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.

    x.m = 1;  lam = 5
    y = rexp(1000, lam)
         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)

      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")

enter image description here

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

  • $\begingroup$ So briefly, does that mean or does that not mean the exponent of an exponentially distributed random variable is Pareto distributed? $\endgroup$ Nov 30 '21 at 12:49
  • 2
    $\begingroup$ @RichardHardy I suspect it is trying to say yes, with minimum value $1$ (and shape parameter $\alpha=\lambda$) $\endgroup$
    – Henry
    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.

  • $\begingroup$ This is a Pareto distribution. $\endgroup$
    – Yves
    Jan 1 at 10:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.