Poisson Closure Under Independent Addition

Why Independent Poisson Counts Add

A Poisson random variable models a count over a specified scope, such as the number of events from one source during a fixed interval. In many probability problems, the count of interest combines several sources: total service requests received by two independent systems, for example, or total defects recorded by separate inspection lines. When the component counts are independent and each is Poisson, their sum is also Poisson. Its parameter is the sum of the component parameters.

The component parameters need not be equal. The essential condition is independence; the rates or expected counts must also refer to the scopes being combined. If the parameters refer to different time intervals or units, first convert them to the same scope. This result is aligned with the SOA Exam P (Probability) syllabus.

Definition: A random variable $X$ has a Poisson distribution with parameter $\lambda$, written $X\sim\operatorname{Poisson}(\lambda)$, if $\lambda\ge 0$ and, for integers $k\ge 0$, $$ P(X=k)=e^{-\lambda}\frac{\lambda^k}{k!}. $$ The parameter $\lambda$ is both the mean and the variance of $X$. When $\lambda=0$, $X=0$ with probability 1.

For a count over a fixed time interval, $\lambda$ is the expected number of events during that interval. In a constant-rate model with rate $r$ events per unit of time, an interval of length $t$ has parameter $\lambda=rt$. This conversion is valid when that rate model applies to the interval in question.

The Closure Result

Theorem: If $X_1,\ldots,X_m$ are mutually independent and $X_j\sim\operatorname{Poisson}(\lambda_j)$ for each $j$, then $$ S=X_1+\cdots+X_m\sim\operatorname{Poisson}(\Lambda), \qquad \Lambda=\lambda_1+\cdots+\lambda_m. $$ The conclusion requires independence. The parameters may differ, and each parameter must describe the count being combined.

The result is not a claim that any two counts with Poisson marginal distributions have a Poisson sum. If the counts are dependent, the closure theorem cannot be applied just by adding their parameters. Independence is also not a consequence of having different event sources; it must follow from the model or be given in the problem.

Why the Parameters Add: Convolution

For two independent Poisson variables, the probability that their sum equals $n$ is found by considering every way the total $n$ could be divided between them. Independence lets us multiply the probabilities for each pair of component counts.

\begin{aligned} P(X+Y=n) &=\sum_{k=0}^{n}P(X=k)P(Y=n-k)\\ &=\sum_{k=0}^{n} e^{-\lambda_1}\frac{\lambda_1^k}{k!} e^{-\lambda_2}\frac{\lambda_2^{n-k}}{(n-k)!}\\ &=e^{-(\lambda_1+\lambda_2)} \frac{1}{n!} \sum_{k=0}^{n}\binom{n}{k} \lambda_1^k\lambda_2^{n-k}\\ &=e^{-(\lambda_1+\lambda_2)} \frac{(\lambda_1+\lambda_2)^n}{n!}. \end{aligned}

The last equality is the binomial theorem. The resulting expression is exactly the Poisson probability formula with parameter $\lambda_1+\lambda_2$. The argument covers every integer $n\ge0$, so it establishes the full distribution, not just its mean.

Confirmation Using Moment Generating Functions

The moment generating function of $X\sim\operatorname{Poisson}(\lambda)$ follows directly from its probability mass function:

\begin{aligned} M_X(t) &=E[e^{tX}] =\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}

For independent variables, the moment generating function of their sum is the product of their moment generating functions. Thus, for independent $X_j\sim\operatorname{Poisson}(\lambda_j)$,

M_S(t)=\prod_{j=1}^{m}\exp\!\bigl(\lambda_j(e^t-1)\bigr) =\exp\!\bigl((\lambda_1+\cdots+\lambda_m)(e^t-1)\bigr).

This is the moment generating function of a Poisson variable with parameter $\Lambda=\lambda_1+\cdots+\lambda_m$. Since the moment generating function exists in a neighborhood of zero and identifies the distribution, this also establishes the closure result.

Using the Result in Probability Calculations

Once a sum has been recognized as Poisson, calculate the requested probability from its single combined parameter. For an exact count $n$, use $e^{-\Lambda}\Lambda^n/n!$. For a range of counts, add the relevant probability mass function terms. For a complement such as $P(S\ge c)$, it is often shorter to subtract the probabilities below $c$ from 1.

Worked Example: Combining Two Independent Counts

Hypothetical scenario. During a specified shift, a dispatch desk receives $X\sim\operatorname{Poisson}(2.4)$ urgent messages from one independent channel and $Y\sim\operatorname{Poisson}(1.6)$ from another. Assume the counts are independent. Find (a) the probability that exactly 3 messages arrive in total and (b) the probability that at most 1 arrives.

Method and parameter. Apply Poisson closure. The combined count is $S=X+Y$, with distribution $\operatorname{Poisson}(\Lambda)$, where $$ \Lambda=2.4+1.6=4. $$ The common mistake avoided here is calculating separate probabilities for the two channels when the event concerns only the total.

(a) Exactly 3. Substitute $n=3$ and $\Lambda=4$: $$ P(S=3)=e^{-4}\frac{4^3}{3!} =e^{-4}\frac{64}{6} =10.6666667e^{-4} \approx 0.1954. $$

(b) At most 1. This event includes $S=0$ and $S=1$: $$ \begin{aligned} P(S\le 1) &=P(S=0)+P(S=1)\\ &=e^{-4}\frac{4^0}{0!}+e^{-4}\frac{4^1}{1!}\\ &=e^{-4}(1+4)=5e^{-4} \approx 0.0916. \end{aligned} $$

The probabilities are reported to 4 decimal places. As a check, the probabilities for $S=0$ and $S=1$ are $e^{-4}$ and $4e^{-4}$, respectively, so their sum is $5e^{-4}$, as used above.

Worked Example: Three Sources and a Complement

Hypothetical scenario. Three independent data-processing queues have Poisson job counts during the same one-hour window, with parameters $1.2$, $2.3$, and $0.5$, respectively. Find the probability that at least 3 jobs arrive in total.

Method and combined distribution. Since the three counts are independent and refer to the same hour, their sum is Poisson with parameter $$ \Lambda=1.2+2.3+0.5=4. $$ Thus $S\sim\operatorname{Poisson}(4)$. The common mistake avoided is adding probabilities of events such as “at least 3” for individual queues; the requested event applies to the total.

It is simpler to use the complement: $$ \begin{aligned} P(S\ge 3) &=1-P(S=0)-P(S=1)-P(S=2)\\ &=1-e^{-4}\left(\frac{4^0}{0!} +\frac{4^1}{1!}+\frac{4^2}{2!}\right)\\ &=1-e^{-4}(1+4+8)\\ &=1-13e^{-4}\\ &\approx 1-13(0.0183156389)\\ &\approx 0.7619. \end{aligned} $$

The answer is reported to 4 decimal places. The excluded cases are exactly the nonnegative integer totals below 3: 0, 1, and 2.

Worked Example: Aligning Rates to the Same Time Interval

Hypothetical scenario. Two independent sensors record events. Their constant rates are $0.08$ and $0.12$ events per minute. Find the probability that the total number recorded by both sensors in a 10-minute window is between 2 and 4 inclusive. Assume each sensor’s count in that window is Poisson with parameter equal to its rate times the window length.

Method and time conversion. Convert both rates to expected counts over 10 minutes: $$ \lambda_1=(0.08)(10)=0.8,\qquad \lambda_2=(0.12)(10)=1.2. $$ By independence, the total $S$ is Poisson with parameter $$ \Lambda=0.8+1.2=2. $$ The common mistake avoided is adding the per-minute rates and using $0.20$ as the Poisson parameter for the whole 10-minute interval. That value is a rate per minute, not the expected count over the requested interval.

The requested inclusive range contains 2, 3, and 4: $$ \begin{aligned} P(2\le S\le 4) &=e^{-2}\left(\frac{2^2}{2!} +\frac{2^3}{3!}+\frac{2^4}{4!}\right)\\ &=e^{-2}\left(2+\frac{4}{3}+\frac{2}{3}\right)\\ &=4e^{-2}\\ &\approx 0.5413. \end{aligned} $$

The answer is reported to 4 decimal places. The three terms sum to $4e^{-2}$ because $4/3+2/3=2$.

Conditions, Checks, and Common Traps

Before using Poisson closure, check that each component is actually Poisson, that the component counts are independent, and that they refer to the same event definition and time interval (or other scope). If each count is specified directly by a Poisson parameter, no constant-rate interpretation is needed; simply add those parameters. If rates are given, convert each to an expected count for the relevant interval first.

Ignoring dependence. Poisson marginal distributions alone do not justify adding parameters. If two counts share events or are otherwise dependent, the closure theorem does not apply.
Mixing rates and expected counts. A rate such as events per hour is not itself the parameter for a 10-minute count. Multiply by the interval length expressed in hours.
Using the wrong event. For a total count, calculate the Poisson probability for the combined variable; do not combine probabilities for separate component events as if they were the total.
Misreading an inclusive range. $2\le S\le4$ includes the three values 2, 3, and 4. Write out the included integer values before summing.

A quick check is to compare the parameter of the total with the sum of the component means. Since $E[X_j]=\lambda_j$, linearity of expectation gives $E[S]=\sum_j\lambda_j=\Lambda$, agreeing with the mean of the derived Poisson distribution. This check does not prove closure, but it can reveal an omitted source or a time-unit error.

Key takeaway: For mutually independent Poisson counts on compatible scopes, add the parameters first. The total is Poisson with that combined parameter, regardless of whether the individual parameters are equal.

Check Your Understanding

For each problem, verify the independence and scope conditions before combining parameters.

Two independent instruments record counts with distributions $\operatorname{Poisson}(1.5)$ and $\operatorname{Poisson}(2.5)$ during the same test. Find the probability of exactly 2 total records. (Answer: $0.1465$, to 4 decimal places.)
Three independent counters have Poisson counts with parameters $0.7$, $1.3$, and $2.0$ over the same interval. Find the probability that the total is at most 1. (Answer: $5e^{-4}\approx0.0916$, to 4 decimal places.)
Two independent alert sources have constant rates of $2.4$ and $1.6$ alerts per hour. Under the Poisson count model, find the probability of at least one alert from either source during 30 minutes. (Answer: $1-e^{-2}\approx0.8647$, to 4 decimal places.)
Four independent stations record counts over a common interval with Poisson parameters $1.2$, $0.8$, $2.5$, and $2.5$. Find the probability that the total count is exactly 6. (Answer: $e^{-7}7^6/6!\approx0.1490$, to 4 decimal places.)
Suppose $X\sim\operatorname{Poisson}(2)$ and a second count is defined by $Y=X$. Can the independent-Poisson closure theorem be used to conclude that $X+Y\sim\operatorname{Poisson}(4)$? (Answer: No. $X$ and $Y$ are dependent because $Y=X$; in fact $X+Y=2X$, which takes only even values and is not $\operatorname{Poisson}(4)$.)

What You'll Learn