Expectation of Exponential Distribution

Theorem
Let $X$ be a continuous random variable of the exponential distribution with parameter $\beta$.

Then the expectation of $X$ is given by:
 * $E \left({X}\right) = \beta$

Proof
The expectation is:


 * $\displaystyle E \left({X}\right) := \int_{x \mathop \in \Omega_X} x \ f_X \left({x}\right) \ \mathrm d x$

which, for the exponential distribution: is:


 * $\displaystyle E \left({X}\right) = \int_0^\infty x \frac 1 \beta \exp{\left(- \frac x \beta \right)} \ \mathrm d x$

where $\exp$ is the exponential function. Substituting $u = \dfrac x \beta$:


 * $\displaystyle E \left({X}\right) = \beta\int_0^ \infty u \exp{\left(-u\right)} \ \mathrm d u$

The integral evaluates to:


 * $\displaystyle E \left({X}\right) = \left.{-\beta \left({u+1}\right) \exp{\left(-u\right)} }\right|_0^\infty$

This evaluation is the limit
 * $\displaystyle E \left({X}\right) = \beta - \beta \lim_{u \to \infty} {\frac {u+1} {\exp u}}$

The exponential function tends to infinity and therefore the limit is indeterminate ($\frac \infty \infty$) and satisfies L'Hôpital's Rule, Corollary 2. This allows us to take the derivatives of the numerator and denominator and achieve the same limit. This changes the limit to:
 * $\displaystyle E \left({X}\right) = \beta - \beta \lim_{u \to \infty} {\frac 1 {\exp u}}$

This limit tends to zero and therefore
 * $E \left({X}\right) = \beta$