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.