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).
 A: Sample from a Pareto distribution. If $Y\sim\mathsf{Exp}(\mathrm{rate}=\lambda),$ then $X = x_m\exp(Y)$ has a Pareto distribution with density function $f_X(x) = \frac{\lambda x_m^\lambda}{x^{\lambda+1}}$
and CDF $F_X(x) = 1-\left(\frac{x_m}{x}\right)^\lambda,$
for $x\ge x_m > 0.$ The minimum value $x_m > 0$ is necessary
for the integral of the density to exist.
Consider the random sample y of $n = 1000$ observations
from $\mathsf{Exp}(\mathrm{rate}=\lambda=5)$ along with
the Pareto sample y resulting from the transformation above.
    set.seed(1128)
    x.m = 1;  lam = 5
    y = rexp(1000, lam)
    summary(y)
 
         Min.   1st Qu.    Median      Mean   3rd Qu.      Max. 
    0.0001314 0.0519039 0.1298572 0.1946130 0.2743406 1.9046195 

    x = x.m*exp(y)
    summary(x)

      Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
     1.000   1.053   1.139   1.245   1.316   6.717 

Below is the empirical CDF (ECDF) of Pareto sample x along
with the CDF (dotted orange) of the distribution from which it was sampled. Tick marks along the horizontal axis show individual
values of x.
    plot(ecdf(x), main="ECDF of Pareto Sample")
     curve(1 - (x.m/x)^lam, add=T, 1, 4, 
           lwd=3, col="orange", lty="dotted")
     rug(x)


Ref: See the Wikipedia page on Pareto distributions, under the heading for relationship to exponential.
A: 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.
