| Exam Board | Edexcel |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2013 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Central limit theorem |
| Type | Estimator properties and bias |
| Difficulty | Challenging +1.2 This is a Further Maths S4 question on estimator properties requiring standard proofs of expectation, consistency, and bias. While it involves multiple parts and algebraic manipulation, the techniques are textbook applications of E(X), Var(X) definitions and linearity of expectation—no novel insight required. The consistency proof and bias calculation follow directly from standard results, making it moderately above average difficulty but well within reach for prepared FM students. |
| Spec | 5.05b Unbiased estimates: of population mean and variance |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(E\left(\sum_{i=1}^{n} W_i\right) = n\mu\) | B1 | Correct |
| \(E(W_i^2) = \text{Var}(W_i) + [E(W_i)]^2 = \sigma^2 + \mu^2\) | M1 | Using \(\text{Var}(W) = E(W^2) - [E(W)]^2\) |
| \(E\left(\sum_{i=1}^{n} W_i^2\right) = \sum_{i=1}^{n} E(W_i^2) = n(\sigma^2 + \mu^2)\) | M1 A1 | Summing correctly |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(E(\bar{X}) = \mu\) (unbiased) | B1 | |
| \(\text{Var}(\bar{X}) = \frac{\sigma^2}{n} \to 0\) as \(n \to \infty\) | M1 | |
| Since \(E(\bar{X}) = \mu\) and \(\text{Var}(\bar{X}) \to 0\), \(\bar{X}\) is consistent | A1 | Both conditions required |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(E(U) = \frac{1}{n} \cdot n(\sigma^2 + \mu^2) - \frac{1}{n^2} \cdot [n\sigma^2 + n^2\mu^2 + \text{...}]\) | M1 | Using results from (a) |
| \(E\left(\frac{1}{n}\sum W_i^2\right) = \sigma^2 + \mu^2\) | A1 | |
| \(E\left(\left(\frac{1}{n}\sum W_i\right)^2\right) = \frac{\sigma^2}{n} + \mu^2\) | A1 | |
| \(E(U) = \sigma^2 + \mu^2 - \frac{\sigma^2}{n} - \mu^2 = \frac{(n-1)\sigma^2}{n}\) | A1 | Bias \(= -\frac{\sigma^2}{n}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Unbiased estimator \(= \frac{n}{n-1}U\), so \(k = \frac{n}{n-1}\) | B1 |
# Question 8:
## Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $E\left(\sum_{i=1}^{n} W_i\right) = n\mu$ | B1 | Correct |
| $E(W_i^2) = \text{Var}(W_i) + [E(W_i)]^2 = \sigma^2 + \mu^2$ | M1 | Using $\text{Var}(W) = E(W^2) - [E(W)]^2$ |
| $E\left(\sum_{i=1}^{n} W_i^2\right) = \sum_{i=1}^{n} E(W_i^2) = n(\sigma^2 + \mu^2)$ | M1 A1 | Summing correctly |
## Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $E(\bar{X}) = \mu$ (unbiased) | B1 | |
| $\text{Var}(\bar{X}) = \frac{\sigma^2}{n} \to 0$ as $n \to \infty$ | M1 | |
| Since $E(\bar{X}) = \mu$ and $\text{Var}(\bar{X}) \to 0$, $\bar{X}$ is consistent | A1 | Both conditions required |
## Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $E(U) = \frac{1}{n} \cdot n(\sigma^2 + \mu^2) - \frac{1}{n^2} \cdot [n\sigma^2 + n^2\mu^2 + \text{...}]$ | M1 | Using results from (a) |
| $E\left(\frac{1}{n}\sum W_i^2\right) = \sigma^2 + \mu^2$ | A1 | |
| $E\left(\left(\frac{1}{n}\sum W_i\right)^2\right) = \frac{\sigma^2}{n} + \mu^2$ | A1 | |
| $E(U) = \sigma^2 + \mu^2 - \frac{\sigma^2}{n} - \mu^2 = \frac{(n-1)\sigma^2}{n}$ | A1 | Bias $= -\frac{\sigma^2}{n}$ |
## Part (d):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Unbiased estimator $= \frac{n}{n-1}U$, so $k = \frac{n}{n-1}$ | B1 | |8. A random sample $W _ { 1 } , W _ { 2 } \ldots , W _ { n }$ is taken from a distribution with mean $\mu$ and variance $\sigma ^ { 2 }$
\begin{enumerate}[label=(\alph*)]
\item Write down $\mathrm { E } \left( \sum _ { i = 1 } ^ { n } W _ { i } \right)$ and show that $\mathrm { E } \left( \sum _ { i = 1 } ^ { n } W _ { i } ^ { 2 } \right) = n \left( \sigma ^ { 2 } + \mu ^ { 2 } \right)$
An estimator for $\mu$ is
$$\bar { X } = \frac { 1 } { n } \sum _ { i = 1 } ^ { n } W _ { i }$$
\item Show that $\bar { X }$ is a consistent estimator for $\mu$.
An estimator of $\sigma ^ { 2 }$ is
$$U = \frac { 1 } { n } \sum _ { i = 1 } ^ { n } W _ { i } ^ { 2 } - \left( \frac { 1 } { n } \sum _ { i = 1 } ^ { n } W _ { i } \right) ^ { 2 }$$
\item Find the bias of $U$.
\item Write down an unbiased estimator of $\sigma ^ { 2 }$ in the form $k U$, where $k$ is in terms of $n$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S4 2013 Q8 [12]}}