91Ó°ÊÓ

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

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

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

Therefore, the PDF of X is expressed as,

\({f_x} = \left\{ \begin{array}{l}\frac{1}{{1 - 0}} = 1\;\;\;\;\;\;\;\;\;\;0 \le x \le 1\\0;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;otherwise\end{array} \right.\)

The CDF of a uniform distribution is obtained by using the formula:

\(\begin{aligned}{F_X}\left( x \right) &= P\left( {X \le x} \right)\\ &= \frac{{x - a}}{{b - a}}\\ &= \frac{{x - 0}}{{1 - 0}}\\ &= x\end{aligned}\)

Thus, the CDF is,

\({F_X}\left( x \right) = \left\{ \begin{array}{l}0\;;\;\;\;\;\;\;\;\;\;\;\;x < 0\\x\;;\;\;\;\;\;\;\;\;\;\;\;0 \le x \le 1\\1\;;\;\;\;\;\;\;\;\;\;\;\;\;x > b\end{array} \right.\)

a.

Define, \(Y = {X^2}\)

Consider the limits as,

\(\begin{array}{l}x = 0 \Rightarrow y = 0\\x = 1 \Rightarrow y = 1\end{array}\)

The PDF of the variable Y is obtained using the CDF approach, as shown below,

\(\begin{aligned}{F_Y}\left( y \right) &= P\left( {Y \le y} \right)\\ &= P\left( {{X^2} \le y} \right)\\ &= P\left( {\left| X \right| \le \sqrt y } \right)\\ &= P\left( { - \sqrt y \le X \le \sqrt y } \right)\\ &= {F_X}\left( {\sqrt y } \right) - F\left( { - \sqrt y } \right)\\ &= \sqrt y \end{aligned}\)

Therefore, the CDF of Y is \(\sqrt {\bf{y}} \).

The pdf is obtained from CDF by differentiating it with respect to the variable, as shown below,

\(\begin{aligned}{f_y} &= \frac{d}{{dy}}\left( {{F_Y}\left( y \right)} \right)\\ &= \frac{d}{{dy}}\left( {\sqrt y } \right)\\ &= \frac{1}{{2\sqrt y }}\end{aligned}\)

Therefore, the PDF of Y is:

\({f_y} = \left\{ \begin{array}{l}\frac{1}{{2\sqrt y }},\;\;\;\;0 < y < 1\\0,\;\;\;\;\;\;\;\;\;o.w\end{array} \right.\)

b.

Consider, \(Y = - {X^3}\)

\(\begin{array}{l}x = 0 \Rightarrow y = 0\\x = 1 \Rightarrow y = - 1\end{array}\)

The CDF of random variable Y is,

\(\begin{aligned}{F_Y}\left( y \right) &= P\left( {Y \le y} \right)\\ &= P\left( { - {X^3} \le y} \right)\\ &= P\left( {{X^3} \ge - y} \right)\end{aligned}\)

Further,

\(\begin{aligned}{F_Y}\left( Y \right) &= 1 - P\left( {X \le \sqrt[3]{y}} \right)\\ &= 1 - {F_X}\left( {{y^{\frac{1}{3}}}} \right)\\ &= 1 - {y^{\frac{1}{3}}}\end{aligned}\)

Therefore, the CDF is \({\bf{1 - }}{{\bf{y}}^{\frac{{\bf{1}}}{{\bf{3}}}}}\).

The PDF of Random Variable Y is,

\(\begin{aligned}{f_y}\left( y \right) &= \frac{d}{{dy}}\left( {{F_Y}\left( y \right)} \right)\\ &= \frac{d}{{dy}}\left( {1 - {y^{\frac{1}{3}}}} \right)\\ &= - \frac{1}{{3{{\left| y \right|}^{^{\frac{2}{3}}}}}},\;\;\;\;\;\;\;\; - 1 < y < 0\end{aligned}\)

Therefore, the PDF is,

\({f_Y}\left( y \right) = \left\{ \begin{array}{l} - \frac{1}{{3{{\left| y \right|}^{\frac{2}{3}}}}},\;\;\;\;\;\;\; - 1 < y < 0\\0,\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;o.w\end{array} \right.\)

c.

Consider \(Y = {X^{\frac{1}{2}}}\).

\(\begin{aligned}x = 0 \Rightarrow y = 0\\x = 1 \Rightarrow y = 1\end{aligned}\)

The CDF of random variable Y is,

\(\begin{aligned}{F_Y}\left( y \right) &= P\left( {Y \le y} \right)\\ &= P\left( {{X^{0.5}} \le y} \right)\\ &= P\left( {X \le {y^2}} \right)\\ &= {F_X}\left( {{y^2}} \right)\\ &= {y^2}\end{aligned}\)

Therefore, the CDF is\({{\bf{y}}^2}\).

The PDF of Random Variable Y is,

\(\begin{aligned}{f_y} &= \frac{d}{{dy}}\left( {{F_Y}\left( y \right)} \right)\\ &= \frac{d}{{dy}}\left( {{y^2}} \right)\\ &= 2y\end{aligned}\)

Therefore, the PDF is,

\({f_Y}\left( y \right) = \left\{ \begin{array}{l}2y,\;\;\;0 < y < 1\\0,\,\;\;\;\;o.w\end{array} \right.\)