Finding the PDF of a function if X is exponentially distributed with a given parameter $\lambda$ = 3 Find the pdf of Y = $e^{\frac{1}{2}x}$ if x is exponentially distributed with parameter $\lambda =$ 3.
This question was given to me as a review for an upcoming exam, I am not really sure where to start.
I'm thinking :
$F_Y(y) =$ 
\begin{cases} 
      \lambda e^{-\lambda x} & x\geq 0 \\
      0 & otherwise
   \end{cases}
as a start.
Update Attempt:
$P(Y \leq y)$
$= e^{\frac{X}{2}} \leq y$
$X \leq 2ln(y)$
$\frac{d}{dy}(1-e^{-2 \lambda ln(y)})$
$= \frac {2 \lambda e^{-2\lambda ln(y)}}{y}$
$F_Y(y)=$
\begin{cases} 
      \frac {2 \lambda e^{-2\lambda ln(y)}}{y} & y\geq 1 \\
      0 & otherwise
   \end{cases}$F_y(Y)=$
\begin{cases} 
      \frac {2 (3) e^{-2(3) ln(y)}}{y} & y\geq 1 \\
      0 & otherwise
   \end{cases}
$F_Y(y)=$
\begin{cases} 
      \frac {6 e^{-6 ln(y)}}{y} & y\geq 1 \\
      0 & otherwise
   \end{cases}
$F_Y(y)=$
\begin{cases} 
      6y^{-7} & y\geq 1 \\
      0 & otherwise
   \end{cases}
 A: The first thing you have to do is master case sensitivity, otherwise you are just going to confuse yourself.   Don't mix up when you use random variable $Y$ and term $y$.
Also, your final answer should not contain any term $x$.   $F_Y(y)$ will clearly be a function of term $y$.
But moving on...

If $X$ is exponential with parameter $\lambda:=3$, and $Y=\mathrm e^{X/2}$, then the pdf for $Y$ is supported over $[1;\infty)$ and ....$$\begin{split}\mathsf P(Y\leq y)&=\mathsf P(\mathrm e^{X/2}\leq y)\\[1ex]&=\mathsf P(X\leq 2\ln y)\\[2ex]F_Y(y)&=F_X(2\ln y)\\[2ex]f_Y(y)&=\begin{vmatrix}\dfrac{\mathrm d F_X(2\ln y)}{\mathrm d y\qquad\quad}\end{vmatrix}\\[1ex]&=\begin{vmatrix}\dfrac{\mathrm d (2\ln y)}{\mathrm d y\qquad~~~}\end{vmatrix}f_X(2\ln y)\\&~~\vdots\end{split}$$
A: From your attempt, you have written $P(Y\le y)= 1-e^{-\lambda y}$. This would be true if $Y$ were $Expo(\lambda)$, but we are not told $Y$ is (we are told $X$ is). Instead, note that for $y> 0$, we have $$Y\le y \iff e^{X/2}\le y\iff X \le 2\ln y.$$
Thus $P(Y\le y) = P(X\le 2\ln y)$. See if you can finish from here. 
