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:

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

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

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

So:

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

where $\exp$ is the exponential function.

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

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

The integral evaluates to:

$\displaystyle \expect X = \left.{ -\beta \paren {u+1} \map \exp {-u} }\right|_0^\infty$

So:

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

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 of $X$, $M_X$, is given by:

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

We have:

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

Setting $t = 0$ gives:

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

$\blacksquare$