S4 June 2011 Q6

EdexcelOld spec16 marksContinuous Random Variables

6. A random sample \(X_1, X_2, \ldots, X_n\) is taken from a population where each of the \(X_i\) have a continuous uniform distribution over the interval \([0, \beta]\).
The random variable \(Y = \max\{X_1, X_2, \ldots, X_n\}\).
The probability density function of \(Y\) is given by

\[\mathrm{f}(y) = \begin{cases} \dfrac{n}{\beta^n}\,y^{n-1} & 0 \leqslant y \leqslant \beta \\ 0 & \text{otherwise} \end{cases}\]

(a) Show that \(\mathrm{E}(Y^m) = \dfrac{n}{n+m}\beta^m\). (3)

(b) Write down \(\mathrm{E}(Y)\). (1)

(c) Using your answers to parts (a) and (b), or otherwise, show that \[\mathrm{Var}(Y) = \dfrac{n}{(n+1)^2(n+2)}\beta^2\] (3)

(d) State, giving your reasons, whether or not \(Y\) is a consistent estimator of \(\beta\). (3)

The random variables \(M = 2\bar{X}\), where \(\bar{X} = \dfrac{1}{n}(X_1 + X_2 + \ldots + X_n)\), and \(S = kY\), where \(k\) is a constant, are both unbiased estimators of \(\beta\).

(e) Find the value of \(k\) in terms of \(n\). (1)

(f) State, giving your reasons, which of \(M\) and \(S\) is the better estimator of \(\beta\) in this case. (3)

Five observations of \(X\) are: 8.5 6.3 5.4 9.1 7.6

(g) Calculate the better estimate of \(\beta\). (2)

Mark scheme (a)

Scheme	Marks
\(\mathrm{E}(Y^m) = \dfrac{n}{\beta^n}\displaystyle\int y^m \times y^{n-1}\,\mathrm{d}y \ =, \ \left[\dfrac{n}{\beta^n} \times \dfrac{1}{m+n} \times y^{m+n}\right]_0^{\beta}\)	M1, A1
\(= \dfrac{n}{\cancel{\beta^n}} \times \dfrac{1}{m+n} \times \beta^{m+\cancel{n}} = \dfrac{n}{m+n}\beta^m \quad (*)\)	A1cso
	(3)

Notes

M1 for attempt to integrate \(y^m\mathrm{f}(m)\)
1^st A1 for correct integration (limits not needed yet)
2^nd A1 for use of correct limits and proceeding to printed answer. No incorrect working seen.

Scheme	Marks
\(\mathrm{E}(Y) = \dfrac{n}{n+1}\beta\)	B1
	(1)

Mark scheme (c)

Scheme	Marks
\(\mathrm{E}(Y^2) = \dfrac{n}{n+2}\beta^2, \quad \mathrm{Var}(Y) = \mathrm{E}(Y^2) - [\mathrm{E}(Y)]^2\)	B1,M1
\(\mathrm{Var}(Y) = \dfrac{n}{n+2}\beta^2 - \dfrac{n^2}{(n+1)^2}\beta^2 = \dfrac{n}{(n+1)^2(n+2)}\beta^2 \quad (*)\)	A1cso
	(3)

Notes

M1 for use of their \(\mathrm{E}(Y)\) and \(\mathrm{E}(Y^2)\) in a correct formula for \(\mathrm{Var}(Y)\)

Mark scheme (d)

Scheme	Marks
As \(n \to \infty\) \(\mathrm{E}(Y) \to \beta\), \(\mathrm{Var}(Y) \to 0\) So \(Y\) is a consistent estimator for \(\beta\).	M1,A1 A1
	(3)

Notes

M1 for examining both \(\mathrm{E}(Y)\) and \(\mathrm{Var}(Y)\) for \(n \to \infty\)
1^st A1 for correct limits for both the above
2^nd A1 for a correct statement following correct working

Scheme	Marks
\(k = \dfrac{n+1}{n}\)	B1
	(1)

Mark scheme (f)

Scheme	Marks
\(\mathrm{Var}(M) = 4\mathrm{Var}(\bar{X}) \ = 4\dfrac{\sigma^2}{n} = \dfrac{4}{n} \times \dfrac{\beta^2}{12} = \dfrac{\beta^2}{3n}\)	B1
\(\dfrac{(n+1)^2}{n^2} \times \dfrac{n}{(n+1)^2(n+2)}\beta^2 = \dfrac{\beta^2}{n(n+2)} \lt \dfrac{\beta^2}{3n}\) so \(S\) is better (\(n \gt 1\))	M1A1
	(3)

Notes

M1 for attempting \(\mathrm{Var}(S)\)

Scheme	Marks
Max = 9.1, \(s = \dfrac{6}{5} \times 9.1 = \underline{\mathbf{10.9(2)}}\)	M1A1
	(2)
	(16 marks)