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:


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

Substituting $u = \dfrac x \beta$:


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

The integral evaluates to:


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

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

This limit is indeterminate ($\frac \infty \infty$) and therefore satisfies L'Hôpital's Rule which changes the limit to:
 * $\displaystyle E \left({X}\right) = \beta - \beta \lim_{u \to \infty} {\frac {1} {e^u}}$

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