PDF of e raised to an exponential random variable 
Let $X$ be an exponential random variable with parameter $\lambda > 0$. For $\lambda > 1$, compute the PDF of $\mathrm e^X$.

Is $P(\mathrm e^X \le \mathrm e^x)$ the same as $P(X \le x)$? Does this mean that the PDF of $\mathrm e^X$ is just the PDF of $X$? 
 A: Yes, because the exponential function is increasing $P(e^X \le e^x) = P(X \le x)$.
This means the CDF $F_{e^X}(e^x) = F_X(x)$, or $F_{e^X}(t) = F_X(\ln (t))$ for $t > 0$.  Differentiate to get the PDF.
A: (1) Yes, as $\exp$ is monotone $\exp(X) \le \exp (x)$ is equivalent to $X\le x$.
(2) No, that does not imply that $\exp(X)$ has the same pdf as $X$. We have 
$$ F_{\exp X}(x) = P\bigl(\exp(X) \le x\bigr) = P(X \le \log x) = F_X(\log x) $$
where $F_X$ denotes the cdf of $X$. Taking derivatives, we have for the pdf
$$ f_{\exp X}(x) = f_X(\log x) \cdot \frac 1x. $$
A: The simple transformation formula for transforming an Exponential random variable $X {\raise.17ex\hbox{$\scriptstyle\sim$}}\text{exp}(\lambda)$ to $Y=e^X$, $$f_Y(y) = f_X(x)_{x=\text{ln}y} \left| \begin{array}{c} \frac{dx}{dy} \end{array} \right|_{x=\text{ln}y}$$ yields $$f_Y(y) = f_X(\text{ln}y)\frac{1}{y}, \forall y\ge1$$ which conforms with the other answers. And if you looked at the CDF by integrating the above, you get a similar CDF as the original exponential. $$F_Y(y) = \int_{1}^y\lambda e^{\lambda\text{ln}y'}\frac{1}{y'}\text{d}y'=\int_0^{\text{ln} y}\lambda e^{\lambda x}\text{d}x$$using change of variables which is equal to $F_X(\text{ln}y)$.
