \left[\frac{-e^{-\frac{1}{2}x(1-2t)}}{(1-2t)^2}\right]_0^{\infty}$ | A1 | $= \left[\frac{-e^{-\frac{1}{2}x(1-2t)}}{4(t-\frac{1}{2})^2}\right]_0^{\infty}$. Allow without limits |
| $= \text{AG}\ (1-2t)^{-2}$ | A1 | With evidence, cwo |
| Requires $1-2t > 0$ for correct limits | B1 | Or for convergence of the integral |
| **[6]** | | |
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $M'(t) = 4(1-2t)^{-3}$, $E(X) = 4$ cwo | B1 | or from $1+4t$ |
| $M''(t) = 24(1-2t)^{-4}$, $E(X^2) = 24$ cwo | B1 | $+12t^2$ |
| $\text{Var} = 24 - 16 = 8$ | B1FT | provided $\text{Var} > 0$ |
| **[3]** | | |
---3 The continuous random variable $X$ has probability density function given by
$$\mathrm { f } ( x ) = \begin{cases} \frac { 1 } { 4 } x \mathrm { e } ^ { - \frac { 1 } { 2 } x } & x \geqslant 0 \\ 0 & \text { otherwise } . \end{cases}$$
(i) Show that the moment generating function of $X$ is $( 1 - 2 t ) ^ { - 2 }$ for $t < \frac { 1 } { 2 }$, and state why the condition $t < \frac { 1 } { 2 }$ is necessary.\\
(ii) Use the moment generating function to find $\operatorname { Var } ( X )$.
\hfill \mbox{\textit{OCR S4 2013 Q3 [9]}}