How to Find the Mean and Variance of an Exponential Distribution

The mean and variance of an exponential distribution depend on how its parameter is defined. The most common convention uses a rate parameter, usually written as $\lambda$. Under that convention, the mean is $1/\lambda$ and the variance is $1/\lambda^2$. The derivation below starts from the probability density function, then works through a numerical example and explains the alternative scale convention.

Exponential distribution using the rate parameter

A continuous random variable $X$ has an exponential distribution with rate $\lambda>0$ if its density is

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

and $f(x)=0$ for $x<0$. The rate describes how quickly events occur, so its units are the inverse of the units used for $X$. For example, if $X$ is measured in minutes, $\lambda$ is measured in events per minute (or, more precisely, inverse minutes).

For a continuous random variable, the expectation is the integral of the value times its density:

$$\mathbb{E}[X]=\int_0^\infty x f(x)\,dx.$$

The second moment is calculated similarly, replacing $x$ with $x^2$:

$$\mathbb{E}[X^2]=\int_0^\infty x^2 f(x)\,dx.$$

Once these two moments are known, use $\operatorname{Var}(X)=\mathbb{E}[X^2]-(\mathbb{E}[X])^2$.

Deriving the mean

Substitute the exponential density into the expectation integral:

$$\begin{aligned} \mathbb{E}[X] &=\int_0^\infty x\lambda e^{-\lambda x}\,dx\\ &=\left[-xe^{-\lambda x}\right]_0^\infty+\int_0^\infty e^{-\lambda x}\,dx\\ &=0+\frac{1}{\lambda}\\ &=\frac{1}{\lambda}. \end{aligned}$$

The second line follows from integration by parts. The boundary term is zero: at $x=0$ it is zero, and as $x$ tends to infinity, $xe^{-\lambda x}$ tends to zero. The remaining integral evaluates to $1/\lambda$.

Deriving the variance

First compute the second moment:

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

The second line is another integration by parts; the boundary term vanishes, just as in the mean calculation. Now subtract the square of the mean:

$$\begin{aligned} \operatorname{Var}(X) &=\mathbb{E}[X^2]-(\mathbb{E}[X])^2\\ &=\frac{2}{\lambda^2}-\left(\frac{1}{\lambda}\right)^2\\ &=\frac{1}{\lambda^2}. \end{aligned}$$

Therefore, for the rate parameterization,

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

The standard deviation is the square root of the variance, so it is also $1/\lambda$. The mean and standard deviation have the same units as $X$, while the variance has squared units.

Worked example

Suppose a waiting time $X$, measured in minutes, follows an exponential distribution with rate $\lambda=0.4$ per minute. Then

$$\mathbb{E}[X]=\frac{1}{0.4}=2.5\text{ minutes}, \qquad \operatorname{Var}(X)=\frac{1}{0.4^2}=6.25\text{ minutes}^2.$$

The standard deviation is $\sqrt{6.25}=2.5$ minutes. As a related check, the probability of waiting longer than five minutes is

$$\Pr(X>5)=e^{-\lambda\cdot 5}=e^{-2}\approx 0.1353.$$

This uses the exponential survival function, $\Pr(X>t)=e^{-\lambda t}$ for $t\ge0$. It also illustrates why the rate’s units matter: multiplying a rate in inverse minutes by a time in minutes gives a unitless exponent.

If the parameter is a scale instead

Some texts define the exponential distribution with a scale parameter $\theta>0$ and write its density as

$$f(x)=\frac{1}{\theta}e^{-x/\theta},\qquad x\ge0.$$

Since $\theta=1/\lambda$, the same results become

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

Check the density or the author’s parameter definition before using a formula. A parameter named $\theta$ is not automatically a scale: notation varies, and the defining density is what settles the convention.

Common mistakes

• Mixing up rate and scale. With a rate $\lambda$, the mean is its reciprocal. With a scale $\theta$, the mean equals the scale.
• Calling the standard deviation the variance. The variance is $1/\lambda^2$; the standard deviation is $1/\lambda$.
• Dropping the squared units. If time is measured in minutes, variance is in minutes squared, not minutes.
• Using the formula outside the exponential model. These results require a nonnegative continuous variable with the stated exponential density; other distributions can have different moments.