S4 June 2008 Q1
1. A random sample \(X_1, X_2, \ldots, X_{10}\) is taken from a population with mean \(\mu\) and variance \(\sigma^2\).
(a) Determine the bias, if any, of each of the following estimators of \(\mu\).\[\theta_1 = \frac{X_3 + X_4 + X_5}{3},\]\[\theta_2 = \frac{X_{10} - X_1}{3},\]\[\theta_3 = \frac{3X_1 + 2X_2 + X_{10}}{6}.\] (4)
(b) Find the variance of each of these estimators. (5)
(c) State, giving reasons, which of these three estimators for \(\mu\) is
(i) the best estimator,
(ii) the worst estimator. (4)
| Scheme | Marks |
|---|---|
| \(\mathrm{E}(\theta_1) = \dfrac{\mathrm{E}(X_3) + \mathrm{E}(X_4) + \mathrm{E}(X_5)}{3}\) \(= \dfrac{3\mu}{3}\) \(= \mu \qquad \text{Bias} = 0\) allow unbiased | B1 |
| \(\mathrm{E}(\theta_2) = \dfrac{\mathrm{E}(X_{10}) - \mathrm{E}(X_1)}{3}\) \(= 1/3(\mu - \mu)\) \(= 0 \qquad \text{Bias} = -\mu\) allow \(\pm\mu\) | B1, B1 |
| \(\mathrm{E}(\theta_3) = \dfrac{3\mathrm{E}(X_1) + 2\mathrm{E}(X_2) + \mathrm{E}(X_{10})}{6}\) \(= \dfrac{3\mu + 2\mu + \mu}{6}\) \(= \mu \qquad \text{Bias} = 0\) allow unbiased | B1 |
| (4) |
| Scheme | Marks |
|---|---|
| \(\mathrm{Var}(\theta_1) = \dfrac{1}{9}\{\mathrm{Var}(X_3) + \mathrm{Var}(X_4) + \mathrm{Var}(X_5)\}\) | M1 |
| \(= \dfrac{1}{9}\{\sigma^2 + \sigma^2 + \sigma^2\}\) \(= \dfrac{1}{3}\sigma^2\) | A1 |
| \(\mathrm{Var}(\theta_2) = \dfrac{2}{9}\sigma^2\) | B1 |
| \(\mathrm{Var}(\theta_3) = \dfrac{1}{36}\{9\sigma^2 + 4\sigma^2 + \sigma^2\}\) | M1 |
| \(= \dfrac{7}{18}\sigma^2\) | A1 |
| (5) |
Notes
(corrected from the printed mark scheme: the scheme prints \(\mathrm{Var}(\theta_1) = \frac{1}{9}\{\mathrm{Var}\,X_2 + \mathrm{Var}(X_3) + \mathrm{Var}(X_4)\}\); \(\theta_1\) uses \(X_3, X_4, X_5\))
| Scheme | Marks |
|---|---|
| (i) \(\theta_1\) is the better estimator. It has a lower var. and no bias | B1 depB1 |
| (ii) \(\theta_2\) is the worst estimator. It is biased | B1 depB1 |
| (4) |