Question-87
If \(X\sim \text{Exp}( \lambda )\), then find the PDF of \(Y=X^{2}\).
The PDF of \(X\) is:
\[ f_{X}( x) =\begin{cases} \lambda e^{-\lambda x} , & x\geqslant 0\\ 0, & \text{otherwise} \end{cases} \]
Its CDF is:
\[ F_{X}( x) =\begin{cases} 1-e^{-\lambda x} , & x\geqslant 0\\ 0, & \text{otherwise} \end{cases} \]
The CDF of \(Y=X^{2}\) is:
\[ \begin{aligned} F_{Y}( y) & =P( Y\leqslant y)\\ & =P\left( X^{2} \leqslant y\right) \end{aligned} \]
If \(y\leqslant 0\), then \(P\left( X^{2} \leqslant y\right) =0\), since \(X^{2}\) is a non-negative random variable. Therefore, we can safely assume that \(y >0\). Now:
\[ \begin{aligned} P\left( X^{2} \leqslant y\right) & =P\left( -\sqrt{y} \leqslant X\leqslant \sqrt{y}\right)\\ & =F_{X}\left(\sqrt{y}\right) -F_{X}\left( -\sqrt{y}\right)\\ & =F_{X}\left(\sqrt{y}\right)\\ & =1-e^{-\lambda \sqrt{y}} \end{aligned} \]
Therefore, we can express the CDF of \(Y\) as:
\[ \begin{aligned} F_{Y}( y) & =\begin{cases} 1-e^{-\lambda \sqrt{y}} , & y\geqslant 0\\ 0, & \text{otherwise} \end{cases} \end{aligned} \]
Differentiating this with respect to \(y\) gives us the PDF of \(Y\):
\[ \begin{aligned} f_{Y}( y) & =\begin{cases} \frac{\lambda }{2\sqrt{y}} e^{-\lambda \sqrt{y}} , & y\geqslant 0\\ 0, & \text{otherwise} \end{cases} \end{aligned} \]