| Exam Board | Edexcel |
|---|---|
| Module | FS2 (Further Statistics 2) |
| Year | 2021 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Probability Generating Functions |
| Type | Moment generating function problems |
| Difficulty | Challenging +1.2 This is a standard Further Statistics 2 question on estimator properties requiring calculation of bias and variance for given linear estimators. While it involves multiple parts and careful algebraic manipulation, the techniques are routine for FS2: finding E(estimator) and Var(estimator) using linearity of expectation and independence. The uniform distribution makes calculations straightforward, and no novel insight is required—just systematic application of learned methods. |
| Spec | 5.05b Unbiased estimates: of population mean and variance |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \([E(X) = 2\beta]\); \(E(A) = E\!\left(\frac{X_1+X_2}{2}\right) = \frac{1}{2}[E(X)+E(X)]\); \(E(B) = \frac{1}{8}[E(X)+2E(X)+3E(X)]\); \(E(C) = \frac{1}{8}[E(X)+2E(X)-E(X)]\) | M1 | Using independence to calculate \(E(A)\), \(E(B)\) or \(E(C)\); Use of bias \(= E(X) - \beta\) |
| Bias for \(A = E(A) - \beta = 2\beta - \beta = \beta\) | A1 | Correct bias for \(A\) |
| Bias for \(B = E(B) - \beta = 1.5\beta - \beta = 0.5\beta\) | A1 | Correct bias for \(B\) |
| Bias for \(C = E(C) - \beta = 0.5\beta - \beta = -0.5\beta\) | A1 | Correct bias for \(C\) [allow \(+0.5\beta\)] |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\left[\text{Var}(X) = \frac{4}{3}\beta^2\right]\); \(\text{Var}(B) = \text{Var}\!\left(\frac{X_1+2X_2+3X_3}{8}\right) = \frac{1}{64}[\text{Var}(X)+4\text{Var}(X)+9\text{Var}(X)]\) | M1 | Realising variances need to be compared and attempt at linear combination of variances for \(B\) or \(C\) |
| \(\text{Var}(C) = \frac{1}{64}[\text{Var}(X)+4\text{Var}(X)+\text{Var}(X)]\) | (M1 continued) | |
| \(\text{Var}(B) = \frac{7}{32}\text{Var}(X) = \frac{7}{24}\beta^2\) | A1 | Correct \(\text{Var}(B)\) |
| \(\text{Var}(C) = \frac{3}{32}\text{Var}(X) = \frac{1}{8}\beta^2\) | A1 | Correct \(\text{Var}(C)\) |
| Since both have same bias and \(\text{Var}(C) < \text{Var}(B)\), therefore \(C\) is the better estimator | B1ft | Correct comparison and deduction that \(C\) is better than \(A\) and \(B\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| Any unbiased estimator, e.g. \(\frac{X_1+X_2+X_3}{6}\) | B1 | Allow any unbiased estimator, e.g. \(\frac{X_1}{2}\) |
# Question 5:
## Part (a):
| Working | Mark | Guidance |
|---------|------|----------|
| $[E(X) = 2\beta]$; $E(A) = E\!\left(\frac{X_1+X_2}{2}\right) = \frac{1}{2}[E(X)+E(X)]$; $E(B) = \frac{1}{8}[E(X)+2E(X)+3E(X)]$; $E(C) = \frac{1}{8}[E(X)+2E(X)-E(X)]$ | M1 | Using independence to calculate $E(A)$, $E(B)$ or $E(C)$; Use of bias $= E(X) - \beta$ |
| Bias for $A = E(A) - \beta = 2\beta - \beta = \beta$ | A1 | Correct bias for $A$ |
| Bias for $B = E(B) - \beta = 1.5\beta - \beta = 0.5\beta$ | A1 | Correct bias for $B$ |
| Bias for $C = E(C) - \beta = 0.5\beta - \beta = -0.5\beta$ | A1 | Correct bias for $C$ [allow $+0.5\beta$] |
## Part (b):
| Working | Mark | Guidance |
|---------|------|----------|
| $\left[\text{Var}(X) = \frac{4}{3}\beta^2\right]$; $\text{Var}(B) = \text{Var}\!\left(\frac{X_1+2X_2+3X_3}{8}\right) = \frac{1}{64}[\text{Var}(X)+4\text{Var}(X)+9\text{Var}(X)]$ | M1 | Realising variances need to be compared and attempt at linear combination of variances for $B$ or $C$ |
| $\text{Var}(C) = \frac{1}{64}[\text{Var}(X)+4\text{Var}(X)+\text{Var}(X)]$ | (M1 continued) | |
| $\text{Var}(B) = \frac{7}{32}\text{Var}(X) = \frac{7}{24}\beta^2$ | A1 | Correct $\text{Var}(B)$ |
| $\text{Var}(C) = \frac{3}{32}\text{Var}(X) = \frac{1}{8}\beta^2$ | A1 | Correct $\text{Var}(C)$ |
| Since both have same bias and $\text{Var}(C) < \text{Var}(B)$, therefore $C$ is the better estimator | B1ft | Correct comparison and deduction that $C$ is better than $A$ and $B$ |
## Part (c):
| Working | Mark | Guidance |
|---------|------|----------|
| Any unbiased estimator, e.g. $\frac{X_1+X_2+X_3}{6}$ | B1 | Allow any unbiased estimator, e.g. $\frac{X_1}{2}$ |
---\begin{enumerate}
\item The continuous random variable $X$ is uniformly distributed over the interval $[ 0,4 \beta ]$, where $\beta$ is an unknown constant.
\end{enumerate}
Three independent observations, $X _ { 1 } , X _ { 2 }$ and $X _ { 3 }$, are taken of $X$ and the following estimators for $\beta$ are proposed
$$\begin{aligned}
& A = \frac { X _ { 1 } + X _ { 2 } } { 2 } \\
& B = \frac { X _ { 1 } + 2 X _ { 2 } + 3 X _ { 3 } } { 8 } \\
& C = \frac { X _ { 1 } + 2 X _ { 2 } - X _ { 3 } } { 8 }
\end{aligned}$$
(a) Calculate the bias of $A$, the bias of $B$ and the bias of $C$\\
(b) By calculating the variances, explain which of $B$ or $C$ is the better estimator for $\beta$\\
(c) Find an unbiased estimator for $\beta$
\hfill \mbox{\textit{Edexcel FS2 2021 Q5 [10]}}