Mean and Variance of the Exponential Distribution

Let $X$ follow an exponential distribution with rate parameter $\lambda > 0,$ so its pdf is

$$f(x) = \lambda e^{-\lambda x}, \quad x \geq 0.$$

The mean and variance of $X$ are

$$E[X] = \dfrac{1}{\lambda}, \qquad \text{Var}(X) = \dfrac{1}{\lambda^2}.$$

Taking the square root of the variance, the standard deviation is

$$\sigma = \sqrt{\text{Var}(X)} = \dfrac{1}{\lambda}.$$

Notice that the mean and standard deviation of an exponential random variable are equal.

For example, if $\lambda = 5,$ then

$$\begin{align*} E[X] &= \dfrac{1}{5} = 0.2, \\[3pt] \text{Var}(X) &= \dfrac{1}{5^2} = \dfrac{1}{25} = 0.04. \end{align*}$$