S3 January 2010 4 OCR

(i)

Note that since V \ge 1 we have \displaystyle 0 <Y = \frac 1 {1 + V} \le \frac 1 2.

So for 0 < y \le \dfrac 1 2 the CDF is given by:

\begin{align*}\mathrm P(Y \le y) & = \mathrm P \left(\frac 1 {1 + V} \le y\right) \\ & = \mathrm P\left(\frac 1 y - 1 \le V\right) \\ & = 1 - \mathrm P \left(V \le \frac 1 y - 1\right) \\ & = 1 - F \left(\frac 1 y - 1\right) \\ & = 1 - \left(1 - \frac 8 {\left(1 + \frac 1 y - 1\right)^3}\right) \\ & = \frac 8 {\frac 1 {y^3}} \\ & = 8y^3 \end{align*}

Differentiating, the PDF of Y is then 8 \times 3y^2 = 24y^2 if 0 < y \le \dfrac 1 2 and 0 otherwise.

(ii)

We have:

\begin{align*}\mathrm E \left(\frac 1 {Y^2}\right) & = \int_0^{\frac 1 2} \frac {24y^2} {y^2} \mathrm dy \\ & = 24 \int_0^{\frac 1 2} \mathrm dy \\ & = 24 \times \frac 1 2 \\ & = 12\end{align*}