| Exam Board | Edexcel |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2017 |
| Session | June |
| Marks | 19 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Random Variables |
| Type | Finding unbiased estimator constraints |
| Difficulty | Challenging +1.2 This is a structured S4 question on unbiased estimators with clear scaffolding through parts (a)-(e). Parts (a)-(c) involve standard expectation/variance algebra and basic calculus for optimization. Parts (d)-(e) require understanding bias and manipulating E[X(X-1)], which is a known technique in S4. While it's Further Maths content (making it inherently harder than core), the heavy scaffolding and standard techniques place it slightly above average difficulty overall. |
| Spec | 5.02b Expectation and variance: discrete random variables5.02c Linear coding: effects on mean and variance5.02d Binomial: mean np and variance np(1-p)5.05b Unbiased estimates: of population mean and variance |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(E\!\left(\dfrac{aX_1 + bX_2}{n}\right) = \dfrac{anp + bnp}{n} = ap + bp = (a+b)p\) | M1 | |
| \(a + b = 1\) | A1* cso |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\text{Var}\!\left(\dfrac{aX_1+bX_2}{n}\right) = \dfrac{1}{n^2}\!\left(a^2np(1-p)+b^2np(1-p)\right)\) | M1 A1 | |
| \(= \dfrac{p(1-p)(a^2+b^2)}{n}\) | ||
| \(= \dfrac{p(1-p)(a^2+(1-a)^2)}{n}\) | M1d | Using \(b = 1-a\) |
| \(= \dfrac{(2a^2-2a+1)p(1-p)}{n}\) | A1* cso |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Min value when \(\dfrac{(4a-2)p(1-p)}{n} = 0\) | M1A1 | |
| \(\Rightarrow 4a - 2 = 0\) | ||
| \(a = \dfrac{1}{2},\ b = \dfrac{1}{2}\) | A1A1ft | |
| \(\dfrac{d^2\text{Var}(\hat{p})}{da^2} = \dfrac{4p(1-p)}{n} > 0\) or quadratic with positive \(x^2\) \(\therefore\) minimum point or sketch | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(E\!\left(\dfrac{aX_1+bX_2}{n}\right)^2 = E\!\left(\dfrac{a^2X_1^2+b^2X_2^2+2abX_1X_2}{n^2}\right)\) | M1 | |
| \(= \dfrac{1}{n^2}\!\left(a^2np(1-p)+a^2n^2p^2+b^2np(1-p)+b^2n^2p^2+2abn^2p^2\right)\) | M1d | |
| \(= \dfrac{(a^2+b^2)p(1-p)}{n} + p^2(a+b)^2\) | ||
| \(= \dfrac{(a^2+b^2)p(1-p)}{n} + p^2\); \(> p^2\) since \(\dfrac{(a^2+b^2)p(1-p)}{n} > 0\) \(\therefore\) biased | A1; A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| As \(n \to \infty\), \(E(\hat{p}^2) \to p^2\), therefore bias \(\to 0\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(E(X_1(X_1-1)) = E(X_1^2) - E(X_1)\) | M1 | |
| \(= np(1-p) + n^2p^2 - np\) | ||
| \(= np - np^2 + n^2p^2 - np = np^2(n-1)\) | A1 | |
| Unbiased estimator \(= \dfrac{X_1(X_1-1)}{n(n-1)}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Using \(\frac{aE(X_1) + bE(X_2)}{n}\) and substituting \(E(X_1) = np\) and \(E(X_2) = np\) | M1 | |
| Need \(p(a+b) = p\) and statement \(a + b = 1\) with no errors | A1* | Answer given (cso) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Using \(\frac{a^2\text{Var}(X_1) + b^2\text{Var}(X_2)}{n^2}\) and substituting \(\text{Var}(X_1) = np(1-p)\) | M1 | May be implied by \(\frac{1}{n^2}(a^2np(1-p) + b^2np(1-p))\) |
| Correct answer in any form | A1 | |
| Substitute \(b = 1 - a\) | M1d | Dependent on 1st M1 |
| Method must be shown, no errors | A1* | cso |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\frac{d}{da}(\text{Var})\) | M1 | Must differentiate with respect to \(a\), or attempt to complete the square |
| Correct diff \(= 0\) or \(2\left(a - \frac{1}{2}\right)^2 + \frac{1}{2}\) | A1 | |
| \(a = 0.5\) | A1 | |
| \(b = 1 - a\) | A1 | Follow through |
| Reason why minimum | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Multiplying out and using \(E(aX) = aE(X)\) | M1 | May use their values of \(a\) and \(b\) |
| Using \(E(X^2) = \text{Var}(X) + [E(X)]^2\) | M1d | Dependent on previous M mark |
| \(\frac{(2a^2 - 2a+1)p(1-p)}{n} + p^2\) or \(\frac{(a^2+b^2)p(1-p)}{n} + p^2\) | A1 | Must be of form \(p^2\) + a single term |
| Reason why not equal to \(p^2\), plus statement so biased | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Follow on from expression \(p^2 + \ldots\) with \(a\) and \(b\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Multiplying out correctly, substituting \(np\) for \(E(X)\), or using \(E(\hat{p})^2 = \text{Var}(\hat{p}) + [E(\hat{p})]^2\) | M1 | |
| Allow \(= \frac{(2a^2 - 2a+1)p(1-p)}{n} + p^2\) | ||
| \(np^2(n-1)\) | A1 | |
| \(\frac{X_1(X_1-1)}{n(n-1)}\) | A1 | |
| NB \(\frac{X_1(X_1-1)}{n(n-1)}\) gains all 3 marks |
# Question 6:
## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $E\!\left(\dfrac{aX_1 + bX_2}{n}\right) = \dfrac{anp + bnp}{n} = ap + bp = (a+b)p$ | M1 | |
| $a + b = 1$ | A1* cso | |
## Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\text{Var}\!\left(\dfrac{aX_1+bX_2}{n}\right) = \dfrac{1}{n^2}\!\left(a^2np(1-p)+b^2np(1-p)\right)$ | M1 A1 | |
| $= \dfrac{p(1-p)(a^2+b^2)}{n}$ | | |
| $= \dfrac{p(1-p)(a^2+(1-a)^2)}{n}$ | M1d | Using $b = 1-a$ |
| $= \dfrac{(2a^2-2a+1)p(1-p)}{n}$ | A1* cso | |
## Part (c)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Min value when $\dfrac{(4a-2)p(1-p)}{n} = 0$ | M1A1 | |
| $\Rightarrow 4a - 2 = 0$ | | |
| $a = \dfrac{1}{2},\ b = \dfrac{1}{2}$ | A1A1ft | |
| $\dfrac{d^2\text{Var}(\hat{p})}{da^2} = \dfrac{4p(1-p)}{n} > 0$ or quadratic with positive $x^2$ $\therefore$ minimum point or sketch | B1 | |
## Part (d)(i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $E\!\left(\dfrac{aX_1+bX_2}{n}\right)^2 = E\!\left(\dfrac{a^2X_1^2+b^2X_2^2+2abX_1X_2}{n^2}\right)$ | M1 | |
| $= \dfrac{1}{n^2}\!\left(a^2np(1-p)+a^2n^2p^2+b^2np(1-p)+b^2n^2p^2+2abn^2p^2\right)$ | M1d | |
| $= \dfrac{(a^2+b^2)p(1-p)}{n} + p^2(a+b)^2$ | | |
| $= \dfrac{(a^2+b^2)p(1-p)}{n} + p^2$; $> p^2$ since $\dfrac{(a^2+b^2)p(1-p)}{n} > 0$ $\therefore$ biased | A1; A1 | |
## Part (d)(ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| As $n \to \infty$, $E(\hat{p}^2) \to p^2$, therefore bias $\to 0$ | B1 | |
## Part (e)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $E(X_1(X_1-1)) = E(X_1^2) - E(X_1)$ | M1 | |
| $= np(1-p) + n^2p^2 - np$ | | |
| $= np - np^2 + n^2p^2 - np = np^2(n-1)$ | A1 | |
| Unbiased estimator $= \dfrac{X_1(X_1-1)}{n(n-1)}$ | A1 | |
## Question (a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Using $\frac{aE(X_1) + bE(X_2)}{n}$ and substituting $E(X_1) = np$ and $E(X_2) = np$ | M1 | |
| Need $p(a+b) = p$ and statement $a + b = 1$ with no errors | A1* | Answer given (cso) |
---
## Question (b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Using $\frac{a^2\text{Var}(X_1) + b^2\text{Var}(X_2)}{n^2}$ and substituting $\text{Var}(X_1) = np(1-p)$ | M1 | May be implied by $\frac{1}{n^2}(a^2np(1-p) + b^2np(1-p))$ |
| Correct answer in any form | A1 | |
| Substitute $b = 1 - a$ | M1d | Dependent on 1st M1 |
| Method must be shown, no errors | A1* | cso |
---
## Question (c):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\frac{d}{da}(\text{Var})$ | M1 | Must differentiate with respect to $a$, or attempt to complete the square |
| Correct diff $= 0$ or $2\left(a - \frac{1}{2}\right)^2 + \frac{1}{2}$ | A1 | |
| $a = 0.5$ | A1 | |
| $b = 1 - a$ | A1 | Follow through |
| Reason why minimum | B1 | |
---
## Question (d)(i):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Multiplying out and using $E(aX) = aE(X)$ | M1 | May use their values of $a$ and $b$ |
| Using $E(X^2) = \text{Var}(X) + [E(X)]^2$ | M1d | Dependent on previous M mark |
| $\frac{(2a^2 - 2a+1)p(1-p)}{n} + p^2$ or $\frac{(a^2+b^2)p(1-p)}{n} + p^2$ | A1 | Must be of form $p^2$ + a single term |
| Reason why not equal to $p^2$, plus statement so biased | A1 | |
## Question (d)(ii):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Follow on from expression $p^2 + \ldots$ with $a$ and $b$ | B1 | |
---
## Question (e):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Multiplying out correctly, substituting $np$ for $E(X)$, or using $E(\hat{p})^2 = \text{Var}(\hat{p}) + [E(\hat{p})]^2$ | M1 | |
| Allow $= \frac{(2a^2 - 2a+1)p(1-p)}{n} + p^2$ | | |
| $np^2(n-1)$ | A1 | |
| $\frac{X_1(X_1-1)}{n(n-1)}$ | A1 | |
| NB $\frac{X_1(X_1-1)}{n(n-1)}$ gains all 3 marks | | |6. The independent random variables $X _ { 1 }$ and $X _ { 2 }$ are each distributed $\mathrm { B } ( n , p )$, where $n > 1$ An unbiased estimator for $p$ is given by
$$\hat { p } = \frac { a X _ { 1 } + b X _ { 2 } } { n }$$
where $a$ and $b$ are constants.\\[0pt]
[You may assume that if $X _ { 1 }$ and $X _ { 2 }$ are independent then $\mathrm { E } \left( X _ { 1 } X _ { 2 } \right) = \mathrm { E } \left( X _ { 1 } \right) \mathrm { E } \left( X _ { 2 } \right)$ ]
\begin{enumerate}[label=(\alph*)]
\item Show that $a + b = 1$
\item Show that $\operatorname { Var } ( \hat { p } ) = \frac { \left( 2 a ^ { 2 } - 2 a + 1 \right) p ( 1 - p ) } { n }$
\item Hence, justifying your answer, determine the value of $a$ and the value of $b$ for which $\hat { p }$ has minimum variance.
\item \begin{enumerate}[label=(\roman*)]
\item Show that $\hat { p } ^ { 2 }$ is a biased estimator for $p ^ { 2 }$
\item Show that the bias $\rightarrow 0$ as $n \rightarrow \infty$
\end{enumerate}\item By considering $\mathrm { E } \left[ X _ { 1 } \left( X _ { 1 } - 1 \right) \right]$ find an unbiased estimator for $p ^ { 2 }$
\end{enumerate}
\hfill \mbox{\textit{Edexcel S4 2017 Q6 [19]}}