How to find the distribution of $Y = {e^{aX}}$ when $X$ is an exponential random variable?

Question

How to find the distribution of $Y = {e^{aX}}$ when $X$ is an exponential random variable with ${f_X}\left( x \right) = \frac{1}{\lambda }\exp \left( { - \frac{x}{\lambda }} \right)$ and $a>0$ ?

First attemp with $a=1$ by using Mathematica return a Pareto Distribution However, for $a=3$ or in general $a>0$, Mathematica cannot do anything: Please help me with this ! Thank you very much !

Answer 1

If you are uncomfortable with a manual change of variables on the density, then use the cumulative distribution functions

• $\mathbb P(X \le x) = 1-\exp \left( { - \frac{x}{\lambda }} \right)$ if $x \ge 0$, and $=0$ if $x \le 0$

• $\mathbb P(Y \le e^{ax}) = 1-\exp \left( { - \frac{x}{\lambda }} \right)$ if $x \ge 0$, and $=0$ if $x \lt 0$

• $\mathbb P(Y \le y) = 1-\exp \left( { - \frac{\frac1a \log_e(y)}{\lambda }} \right) =1-y^{-\frac{1}{a\lambda}}$ if $y \ge 1$, and $=0$ if $y \lt 1$

• $f_Y(y) = \frac{1}{a\lambda}y^{-\frac{1}{a\lambda}-1}$ if $y \ge 1$, and $=0$ if $y \lt 1$

which is indeed a Pareto distribution with minimum $1$

Answer 2

As an alternative, if you want to directly derive $f_Y(y)$, consider that

$$y=e^{ax}$$

$$x=\frac{1}{a}\log y$$

$$|x'|=\frac{1}{ay}$$

thus

$$f_Y(y)=\frac{1}{ay\lambda}\left[e^{\log(1/y)}\right]^{1/(a\lambda)}=\frac{1}{a\lambda y^{1/(a\lambda)+1}}$$

This is your density setting $a=3$ and $\lambda=1$

Answer 3

Because $X$ is exponential, we know $P(X \geq 0) = 1$, which means $P(Y \geq 1) = 1$, so the PDF of $Y$ will be zero outside $[1, \infty)$.

We know that the CDF of $X$ is $$ P(X \leq z) = \int_{0}^z \lambda^{-1} e^{-\lambda^{-1} x} \, dx = 1 - e^{-\lambda^{-1} z} $$

Then write down the CDF of $Y$ and use what you know: for $y \geq 1$, \begin{align*} P(Y \leq y) &= P( e^{a X} \leq y ) \\ &= P( a X \leq \log y ) \\ &= P( X \leq a^{-1} \log y ) \\ &= 1 - e^{-(a \lambda)^{-1} \log y} \\ &= 1 - e^{\log y^{-(a \lambda)^{-1}}} \\ &= 1 - y^{-(a \lambda)^{-1}} \end{align*}

Let $\beta = (a \lambda)^{-1}$. Now differentiate with respect to $y$ on both sides to obtain, for $y \geq 1$, $$ f_Y(y) = \beta y^{-\beta - 1} $$ If you want, you can write out the PDF in full: $$ f_Y(y) = \begin{cases} 0 & y < 1 \\ \frac{\beta}{y^{\beta + 1}} & y \geq 1. \end{cases} $$ This is the PDF of a Pareto random variable with parameter $\beta$ with $\beta = (a \lambda)^{-1}$ as above. In Mathematica's notation, it would be ParetoDistribution[1,$\beta$].