Deriving the Mean and Variance of a Poisson Count

Why Derive the Mean and Variance?

A Poisson model describes a count over a specified period or amount of exposure. Its parameter determines both the expected count and how much the count varies around that expectation. Deriving those facts from the probability mass function makes the formulas easier to remember and helps prevent a common error: confusing the second factorial moment with the second raw moment.

The results are especially useful in actuarial count models. If a hypothetical portfolio has a Poisson-distributed number of claims, the mean gives the expected claim count and the variance quantifies count uncertainty under the model. If the count is divided by exposure to form a frequency, its variance changes as well; it is not the variance of the original count.

Definition: A random variable $X$ has a Poisson distribution with parameter $\lambda$, written $X\sim\operatorname{Poisson}(\lambda)$, if $\lambda\geq 0$, $X$ takes nonnegative integer values, and $$ P(X=x)=e^{-\lambda}\frac{\lambda^x}{x!},\qquad x=0,1,2,\ldots. $$ For $\lambda=0$, $X=0$ with probability 1. The derivations below use $\lambda>0$; their final formulas also hold at $\lambda=0$.

The parameter $\lambda$ is the mean count implied by the model, but here it will be derived rather than assumed. For a count over exposure $t$, if the assumed constant rate is $r$ per unit of exposure, the Poisson parameter is $\lambda=rt$. This use of a rate requires that the count model and exposure definition are appropriate for the situation; it does not establish that a Poisson model is appropriate for every real count process.

Factorial Moments from the Probability Mass Function

The first factorial moment is $E[X]$. The second factorial moment is $E[X(X-1)]$, which differs from the second raw moment $E[X^2]$. Factorial terms are convenient for Poisson calculations because the factors $x$ and $x(x-1)$ cancel corresponding parts of $x!$.

For $\lambda>0$, start with the expectation definition and substitute the Poisson probability mass function:

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

The $x=0$ term vanishes because it is multiplied by zero. In the remaining sum, let $y=x-1$. As $x$ ranges from 1 upward, $y$ ranges from 0 upward. The exponential series gives $\sum_{y=0}^{\infty}\lambda^y/y!=e^\lambda$, so

E[X]=\lambda e^{-\lambda}e^\lambda=\lambda.

Now calculate the second factorial moment. The terms for $x=0$ and $x=1$ vanish, so the sum begins at 2. For $x\geq 2$, $x(x-1)/x!=1/(x-2)!$:

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

To obtain the variance, first relate the factorial moment to the second raw moment. Since $X^2=X(X-1)+X$, linearity of expectation gives $E[X^2]=E[X(X-1)]+E[X]$. Then apply the definition $\operatorname{Var}(X)=E[X^2]-(E[X])^2$.

Theorem: If $X\sim\operatorname{Poisson}(\lambda)$, then $$ E[X]=\lambda,\qquad E[X(X-1)]=\lambda^2,\qquad E[X^2]=\lambda^2+\lambda,\qquad \operatorname{Var}(X)=\lambda. $$ The standard deviation is therefore $\sqrt{\lambda}$, not $\lambda$.

Worked Applications

Worked Example: Monthly Service Requests

A hypothetical service desk models the number $X$ of requests received in one month as $\operatorname{Poisson}(2.5)$. Find the expected number of requests, the second factorial moment, the second raw moment, and the variance. Check the variance using the raw-moment formula.

Method and model. The count is assumed to have a Poisson distribution with parameter $\lambda=2.5$ requests per month. Apply the factorial-moment derivation.

Mean and second factorial moment.

E[X]=\lambda=2.5,\qquad E[X(X-1)]=\lambda^2=(2.5)^2=6.25.

Second raw moment. Use $X^2=X(X-1)+X$:

$$ E[X^2]=E[X(X-1)]+E[X]=6.25+2.5=8.75. $$

Variance and check.

\operatorname{Var}(X)=E[X^2]-(E[X])^2 =8.75-(2.5)^2=8.75-6.25=2.5.

Thus the expected count is 2.5 requests, and the variance is 2.5 requests-squared. The check agrees with $\operatorname{Var}(X)=\lambda$. A common mistake is to report 6.25 as $E[X^2]$; that is the second factorial moment, not the second raw moment. Adding $E[X]=2.5$ gives the correct raw moment, 8.75.

Worked Example: Claim Count Across Policy Exposure

For a hypothetical coverage, assume claim counts follow a Poisson model with a constant rate of $0.30$ claims per policy-year. A group is observed for 20 policy-years of total exposure. Find the model parameter, expected total count, variance, and standard deviation of the total count.

Method and model. The assumed rate is $r=0.30$ claims per policy-year and the total exposure is $t=20$ policy-years. Therefore the Poisson parameter for the total count $N$ is

\lambda=rt=(0.30)(20)=6,\qquad N\sim\operatorname{Poisson}(6).

The unit policy-years cancels in the rate-times-exposure calculation, leaving a mean count parameter of 6 claims. Apply the derived formulas:

E[N]=6,\qquad \operatorname{Var}(N)=6,\qquad \operatorname{SD}(N)=\sqrt{6}\approx 2.4495.

The mean and variance are both 6; the standard deviation is approximately 2.4495 claims. These are distinct summaries: the variance is measured in squared count units, while the standard deviation is measured in count units.

Check. The second factorial moment is $E[N(N-1)]=6^2=36$, so $E[N^2]=36+6=42$. Consequently,

\operatorname{Var}(N)=42-6^2=42-36=6.

The result agrees with the direct formula. A common error is to use the rate, $0.30$, as the parameter for the whole group. That would ignore the 20 policy-years of exposure; the parameter is $0.30\times20=6$, not $0.30$.

Worked Example: Variance of a Claim Frequency

A hypothetical portfolio has 50 policy-years of exposure. Assume its claim count $N$ is Poisson with a constant rate of $0.08$ claims per policy-year. Find the expected claim frequency $N/50$ and its variance. Also calculate the variance of the total count to distinguish the two quantities.

Method and model. The count parameter is the rate multiplied by exposure:

\lambda=(0.08)(50)=4,\qquad N\sim\operatorname{Poisson}(4).

It follows that $E[N]=4$ and $\operatorname{Var}(N)=4$. Define the frequency $R=N/50$, measured in claims per policy-year. By the scaling rules for expectation and variance,

E[R]=E\left[\frac{N}{50}\right]=\frac{E[N]}{50} =\frac{4}{50}=0.08,

\operatorname{Var}(R)=\operatorname{Var}\left(\frac{N}{50}\right) =\frac{\operatorname{Var}(N)}{50^2} =\frac{4}{2500}=0.0016.

Thus the expected frequency is $0.08$ claims per policy-year, the variance of the total count is 4, and the variance of the frequency is 0.0016. The frequency variance is divided by the square of the exposure, not by the exposure itself. Using $4/50=0.08$ as the frequency variance would confuse the mean frequency with its variance.

Common Errors and What the Results Mean

The derivation uses the Poisson probability mass function and the exponential-series identity. It is valid for a Poisson-distributed count; the equality of mean and variance is not a general rule for all discrete random variables. Before applying the result, identify the count, its exposure, and the Poisson parameter for that count.

Confusing factorial and raw moments. The second factorial moment is $E[X(X-1)]=\lambda^2$, whereas $E[X^2]=\lambda^2+\lambda$.
Stopping before calculating variance. Knowing $E[X(X-1)]$ is not yet knowing $\operatorname{Var}(X)$. Use $X^2=X(X-1)+X$, then subtract $(E[X])^2$.
Using the rate instead of the count parameter. For rate $r$ per exposure unit and exposure $t$, calculate $\lambda=rt$ before finding the count mean and variance.
Reporting variance as standard deviation. For a Poisson count, the variance is $\lambda$, but the standard deviation is $\sqrt{\lambda}$.
Scaling a variance incorrectly. If a count is divided by exposure $t$, its variance is divided by $t^2$, because $\operatorname{Var}(cX)=c^2\operatorname{Var}(X)$.

Key takeaway: The Poisson probability mass function gives $E[X]=\lambda$ and $E[X(X-1)]=\lambda^2$. Combining these with $X^2=X(X-1)+X$ yields $E[X^2]=\lambda^2+\lambda$ and $\operatorname{Var}(X)=\lambda$.

Check Your Understanding

For each problem, state the relevant Poisson parameter before calculating the requested quantities.

A hypothetical monitoring system records $X\sim\operatorname{Poisson}(3.5)$ alerts in a fixed interval. Find $E[X]$ and $\operatorname{Var}(X)$. (Answer: mean $=3.5$, variance $=3.5$)
A hypothetical repair program has a constant assumed rate of $0.12$ repairs per equipment-year and observes 25 equipment-years. Find the expected total count and its variance. (Answer: $\lambda=3$, mean $=3$, variance $=3$)
If $Y\sim\operatorname{Poisson}(5)$, find $E[Y(Y-1)]$ and $E[Y^2]$. (Answer: $25$ and $30$, respectively)
A hypothetical portfolio has 40 policy-years of exposure and an assumed rate of $0.06$ claims per policy-year. Let $N$ be the total claim count and $R=N/40$. Find $E[R]$ and $\operatorname{Var}(R)$. (Answer: $E[R]=0.06$, $\operatorname{Var}(R)=0.0015$)
A hypothetical monthly count $Z$ is modeled as $\operatorname{Poisson}(7)$. Calculate $E[Z^2]$, then use it to verify the variance. (Answer: $E[Z^2]=56$, and $\operatorname{Var}(Z)=56-49=7$)

What You'll Learn