# 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).

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)


• @RichardHardy I suspect it is trying to say yes, with minimum value $1$ (and shape parameter $\alpha=\lambda$) Nov 30 '21 at 13:52
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 $$f_Y(t) = \lambda t^{-\lambda -1},\quad t>1.$$
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.