6 The continuous random variable \(X\) has (cumulative) distribution function given by
$$\mathrm { F } ( x ) = \begin{cases} 0 & x < \frac { 1 } { 2 }
\frac { 2 x - 1 } { x + 1 } & \frac { 1 } { 2 } \leqslant x \leqslant 2
1 & x > 2 . \end{cases}$$
- Given that \(Y = \frac { 1 } { X }\), find the (cumulative) distribution function of \(Y\), and deduce that \(Y\) and \(X\) have identical distributions.
- Find \(\mathrm { E } ( X + 1 )\) and deduce the value of \(\mathrm { E } \left( \frac { 1 } { X } \right)\).