Poisson Sums Through MGFs

Why Poisson Sums Are Useful

A Poisson random variable models a count of events in a specified observation period or exposure, when its parameter gives the expected count in that period or exposure. In practice, a total count may combine events from several independent sources: for example, counts from separate hypothetical service channels or separate categories of insured events. The moment generating function (MGF) gives a direct way to establish the distribution of that total.

The key result is simple but depends on an important condition: the Poisson random variables being added must be independent. Under that condition, their MGFs multiply, and the product has the form of a Poisson MGF with parameter equal to the sum of the individual parameters. This lets us identify the distribution of the total before calculating probabilities.

Definition: A random variable $X$ has a Poisson distribution with parameter $\lambda>0$, written $X\sim\operatorname{Poisson}(\lambda)$, if $$ P(X=k)=e^{-\lambda}\frac{\lambda^k}{k!},\qquad k=0,1,2,\ldots. $$ The parameter $\lambda$ is the expected count for the specified observation period or exposure. A Poisson variable has $E[X]=\lambda$ and $\operatorname{Var}(X)=\lambda$.

The Poisson MGF

The MGF of a random variable $X$ is $M_X(t)=E[e^{tX}]$, for values of $t$ for which this expectation is finite. For a Poisson variable, substitute its probability mass function into the expectation. The exponential series then gives the MGF explicitly.

\begin{aligned} M_X(t) &=\sum_{k=0}^{\infty}e^{tk}e^{-\lambda}\frac{\lambda^k}{k!}\\ &=e^{-\lambda}\sum_{k=0}^{\infty}\frac{(\lambda e^t)^k}{k!}\\ &=e^{-\lambda}e^{\lambda e^t}\\ &=\exp\!\bigl(\lambda(e^t-1)\bigr). \end{aligned}

The exponential series converges for every real value of $t$, so this MGF is finite for all real $t$. Its form is especially useful: the Poisson parameter appears as a coefficient multiplying $e^t-1$. When independent Poisson MGFs are multiplied, those coefficients add.

Theorem: If $X_1,\ldots,X_n$ are independent and $X_j\sim\operatorname{Poisson}(\lambda_j)$ for each $j$, then $$ S=\sum_{j=1}^{n}X_j\sim\operatorname{Poisson}\!\left(\sum_{j=1}^{n}\lambda_j\right). $$ Independence is required for the MGF product step. The parameters must describe counts on the specified periods or exposures used in the model.

Why the Parameters Add

The product rule for independent MGFs states that if $X$ and $Y$ are independent, then $M_{X+Y}(t)=M_X(t)M_Y(t)$, wherever the MGFs exist. For independent Poisson variables, substitute the individual MGF formulas into that rule.

\begin{aligned} M_{X+Y}(t) &=M_X(t)M_Y(t)\\ &=\exp\!\bigl(\lambda_1(e^t-1)\bigr) \exp\!\bigl(\lambda_2(e^t-1)\bigr)\\ &=\exp\!\bigl((\lambda_1+\lambda_2)(e^t-1)\bigr). \end{aligned}

The last expression is the MGF of a Poisson variable with parameter $\lambda_1+\lambda_2$. Since the MGFs agree for all real $t$, the uniqueness theorem for MGFs identifies the distribution of the sum. For $n$ independent variables, the same argument multiplies $n$ exponential terms, adding all their parameters.

This also provides a useful check. The mean and variance of the sum are both $\sum_j\lambda_j$: independence gives the sum of the means and the sum of the variances, respectively. These checks support the MGF identification, although matching the mean and variance alone is not enough to establish a distribution.

State the model.
Write down each Poisson distribution, its parameter, the observation period or exposure, and the independence assumption.

Apply the product rule.
Multiply the individual MGFs. Combine their exponents to obtain a single coefficient of $e^t-1$.

Identify the total.
Match the resulting MGF to the Poisson MGF, then use the summed parameter for any requested probability.

Check the result.
Confirm that the total parameter agrees with the sum of the component means and variances.

Worked Applications

Worked Example: Combining Two Independent Event Counts

Model and method. In a hypothetical one-week monitoring period, let $X$ count events from one source and $Y$ count events from a separate source. Assume $X\sim\operatorname{Poisson}(2.4)$, $Y\sim\operatorname{Poisson}(3.6)$, and $X$ and $Y$ are independent. Find the distribution of the total $S=X+Y$, then calculate $P(S\geq 2)$. Use the product rule and Poisson probability formula.

The component MGFs are $$ M_X(t)=\exp\!\bigl(2.4(e^t-1)\bigr), \qquad M_Y(t)=\exp\!\bigl(3.6(e^t-1)\bigr). $$ By independence, $$ \begin{aligned} M_S(t) &=M_X(t)M_Y(t)\\ &=\exp\!\bigl((2.4+3.6)(e^t-1)\bigr)\\ &=\exp\!\bigl(6(e^t-1)\bigr). \end{aligned} $$

This is the MGF of $\operatorname{Poisson}(6)$, so $S\sim\operatorname{Poisson}(6)$. The requested event is the complement of $S=0$ and $S=1$: $$ \begin{aligned} P(S\geq 2) &=1-P(S=0)-P(S=1)\\ &=1-e^{-6}\frac{6^0}{0!}-e^{-6}\frac{6^1}{1!}\\ &=1-e^{-6}(1+6)\\ &=1-7(0.002478752)\\ &=1-0.017351264\\ &=0.982648736. \end{aligned} $$

Rounded to four decimal places, $P(S\geq2)=0.9826$. As a check, $E[S]=2.4+3.6=6$ and, by independence, $\operatorname{Var}(S)=2.4+3.6=6$, consistent with the identified Poisson distribution. The method avoids calculating the total probability by separately enumerating every pair $(X,Y)$ that could produce a total of at least two.

Worked Example: Finding an Exact Total from Three Sources

Model and method. For a hypothetical four-day period, let independent counts from three sources be $X_1\sim\operatorname{Poisson}(1.25)$, $X_2\sim\operatorname{Poisson}(2.00)$, and $X_3\sim\operatorname{Poisson}(0.75)$. Find the distribution of $S=X_1+X_2+X_3$, and calculate the probability that the total equals four. Use repeated application of the product rule.

The total MGF is the product of the three MGFs: $$ \begin{aligned} M_S(t) &=\prod_{j=1}^{3}M_{X_j}(t)\\ &=\exp\!\bigl(1.25(e^t-1)\bigr) \exp\!\bigl(2.00(e^t-1)\bigr) \exp\!\bigl(0.75(e^t-1)\bigr)\\ &=\exp\!\bigl((1.25+2.00+0.75)(e^t-1)\bigr)\\ &=\exp\!\bigl(4(e^t-1)\bigr). \end{aligned} $$ Thus $S\sim\operatorname{Poisson}(4)$. Applying the Poisson probability formula, $$ \begin{aligned} P(S=4) &=e^{-4}\frac{4^4}{4!}\\ &=e^{-4}\frac{256}{24}\\ &=e^{-4}(10.6666667)\\ &=(0.018315639)(10.6666667)\\ &=0.195366816. \end{aligned} $$

Rounded to four decimal places, $P(S=4)=0.1954$. The parameter check is $1.25+2.00+0.75=4$; consequently, both the mean and variance of the total are four. A common mistake is to multiply the three parameters. That would give $1.25(2.00)(0.75)=1.875$, which is not the coefficient in the product MGF and also conflicts with the total mean of four.

Worked Example: Using a Total Mean to Determine a Component Parameter

Model and method. In a hypothetical observation period, independent counts $X$ and $Y$ satisfy $X\sim\operatorname{Poisson}(\theta)$ and $Y\sim\operatorname{Poisson}(\theta+2)$, where $\theta>0$. The expected total $S=X+Y$ is eight. Find $\theta$, identify the distribution of $S$ by its MGF, and calculate $P(S\leq1)$.

First, use the mean of a Poisson variable and linearity of expectation: $$ E[S]=E[X]+E[Y]=\theta+(\theta+2)=2\theta+2. $$ Since $E[S]=8$, $$ 2\theta+2=8,\qquad 2\theta=6,\qquad \theta=3. $$ Therefore, the component parameters are three and five. Independence gives $$ \begin{aligned} M_S(t) &=\exp\!\bigl(3(e^t-1)\bigr)\exp\!\bigl(5(e^t-1)\bigr)\\ &=\exp\!\bigl(8(e^t-1)\bigr). \end{aligned} $$ Thus $S\sim\operatorname{Poisson}(8)$, in agreement with the given total mean. Now calculate $$ \begin{aligned} P(S\leq1) &=P(S=0)+P(S=1)\\ &=e^{-8}\frac{8^0}{0!}+e^{-8}\frac{8^1}{1!}\\ &=e^{-8}(1+8)\\ &=9(0.000335463)\\ &=0.003019167. \end{aligned} $$

Rounded to four decimal places, $P(S\leq1)=0.0030$. This method first solves for the unknown component parameter, then verifies the sum with its MGF. It avoids treating the given total mean as the parameter of either individual component; the individual parameters are three and five, while the total parameter is eight.

Common Mistakes and Scope of the Result

The rate-addition rule is short, so errors often come from overlooking what its parameters mean or when its assumptions hold.

Multiplying parameters instead of adding them. Independence multiplies the MGFs, but the MGF exponents contain the parameters as coefficients of the same term $e^t-1$. Combining those exponents adds the parameters.
Using a parameter from the wrong observation period. Each $\lambda_j$ must refer to the period or exposure for the count being added. If a model gives counts over different durations, first ensure that the parameters correspond to the actual intervals being combined.
Forgetting independence. The product rule used here requires independent counts. If they are dependent, the MGF of the sum cannot in general be obtained by multiplying the individual MGFs; this Poisson-sum result does not follow from the individual distributions alone.
Stopping after finding the distribution. Once the total is identified as $\operatorname{Poisson}(\lambda_1+\cdots+\lambda_n)$, use that total parameter in the probability formula. Do not use just one component parameter.

The result matters because it reduces a potentially long sum over combinations of component counts to one Poisson calculation. It also provides immediate checks: the total mean and variance must both equal the sum of the component parameters. Those checks can reveal an arithmetic or modeling error, but they do not replace the independence assumption or the MGF argument.

Key takeaway: For independent Poisson counts on the specified observation periods, multiply their MGFs and combine the exponents. The sum is Poisson, with parameter equal to the sum of the component parameters.

Check Your Understanding

For each problem, state or use the distribution of the total before evaluating any requested probability. Assume independence wherever it is specified.

Independent counts have distributions $\operatorname{Poisson}(1.7)$ and $\operatorname{Poisson}(2.3)$. Identify the total distribution and find the probability of no events in the total. (Answer: $S\sim\operatorname{Poisson}(4)$, $P(S=0)=e^{-4}=0.0183$.)
Three independent counts have parameters $0.8$, $1.2$, and $2.0$. Find the probability their total is exactly one. (Answer: $S\sim\operatorname{Poisson}(4)$, $P(S=1)=4e^{-4}=0.0733$.)
Three independent hypothetical event counts over a common period have parameters $3.5$, $1.5$, and $2.0$. Find the probability that their total is at most one. (Answer: $S\sim\operatorname{Poisson}(7)$, $P(S\leq1)=8e^{-7}=0.0073$.)
Independent counts have distributions $\operatorname{Poisson}(\theta)$ and $\operatorname{Poisson}(3\theta)$. If their total has mean 12, find $\theta$, the total distribution, and the total variance. (Answer: $\theta=3$, $S\sim\operatorname{Poisson}(12)$, and $\operatorname{Var}(S)=12$.)
Independent counts have MGFs $\exp(1.4(e^t-1))$ and $\exp(2.6(e^t-1))$. Identify the total distribution and find the probability that the total is at least two. (Answer: $S\sim\operatorname{Poisson}(4)$, $P(S\geq2)=1-5e^{-4}=0.9084$.)

What You'll Learn