| Exam Board | Edexcel |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2003 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moment generating functions |
| Type | Construct combined estimator |
| Difficulty | Standard +0.3 This is a standard S4 question on unbiased estimators and minimum variance. Part (a) requires routine verification using linearity of expectation. Part (b) involves straightforward application of variance properties with the constraint from unbiasedness. Part (c) is basic calculus (differentiation to find minimum). All steps are mechanical applications of well-practiced techniques with no novel insight required, making it slightly easier than average. |
| Spec | 5.02b Expectation and variance: discrete random variables5.02c Linear coding: effects on mean and variance |
| Answer | Marks | Guidance |
|---|---|---|
| \(E\left(\frac{2}{3}X_1 - \frac{1}{2}X_2 + \frac{5}{6}X_3\right) = \frac{2}{3}\mu - \frac{1}{2}\mu + \frac{5}{6}\mu = \mu\) | M1A1 | |
| \(E(Y) = \mu \Rightarrow\) unbiased | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(E(aX_1 + bX_2) = a\mu + b\mu = \mu\) | M1 | |
| \(a + b = 1\) | A1 | |
| \(\text{Var}(aX_1 + bX_2) = a^2\sigma^2 + b^2\sigma^2\) | M1A1 | |
| \(= a^2\sigma^2 + (1-a)^2\sigma^2\) | M1 | |
| \(= (2a^2 - 2a + 1)\sigma^2\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Min value when \((4a - 2)\sigma^2 = 0\) | M1A1 | |
| \(\Rightarrow 4a - 2 = 0\) | A1 | |
| \(a = \frac{1}{2}, b = \frac{1}{2}\) | \(\frac{d}{da}(\text{Var}) = 0\) all correct | A1A1 \(\mathcal{f}\) |
## Part (a)
$E\left(\frac{2}{3}X_1 - \frac{1}{2}X_2 + \frac{5}{6}X_3\right) = \frac{2}{3}\mu - \frac{1}{2}\mu + \frac{5}{6}\mu = \mu$ | | M1A1 |
$E(Y) = \mu \Rightarrow$ unbiased | | B1 | (3)
## Part (b)
$E(aX_1 + bX_2) = a\mu + b\mu = \mu$ | | M1 |
$a + b = 1$ | | A1 |
$\text{Var}(aX_1 + bX_2) = a^2\sigma^2 + b^2\sigma^2$ | | M1A1 |
$= a^2\sigma^2 + (1-a)^2\sigma^2$ | | M1 |
$= (2a^2 - 2a + 1)\sigma^2$ | | A1 | (6)
## Part (c)
Min value when $(4a - 2)\sigma^2 = 0$ | | M1A1 |
$\Rightarrow 4a - 2 = 0$ | | A1 |
$a = \frac{1}{2}, b = \frac{1}{2}$ | $\frac{d}{da}(\text{Var}) = 0$ all correct | A1A1 $\mathcal{f}$ | (5)
---A random sample of three independent variables $X_1$, $X_2$ and $X_3$ is taken from a distribution with mean $\mu$ and variance $\sigma^2$.
\begin{enumerate}[label=(\alph*)]
\item Show that $\frac{2}{5}X_1 - \frac{1}{5}X_2 + \frac{4}{5}X_3$ is an unbiased estimator for $\mu$. [3]
\end{enumerate}
An unbiased estimator for $\mu$ is given by $\hat{\mu} = aX_1 + bX_2$ where $a$ and $b$ are constants.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Show that Var($\hat{\mu}$) = $(2a^2 - 2a + 1)\sigma^2$. [6]
\item Hence determine the value of $a$ and the value of $b$ for which $\hat{\mu}$ has minimum variance. [5]
\end{enumerate}
\hfill \mbox{\textit{Edexcel S4 2003 Q6 [14]}}