# Expectation of Exponential Distribution

## Theorem

Let $X$ be a continuous random variable of the exponential distribution with parameter $\beta$ for some $\beta \in \R_{> 0}$

Then the expectation of $X$ is given by:

$\expect X = \beta$

## Proof 1

The expectation of a continuous random variable $X$ with sample space $\Omega_X$ is given by:

$\ds \expect X := \int_{x \mathop \in \Omega_X} x \map {f_X} x \rd x$

For the exponential distribution, we have $\Omega_X = \hointr 0 \infty$, and:

$\ds \map {f_X} x = \frac 1 \beta \map \exp {- \frac x \beta}$

So:

$\ds \expect X = \int_0^\infty x \frac 1 \beta \map \exp {- \frac x \beta} \rd x$

where $\exp$ is the exponential function.

Substituting $u = \dfrac x \beta$, we have::

$\ds \expect X = \beta \int_0^\infty u \map \exp {-u} \rd u$

The integral evaluates to:

$\ds \expect X = \bigintlimits {-\beta \paren {u + 1} \map \exp {-u} } 0 \infty$

So:

$\ds \expect X = \beta - \beta \lim_{u \mathop \to \infty} \frac {u + 1} {\exp u}$

By Limit at Infinity of Polynomial over Complex Exponential, it follows that this limit is zero, so that:

$\expect X = \beta$

$\blacksquare$

## Proof 2

From Moment Generating Function of Exponential Distribution, the moment generating function $M_X$ of $X$, is given by:

$\map {M_X} t = \dfrac 1 {1 - \beta t}$
$\expect X = \map {M_X'} 0$

We have:

 $\ds \map {M_X'} t$ $=$ $\ds \map {\frac \d {\d t} } {\frac 1 {1 - \beta t} }$ $\ds$ $=$ $\ds \frac {-\beta} {-1} \frac 1 {\paren {1 - \beta t}^2}$ Chain Rule for Derivatives, Derivative of Power $\ds$ $=$ $\ds \frac \beta {\paren {1 - \beta t}^2}$

Setting $t = 0$ gives:

 $\ds \expect X$ $=$ $\ds \frac \beta {\paren {1 - 0 \beta}^2}$ $\ds$ $=$ $\ds \beta$

$\blacksquare$