# S3 January 2008 7 OCR

(i)

Integrating, we have:

\displaystyle F(t) = \begin{cases}0 & t \le 0 \\ t^4 & 0 < t \le 1 \\ 1 & t > 1\end{cases}

(ii)

Note that since 0 < T \le 1, we have H \ge 1. So for h \ge 1 we have:

\begin{align*}\mathrm P(H \le h) & = \mathrm P\left(\frac 1 {T^4} \le h\right) \\ & = \mathrm P\left(\frac 1 h \le T^4\right) \\ & = \mathrm P \left(\frac 1 {h^{1/4}} \le T\right) \\ & = 1 - \mathrm P \left(T \le \frac 1 {h^{1/4}}\right) \\ & = 1 - F \left(\frac 1 {h^{1/4}}\right)\end{align*}

This is equal to \displaystyle 1 - \left(\frac 1 {h^{1/4}}\right)^4 = 1 - \frac 1 h for h \ge 1 and 0 for h < 1.

Differentiating, the PDF of H is given by:

\displaystyle g(h) = \frac 1 {h^2}

for h \ge 1.

(iii)

We have:

\begin{align*}\mathrm E(1 + 2H^{-1}) & = \int_1^\infty \left(1 + \frac 2 h\right) \times \frac 1 {h^2} \mathrm dh \\ & = \int_1^\infty \left(\frac 1 {h^2} + \frac 2 {h^3}\right) \mathrm dh \\ & = \left[-\frac 1 h - \frac 1 {h^2}\right]_1^\infty \\ & = -0 - 0 + \frac 1 1 +\frac 1 {1^2} \\ & = 2\end{align*}