What Is the Standard Deviation of an Exponential Distribution?

For an exponential distribution, the standard deviation is the reciprocal of its rate parameter. Since authors sometimes describe the same distribution using a scale parameter instead, first check which parameterization you have. The derivation below also shows why the exponential distribution’s standard deviation equals its mean.

Formula for the standard deviation

If $X$ has an exponential distribution with rate parameter $\lambda>0$, its probability density function is

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

The mean and variance are

$$\mathbb{E}[X]=\frac{1}{\lambda},\qquad \operatorname{Var}(X)=\frac{1}{\lambda^2}.$$

Therefore, the standard deviation is

$$\sigma_X=\sqrt{\operatorname{Var}(X)}=\frac{1}{\lambda}.$$

The rate $\lambda$ measures events per unit of time (or distance, or another measured quantity). Thus $1/\lambda$ has the units of the quantity being measured, as a standard deviation should.

Deriving the result from moments

The variance is $\mathbb{E}[X^2]-(\mathbb{E}[X])^2$. For the exponential density, substitution $u=\lambda x$ gives

$$\begin{aligned} \mathbb{E}[X] &= \int_0^\infty x\lambda e^{-\lambda x}\,dx =\frac{1}{\lambda}\int_0^\infty u e^{-u}\,du =\frac{1}{\lambda},\\ \mathbb{E}[X^2] &= \int_0^\infty x^2\lambda e^{-\lambda x}\,dx =\frac{1}{\lambda^2}\int_0^\infty u^2 e^{-u}\,du =\frac{2}{\lambda^2}. \end{aligned}$$

The last integrals are $1$ and $2$, respectively; they can be evaluated by integration by parts. So

$$\operatorname{Var}(X)=\frac{2}{\lambda^2}-\left(\frac{1}{\lambda}\right)^2=\frac{1}{\lambda^2},$$

and taking the square root yields $\sigma_X=1/\lambda$.

Rate and scale parameterizations

Some books use the scale parameter $\theta=1/\lambda$ instead of the rate. In that convention, the density is

$$f(x)=\frac{1}{\theta}e^{-x/\theta},\qquad x\ge 0,$$

and both the mean and standard deviation are $\theta$. Equivalently, if a problem gives the mean $\mu$ of an exponential distribution, its standard deviation is also $\mu$.

Given parameter	Standard deviation
Rate $\lambda$	$1/\lambda$
Scale $\theta$	$\theta$
Mean $\mu$	$\mu$

The rate and scale are reciprocals, not interchangeable numerical values. For example, a rate of $0.2$ per minute corresponds to a scale of $5$ minutes.

Worked example

Suppose the time between arrivals follows an exponential distribution with rate $\lambda=0.2$ arrivals per minute. Then

$$\sigma_X=\frac{1}{0.2\text{ per minute}}=5\text{ minutes}.$$

The expected waiting time is also $1/0.2=5$ minutes. If the same distribution is described by its scale, the scale is $\theta=5$ minutes, and the standard deviation is $\theta=5$ minutes. The units are a useful check: the answer is in minutes, not arrivals per minute.

Common mistakes

• Using the rate as the standard deviation. The rate has inverse units. Take its reciprocal: $\sigma=1/\lambda$.
• Confusing rate and scale. A rate of $0.2$ per minute is not a scale of $0.2$ minutes; its scale is $5$ minutes.
• Using the variance as the standard deviation. The variance is $1/\lambda^2$; take its square root to obtain $1/\lambda$.
• Treating a sample’s standard deviation as fixed by the model. The formula gives the population or theoretical standard deviation. A finite sample has a sample standard deviation calculated from its observed values, which can differ from the theoretical value.

For a quick check, an exponential distribution with mean 8 seconds has standard deviation 8 seconds. If the stated rate is instead $1/8$ per second, the same result follows by taking its reciprocal.