We can now derive the mean and variance of the exponential distribution.

Exponential mean and variance

If \(X\) has an \(\ExponDistn(\lambda) \) distribution with pdf

\[ f(x) \;\;=\;\; \lambda e^{-\lambda x} \]

then its mean and variance are

\[ E[X] \;\;=\;\; \frac 1 {\lambda} \spaced{and} \Var(X) \;\;=\;\; \frac 1 {\lambda^2} \]

The distribution's mean is

\[ \begin{align} E[X] \;\;&=\;\; \int_0^{\infty} {x \times \lambda e^{-\lambda x}} \; dx \\ &=\;\; \int_0^{\infty} {\frac y {\lambda} \times \lambda e^{-y}} \; \frac {dy}{\lambda} \quad\quad \text{with a change of variable } y = \lambda x \\ &=\;\; \frac{1}{\lambda} \int_0^{\infty} {y e^{-y}} \; dy \\ &=\;\; \frac{1}{\lambda} \times \Gamma(2) \;\;=\;\; \frac{1}{\lambda} \times (1!) \;\;=\;\; \frac{1}{\lambda} \end{align} \]

In a similar way,

\[ \begin{align} E[X^2] \;\;&=\;\; \int_0^{\infty} {x^2 \times \lambda e^{-\lambda x}} \; dx \\ &=\;\; \frac{1}{\lambda^2} \int_0^{\infty} {y^2 e^{-y}} \; dy \\ &=\;\; \frac{1}{\lambda^2} \times \Gamma(3) \;\;=\;\; \frac{1}{\lambda^2} \times (2!) \;\;=\;\; \frac{2}{\lambda^2} \end{align} \]

The variance is

\[ \begin{align} \Var(X) \;\;&=\;\; E[X^2] - \left(E[X]\right)^2 \\ &=\;\; \frac{2}{\lambda^2} - \left(\frac{1}{\lambda}\right)^2 \\ &=\;\; \frac 1 {\lambda^2} \end{align} \]