The continuous random variable $X$ takes values only in the interval [$a$, $b$] and $F$ denotes its cumulative distribution function. Using integration by parts, show that: $$E(X) = b - \int_a^b {F(x){\rm{d}}x}.$$

The continuous random variable $Y$ has probability density function $f$ given by: $$\begin{array}{*{20}{c}}   {f(y) = \cos y,}&{0 \leqslant y \leqslant \frac{\pi }{2}} \\   {f(y) = 0,}&{{\text{elsewhere}}{\text{.}}} \end{array}$$

  (i)     Obtain an expression for the cumulative distribution function of $Y$ , valid for $0 \le y \le \frac{\pi }{2}$ . Use the result in (a) to determine $E(Y)$ .

  (ii)     The random variable $U$ is defined by $U = {Y^n}$ , where $n \in {\mathbb{Z}^ + }$ . Obtain an expression for the cumulative distribution function of $U$ valid for $0 \le u \le {\left( {\frac{\pi }{2}} \right)^n}$ .

  (iii)     The medians of $U$ and $Y$ are denoted respectively by ${m_u}$ and ${m_y}$ . Show that ${m_u} = m_y^n$ .

$E(X) = \int_a^b {xf(x){\rm{d}}x}$     M1

$= \left[ {xF(x)} \right]_a^b - \int_a^b {F(x){\rm{d}}x}$     A1

$= bF(b) - aF(a) - \int_a^b {F(x){\rm{d}}x}$     A1

$= b - \int_a^b {F(x){\rm{d}}x}$ because $F(a) = 0$ and $F(b) = 1$     A1

[4 marks]

(i)     let $G$ denote the cumulative distribution function of $Y$

$G(y) = \int_0^y {\cos t{\rm{d}}t}$     M1

$= \left[ {\sin t} \right]_0^y$     (A1)

$= \sin y$     A1

$E(Y) = \frac{\pi }{2} - \int_0^{\frac{\pi }{2}} {\sin y{\rm{d}}y}$     M1

$= \frac{\pi }{2} + \left[ {\cos y} \right]_0^{\frac{\pi }{2}}$     A1

$= \frac{\pi }{2} - 1$     A1

(ii)     CDF of $U = P(U \le u)$     M1

$= P({Y^n} \le u)$     A1

$= P({Y^{}} \le {u^{\frac{1}{n}}})$     A1

$= G({u^{\frac{1}{n}}})$     (A1)

$= \sin \left( {{u^{\frac{1}{n}}}} \right)$     A1

(iii)     ${m_y}$ satisfies the equation $\sin {m_y} = \frac{1}{2}$     A1

${m_u}$ satisfies the equation $\sin \left( {m_u^{\frac{1}{n}}} \right) = \frac{1}{2}$     A1

therefore ${m_y} = m_u^{\frac{1}{n}}$     A1

${m_u} = m_y^n$     AG

[14 marks]

Solutions to (a) were often unconvincing. Candidates were expected to include in their solution the fact that $F(a) = 0$ and $F(b) = 1$ .

In (b) (i), it was not enough to state that $G(y) = \int {\cos y{\rm{d}}y = \sin y}$ although that, fortuitously, gave the correct answer on this occasion. The correct approach was either to state that $G(y) = \int_0^y {\cos t{\rm{d}}t}  = \sin y$ or that $G(y) = \int {\cos y{\rm{d}}y}  = \sin x + C$ and then show that $C = 0$ because $F(0) = 0$ or $F\left( {\frac{\pi }{2}} \right) = 1$ . Solutions to (b) (ii) and (iii) were often disappointing, giving the impression that many of the candidates were not familiar with dealing with cumulative distribution functions.