91Ó°ÊÓ

The random variable X follows uniform distribution on the interval [0,1], i.e.,\(X \sim U\left[ {0,1} \right]\).

The pdf of a uniform distribution is obtained by using the formula: \(\frac{1}{{b - a}};a \le x \le b\)

Here, \(a = 0,b = 1\)

\({f_x} = \frac{1}{{1 - 0}};0 \le x \le 1\)

\(\begin{aligned}{f_x} &= 1,0 \le x \le 1\\ &= 0,o.w.\end{aligned}\)

And given pdf of Y is

\(\begin{aligned}g\left( y \right) &= \frac{3}{8}{y^2},0 < y < 2\\ &= 0,otherwise. \ldots \left( 1 \right)\end{aligned}\)

By Theorem 3.8.4,

Let X be anongoing randomized variable with PDF f(x). Let \(y = g\left( x \right)\) be an increasing/decreasing function. The density function of a random variable \(y = g\left( x \right)\) is given by

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

\(x = s\left( y \right)\)is the inverse of the function \(g\left( x \right)\)

We know that,

\(f\left[ {s\left( y \right)} \right] = 1\)since\(X \sim U\left[ {0,1} \right]\)

By substituting\(\left( {\rm{1}} \right)\;{\rm{in}}\;\left( {\rm{2}} \right)\)

\(\begin{aligned} &\Rightarrow \frac{3}{8}{y^2} = 1 \cdot \left| {\frac{{ds\left( y \right)}}{{dy}}} \right|\\ &\Rightarrow \left| {\frac{{ds\left( y \right)}}{{dy}}} \right| = \frac{3}{8}{y^2}\end{aligned}\)

Integrating both sides

\(x = \frac{{{y^3}}}{8} \ldots \left( 3 \right)\)

\(y = {\left( {8x} \right)^{\frac{1}{3}}}\) from \(\left( 3 \right)\)

Therefore, the Random Variable \(Y = r\left( X \right)\) is \(y = {\left( {8x} \right)^{\frac{1}{3}}}\)