A2 October 2021 Q5

5. The continuous random variable \(X\) is uniformly distributed over the interval \([0, 4\beta]\), where \(\beta\) is an unknown constant.

Three independent observations, \(X_1\), \(X_2\) and \(X_3\), are taken of \(X\) and the following estimators for \(\beta\) are proposed

\[A = \frac{X_1 + X_2}{2}\]\[B = \frac{X_1 + 2X_2 + 3X_3}{8}\]\[C = \frac{X_1 + 2X_2 - X_3}{8}\]

(a) Calculate the bias of \(A\), the bias of \(B\) and the bias of \(C\) (5)

(b) By calculating the variances, explain which of \(B\) or \(C\) is the better estimator for \(\beta\) (4)

Mark scheme (a)

Scheme	Marks	AO
\([\mathrm{E}(X) = 2\beta]\) \(\mathrm{E}(A) = \mathrm{E}\left(\dfrac{X_1 + X_2}{2}\right) = \tfrac{1}{2}[\mathrm{E}(X) + \mathrm{E}(X)]\) \(\mathrm{E}(B) = \mathrm{E}\left(\dfrac{X_1 + 2X_2 + 3X_3}{8}\right) = \tfrac{1}{8}[\mathrm{E}(X) + 2\mathrm{E}(X) + 3\mathrm{E}(X)]\) \(\mathrm{E}(C) = \mathrm{E}\left(\dfrac{X_1 + 2X_2 - X_3}{8}\right) = \tfrac{1}{8}[\mathrm{E}(X) + 2\mathrm{E}(X) - \mathrm{E}(X)]\)	M1	3.1a
Bias for \(A = \mathrm{E}(A) - \beta = 2\beta - \beta = \beta\) Bias for \(B = \mathrm{E}(B) - \beta = 1.5\beta - \beta = 0.5\beta\) Bias for \(C = \mathrm{E}(C) - \beta = 0.5\beta - \beta = -0.5\beta\)	M1 A1 A1 A1	2.1 1.1b 1.1b 1.1b
	(5)

M1: Using independence to calculate the \(\mathrm{E}(A)\), \(\mathrm{E}(B)\) or \(\mathrm{E}(C)\)

M1: Use of bias = \(\mathrm{E}(X) - \beta\)

A1: Correct bias for \(A\)

A1: Correct bias for \(B\)

A1: Correct bias for \(C\) [allow \(+0.5\beta\)]

Mark scheme (b)

Scheme	Marks	AO
\(\left[\mathrm{Var}(X) = \dfrac{4}{3}\beta^2\right]\) Better estimator would have the smallest bias and the least variance. \(B\) and \(C\) have equal bias, so we select the estimator with the smallest variance \(\mathrm{Var}(B) = \mathrm{Var}\left(\dfrac{X_1 + 2X_2 + 3X_3}{8}\right)\) \(= \tfrac{1}{64}[\mathrm{Var}(X) + 4\mathrm{Var}(X) + 9\mathrm{Var}(X)]\) \(\mathrm{Var}(C) = \mathrm{Var}\left(\dfrac{X_1 + 2X_2 - X_3}{8}\right)\) \(= \tfrac{1}{64}[\mathrm{Var}(X) + 4\mathrm{Var}(X) + \mathrm{Var}(X)]\)	M1	2.1
\(\mathrm{Var}(B) = \tfrac{7}{32}\mathrm{Var}(X)\left[= \tfrac{7}{24}\beta^2\right]\) \(\mathrm{Var}(C) = \tfrac{3}{32}\mathrm{Var}(X)\left[= \tfrac{1}{8}\beta^2\right]\)	A1 A1	1.1b 1.1b
(Since both have same bias and) \(\mathrm{Var}(C) \lt \mathrm{Var}(B)\) therefore \(C\) is the better estimator.	B1ft	2.2a
	(4)

M1: Realising that variances need to be compared and attempt at linear combination of variances for \(B\) or \(C\)

A1: Correct \(\mathrm{Var}(B)\)

A1: Correct \(\mathrm{Var}(C)\)

A1ft: Correct comparison and deduction that \(C\) a better estimator than \(A\) and \(B\).

Scheme	Marks	AO
Any unbiased estimator, e.g. \(\dfrac{X_1 + X_2 + X_3}{6}\)	B1	3.5c
	(1)
	(10 marks)