S4 June 2013 Q4

EdexcelOld spec16 marksContinuous Random Variables

4. A random sample of size 2, \(X_1\) and \(X_2\), is taken from the random variable \(X\) which has a continuous uniform distribution over the interval \([-a, 2a]\), \(a \gt 0\)

(a) Show that \(\bar{X} = \dfrac{X_1 + X_2}{2}\) is a biased estimator of \(a\) and find the bias. (3)

The random variable \(Y = k\bar{X}\) is an unbiased estimator of \(a\).

(b) Write down the value of the constant \(k\). (1)

The random variable \(M\) is the maximum of \(X_1\) and \(X_2\)

The probability density function, \(m(x)\), of \(M\) is given by

\[m(x) = \begin{cases} \dfrac{2(x + a)}{9a^2} & -a \leqslant x \leqslant 2a \\ 0 & \text{otherwise} \end{cases}\]

(d) Show that \(M\) is an unbiased estimator of \(a\). (4)

Given that \(\mathrm{E}(M^2) = \dfrac{3}{2}a^2\)

(e) find \(\mathrm{Var}(M)\). (1)

(f) State, giving a reason, whether you would use \(Y\) or \(M\) as an estimator of \(a\). (2)

A random sample of two values of \(X\) are 5 and −1

(g) Use your answer to part (f) to estimate \(a\). (1)

Mark scheme (a)

Scheme	Marks
\(\mathrm{E}(X) = \mu = \dfrac{2a - a}{2} = \dfrac{a}{2};\qquad \mathrm{E}(\bar{X}) = \mu = \dfrac{a}{2}\) so biased estimator for \(a\)	M1;A1
Bias \(= \dfrac{a}{2} - a = -\dfrac{a}{2}\)	B1(accept \(\pm\))
	(3)

Notes

M1 for use of formula or integration or symmetry to find \(\mathrm{E}(X)\)

Scheme	Marks
\(k = 2\)	B1
	(1)

Mark scheme (c)

Scheme	Marks
\(\mathrm{Var}(X) = \sigma^2 = \dfrac{(2a - -a)^2}{12} = \dfrac{9a^2}{12} = \dfrac{3a^2}{4};\qquad \mathrm{Var}(\bar{X}) = \dfrac{\sigma^2}{2}\)	B1;B1
\(\mathrm{Var}(Y) = k^2\mathrm{Var}(\bar{X}) = 4, \times \dfrac{\sigma^2}{2} = 4 \times \dfrac{3a^2}{4 \times 2} = \dfrac{3}{2}a^2\)	M1,A1
	(4)

Notes

1^st B1 for use of formula for variance
2^nd B1 for use of \(\dfrac{\sigma^2}{n}\) formula
M1 for \(k^2\mathrm{Var}(\bar{X})\) and ft their \(k\)

Mark scheme (d)

Scheme	Marks
\(\mathrm{E}(M) = \displaystyle\int \dfrac{2x(x+a)}{9a^2}\,\mathrm{d}x = \left[\dfrac{2x^3}{27a^2} + \dfrac{ax^2}{9a^2}\right]_{-a}^{2a},\ = \left(\dfrac{16a}{27} + \dfrac{4a}{9}\right) - \left(-\dfrac{2a}{27} + \dfrac{a}{9}\right)\ [= a]\)	M1A1,M1d
So \(\mathrm{E}(M) = a\) and therefore \(M\) is an unbiased estimator for \(a\)	A1cso
	(4)

Notes

1^st M1 for attempt at correct integration of correct expression
1^st A1 for correct integration
2^nd M1d dependent on previous M, for attempting to use correct limits
2^nd A1 need statement that \(M\) is therefore unbiased

Scheme	Marks
\(\mathrm{Var}(M) \lt \mathrm{Var}(Y)\), so \(M\) is the better estimator of \(a\)	M1, A1
	(2)

Scheme	Marks
Maximum value = 5	B1ft
	(1)
	(16 marks)