| Exam Board | Edexcel |
|---|---|
| Module | FS2 (Further Statistics 2) |
| Year | 2022 |
| Session | June |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Probability Generating Functions |
| Type | Moment generating function problems |
| Difficulty | Challenging +1.8 This is a challenging Further Statistics 2 question requiring manipulation of weighted sums, proving unbiasedness, calculating variances of complex estimators, and comparing efficiency. While the techniques are standard (linearity of expectation/variance, independence), the algebraic manipulation is substantial and the weighted estimator K requires careful handling of the sum ∑r. The comparison for large n requires asymptotic analysis. This is harder than typical A-level questions but represents a standard FS2 problem rather than requiring exceptional insight. |
| Spec | 5.02b Expectation and variance: discrete random variables5.02c Linear coding: effects on mean and variance5.05b Unbiased estimates: of population mean and variance |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(E(K) = \frac{2E(X)}{n(n+1)} + \frac{4E(X)}{n(n+1)} + \frac{6E(X)}{n(n+1)} + \cdots + \frac{2nE(X)}{n(n+1)}\) | M1 | Using independence to set up expression for \(E(K)\) |
| \(= \frac{1}{n(n+1)}\left(\sum_{r=1}^{n} 2r\right)E(X) = \frac{1}{n(n+1)}\cdot\frac{(2+2n)n}{2}\cdot E(X) = E(X)\) | M1 | Use of \(\sum_{r=1}^{n} r = \frac{n(n+1)}{2}\) |
| \(= \mu\) (therefore \(K\) is an unbiased estimator of \(\mu\)) | A1cso | Correct conclusion from correct working |
| \(E(L) = \frac{2E(X)}{3} + \frac{(n-2)E(X)}{3(n-2)} = \frac{2E(X)}{3} + \frac{1E(X)}{3} = E(X)\) | M1, A1cso | M1: using independence to set up \(E(L)\); A1cso: correct conclusion |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\text{Var}(K) = \text{Var}\!\left(\frac{2X}{n(n+1)}\right) + \text{Var}\!\left(\frac{4X}{n(n+1)}\right) + \cdots + \text{Var}\!\left(\frac{2nX}{n(n+1)}\right)\) | M1 | Using independence to set up expression for \(\text{Var}(K)\) |
| \(= \frac{2^2}{n^2(n+1)^2}\text{Var}(X) + \frac{4^2}{n^2(n+1)^2}\text{Var}(X) + \cdots + \frac{(2n)^2}{n^2(n+1)^2}\text{Var}(X)\) | M1 | Use of \(\text{Var}(aX) = a^2\text{Var}(X)\) |
| \(= \frac{4\sum_{r=1}^{n} r^2}{n^2(n+1)^2}\text{Var}(X) = \frac{\frac{4}{6}n(n+1)(2n+1)}{n^2(n+1)^2}\sigma^2\) | M1 | Use of \(\sum r^2 = \frac{n(n+1)(2n+1)}{6}\) |
| \(= \frac{2(2n+1)}{3n(n+1)}\sigma^2\) | A1 | Correct simplified answer |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\text{Var}(L) = \text{Var}\!\left(\frac{X_1 + X_2}{3}\right) + \text{Var}\!\left(\frac{X_3 + X_4 + \cdots + X_n}{3(n-2)}\right)\) | M1 | Using independence to set up expression for \(\text{Var}(L)\) |
| \(= \frac{2}{9}\text{Var}(X) + \frac{(n-2)}{9(n-2)^2}\text{Var}(X)\) | M1 | Correct expansion |
| \(= \frac{2n-3}{9(n-2)}\sigma^2\) | A1 | Correct simplified answer |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| For large values of \(n\): \(\text{Var}(K) \to 0\), \(\quad \text{Var}(L) \to \frac{2}{9}\sigma^2\) | M1 | Correct limiting behaviour identified |
| Since both are unbiased, the better estimator is the one with smaller variance; \(0 < \frac{2}{9}\sigma^2\) | M1 | Comparison using smaller variance criterion |
| Therefore \(K\) is the better estimator and Korhan wins the challenge. | A1 | Correct conclusion |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Use of \(\sum_{r=1}^{n} r^2\) | M1 | |
| Correct equivalent expression for \(\text{Var}(K)\) | A1 | |
| Using independence to set up expression for \(\text{Var}(L)\) | M1 | |
| Use of \(\text{Var}(aX) = a^2\text{Var}(X)\) and \(\text{Var}(X_1 + X_2) = 2\text{Var}(X)\) | M1 | Implied by either correct term |
| Correct equivalent expression for \(\text{Var}(L)\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Finding correct limits as \(n\) gets larger for each expression | M1 | Allow ft |
| Solving \(\frac{2n-3}{9(n-2)} = \frac{2(2n+1)}{3n(n+1)} \rightarrow n = -0.53,\ 2.46,\ 4.57\) | Allow comments relating to \(n \geq 5\ (4.57)\) | |
| Correct explanation | M1 | |
| Deducing that Korhan is the winner | A1 | Dependent upon both M marks |
# Question 6:
## Part 6(a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $E(K) = \frac{2E(X)}{n(n+1)} + \frac{4E(X)}{n(n+1)} + \frac{6E(X)}{n(n+1)} + \cdots + \frac{2nE(X)}{n(n+1)}$ | M1 | Using independence to set up expression for $E(K)$ |
| $= \frac{1}{n(n+1)}\left(\sum_{r=1}^{n} 2r\right)E(X) = \frac{1}{n(n+1)}\cdot\frac{(2+2n)n}{2}\cdot E(X) = E(X)$ | M1 | Use of $\sum_{r=1}^{n} r = \frac{n(n+1)}{2}$ |
| $= \mu$ (therefore $K$ is an unbiased estimator of $\mu$) | A1cso | Correct conclusion from correct working |
| $E(L) = \frac{2E(X)}{3} + \frac{(n-2)E(X)}{3(n-2)} = \frac{2E(X)}{3} + \frac{1E(X)}{3} = E(X)$ | M1, A1cso | M1: using independence to set up $E(L)$; A1cso: correct conclusion |
## Part 6(b)(i):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\text{Var}(K) = \text{Var}\!\left(\frac{2X}{n(n+1)}\right) + \text{Var}\!\left(\frac{4X}{n(n+1)}\right) + \cdots + \text{Var}\!\left(\frac{2nX}{n(n+1)}\right)$ | M1 | Using independence to set up expression for $\text{Var}(K)$ |
| $= \frac{2^2}{n^2(n+1)^2}\text{Var}(X) + \frac{4^2}{n^2(n+1)^2}\text{Var}(X) + \cdots + \frac{(2n)^2}{n^2(n+1)^2}\text{Var}(X)$ | M1 | Use of $\text{Var}(aX) = a^2\text{Var}(X)$ |
| $= \frac{4\sum_{r=1}^{n} r^2}{n^2(n+1)^2}\text{Var}(X) = \frac{\frac{4}{6}n(n+1)(2n+1)}{n^2(n+1)^2}\sigma^2$ | M1 | Use of $\sum r^2 = \frac{n(n+1)(2n+1)}{6}$ |
| $= \frac{2(2n+1)}{3n(n+1)}\sigma^2$ | A1 | Correct simplified answer |
## Part 6(b)(ii):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\text{Var}(L) = \text{Var}\!\left(\frac{X_1 + X_2}{3}\right) + \text{Var}\!\left(\frac{X_3 + X_4 + \cdots + X_n}{3(n-2)}\right)$ | M1 | Using independence to set up expression for $\text{Var}(L)$ |
| $= \frac{2}{9}\text{Var}(X) + \frac{(n-2)}{9(n-2)^2}\text{Var}(X)$ | M1 | Correct expansion |
| $= \frac{2n-3}{9(n-2)}\sigma^2$ | A1 | Correct simplified answer |
## Part 6(c):
| Working/Answer | Mark | Guidance |
|---|---|---|
| For large values of $n$: $\text{Var}(K) \to 0$, $\quad \text{Var}(L) \to \frac{2}{9}\sigma^2$ | M1 | Correct limiting behaviour identified |
| Since both are unbiased, the better estimator is the one with smaller variance; $0 < \frac{2}{9}\sigma^2$ | M1 | Comparison using smaller variance criterion |
| Therefore $K$ is the better estimator and Korhan wins the challenge. | A1 | Correct conclusion |
# Question 6 (ii):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Use of $\sum_{r=1}^{n} r^2$ | M1 | |
| Correct equivalent expression for $\text{Var}(K)$ | A1 | |
| Using independence to set up expression for $\text{Var}(L)$ | M1 | |
| Use of $\text{Var}(aX) = a^2\text{Var}(X)$ and $\text{Var}(X_1 + X_2) = 2\text{Var}(X)$ | M1 | Implied by either correct term |
| Correct equivalent expression for $\text{Var}(L)$ | A1 | |
# Question 6 (c):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Finding correct limits as $n$ gets larger for each expression | M1 | Allow ft |
| Solving $\frac{2n-3}{9(n-2)} = \frac{2(2n+1)}{3n(n+1)} \rightarrow n = -0.53,\ 2.46,\ 4.57$ | | Allow comments relating to $n \geq 5\ (4.57)$ |
| Correct explanation | M1 | |
| Deducing that Korhan is the winner | A1 | Dependent upon both M marks |
---\begin{enumerate}
\item Korhan and Louise challenge each other to find an estimator for the mean, $\mu$, of the continuous random variable $X$ which has variance $\sigma ^ { 2 }$\\
$X _ { 1 } , X _ { 2 } , X _ { 3 } , \ldots , X _ { n }$ are $n$ independent observations taken from $X$\\
Korhan's estimator is given by
\end{enumerate}
$$K = \frac { 2 } { n ( n + 1 ) } \sum _ { r = 1 } ^ { n } r X _ { r }$$
Louise's estimator is given by
$$L = \frac { X _ { 1 } + X _ { 2 } } { 3 } + \frac { X _ { 3 } + X _ { 4 } + \ldots + X _ { n } } { 3 ( n - 2 ) }$$
(a) Show that $K$ and $L$ are both unbiased estimators of $\mu$\\
(b) (i) Find $\operatorname { Var } ( K )$\\
(ii) Find $\operatorname { Var } ( L )$
The winner of the challenge is the person who finds the better estimator.\\
(c) Determine the winner of the challenge for large values of $n$.
Give reasons for your answer.
\hfill \mbox{\textit{Edexcel FS2 2022 Q6 [15]}}