Adding Independent Poisson Counts

Why Combine Poisson Counts?

A total count can come from several distinct sources. A clinic might record arrivals from two independently operating referral channels, or a warehouse might count defects produced by several independent machines. If each source has a Poisson count, a natural question is whether the combined total also has a Poisson distribution—and, if so, what its parameter is.

The answer is useful because it turns a calculation involving several sources into one Poisson calculation. The key condition is not merely that each source separately follows a Poisson model. The source counts must also be jointly independent. The result concerns counts over the same defined observation window or exposure; each parameter describes the expected count for that window.

Theorem: Let $X_1,\ldots,X_m$ be jointly independent random variables, with $X_i\sim\operatorname{Poisson}(\lambda_i)$ for each $i$, where $\lambda_i\geq 0$. Then their sum $S=X_1+\cdots+X_m$ has a Poisson distribution with parameter $\lambda_1+\cdots+\lambda_m$: $$ S\sim\operatorname{Poisson}\left(\sum_{i=1}^{m}\lambda_i\right). $$

A Poisson parameter is the expected count for the specified observation window, and the variance of that count is also equal to the parameter. When sources are recorded over the same window, their expected counts add. The theorem says more than that their means add: under joint independence, their entire combined count has a Poisson distribution with this summed parameter.

Deriving the Result for Two Counts

First consider two independent counts, $X\sim\operatorname{Poisson}(\lambda_1)$ and $Y\sim\operatorname{Poisson}(\lambda_2)$, and define $S=X+Y$. To have $S=k$, the first count can be any integer $j$ from 0 to $k$, while the second must equal $k-j$. These are mutually exclusive possibilities. Independence lets us multiply the probabilities for each pair of values, then add across all possible pairs.

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

In the fourth line, $1/(j!(k-j)!)=\binom{k}{j}/k!$. The final equality is the binomial theorem. The resulting expression is the Poisson probability mass function with parameter $\lambda_1+\lambda_2$, for every nonnegative integer $k$. Thus $X+Y$ has that Poisson distribution.

For more than two counts, apply the two-count result repeatedly, but retain the theorem's joint-independence condition. If $X_1,\ldots,X_m$ are jointly independent, the group $(X_1,\ldots,X_{m-1})$ is independent of $X_m$; consequently, any function of the first group, including $X_1+\cdots+X_{m-1}$, is independent of $X_m$. The two-count result can therefore be applied to that partial sum and $X_m$, and then repeated for the remaining counts. This group-independence step is why pairwise independence alone is not enough to justify the argument.

Worked Applications

Worked Example: Combining Two Counts in One Window

Suppose two independent service desks handle requests during the same hour. Assume the first desk's count is Poisson with parameter 2 requests per hour and the second desk's count is Poisson with parameter 3 requests per hour. Find the probability that the two desks handle exactly 4 requests altogether during an hour.

Let $X$ and $Y$ be the two hourly counts. The model states that they are independent Poisson counts, so the sum has parameter $2+3=5$ requests per hour. Therefore $S=X+Y\sim\operatorname{Poisson}(5)$. The probability of exactly 4 total requests is

P(S=4)=e^{-5}\frac{5^4}{4!} =e^{-5}\frac{625}{24} \approx 0.1755.

The assumptions are that both counts concern the same one-hour window, each follows its stated Poisson model, and the two counts are independent. Under those assumptions, the probability of exactly 4 requests across both desks is about 0.1755, or 17.55%.

Worked Example: A Tail Probability for Two Sources

Consider two independent sources of equipment alerts observed over a three-hour period. Suppose their alert rates are 1.2 alerts per hour and 0.8 alerts per hour, respectively, and each source follows a Poisson model with a constant rate over this period. Find the probability of at least 5 alerts altogether.

For the three-hour window, the source parameters are $1.2(3)=3.6$ alerts and $0.8(3)=2.4$ alerts. Their counts are assumed independent, so the total count $S$ is Poisson with parameter $3.6+2.4=6$. The event “at least 5” is the complement of counts 0 through 4:

\begin{aligned} P(S\geq 5) &=1-P(S\leq 4)\\ &=1-e^{-6}\sum_{k=0}^{4}\frac{6^k}{k!}\\ &=1-e^{-6}\left(1+6+18+36+54\right)\\ &=1-115e^{-6}\\ &\approx 0.7149. \end{aligned}

The arithmetic uses the terms $6^0/0!=1$, $6^1/1!=6$, $6^2/2!=18$, $6^3/3!=36$, and $6^4/4!=54$. Thus the model gives about a 71.49% probability of at least 5 alerts during the three hours. This interpretation depends on the rate assumptions and independence of the two source counts.

Worked Example: Combining Three Independent Counts

Suppose three independent production lines have defect counts during one shift that follow Poisson distributions with parameters 0.5, 1.0, and 1.5 defects per shift. Find the probability that there are no more than 2 defects across the three lines.

All three counts refer to the same shift and are stated to be jointly independent, so the total $S$ is Poisson with parameter $0.5+1.0+1.5=3$ defects per shift. The requested probability is the sum of the probabilities for 0, 1, and 2 defects:

\begin{aligned} P(S\leq 2) &=\sum_{k=0}^{2}e^{-3}\frac{3^k}{k!}\\ &=e^{-3}\left(1+3+\frac{9}{2}\right)\\ &=8.5e^{-3}\\ &\approx 0.4232. \end{aligned}

According to this model, the probability of at most 2 defects across all three lines in a shift is about 42.32%. The sum rule allows this probability to be calculated from the combined parameter rather than by listing every way defects could be distributed among the lines.

Combining Rates and Exposure

When source rates are given per unit of exposure, first express each source parameter for the observation window in question. For example, if source $i$ has a constant rate $r_i$ per hour and the window lasts $t$ hours, its Poisson parameter for the window is $r_i t$. If all sources are independent and observed over that same window, the total parameter is

\sum_{i=1}^{m} r_i t =t\sum_{i=1}^{m}r_i.

Thus one can either add the source-specific expected counts for the window or add the rates and multiply by the window length. These are equivalent ways to obtain the parameter for the combined count. Keep the units clear: a rate may be given in events per hour, while a parameter for a three-hour count is an expected number of events in those three hours.

The result is particularly useful when a question asks about the total number from multiple independent sources. Instead of considering all possible combinations of source-specific counts, form the combined parameter and use the Poisson probability mass function or an appropriate cumulative probability.

Conditions and a Common Mistake

The conclusion that a sum is Poisson requires more than knowing that the separate counts are Poisson. In particular, the counts need to be jointly independent and to refer to compatible observation windows or exposures. For several counts, joint independence means that the variables as a collection are independent; pairwise independence only asserts independence for each pair and does not, in general, ensure that one count is independent of a sum of others.

A common mistake is to add the parameters immediately whenever several Poisson counts appear. For example, if two sources share a common cause that can increase both counts at once, their counts may be dependent. Even if each source's marginal count is Poisson, the theorem above then does not establish that the total has a Poisson distribution. One must check the independence assumption, rather than treating the sum rule as a consequence of the individual Poisson models alone.

Key takeaway: For jointly independent Poisson counts measured over the relevant common exposure, add the parameters to obtain the distribution of the total. If the counts are dependent, do not apply this theorem without additional justification.

Check Your Understanding

Use the theorem and its assumptions to answer these questions.

Two independent counts over the same observation window have Poisson parameters 1.7 and 2.3. What is the distribution of their total, and what is its parameter?
For independent Poisson counts with parameters 2 and 4, write an expression for the probability that their total is exactly 3.
Three counts are each Poisson, and every pair is independent. Is that information alone enough to invoke the theorem for their sum? Explain.
An independent source has a constant rate of 0.6 events per hour, and another independent source has a rate of 0.9 events per hour. What is the combined parameter for a two-hour window?
Why is it not sufficient to know that each of two counts follows a Poisson distribution before concluding their sum is Poisson?