Exponential of Negative of Exponential Random Variable has Beta Distribution

Theorem

Let $\beta$ be a positive real number.

Let $X \sim \Exponential \beta$ where $\Exponential \beta$ is the exponential distribution with parameter $\beta$.

Then:

$e^{-X} \sim \BetaDist {\dfrac 1 \beta} 1$

Proof

Note that if:

$Y \sim \BetaDist {\dfrac 1 \beta} 1$

then the probability density function of $Y$, $f_Y$ is given by:

\(\ds \map {f_Y} y\)

\(=\)

\(\ds \frac {y^{\frac 1 \beta - 1} \paren {1 - y}^{1 - 1} } {\map \Beta {\frac 1 \beta, 1} }\)

Definition of Beta Distribution

\(\ds \)

\(=\)

\(\ds \frac {y^{\frac 1 \beta - 1} } {\frac {\map \Gamma {\frac 1 \beta} \map \Gamma 1} {\map \Gamma {\frac 1 \beta + 1} } }\)

Definition of Beta Function

\(\ds \)

\(=\)

\(\ds \frac 1 {\frac {\map \Gamma {\frac 1 \beta} } {\frac 1 \beta \map \Gamma {\frac 1 \beta} } } y^{\frac 1 \beta}\)

Gamma Difference Equation

\(\ds \)

\(=\)

\(\ds \frac 1 \beta y^{\frac 1 \beta}\)

for each $y > 0$.

Let:

$Z = e^{-X}$

It suffices to show that $Z$ has the same probability density function as $Y$.

We have:

\(\ds \map \Pr {Z \le z}\)

\(=\)

\(\ds \map \Pr {e^{-X} \le z}\)

\(\ds \)

\(=\)

\(\ds \map \Pr {-X \le \ln z}\)

\(\ds \)

\(=\)

\(\ds \map \Pr {X \ge -\ln z}\)

\(\ds \)

\(=\)

\(\ds \map \exp {-\frac {-\ln z} {\beta} }\)

Definition of Exponential Distribution

\(\ds \)

\(=\)

\(\ds z^{\frac 1 \beta}\)

By Derivative of Power, the probability density function of $Z$, $f_Z$ is therefore given by:

$\map {f_Z} z = \dfrac 1 \beta z^{\frac 1 \beta - 1}$

for each $z > 0$.

So:

$f_Y = f_Z$

$\blacksquare$