A2 June 2022 Q6
6. Korhan and Louise challenge each other to find an estimator for the mean, \(\mu\), of the continuous random variable \(X\) which has variance \(\sigma^2\)
\(X_1, X_2, X_3, \ldots, X_n\) are \(n\) independent observations taken from \(X\)
Korhan’s estimator is given by
\[K = \frac{2}{n(n+1)}\sum_{r=1}^{n} rX_r\]Louise’s estimator is given by
\[L = \frac{X_1 + X_2}{3} + \frac{X_3 + X_4 + \ldots + X_n}{3(n-2)}\]The winner of the challenge is the person who finds the better estimator.
Give reasons for your answer. (3)
| Scheme | Marks | AO |
|---|---|---|
| \(\mathrm{E}(K) = \dfrac{2\mathrm{E}(X)}{n(n+1)} + \dfrac{4\mathrm{E}(X)}{n(n+1)} + \dfrac{6\mathrm{E}(X)}{n(n+1)} + \ldots + \dfrac{2n\mathrm{E}(X)}{n(n+1)}\) oe | M1 | 3.1a |
| \(= \dfrac{1}{n(n+1)}\left(\displaystyle\sum_{r=1}^{n} 2r\right)\mathrm{E}(X) = \dfrac{1}{n(n+1)}\left(\dfrac{(2+2n)n}{2}\right)\mathrm{E}(X)\ (= \mathrm{E}(X))\) | M1 | 2.1 |
| \(= \mu\) (therefore \(K\) is an unbiased estimator of \(\mu\)) | A1cso | 1.1b |
| \(\mathrm{E}(L) = \dfrac{2\mathrm{E}(X)}{3} + \dfrac{(n-2)\mathrm{E}(X)}{3(n-2)} = \dfrac{2\mathrm{E}(X)}{3} + \dfrac{1\mathrm{E}(X)}{3}\ (= \mathrm{E}(X))\) | M1 | 1.1b |
| \(= \mu\) (therefore \(L\) is an unbiased estimator of \(\mu\)) | A1cso | 1.1b |
| (5) |
Notes
M1: Using independence to set up expression for \(\mathrm{E}(K)\)
M1: Use of \(\displaystyle\sum_{r=1}^{n} r \left(= \frac{n(n+1)}{2}\right)\)
A1cso: Correct conclusion from correct working
M1: Using independence to set up expression for \(\mathrm{E}(L)\)
A1cso: Correct conclusion from correct working
| Scheme | Marks | AO |
|---|---|---|
| (i) \(\mathrm{Var}(K) = \mathrm{Var}\left(\dfrac{2X}{n(n+1)}\right) + \mathrm{Var}\left(\dfrac{4X}{n(n+1)}\right) + \ldots + \mathrm{Var}\left(\dfrac{2n(X)}{n(n+1)}\right)\) | M1 | 3.1a |
| \(= \dfrac{2^2}{n^2(n+1)^2}\mathrm{Var}(X) + \dfrac{4^2}{n^2(n+1)^2}\mathrm{Var}(X) + \ldots + \dfrac{(2n)^2}{n^2(n+1)^2}\mathrm{Var}(X)\) | M1 | 1.1b |
| \(= \dfrac{\sum_{r=1}^{n}(2r)^2}{n^2(n+1)^2}\mathrm{Var}(X) = \dfrac{4\sum_{r=1}^{n} r^2}{n^2(n+1)^2}\mathrm{Var}(X) = \dfrac{\frac{4}{6}(n)(n+1)(2n+1)}{n^2(n+1)^2}\sigma^2\) | M1 | 2.1 |
| \(= \dfrac{2(2n+1)}{3n(n+1)}\sigma^2\) | A1 | 1.1b |
| (ii) \(\mathrm{Var}(L) = \mathrm{Var}\left(\dfrac{X_1 + X_2}{3}\right) + \mathrm{Var}\left(\dfrac{X_3 + X_4 + \ldots + X_n}{3(n-2)}\right)\) | M1 | 1.1b |
| \(= \dfrac{2}{9}\mathrm{Var}(X) + \dfrac{(n-2)}{9(n-2)^2}\mathrm{Var}(X)\) | M1 | 2.1 |
| \(= \dfrac{2n-3}{9(n-2)}\sigma^2\) | A1 | 1.1b |
| (7) |
Notes
(i) M1: Using independence to set up expression for \(\mathrm{Var}(K)\)
M1: Use of \(\mathrm{Var}(aX) = a^2\mathrm{Var}(X)\) Implied by \(\dfrac{2^2}{n^2(n+1)^2}\)
M1: Use of \(\displaystyle\sum_{r=1}^{n} r^2\)
A1: Correct equivalent expression for \(\mathrm{Var}(K)\) oe
(ii) M1: Using independence to set up expression for \(\mathrm{Var}(L)\)
M1: Use of \(\mathrm{Var}(aX) = a^2\mathrm{Var}(X)\) and understanding \(\mathrm{Var}(X_1 + X_2) = 2\mathrm{Var}(X)\)
Implied by either correct term
A1: Correct equivalent expression for \(\mathrm{Var}(L)\) oe
| Scheme | Marks | AO |
|---|---|---|
| For large values of \(n\) \(\mathrm{Var}(K) \rightarrow 0 \qquad \mathrm{Var}(L) \rightarrow \tfrac{2}{9}(\sigma^2)\) | M1 | 2.1 |
| (Since both are unbiased,) the better estimator is the one with the smaller variance or \(0 \lt \tfrac{2}{9}(\sigma^2)\) | M1 | 2.4 |
| Therefore \(K\) is the better estimator and Korhan wins the challenge. | A1 | 2.2a |
| (3) | ||
| (15 marks) |
Notes
M1: Finding correct limits as \(n\) gets larger for each expression allow ft
Note: Solving \(\dfrac{2n-3}{9(n-2)} = \dfrac{2(2n+1)}{3n(n+1)} \rightarrow n = -0.53,\ 2.46,\ 4.57\) so allow comments relating to \(n \geqslant 5\ (4.57)\)
M1: Correct explanation
A1: Deducing that Korhan is the winner (dependent upon both M marks)