S4 June 2015 Q6
6. A random sample \(X_1, X_2, X_3, \ldots, X_{2n}\) is taken from a population with mean \(\dfrac{\mu}{3}\) and variance \(3\sigma^2\). A second random sample \(Y_1, Y_2, Y_3, \ldots, Y_n\) is taken from a population with mean \(\dfrac{\mu}{2}\) and variance \(\dfrac{\sigma^2}{2}\), where the \(X\) and \(Y\) variables are all independent.
\(A\), \(B\) and \(C\) are possible estimators of \(\mu\), where
\[A = \frac{X_1 + X_2 + X_3 + Y_1 + Y_2}{2}\] \[B = \frac{3X_1}{2} + \frac{2Y_1}{3}\] \[C = \frac{3X_1 + 4Y_1}{3}\]The estimator
\[D = \frac{1}{k}\left(\sum_{i=1}^{2n} X_i + \sum_{i=1}^{n} Y_i\right)\]is an unbiased estimator of \(\mu\).
| Scheme | Marks |
|---|---|
| \(\mathrm{E}[A] = \dfrac{1}{2}\left(\mathrm{E}[X_1] + \mathrm{E}[X_2] + \mathrm{E}[X_3] + \mathrm{E}[Y_1] + \mathrm{E}[Y_2]\right) = \dfrac{1}{2}\left(3 \times \dfrac{\mu}{3} + 2 \times \dfrac{\mu}{2}\right) = \mu\) | M1 |
| Therefore \(A\) is an unbiased estimator | A1 |
| \(\mathrm{E}[B] = \dfrac{3\mathrm{E}[X_1]}{2} + \dfrac{2\mathrm{E}[Y_1]}{3} = \dfrac{3}{2} \times \dfrac{\mu}{3} + \dfrac{2}{3} \times \dfrac{\mu}{2} = \dfrac{5\mu}{6}\) | A1 |
| Therefore \(B\) is biased with bias \((-)\dfrac{\mu}{6}\) | B1ft |
| \(\mathrm{E}[C] = \dfrac{1}{3}\left(3\mathrm{E}[X_1] + 4\mathrm{E}[Y_1]\right) = \dfrac{1}{3}\left(\dfrac{3\mu}{3} + \dfrac{4\mu}{2}\right) = \mu\) Therefore \(C\) is an unbiased estimator | A1 |
| (5) |
Notes
M1 for a correct method for E(A) or E(B) or E(C)
A1 for each correct expectation with a correct method
B1ft bias of B, condone missing – sign. Do not allow a bias of 0
| Scheme | Marks |
|---|---|
| Best estimator is unbiased estimator with least variance \(\mathrm{Var}(A) = \dfrac{1}{4}\left(\mathrm{Var}\,X_1 + \mathrm{Var}\,X_2 + \mathrm{Var}\,X_3 + \mathrm{Var}\,Y_1 + \mathrm{Var}\,Y_2\right)\) | M1 |
| \(= \dfrac{1}{4}\left(3 \times 3\sigma^2 + 2 \times \dfrac{\sigma^2}{2}\right) = \dfrac{5\sigma^2}{2}\) | A1 |
| \(\mathrm{Var}(C) = \dfrac{1}{9}\left(9\,\mathrm{Var}\,X_1 + 16\,\mathrm{Var}\,Y_1\right) = \dfrac{1}{9}\left(9 \times 3\sigma^2 + 16 \times \dfrac{\sigma^2}{2}\right) = \dfrac{35\sigma^2}{9}\) | A1 |
| Therefore \(A\) is a better estimator of \(\mu\) (smaller variance) | B1dft |
| (4) |
Notes
M1 Use of \(\mathrm{Var}(aX) = a^2\mathrm{Var}(X)\) and subst \(3\sigma^2\) for \(\mathrm{Var}(X)\) and \(\dfrac{\sigma^2}{2}\) for \(\mathrm{Var}(Y)\)
A1 for each correct variance
B1dft their variances. Dep on m1 being awarded. If no variances given then B0
| Scheme | Marks |
|---|---|
| \(\mathrm{E}[D] = \dfrac{1}{k}\left(2n \times \dfrac{\mu}{3} + n \times \dfrac{\mu}{2}\right) = \mu\) | M1A1 |
| \(k = \dfrac{2n}{3} + \dfrac{n}{2} = \dfrac{7n}{6}\) | A1 |
| (3) |
Notes
M1 attempts E(\(D\)) and puts = to \(\mu\) (may be implied)
A1 for E(\(D\))
| Scheme | Marks |
|---|---|
| \(\mathrm{Var}(D) = \dfrac{1}{k^2}\left(2n \times 3\sigma^2 + n \times \dfrac{\sigma^2}{2}\right) = \dfrac{1}{k^2} \times \dfrac{13n\sigma^2}{2}\) | M1 |
| \(\mathrm{Var}(D) = \dfrac{36}{49n^2} \times \dfrac{13n\sigma^2}{2} = \dfrac{234\sigma^2}{49n}\) | M1d A1 |
| Therefore \(\mathrm{Var}\,D \to 0\) as \(n \to \infty\), therefore \(D\) is a consistent estimator | A1dd |
| (4) |
Notes
M1 for \(\dfrac{1}{k^2}\left(2n \times 3\sigma^2 + n \times \dfrac{\sigma^2}{2}\right)\) or \(\dfrac{1}{k^2} \times \dfrac{13n\sigma^2}{2}\)
M1d for subst in \(k\)
A1 Correct Var (\(D\))
A1dd Need correct reason for being a consistent estimator dep on previous method marks being awarded
| Scheme | Marks |
|---|---|
| Want \(\dfrac{234\sigma^2}{49n} \lt \dfrac{5\sigma^2}{2}\) Therefore \(\dfrac{234}{49} \times \dfrac{2}{5} \lt n\) | M1 |
| \(n \gt 1.910\ldots\) So minimum value is \(n = 2\) | A1cso |
| (2) | |
| (18 marks) |
Notes
M1 for forming an inequality with their Var(\(D\)) < their best estimator leading to \(n\)