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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).

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2 Answers 2

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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)

enter image description here

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

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  • $\begingroup$ So briefly, does that mean or does that not mean the exponent of an exponentially distributed random variable is Pareto distributed? $\endgroup$ Nov 30, 2021 at 12:49
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    $\begingroup$ @RichardHardy I suspect it is trying to say yes, with minimum value $1$ (and shape parameter $\alpha=\lambda$) $\endgroup$
    – Henry
    Nov 30, 2021 at 13:52
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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.

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    $\begingroup$ This is a Pareto distribution. $\endgroup$
    – Yves
    Jan 1, 2022 at 10:37

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