Distribution of the exponential of an exponentially distributed random variable? Let $X$ be an exponentially distributed random variable, that is, with density function $f(x)=\lambda e^{-\lambda x}$ for $x\ge 0$ ($\lambda>0$), and cdf $F_X(x)=1 - e^{-\lambda x}$. What is the distribution of $Y=\exp(X)$?
(Note the similar question Distribution of the exponential of an exponential random variable, but that involves a complex number parameter).

First, note that the range of $\DeclareMathOperator{\P}{\mathbb{P}}  Y$ is $(1, \infty)$. First find the cumulative distribution function of $Y$ in the usual way:
$$\begin{align}  
   F_Y(t) & = \P(Y \leq  t) = \P(e^X \le t)        \\
          & = \P( X \leq \ln(t) )                  \\
          & = F_X( \ln(t) ) = 1-e^{-\lambda \ln(t)} \\
          & = 1- e^{\ln( t^{-\lambda})}             \\
          & = 1-t^{-\lambda}
\end{align}$$ for $t\gt 1$. By differentiation we find the density function
$$ f_Y(t) = \lambda t^{-\lambda -1},\quad t>1. $$
which is a Pareto distribution.

Note that this is suspiciously similar to the density of a beta prime distribution.  Define $U=T-1$, which has density function
$$ f_U(u)= \lambda (u+1)^{-\lambda -1},\quad u>0 $$ which we can rewrite as
$$ f_U(u)=\frac{u^{1-1} (u+1)^{-\lambda-1}}{B(1,\lambda)} $$ which we can see is a beta prime density.
So we can reformulate: $e^X -1$ has a beta prime distribution.
