Example 1.1.2. Let \(X\) be a continuous non-negative random variable with pdf \(f_X(x)\) and \(Y=X^k\), find the pdf of \(Y\) \[Y=X^k,\hspace {0.3cm} g(x)=X^k,\hspace {0.3cm} g^{-1}(y)=y^{\frac {1}{k}}\] \begin {align*} f_Y(y) &=f_X\Big (g^{-1}(y)\Big ).\Big |\frac {d}{dy}g^{-1}(y)\Big |\\ &=f_X\Big (y^{\frac {1}{k}}\Big ).\frac {1}{k}.y^{\frac {1}{k}-1}\\ &=\frac {1}{k}y^{\frac {1}{k}-1}f_X\Big (y^{\frac {1}{k}}\Big ) \end {align*}

e.g if \(k=2\) \[f_Y(y)=\frac {1}{2}y^{-\frac {1}{2}}f_Y\Big (y^{\frac {1}{2}}\Big )\] \[f_Y(y)=\frac {1}{2\sqrt {y}}f_X\Big (\sqrt {y}\Big ),\hspace {0.3cm} y\in g(I),\hspace {0.3cm} \text {where}\hspace {0.3cm} f_X(x)\hspace {0.3cm} \text {is supported on}\hspace {0.3cm} I\]
The cdf technique has been used to prove theorem 1.1.1, however the cdf technique can be used directly to find the density of \(Y=g(x)\).

Example 1.1.3. Let \(X\) be a continuous random variable with pdf \(f_X(x)\), find the pdf of \(Y=|X|\). \begin {align*} F_Y(y) &=P(Y\leq y)=P(|X|\leq y)\\ &=P(-y\leq X\leq y)\\ &=F_X(y)-F_X(-y) \end {align*}

\[\implies \hspace {0.4cm} f_Y(y) =f_X(y)+f_X(-y).\]
\[\text {If}\hspace {0.4cm} X\thicksim N(\mu , \sigma ^2),\hspace {0.4cm}\text {and}\hspace {0.4cm} Y=|X|\] \[f_X(x)=\frac {1}{\sqrt {2\pi \sigma ^2}}e^{-\frac {1}{2}\Big (\frac {x-\mu }{\sigma }\Big )^2}\] \begin {align*} f_Y(y) &=\frac {1}{\sqrt {2\pi \sigma ^2}}e^{-\frac {1}{2}\Big (\frac {y-\mu }{\sigma }\Big )^2}+\frac {1}{\sqrt {2\pi \sigma ^2}}e^{-\frac {1}{2}\Big (\frac {-y-\mu }{\sigma }\Big )^2}\\\\ &=\frac {1}{\sqrt {2\pi \sigma ^2}}\Bigg \{e^{-\frac {1}{2\sigma ^2}(y^2-2\mu y+\mu ^2)}+e^{-\frac {1}{2\sigma ^2}(y^2+2\mu y+\mu ^2)}\Bigg \}\\ &=\frac {1}{\sqrt {2\pi \sigma ^2}}e^{-\frac {1}{2\sigma ^2}(y^2+\mu ^2)}\Bigg \{e^{\frac {1}{2}2\mu y}+e^{-\frac {1}{2}2\mu y}\Bigg \}\\ &=\frac {1}{\sqrt {2\pi \sigma ^2}}e^{-\frac {1}{2\sigma ^2}(y^2+\mu ^2)}\Bigg \{e^{\frac {\mu y}{\sigma ^2}}+e^{-\frac {\mu y}{\sigma ^2}}\Bigg \},\hspace {0.3cm} y>0.\\\\ \end {align*}