| Exam Board | Edexcel |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2011 |
| Session | June |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Central limit theorem |
| Type | Estimator properties and bias |
| Difficulty | Challenging +1.2 This is a structured S4 question on estimator properties with given pdf. Parts (a)-(c) involve standard integration and variance formula application. Parts (d)-(f) test understanding of consistency and efficiency concepts with straightforward reasoning. Part (g) is simple arithmetic. While it covers advanced statistics topics (order statistics, estimator comparison), the question provides the pdf and guides students through each step methodically, making it moderately above average difficulty but not requiring novel insight. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.05b Unbiased estimates: of population mean and variance |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(E(Y^m) = \frac{n}{\beta^n} \int y^m \times y^{n-1}\, dy = \left[\frac{n}{\beta^n} \times \frac{1}{m+n} \times y^{m+n}\right]_0^\beta\) | M1, A1 | M1 for attempt to integrate \(y^m f(m)\); 1st A1 for correct integration (limits not needed yet) |
| \(= \frac{n}{\beta^n} \times \frac{1}{m+n} \times \beta^{m+n} = \frac{n}{m+n}\beta^m\) | A1cso | 2nd A1 for use of correct limits and proceeding to printed answer; no incorrect working seen |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(E(Y) = \frac{n}{n+1}\beta\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(E(Y^2) = \frac{n}{n+2}\beta^2\), \(\quad \text{Var}(Y) = E(Y^2) - [E(Y)]^2\) | B1, M1 | M1 for use of their \(E(Y)\) and \(E(Y^2)\) in a correct formula for \(\text{Var}(Y)\) |
| \(\text{Var}(Y) = \frac{n}{n+2}\beta^2 - \frac{n^2}{(n+1)^2}\beta^2 = \frac{n}{(n+1)^2(n+2)}\beta^2\) | A1cso |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| As \(n \to \infty\), \(E(Y) \to \beta\), \(\text{Var}(Y) \to 0\) | M1, A1 | M1 for examining both \(E(Y)\) and \(\text{Var}(Y)\) for \(n \to \infty\); 1st A1 for correct limits for both |
| So \(Y\) is a consistent estimator for \(\beta\) | A1 | 2nd A1 for a correct statement following correct working |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(k = \frac{n+1}{n}\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\text{Var}(M) = 4\text{Var}(\bar{X}) = 4\frac{\sigma^2}{n} = \frac{4}{n} \times \frac{\beta^2}{12} = \frac{\beta^2}{3n}\) | B1 | |
| \(\frac{(n+1)^2}{n^2} \times \frac{n}{(n+1)^2(n+2)}\beta^2 = \frac{\beta^2}{n(n+2)} < \frac{\beta^2}{3n}\) so \(S\) is better \((n>1)\) | M1A1 | M1 for attempting \(\text{Var}(S)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\text{Max} = 9.1,\quad s = \frac{6}{5} \times 9.1 = \mathbf{10.9(2)}\) | M1A1 | M1 for correct use of \(S\) to find estimate |
# Question 6:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $E(Y^m) = \frac{n}{\beta^n} \int y^m \times y^{n-1}\, dy = \left[\frac{n}{\beta^n} \times \frac{1}{m+n} \times y^{m+n}\right]_0^\beta$ | M1, A1 | M1 for attempt to integrate $y^m f(m)$; 1st A1 for correct integration (limits not needed yet) |
| $= \frac{n}{\beta^n} \times \frac{1}{m+n} \times \beta^{m+n} = \frac{n}{m+n}\beta^m$ | A1cso | 2nd A1 for use of correct limits and proceeding to printed answer; no incorrect working seen |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $E(Y) = \frac{n}{n+1}\beta$ | B1 | |
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $E(Y^2) = \frac{n}{n+2}\beta^2$, $\quad \text{Var}(Y) = E(Y^2) - [E(Y)]^2$ | B1, M1 | M1 for use of their $E(Y)$ and $E(Y^2)$ in a correct formula for $\text{Var}(Y)$ |
| $\text{Var}(Y) = \frac{n}{n+2}\beta^2 - \frac{n^2}{(n+1)^2}\beta^2 = \frac{n}{(n+1)^2(n+2)}\beta^2$ | A1cso | |
## Part (d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| As $n \to \infty$, $E(Y) \to \beta$, $\text{Var}(Y) \to 0$ | M1, A1 | M1 for examining both $E(Y)$ and $\text{Var}(Y)$ for $n \to \infty$; 1st A1 for correct limits for both |
| So $Y$ is a consistent estimator for $\beta$ | A1 | 2nd A1 for a correct statement following correct working |
## Part (e):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $k = \frac{n+1}{n}$ | B1 | |
## Part (f):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\text{Var}(M) = 4\text{Var}(\bar{X}) = 4\frac{\sigma^2}{n} = \frac{4}{n} \times \frac{\beta^2}{12} = \frac{\beta^2}{3n}$ | B1 | |
| $\frac{(n+1)^2}{n^2} \times \frac{n}{(n+1)^2(n+2)}\beta^2 = \frac{\beta^2}{n(n+2)} < \frac{\beta^2}{3n}$ so $S$ is better $(n>1)$ | M1A1 | M1 for attempting $\text{Var}(S)$ |
## Part (g):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\text{Max} = 9.1,\quad s = \frac{6}{5} \times 9.1 = \mathbf{10.9(2)}$ | M1A1 | M1 for correct use of $S$ to find estimate |
---\begin{enumerate}
\item A random sample $X _ { 1 } , X _ { 2 } , \ldots , X _ { n }$ is taken from a population where each of the $X _ { i }$ have a continuous uniform distribution over the interval $[ 0 , \beta ]$.\\
The random variable $Y = \max \left\{ X _ { 1 } , X _ { 2 } , \ldots , X _ { n } \right\}$.\\
The probability density function of $Y$ is given by
\end{enumerate}
$$f ( y ) = \left\{ \begin{array} { c c }
\frac { n } { \beta ^ { n } } y ^ { n - 1 } & 0 \leqslant y \leqslant \beta \\
0 & \text { otherwise }
\end{array} \right.$$
(a) Show that $\mathrm { E } \left( Y ^ { m } \right) = \frac { n } { n + m } \beta ^ { m }$.\\
(b) Write down $\mathrm { E } ( Y )$.\\
(c) Using your answers to parts (a) and (b), or otherwise, show that
$$\operatorname { Var } ( Y ) = \frac { n } { ( n + 1 ) ^ { 2 } ( n + 2 ) } \beta ^ { 2 }$$
(d) State, giving your reasons, whether or not $Y$ is a consistent estimator of $\beta$.
The random variables $M = 2 \bar { X }$, where $\bar { X } = \frac { 1 } { n } \left( X _ { 1 } + X _ { 2 } + \ldots + X _ { n } \right)$, and $S = k Y$, where $k$ is a constant, are both unbiased estimators of $\beta$.\\
(e) Find the value of $k$ in terms of $n$.\\
(f) State, giving your reasons, which of $M$ and $S$ is the better estimator of $\beta$ in this case.
Five observations of $X$ are: $\quad \begin{array} { l l l l l } 8.5 & 6.3 & 5.4 & 9.1 & 7.6 \end{array}$\\
(g) Calculate the better estimate of $\beta$.\\
\hfill \mbox{\textit{Edexcel S4 2011 Q6 [16]}}