S4 June 2009 Q6

6. A continuous uniform distribution on the interval \([0, k]\) has mean \(\dfrac{k}{2}\) and variance \(\dfrac{k^2}{12}\).

A random sample of three independent variables \(X_1\), \(X_2\) and \(X_3\) is taken from this distribution.

(a) Show that \(\dfrac{2}{3}X_1 + \dfrac{1}{2}X_2 + \dfrac{5}{6}X_3\) is an unbiased estimator for \(k\). (3)

An unbiased estimator for \(k\) is given by \(\hat{k} = aX_1 + bX_2\) where \(a\) and \(b\) are constants.

(b) Show that \(\mathrm{Var}(\hat{k}) = (a^2 - 2a + 2)\dfrac{k^2}{6}\) (6)

(c) Hence determine the value of \(a\) and the value of \(b\) for which \(\hat{k}\) has minimum variance, and calculate this minimum variance. (6)

Mark scheme (a)

Scheme	Marks
\(\mathrm{E}\left(\tfrac{2}{3}X_1 + \tfrac{1}{2}X_2 + \tfrac{5}{6}X_3\right) = \tfrac{2}{3} \times \tfrac{k}{2} + \tfrac{1}{2} \times \tfrac{k}{2} + \tfrac{5}{6} \times \tfrac{k}{2} = k\)	M1 A1
\(\mathrm{E}\left(\tfrac{2}{3}X_1 + \tfrac{1}{2}X_2 + \tfrac{5}{6}X_3\right) = k \Rightarrow\) unbiased	B1
	(3)

(corrected from the printed mark scheme: the second line is printed as “E( X₁ + X₂ + X₃) = k ⇒ unbiased” with the coefficients missing)

Mark scheme (b)

Scheme	Marks
\(\mathrm{E}(aX_1 + bX_2) = a\dfrac{k}{2} + b\dfrac{k}{2} = k\)	M1
\(a + b = 2\)	A1
\(\mathrm{Var}(aX_1 + bX_2) = a^2\dfrac{k^2}{12} + b^2\dfrac{k^2}{12}\)	M1A1
\(= a^2\dfrac{k^2}{12} + (2-a)^2\dfrac{k^2}{12}\)	M1
\(= (2a^2 - 4a + 4)\dfrac{k^2}{12}\) \(= (a^2 - 2a + 2)\dfrac{k^2}{6}\) (*) since answer given	A1 cso
	(6)

Mark scheme (c)

Scheme	Marks
Min value when \((2a - 2)\dfrac{k^2}{6} = 0\) \(\dfrac{\mathrm{d}}{\mathrm{d}a}(\mathrm{Var}) = 0\), all correct, condone missing \(\dfrac{k^2}{6}\)	M1A1
\(\Rightarrow 2a - 2 = 0\) \(a = 1,\ b = 1.\)	A1A1
\(\dfrac{\mathrm{d}^2(\mathrm{Var})}{\mathrm{d}a^2} = \dfrac{2k^2}{6} \gt 0\) since \(k^2 \gt 0\) therefore it is a minimum	M1
min variance \(= (1 - 2 + 2)\dfrac{k^2}{6}\) \(= \dfrac{k^2}{6}\)	B1
	(6)