| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/2 (Pre-U Further Mathematics Paper 2) |
| Year | 2020 |
| Session | Specimen |
| Marks | 3 |
| Topic | Linear combinations of normal random variables |
| Type | Pure expectation and variance calculation |
| Difficulty | Standard +0.3 This is a straightforward application of standard results for expectation and variance of linear combinations of independent normal variables. Part (a) requires simple algebra with E(aX̄+bȲ), part (b) uses Var(aX̄+bȲ)=a²Var(X̄)+b²Var(Ȳ) with substitution, and part (c) is basic calculus (differentiation to find minimum). All steps are routine for Further Maths students with no novel insight required, making it slightly easier than average. |
| Spec | 5.04a Linear combinations: E(aX+bY), Var(aX+bY)5.05b Unbiased estimates: of population mean and variance |
**Question 4(a):**
$E(a\bar{X} + b\bar{Y}) = \mu$ [M1]
$E(a\bar{X} + b\bar{Y}) = aE(\bar{X}) + bE(\bar{Y})$ [M1]
$\Rightarrow a\mu + 3b\mu = \mu \Rightarrow a + 3b = 1$ [A1]
**Total: 3 marks**
**Question 4(b):**
$\text{Var}(a\bar{X} + b\bar{Y}) = a^2\text{Var}(\bar{X}) + b^2\text{Var}(\bar{Y}) = a^2\dfrac{\sigma^2}{n} + 4b^2\dfrac{\sigma^2}{n}$ [M1]
$= \dfrac{\sigma^2}{n}(a^2 + 4b^2) = \dfrac{\sigma^2}{n}(1 - 6b + 9b^2 + 4b^2) = \dfrac{\sigma^2}{n}(1 - 6b + 13b^2)$ **AG** [M1A1]
**Total: 3 marks**
**Question 4(c):**
$\dfrac{\text{d}}{\text{d}b}\text{Var}(a\bar{X} + b\bar{Y}) = -6 + 26b = 0 \Rightarrow b = \dfrac{3}{13}$ [M1A1]
$\Rightarrow \text{Var}_{\min}(a\bar{X} + b\bar{Y}) = \dfrac{\sigma^2}{n}\left(1 - 6 \times \dfrac{3}{13} + 13 \times \dfrac{9}{169}\right) = \dfrac{4\sigma^2}{13n}$ [A1]
**Total: 3 marks**4 The independent random variables $X$ and $Y$ have normal distributions where $X \sim \mathrm {~N} \left( \mu , \sigma ^ { 2 } \right)$ and $Y \sim \mathrm {~N} \left( 3 \mu , 4 \sigma ^ { 2 } \right)$. Two random samples each of size $n$ are taken, one from each of these normal populations.
\begin{enumerate}[label=(\alph*)]
\item Show that $a \bar { X } + b \bar { Y }$ is an unbiased estimator of $\mu$ provided that $a + 3 b = 1$, where $a$ and $b$ are constants and $\bar { X }$ and $\bar { Y }$ are the respective sample means.
In the remainder of the question assume that $a \bar { X } + b \bar { Y }$ is an unbiased estimator of $\mu$.
\item Show that $\operatorname { Var } ( a \bar { X } + b \bar { Y } )$ can be written as $\frac { \sigma ^ { 2 } } { n } \left( 1 - 6 b + 13 b ^ { 2 } \right)$.
\item The value of the constant $b$ can be varied. Find the value of $b$ that gives the minimum of $\operatorname { Var } ( a \bar { X } + b \bar { Y } )$, and hence find the minimum of $\operatorname { Var } ( a \bar { X } + b \bar { Y } )$ in terms of $\sigma$ and $n$.
\end{enumerate}
\hfill \mbox{\textit{Pre-U Pre-U 9795/2 2020 Q4 [3]}}