S3 January 2007 6 OCR

(i)

We have:

\displaystyle \int_1^t \frac 3 {x^4} \mathrm dx = \left[-\frac 1 {x^3}\right]_1^t = 1 - \frac 1 {t^3}

so the CDF of T is given by:

\displaystyle F(t) = \begin{cases}0 & t < 1 \\ 1 - \frac 1 {t^3} & t \ge 1\end{cases}

(ii)

Since T \ge 1, we must have Y = T^3 \ge 1. So for y \ge 1 the CDF of Y is given by:

\begin{align*}\mathrm P(Y \le y) & = \mathrm P(T^3 \le y) \\ & = \mathrm P(T \le y^{1/3}) \\ & = 1 - \frac 1 {(y^{1/3})^3} \\ & = 1 - \frac 1 y\end{align*}

differentiating gives the PDF as:

\displaystyle f_Y(y) = \frac 1 {y^2}

for y \ge 1.

(iii)

We have:

\begin{align*}\mathrm E(\sqrt Y) & = \int_1^\infty \frac {\sqrt y} {y^2} \mathrm dy \\ & = \int_1^\infty y^{-3/2} \mathrm dy \\ & = \left[-\frac 2 {\sqrt y}\right]_1^\infty \\ & = \frac 2 {\sqrt 1} \\ & = 2\end{align*}