Q2Single correct3 Marks13 Apr 2025If X∼Exp(λ)X \sim \text{Exp}(\lambda)X∼Exp(λ), then find the PDF of Y=XY = \sqrt{X}Y=X.AfY(y)={2λye−λy2,y>00,otherwisef_Y(y) = \begin{cases} 2\lambda y e^{-\lambda y^2}, & y > 0 \\ 0, & \text{otherwise} \end{cases}fY(y)={2λye−λy2,0,y>0otherwiseBfY(y)={λe−λy,y>00,otherwisef_Y(y) = \begin{cases} \lambda e^{-\lambda y}, & y > 0 \\ 0, & \text{otherwise} \end{cases}fY(y)={λe−λy,0,y>0otherwiseCfY(y)={λe−λy22,y>00,otherwisef_Y(y) = \begin{cases} \lambda e^{-\frac{\lambda y^2}{2}}, & y > 0 \\ 0, & \text{otherwise} \end{cases}fY(y)={λe−2λy2,0,y>0otherwiseDfY(y)={λye−λy2,y>00,otherwisef_Y(y) = \begin{cases} \lambda y e^{-\lambda y^2}, & y > 0 \\ 0, & \text{otherwise} \end{cases}fY(y)={λye−λy2,0,y>0otherwise