Poisson Probabilities, Mean, and Variance

From a Rate to a Probability Distribution

A rate parameter becomes useful for probability calculations only after it specifies a valid distribution. For a Poisson count, the rate determines the probability of every possible nonnegative integer count. The same parameter also has a direct interpretation in the distribution’s mean and variance: both equal the rate. These facts follow from the defining probability series, rather than needing to be assumed as part of the model.

Throughout, let $X$ be a count of events in a fixed interval or region, and let $\lambda>0$ denote its Poisson rate for that exposure. The possible values of $X$ are $0,1,2,\ldots$. The rate is the expected count for the specified exposure; changing the exposure generally changes the rate. The calculations below establish the corresponding probability and moment formulas.

Definition: A random variable $X$ has a Poisson distribution with rate $\lambda$, written $X\sim\operatorname{Poisson}(\lambda)$, if $\lambda>0$ and its probability mass function is $$ P(X=k)=f(k)=e^{-\lambda}\frac{\lambda^k}{k!},\qquad k=0,1,2,\ldots. $$ Here $0!=1$. The rate $\lambda$ is a nonnegative real number; when $\lambda=0$, the distribution is understood as the degenerate case $P(X=0)=1$.

The factorial in the denominator is essential: it ensures that the probabilities over all possible counts add to one. To see why, recall the exponential series $e^\lambda=\sum_{k=0}^{\infty}\lambda^k/k!$, which converges for every finite real $\lambda$. The same series will let us calculate the moments.

Normalization: The Probabilities Sum to One

For a discrete random variable to have a probability mass function, each assigned probability must be nonnegative and their sum over the support must equal one. For $\lambda>0$, every Poisson probability is nonnegative because $e^{-\lambda}>0$, $\lambda^k>0$, and $k!>0$. Summing over the nonnegative integers gives the required total.

Theorem: For every $\lambda>0$, the Poisson mass function is normalized: $$ \sum_{k=0}^{\infty} e^{-\lambda}\frac{\lambda^k}{k!}=1. $$

Proof. The terms $\lambda^k/k!$ are nonnegative, and the exponential series converges to $e^\lambda$. Multiplication of its sum by the fixed positive constant $e^{-\lambda}$ therefore gives

\sum_{k=0}^{\infty}e^{-\lambda}\frac{\lambda^k}{k!} =e^{-\lambda}\sum_{k=0}^{\infty}\frac{\lambda^k}{k!} =e^{-\lambda}e^\lambda =1.

This verifies the total probability. In the degenerate case $\lambda=0$, the stated mass is one at zero and zero at every positive integer, so it is normalized as well. $\square$

Worked Example: Checking Probabilities at Rate Two

Suppose $X\sim\operatorname{Poisson}(2)$. Find the probabilities of zero, one, and two events, and explain why these three probabilities do not need to sum to one.

Substitute $\lambda=2$ into the mass function. For zero events, $$ P(X=0)=e^{-2}\frac{2^0}{0!}=e^{-2}\approx 0.135335. $$ For one event, $$ P(X=1)=e^{-2}\frac{2^1}{1!}=2e^{-2}\approx 0.270671. $$ For two events, $$ P(X=2)=e^{-2}\frac{2^2}{2!}=2e^{-2}\approx 0.270671. $$

Their sum is $5e^{-2}\approx 0.676676$, which is less than one because it includes only the outcomes $0,1,2$. The remaining probability belongs to counts $3,4,\ldots$. The normalization result applies when the mass is summed over the entire support, not just these three values.

Deriving the Mean from the Defining Series

For a nonnegative integer-valued random variable, the expectation is the sum of each possible value multiplied by its probability, provided that sum is finite. In the Poisson case, the term for $k=0$ contributes zero, and the other terms can be simplified by canceling one factor of $k$ against $k!$. The resulting series is again an exponential series.

Theorem: If $X\sim\operatorname{Poisson}(\lambda)$ with $0<\lambda<\infty$, then $E[X]=\lambda$.

Proof. Since all terms in the expectation sum are nonnegative, the expectation is defined by that sum, possibly initially as an infinite value. For $k\geq 1$, $k/k!=1/(k-1)!$. Thus

\begin{aligned} E[X] &=\sum_{k=0}^{\infty} k e^{-\lambda}\frac{\lambda^k}{k!}\\ &=e^{-\lambda}\sum_{k=1}^{\infty}\frac{\lambda^k}{(k-1)!}\\ &=\lambda e^{-\lambda}\sum_{j=0}^{\infty}\frac{\lambda^j}{j!}\\ &=\lambda e^{-\lambda}e^\lambda\\ &=\lambda. \end{aligned}

In the third line, $j=k-1$; as $k$ runs from 1 upward, $j$ runs from 0 upward. The final value is finite, so the expectation exists. For $\lambda=0$, $X=0$ with probability one and $E[X]=0=\lambda$. $\square$

In practical terms, if a Poisson model has rate $\lambda=4$ events per specified exposure, its mean count over that exposure is four. This statement concerns an average over repeated observations; it does not mean an individual count must equal four.

Second Moment and Variance

The variance is $E[(X-E[X])^2]$, which can be computed as $E[X^2]-(E[X])^2$ when the second moment is finite. A direct calculation of $E[X^2]$ is possible, but it is especially convenient first to calculate $E[X(X-1)]$. The identity $X^2=X(X-1)+X$ then recovers the second moment.

Theorem: If $X\sim\operatorname{Poisson}(\lambda)$ with $0<\lambda<\infty$, then $$ E[X(X-1)]=\lambda^2,\qquad E[X^2]=\lambda^2+\lambda,\qquad \operatorname{Var}(X)=\lambda. $$

Proof. The $k=0$ and $k=1$ terms of $E[X(X-1)]$ are zero. For $k\geq2$, $k(k-1)/k!=1/(k-2)!$. Reindex with $j=k-2$:

\begin{aligned} E[X(X-1)] &=e^{-\lambda}\sum_{k=2}^{\infty}\frac{k(k-1)\lambda^k}{k!}\\ &=e^{-\lambda}\sum_{k=2}^{\infty}\frac{\lambda^k}{(k-2)!}\\ &=\lambda^2 e^{-\lambda}\sum_{j=0}^{\infty}\frac{\lambda^j}{j!}\\ &=\lambda^2. \end{aligned}

Because $X^2=X(X-1)+X$, linearity of expectation and the mean calculation give

E[X^2]=E[X(X-1)]+E[X]=\lambda^2+\lambda.

In particular, the second moment is finite. Therefore the variance identity applies, and

\operatorname{Var}(X)=E[X^2]-(E[X])^2=(\lambda^2+\lambda)-\lambda^2=\lambda.

When $\lambda=0$, the count is constant at zero, so its variance is zero, again equal to $\lambda$. $\square$

Worked Example: Mean and Variance at Rate Four

Let $X\sim\operatorname{Poisson}(4)$. Find its mean and variance, calculate its second moment, and find the probability of exactly three events.

The mean and variance results give $$ E[X]=4,\qquad \operatorname{Var}(X)=4. $$ The second moment is not the variance: using $E[X^2]=\lambda^2+\lambda$, $$ E[X^2]=4^2+4=20. $$ Indeed, $E[X^2]-(E[X])^2=20-16=4$, as required.

For exactly three events, substitute $k=3$ and $\lambda=4$: $$ P(X=3)=e^{-4}\frac{4^3}{3!} =e^{-4}\frac{64}{6} =\frac{32}{3}e^{-4} \approx 0.195367. $$ This is a single point probability, whereas the mean and variance summarize the distribution across all possible counts.

Worked Example: A Fractional Rate

Suppose the number of calls received in a short, specified time interval follows a Poisson model with rate $0.8$. Find the probabilities of zero, one, and two calls, and state the mean and variance in the context of the model.

For zero calls, $$ P(X=0)=e^{-0.8}\frac{0.8^0}{0!}=e^{-0.8}\approx 0.449329. $$ For one call, $$ P(X=1)=e^{-0.8}\frac{0.8^1}{1!}=0.8e^{-0.8}\approx 0.359463. $$ For two calls, $$ P(X=2)=e^{-0.8}\frac{0.8^2}{2!}=0.32e^{-0.8}\approx 0.143785. $$

The mean number of calls in this interval is $E[X]=0.8$, and the variance of the count is $\operatorname{Var}(X)=0.8$. A non-integer mean is entirely consistent with an integer-valued count: the mean describes an average across repetitions, not a possible value that every observation must take.

Interpreting the Equality of Mean and Variance

For a Poisson count, the mean and variance share the numerical value $\lambda$, but they describe different features. The mean gives the distribution’s expected count; the variance measures the expected squared distance from that mean. Their equality is a derived property of this distribution, not a general rule for count variables. In particular, one should not infer the variance of an arbitrary count merely from its mean.

The rate must also be understood relative to the exposure used to define the count. A rate for a short interval and a rate for a longer interval are not interchangeable without adjusting the exposure. Once the model and exposure are specified, $\lambda$ simultaneously determines each point probability and both moments. The defining series makes those relationships explicit.

Common Mistake: Confusing the Second Moment with Variance

A tempting but incorrect calculation is to find $E[X^2]$ and report it as $\operatorname{Var}(X)$. These quantities are not the same: the variance subtracts the square of the mean. For a Poisson variable, $$ E[X^2]=\lambda^2+\lambda,\qquad \operatorname{Var}(X)=E[X^2]-(E[X])^2 =(\lambda^2+\lambda)-\lambda^2 =\lambda. $$

For example, at rate four, $E[X^2]=20$ but $\operatorname{Var}(X)=4$. Another error is to treat $\lambda$ as the probability of an event: $\lambda$ is the rate for the specified exposure, while individual probabilities are given by $e^{-\lambda}\lambda^k/k!$. The probability of exactly one event, for instance, is $\lambda e^{-\lambda}$, not $\lambda$.

Key takeaway: The exponential series normalizes the Poisson mass function. Reindexing the defining sums gives $E[X]=\lambda$ and $E[X(X-1)]=\lambda^2$; since $X^2=X(X-1)+X$, it follows that $\operatorname{Var}(X)=\lambda$.

Check Your Understanding

Use the mass function and derivations above to answer these questions.

For $X\sim\operatorname{Poisson}(1.5)$, write the expression for $P(X=2)$. What are $E[X]$ and $\operatorname{Var}(X)$?
In the normalization calculation, which exponential series is used, and what constant multiplies it?
Why does the $k=0$ term contribute nothing to the mean, and why do the $k=0$ and $k=1$ terms contribute nothing to $E[X(X-1)]$?
If $\lambda=3$, calculate $E[X^2]$ and then verify the variance by subtracting $(E[X])^2$.
Explain why a Poisson variable can have mean $0.8$ even though its possible values are integers.