| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/2 (Pre-U Further Mathematics Paper 2) |
| Session | Specimen |
| Marks | 9 |
| Topic | Moment generating functions |
| Type | Show unbiased estimator |
| Difficulty | Standard +0.3 Part (i) is a straightforward algebraic proof using linearity of expectation (2 marks). Part (ii) requires computing E(X) by integration of a triangular pdf and applying part (i), but the integration is routine and the connection is direct. This is a standard textbook exercise on unbiased estimators, slightly easier than average A-level due to its mechanical nature. |
| Spec | 5.03c Calculate mean/variance: by integration5.05b Unbiased estimates: of population mean and variance |
| Answer | Marks | Guidance |
|---|---|---|
| \(= \frac{1}{a}(a\theta + b) - \frac{b}{a} = \theta\) | M1, A1 | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| \(E(X) = \int_\theta^{\theta+4}\left(\frac{(\theta+4)x-x^2}{8}\right)dx\) | M1 | |
| \(= \left[\frac{(\theta+4)x^2}{16} - \frac{x^3}{24}\right]_\theta^{\theta+4}\) | M1, A1 | |
| \(= \left[\frac{(\theta+4)^3}{48}\right] - \left[\frac{(\theta+4)^2\theta^3}{16} - \frac{\theta^3}{24}\right]\) | M1 | |
| \(= \frac{\theta^3 + 12\theta^2 + 48\theta + 64}{48} - \frac{\theta^3 + 12\theta^2}{48}\) | A1 | |
| \(= \theta + \frac{4}{3}\) | A1 | |
| Hence unbiased estimator is \(X - \frac{4}{3}\) | ft, B1 | 7 |
### (i)
$E\left(\frac{X-b}{a}\right) = \frac{1}{a}E(X) - \frac{b}{a}$
$= \frac{1}{a}(a\theta + b) - \frac{b}{a} = \theta$ | M1, A1 | **2**
### (ii)
$E(X) = \int_\theta^{\theta+4}\left(\frac{(\theta+4)x-x^2}{8}\right)dx$ | M1 |
$= \left[\frac{(\theta+4)x^2}{16} - \frac{x^3}{24}\right]_\theta^{\theta+4}$ | M1, A1 |
$= \left[\frac{(\theta+4)^3}{48}\right] - \left[\frac{(\theta+4)^2\theta^3}{16} - \frac{\theta^3}{24}\right]$ | M1 |
$= \frac{\theta^3 + 12\theta^2 + 48\theta + 64}{48} - \frac{\theta^3 + 12\theta^2}{48}$ | A1 |
$= \theta + \frac{4}{3}$ | A1 |
Hence unbiased estimator is $X - \frac{4}{3}$ | ft, B1 | **7**\begin{enumerate}[label=(\roman*)]
\item The random variable $X$ is such that $\text{E}(X) = a\theta + b$, where $a$ and $b$ are constants and $\theta$ is a parameter.
Show that $\frac{X - b}{a}$ is an unbiased estimator of $\theta$. [2]
\item The continuous random variable $X$ has probability density function given by
$$f(x) = \begin{cases}
\frac{1}{8}(\theta + 4 - x) & \theta \leq x \leq \theta + 4, \\
0 & \text{otherwise}.
\end{cases}$$
Find $\text{E}(X)$ and hence find an unbiased estimator of $\theta$. [7]
\end{enumerate}
\hfill \mbox{\textit{Pre-U Pre-U 9795/2 Q8 [9]}}