91Ó°ÊÓ

The pdf referred to for random variable X is:

\(\begin{aligned}f\left( x \right) &= \frac{1}{2}x,0 < x < 2\\ &= 0,{\rm{otherwise}}.\end{aligned}\)

And the pdf referred to for random variable Y is:

\(\begin{aligned}g\left( y \right) &= \frac{3}{8}{y^2},0 < y < 2\\ &= 0,{\rm{otherwise}}.\end{aligned}\)

\(\begin{aligned}{X_1} &= \frac{X}{2} \sim {\rm{Beta}}\left( {2,1} \right)\\{X_2} &= \frac{Y}{2} \sim {\rm{Beta}}\left( {3,1} \right)\end{aligned}\)

Consider the problem,

\(\begin{array}{l}{X_1} \sim {\rm{Beta}}\left( {2,1} \right)\\ \Rightarrow f\left( {{x_1}} \right) = \left\{ \begin{array}{l}2x,0 < x < 1\\0,{\rm{otherwise}}\end{array} \right.\\{X_2} \sim {\rm{Beta}}\left( {3,1} \right)\\ \Rightarrow f\left( {{x_2}} \right) = \left\{ \begin{array}{l}3{x^2},0 < x < 1\\0,{\rm{otherwise}}\end{array} \right.\end{array}\)

By Theorem 3.8.4,

\(g\left( y \right) = f\left[ {s\left( y \right)} \right]\left| {\frac{{ds\left( y \right)}}{{dy}}} \right| \ldots \left( 2 \right)\)

By backward calculation,

\(\left. \begin{aligned} &= \Pr \left\{ {{X_1} < {t^{\frac{3}{2}}}} \right\}\\ &= {\left[ {{t^{\frac{3}{2}}}} \right]^2}\\ &= {t^3}\end{aligned} \right\}{\rm{using}}\;{\rm{this}}\;{\rm{we}}\,{\rm{can}}\,{\rm{tak}}e\,{X_2} = {X_1}^{\frac{2}{3}}\)

Now,

\(\begin{aligned}{F_{{X_2}}}\left( t \right) &= \Pr \left\{ {{X_2} \le t} \right\}\\ &= \Pr \left\{ {{X_1} \le {t^{\frac{3}{2}}}} \right\}\\ &= {t^3}\end{aligned}\)

Differentiating the CDF to get the PDF

\(\begin{array}{l}{f_{{X_2}}}\left( t \right) = \left\{ \begin{array}{l}3{t^2},0 < t < 1\\0,{\rm{otherwise}}\end{array} \right.\\\therefore {X_2} \sim {\rm{Beta}}\left( {3,1} \right)\end{array}\)