What Is the Expected Value of a Poisson Distribution?

A Poisson random variable counts how many times an event occurs in a fixed interval, area, or other exposure. Its expected value is determined directly by its parameter. Below is the result, a short derivation from the probability mass function, and an example showing how to use an event rate over a longer interval.

Expected value of a Poisson random variable

If $X$ has a Poisson distribution with parameter $\lambda$, written $X\sim\operatorname{Poisson}(\lambda)$, then

Here $\lambda\geq 0$ is the average number of events in the interval or region being modeled. In some applications it is called the mean parameter; it is not the event count observed in any particular interval.

The probability mass function is

The formula applies when $\lambda>0$. At $\lambda=0$, the distribution is concentrated at $X=0$, so its expected value is also zero.

Deriving the mean from the probability mass function

By the definition of expectation for a discrete random variable, multiply each possible value $k$ by its probability and sum over all nonnegative integers:

The $k=0$ term is zero, which is why the final sum starts at $k=1$. Now set $j=k-1$. As $k$ runs from 1 upward, $j$ runs from 0 upward:

The sum is the power-series expansion of $e^{\lambda}$. This calculation shows why the parameter itself is the mean.

Worked example: events per hour

Suppose a help desk receives requests according to a Poisson process at an average rate of 2.5 requests per hour. For six hours, the Poisson parameter is the rate multiplied by the exposure time:

Therefore, the expected number of requests in six hours is 15. This does not mean that exactly 15 requests will arrive in every six-hour period. A particular period might have fewer or more; 15 is the long-run average across many periods of the same length.

The multiplication by time assumes the stated rate applies throughout the interval. More generally, if events follow a homogeneous Poisson process with constant rate $r$ per unit of exposure for duration $t$, then $N(t)\sim\operatorname{Poisson}(rt)$ and $\mathbb{E}[N(t)]=rt$.

What the expected value tells you

An expected value is a probability-weighted average, not a promise about an individual outcome. Since a count can only be a whole number, a Poisson variable never takes the value 2.5, for example. But its mean can be 2.5 because it averages possible whole-number outcomes over repeated observations.

For a Poisson distribution, the variance is also $\lambda$, so its standard deviation is $\sqrt{\lambda}$. This is a separate fact from the expected-value result, but it helps describe how much counts tend to vary around their average. For instance, when $\lambda=15$, the mean is 15 and the standard deviation is $\sqrt{15}\approx3.87$ events.

Common mistakes

• Treating $\lambda$ as an observed count. It is the model’s average count for the specified exposure. An individual observation may be different.
• Assuming the expected value must be an integer. A Poisson outcome is an integer, but its expected value need not be.
• Forgetting to adjust for the interval. If a rate is given per hour but the question concerns six hours, use six hours of exposure: $\lambda=rt$.
• Confusing a Poisson mean with a rate. The mean is a count for a specified interval. A rate has units such as events per hour; multiply it by the duration to obtain the Poisson parameter.
• Changing conventions without checking the model. In the standard notation $\operatorname{Poisson}(\lambda)$, $\lambda$ is the mean. In a word problem, confirm whether the supplied number is already the expected count for the target interval or is instead a rate that needs scaling.

For a quick check, ask what interval the parameter describes and whether it is a count or a rate. Once the Poisson parameter for the interval is identified, that number is the expected value.