Additivity of Independent Poisson Variables

Why Poisson Counts Can Be Combined

A Poisson model describes a count of events over a specified interval or exposure. In practice, a total count may combine events from several sources: for example, separate service channels or distinct types of incidents. If the component counts are independent Poisson random variables measured over the same interval, their sum also has a Poisson distribution. Its parameter is the sum of the component parameters.

The independence condition matters. Knowing that each source has a Poisson count does not, by itself, show that the combined count is Poisson. This tutorial derives the result from the discrete convolution formula and applies it to event-count probabilities. All operational scenarios below are hypothetical.

Definition: A random variable $X$ has a Poisson distribution with parameter $\lambda$, written $X\sim\operatorname{Poisson}(\lambda)$, if $\lambda\geq 0$ and $$ P(X=k)=e^{-\lambda}\frac{\lambda^k}{k!},\qquad k=0,1,2,\ldots. $$ The parameter $\lambda$ is the expected count for the interval or exposure represented by the model. When $\lambda=0$, $X=0$ with probability 1.

For two sources, suppose $X\sim\operatorname{Poisson}(\lambda)$ and $Y\sim\operatorname{Poisson}(\mu)$, and assume $X$ and $Y$ are independent. Both counts must refer to the same specified interval or exposure before they can be added meaningfully. The total count is $S=X+Y$. Since counts are nonnegative integers, $S=k$ can occur through any of the pairs $(X,Y)=(j,k-j)$, where $j=0,\ldots,k$.

Deriving the Distribution by Convolution

For independent discrete random variables, the probability of a sum is found by summing over all pairs of component values that produce that sum. In this case, each possible pair contributes the product of its two probabilities. The following derivation shows why the resulting probability has the Poisson form.

List the pairs giving a total of $k$.
For nonnegative integer counts, $X+Y=k$ exactly when $X=j$ and $Y=k-j$, for one of $j=0,1,\ldots,k$.

Use independence in the convolution.
Independence gives $P(X=j,Y=k-j)=P(X=j)P(Y=k-j)$, so the probability of the total is the sum of these products.

Simplify the finite sum.
Factor out terms common to every summand, then apply the binomial theorem to the remaining sum.

\begin{aligned} P(S=k) &=\sum_{j=0}^{k}P(X=j)P(Y=k-j)\\ &=\sum_{j=0}^{k} \left(e^{-\lambda}\frac{\lambda^j}{j!}\right) \left(e^{-\mu}\frac{\mu^{k-j}}{(k-j)!}\right)\\ &=e^{-(\lambda+\mu)} \frac{1}{k!} \sum_{j=0}^{k}\binom{k}{j}\lambda^j\mu^{k-j}\\ &=e^{-(\lambda+\mu)} \frac{(\lambda+\mu)^k}{k!}. \end{aligned}

In the third line, $1/(j!(k-j)!)=\binom{k}{j}/k!$. The last equality is the binomial theorem. The resulting expression is the Poisson probability with parameter $\lambda+\mu$. Since $S$ can take only nonnegative integer values, this gives its entire probability mass function.

Theorem: If $X\sim\operatorname{Poisson}(\lambda)$ and $Y\sim\operatorname{Poisson}(\mu)$ are independent, then $$ X+Y\sim\operatorname{Poisson}(\lambda+\mu). $$ More generally, if $X_1,\ldots,X_n$ are mutually independent and $X_r\sim\operatorname{Poisson}(\lambda_r)$ for each $r$, then their sum has a Poisson distribution with parameter $\sum_{r=1}^{n}\lambda_r$. This follows by applying the two-variable result repeatedly.

The repeated-sum result requires mutual independence, not merely pairwise independence in every possible setting. For the usual application to independent source counts, confirm that the model specifies the counts as jointly independent. Each parameter must also describe the same interval or exposure; parameters for counts over different durations cannot simply be combined without first bringing them to a common basis.

Using the Combined Parameter

Once the conditions hold, calculate the total parameter and use the ordinary Poisson formula. For an exact total $k$, use $e^{-\Lambda}\Lambda^k/k!$, where $\Lambda$ is the sum of the component parameters. For a cumulative probability, add the relevant exact-count probabilities, or subtract the complement when that requires fewer terms.

When a model gives constant event rates per unit time, a rate must be multiplied by the duration to obtain the Poisson parameter for that duration. For example, a rate of $r$ events per hour over $t$ hours gives parameter $rt$, under the assumed constant-rate Poisson model. Check the time units before adding parameters.

Worked Example: Combining Two Counts for One Interval

A hypothetical monitoring system records two independent types of alerts during a four-hour test. The first count is $X\sim\operatorname{Poisson}(2.4)$, and the second is $Y\sim\operatorname{Poisson}(1.6)$. Find the probability of exactly three alerts in total and the probability of at most one alert.

Method. The counts are independent and refer to the same four-hour interval, so the convolution result applies. The combined parameter is $\Lambda=2.4+1.6=4.0$, and $S=X+Y\sim\operatorname{Poisson}(4)$.

For exactly three alerts, $$ P(S=3)=e^{-4}\frac{4^3}{3!} =e^{-4}\frac{64}{6} =e^{-4}\left(10.6666667\right) \approx 0.1954. $$

For at most one alert, add the probabilities of zero and one: $$ \begin{aligned} P(S\leq 1) &=P(S=0)+P(S=1)\\ &=e^{-4}\frac{4^0}{0!}+e^{-4}\frac{4^1}{1!}\\ &=e^{-4}(1+4)=5e^{-4}\approx 0.0916. \end{aligned} $$

Check. Both probabilities are between 0 and 1, and the probability of at most one alert is small relative to a mean count of 4. The results are rounded to four decimal places. The common mistake avoided here is treating the probability of a total of three as the probability that each source individually records three; the total is modeled using the combined parameter.

Worked Example: Converting Rates to a Common Duration

In a hypothetical service center, independent incoming-request counts from channel A and channel B follow Poisson models with constant rates of 1.2 and 0.8 requests per hour, respectively. Find the probability that exactly five requests arrive across both channels during a three-hour shift.

Method. First convert each rate to a parameter for the three-hour shift. For channel A, $\lambda_A=(1.2)(3)=3.6$. For channel B, $\lambda_B=(0.8)(3)=2.4$. These are independent Poisson counts over the same shift, so the combined count has parameter $\Lambda=3.6+2.4=6$.

Thus the total $S$ has distribution $\operatorname{Poisson}(6)$, and $$ \begin{aligned} P(S=5) &=e^{-6}\frac{6^5}{5!}\\ &=e^{-6}\frac{7776}{120}\\ &=64.8e^{-6}\\ &\approx 0.1606. \end{aligned} $$

Check. The two three-hour expected counts sum to 6, as does the combined parameter. The probability is rounded to four decimal places. The common mistake avoided is adding the hourly rates and then using that one-hour parameter for a three-hour shift; the rates must first be multiplied by the duration.

Worked Example: Combining Three Independent Sources

A hypothetical laboratory classifies independently occurring equipment interruptions into three categories. During a specified test interval, the counts have distributions $X_1\sim\operatorname{Poisson}(0.5)$, $X_2\sim\operatorname{Poisson}(1.5)$, and $X_3\sim\operatorname{Poisson}(2)$, and the three counts are mutually independent. Find the probability of no more than two interruptions in total.

Method. Repeatedly apply the two-variable convolution result. The total $T=X_1+X_2+X_3$ is Poisson with parameter $\Lambda=0.5+1.5+2=4$.

“No more than two” means $T=0$, $T=1$, or $T=2$, so $$ \begin{aligned} P(T\leq 2) &=e^{-4}\left(\frac{4^0}{0!}+\frac{4^1}{1!}+\frac{4^2}{2!}\right)\\ &=e^{-4}(1+4+8)\\ &=13e^{-4}\\ &\approx 0.2381. \end{aligned} $$

Check. The three parameters sum to 4, and the probability includes exactly the three possible totals 0, 1, and 2. The result is rounded to four decimal places. The common mistake avoided is adding only two of the three category parameters or forgetting the $2!$ in the probability for exactly two events.

When the Result Does Not Apply

The theorem is not a general rule that any two Poisson-distributed variables have a Poisson sum. The convolution proof used independence when replacing a joint probability by a product of marginal probabilities. If the counts are dependent, that step may be false, and the resulting sum need not have a Poisson distribution. Adding the component means alone is therefore not enough to identify the distribution of a total.

For example, let $X\sim\operatorname{Poisson}(2)$ and set $Y=X$. Each of $X$ and $Y$ has a Poisson distribution, but the two variables are perfectly dependent. Their sum is $S=2X$, which can take only even values. In particular, $P(S=1)=0$. If one incorrectly applied the independent-sum theorem, one would obtain a Poisson distribution with parameter 4 and calculate $$ P(S=1)=e^{-4}\frac{4^1}{1!}\approx 0.0733, $$ which contradicts the actual probability of zero. The error is the unjustified assumption of independence.

Exam check: Before combining Poisson counts, verify that each component is Poisson, the counts are independent, and the parameters refer to compatible intervals or exposures. Only then add the parameters and calculate the requested probability.

This result is useful whenever a total is assembled from separate independent event sources. It can replace a longer term-by-term convolution calculation with a single Poisson probability calculation, but the simplification is justified only after checking the model assumptions. Careful attention to independence, duration, and the meaning of a cumulative event such as “at most” or “at least” prevents the most common errors.

Check Your Understanding

For each problem, identify the combined parameter before calculating. Give probabilities to four decimal places.

Independent counts have distributions $\operatorname{Poisson}(0.9)$ and $\operatorname{Poisson}(1.1)$ over the same interval. What is the probability that the combined count is zero? (Answer: $e^{-2}\approx 0.1353$)
Two independent hypothetical event sources have constant rates of 2.5 and 1.5 events per day. What is the probability of exactly six events from both sources over two days? (Answer: $e^{-8}8^6/6!\approx 0.1221$)
Independent counts have distributions $\operatorname{Poisson}(1.4)$ and $\operatorname{Poisson}(2.6)$. What is the probability that their sum is at most two? (Answer: $13e^{-4}\approx 0.2381$)
Three mutually independent counts over the same exposure have distributions $\operatorname{Poisson}(0.5)$, $\operatorname{Poisson}(1.5)$, and $\operatorname{Poisson}(2)$. What is the probability that their sum is exactly three? (Answer: $e^{-4}4^3/3!\approx 0.1954$)
Let $X$ and $Y$ be independent with $X\sim\operatorname{Poisson}(1)$ and $Y\sim\operatorname{Poisson}(1)$. What is $P(X+Y\geq 1)$? (Answer: $1-e^{-2}\approx 0.8647$)

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