S4 June 2018 Q6
6. The continuous random variable \(X\) has probability density function \(\mathrm{f}(x)\)
\[\mathrm{f}(x) = \begin{cases} \dfrac{x}{2\theta^2} & 0 \leqslant x \leqslant 2\theta \\ 0 & \text{otherwise} \end{cases}\]where \(\theta\) is a constant.
A random sample \(X_1, X_2, \ldots, X_n\) where \(n \geqslant 2\) is taken to estimate the value of \(\theta\)
The random variable \(S_1 = q\bar{X}\) is an unbiased estimator of \(\theta\)
The continuous random variable \(Y\) is independent of \(X\) and is uniformly distributed over the interval \(\left[0, \dfrac{2\theta}{3}\right]\), where \(\theta\) is the same unknown constant as in \(\mathrm{f}(x)\).
The random variable \(S_2 = aX + bY\) is an unbiased estimator of \(\theta\) and is based on one observation of \(X\) and one observation of \(Y\).
| Scheme | Marks |
|---|---|
| \(\mathrm{E}(X^N) = \displaystyle\int_0^{2\theta} \dfrac{x^{N+1}}{2\theta^2}\,\mathrm{d}x\) | M1 |
| \(= \left[\dfrac{x^{N+2}}{2(N + 2)\theta^2}\right]_0^{2\theta}\) | A1 |
| \(= \dfrac{(2\theta)^{N+2}}{2(N + 2)\theta^2}\) | |
| \(= \dfrac{2^{N+1}}{N + 2}\theta^N\ \ (*)\) | A1cso |
| (3) |
Notes
M1 attempting to integrate \(\dfrac{x^{N+1}}{2\theta^2}\), \(x^{N+1} \to x^{N+2}\) condone missing limits
A1 correct integration
A1 fully correct solution – must see substitution of \(2\theta\)
| Scheme | Marks |
|---|---|
| \(\mathrm{E}(X) = \dfrac{4\theta}{3}\) | B1 |
| \(\mathrm{Var}(X) = 2\theta^2 - \left(\text{"}\dfrac{4\theta}{3}\text{"}\right)^2 \qquad = \dfrac{2\theta^2}{9}\) | M1A1 |
| (3) |
Notes
B1 must have \(\mathrm{E}(X) =\)
M1 allow their \(\mathrm{E}(X)\) if one has been given otherwise must be correct in here
A1 must be using part (a), do not allow if integrated from scratch.
| Scheme | Marks |
|---|---|
| \(q = \dfrac{3}{4}\) | B1 |
| \(\mathrm{Var}(S_1) = \dfrac{9}{16} \times \dfrac{\text{"}2\theta^2\text{"}}{9n}\) | M1 |
| \(= \dfrac{\theta^2}{8n}\) as \(n \to \infty\) \(\mathrm{Var}(S) \to 0\ \therefore\) s[ince it is unbiased] it is a consistent estimator | A1cso |
| (3) |
Notes
M1 for \(\dfrac{9}{16} \times \dfrac{\text{their Var}(X)}{n}\)
A1 cso and for as \(n \to \infty\) \(\mathrm{Var}(S) \to 0\ \therefore\) since it is unbiased it is a consistent estimator
| Scheme | Marks |
|---|---|
| \(\mathrm{E}(S_2) = a \times \text{"}\dfrac{4\theta}{3}\text{"} + b \times \dfrac{\theta}{3}\) | M1 |
| \(a \times \dfrac{4\theta}{3} + b \times \dfrac{\theta}{3} = \theta\) or \(4a + b = 3\) | A1 |
| \(\mathrm{Var}(S_2) = a^2 \times \text{"}\dfrac{2\theta^2}{9}\text{"} + b^2 \times \dfrac{\theta^2}{27}\) | M1 |
| \(\mathrm{Var}(S_2) = a^2 \times \dfrac{2\theta^2}{9} + (3 - 4a)^2 \times \dfrac{\theta^2}{27}\) or \(\mathrm{Var}(S_2) = \left(\dfrac{3 - b}{4}\right)^2 \times \dfrac{2\theta^2}{9} + b^2 \times \dfrac{\theta^2}{27}\) | M1 |
| \(\dfrac{\mathrm{d}\mathrm{Var}(S_2)}{\mathrm{d}a} = \dfrac{4a\theta^2}{9} - \dfrac{8(3 - 4a)\theta^2}{27}\) or \(\dfrac{-(3 - b)\theta^2}{36} + \dfrac{2b\theta^2}{27}\) | M1 |
| \(\dfrac{4a\theta^2}{9} - \dfrac{8(3 - 4a)\theta^2}{27} = 0\) or \(\dfrac{-(3 - b)\theta^2}{36} + \dfrac{2b\theta^2}{27} = 0\) | M1 |
| \(\dfrac{44a}{27} = \dfrac{24}{27}\) or \(\dfrac{11}{108}b = \dfrac{1}{12}\) | |
| \(a = \dfrac{6}{11},\ \ b = \dfrac{9}{11}\) | A1 |
| (7) |
Notes
M1 for \(a \times \text{their } \mathrm{E}(X) + b \times \dfrac{\theta}{3}\)
A1 a correct equation with no \(\theta\)
M1 \(a^2 \times \text{their Var}(X) + b^2 \times \dfrac{\theta^2}{27}\)
M1 subst in for \(a\) or \(b\)
M1 differentiating with respect to \(a\) or \(b\)
M1 putting dVar/d\(a\) = 0 and solving leading to \(a = \ldots\) or \(b = \ldots\)
A1 allow awrt 0.545 and awrt 0.818
| Scheme | Marks |
|---|---|
| \(\mathrm{Var}(S_2) = \left(\text{"}\dfrac{6}{11}\text{"}\right)^2 \times \dfrac{2\theta^2}{9} + \left(\text{"}\dfrac{9}{11}\text{"}\right)^2 \times \dfrac{\theta^2}{27}\) | M1 |
| \(\mathrm{Var}(S_2) = \dfrac{\theta^2}{11}\) | |
| (1) |
Notes
M1 subst \(a\) and \(b\) in to find \(\mathrm{Var}(S_2)\)
| Scheme | Marks |
|---|---|
| \(S_1\) is the better estimator when \(\dfrac{\theta^2}{8n} \lt \dfrac{\theta^2}{11} \Rightarrow n \gt \dfrac{11}{8}\) | M1 |
| \(S_2\) is the better estimator when \(n \lt \dfrac{11}{8}\) | |
| Therefore \(S_1\) is the better estimator since \(n \geqslant 2\) | A1cso |
| (2) | |
| (19 marks) |
Notes
M1 for reason \(\dfrac{\theta^2}{8n} \lt \dfrac{\theta^2}{11} \Rightarrow n \gt \dfrac{11}{8}\) or \(\mathrm{Var}(S_1) \leqslant \dfrac{\theta^2}{16} \lt \dfrac{\theta^2}{11}\)
A1cso correct selection