Consider the random variable X∼exp(3). And let $Y=e^{2X}$.
Find the PDF of Y.
I know by properties of exp distribution that the PDF of $f_X(x)$ is given by $$f_X(x) = \begin{cases} 3e^{-3x}; & \text{ if, } x > 0 \\ 0; & \text{ otherwise } \end{cases}$$
The way I approched the problem is by first trying to find the CDF of Y and then taking the derivative to get the PDF of Y.
This is a exercise on a previous exam so I know the result and I'm getting a wrong one. I think I'm having troubles finding the CDF. This is what I've tried
$F_Y(y) = P(Y\le y)=P(e^{2X}\le y)=P(X\le \frac{ln(y)}{2})$
So I have the upper limit and the lower limit of the integral is 1, as $e^{2\cdot 0}$ is 1 and that is the lowest that $x$ gets.
I then integrate the the whole thing to find the CDF.
$\int_{1}^{\frac{ln(y}{2}} 3e^{-3X} \ dx$
But I'm doing something wrong, because it's suppose to give me the result of $\frac{3}{2}y^\frac{-5}{2}$, for $y>1$
Actually $X$ can go to zero, so the correct lower bound for $x$ is $0$ instead of $1$, giving us: $$F_Y(y) = \int_{0}^{(\ln y)/2} 3e^{-3x} \ \mathrm dx = 1 - y^{-3/2}$$
Therefore: $$f_Y(y) = F_Y'(y) = \frac32y^{-5/2}$$