Addition of Independent Poisson Variables

Why Independent Poisson Counts Add

A Poisson variable often represents the number of events observed in a specified interval or region. If two independent sources contribute events to the same total, it is useful to know the distribution of the combined count without listing every possible split between the sources. The discrete convolution formula gives that distribution directly: the probability of a total of $m$ is obtained by summing over all ways the component counts can add to $m$.

For Poisson variables, this sum simplifies exactly. The total is again Poisson, with rate equal to the sum of the component rates. The key calculation is a finite sum of factorial terms; independence supplies the product of the component probabilities.

Definition: A random variable $X$ has a Poisson distribution with rate $\lambda$, written $X\sim\operatorname{Poisson}(\lambda)$, when $\lambda\geq 0$, $X$ takes values in $\{0,1,2,\ldots\}$, and $$ P(X=k)=e^{-\lambda}\frac{\lambda^k}{k!},\qquad k=0,1,2,\ldots. $$ When $\lambda=0$, the distribution is concentrated at zero: $P(X=0)=1$. For $\lambda>0$, the displayed formula applies in the usual way, including $0^0=1$ when the probability at $k=0$ is evaluated.

Deriving the Two-Variable Sum Law

Let $X$ and $Y$ be independent Poisson variables with rates $\lambda$ and $\mu$, respectively, where both rates are nonnegative. Their sum $S=X+Y$ can only take nonnegative integer values. For a fixed integer $m\geq 0$, the event $S=m$ is the disjoint union of the events $X=k,\;Y=m-k$, for $k=0,\ldots,m$. By independence, each joint probability is the product of its marginal probabilities. This is precisely the discrete convolution formula.

Theorem: If $X\sim\operatorname{Poisson}(\lambda)$ and $Y\sim\operatorname{Poisson}(\mu)$ are independent, then $X+Y\sim\operatorname{Poisson}(\lambda+\mu)$.

Proof. For every integer $m\geq 0$, use the disjoint split of the event $X+Y=m$ and independence:

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

The penultimate line uses $\binom{m}{k}=m!/[k!(m-k)!]$. The last line is the binomial theorem. The resulting expression is the Poisson probability mass function with rate $\lambda+\mu$, for every possible value $m$. Therefore $X+Y\sim\operatorname{Poisson}(\lambda+\mu)$. If either rate is zero, the same calculation remains valid, or the zero-rate variable can be seen to equal zero with probability one. $\square$

Worked Example: A Convolution at a Particular Total

Suppose $X$ and $Y$ are independent with $X\sim\operatorname{Poisson}(2)$ and $Y\sim\operatorname{Poisson}(3)$. Calculate $P(X+Y=4)$ by listing the possible splits of 4.

The splits are $(0,4),(1,3),(2,2),(3,1),(4,0)$. Applying independence gives

\begin{aligned} P(X+Y=4) &=\sum_{k=0}^{4}P(X=k)P(Y=4-k)\\ &=e^{-5}\left( \frac{3^4}{4!} +\frac{2\cdot3^3}{1!3!} +\frac{2^2\cdot3^2}{2!2!} +\frac{2^3\cdot3}{3!1!} +\frac{2^4}{4!} \right)\\ &=e^{-5}\left( \frac{81}{24}+9+9+4+\frac{16}{24} \right)\\ &=e^{-5}\frac{625}{24}. \end{aligned}

The rate-sum result gives $X+Y\sim\operatorname{Poisson}(5)$, so its probability at 4 is $e^{-5}5^4/4!=e^{-5}(625/24)$, agreeing with the explicit convolution.

Extending the Result to a Finite Collection

The two-variable result suggests the general rule, but it is also possible to derive it directly by summing over all component counts that produce a given total. The rates need not be equal. Write $\Lambda=\sum_{i=1}^{r}\lambda_i$, where $r$ is a positive integer and each $\lambda_i\geq 0$.

Theorem: Let $X_1,\ldots,X_r$ be mutually independent, with $X_i\sim\operatorname{Poisson}(\lambda_i)$ for $i=1,\ldots,r$. Then $$ \sum_{i=1}^{r}X_i\sim\operatorname{Poisson}\left(\sum_{i=1}^{r}\lambda_i\right). $$

Proof. Fix an integer $m\geq0$. Each way to obtain a total of $m$ is an $r$-tuple of nonnegative integers $(k_1,\ldots,k_r)$ satisfying $k_1+\cdots+k_r=m$. These events are disjoint and exhaust the event that the sum equals $m$. Mutual independence means the probability for any such tuple factors into the product of its $r$ Poisson probabilities. Therefore

\begin{aligned} P\left(\sum_{i=1}^{r}X_i=m\right) &=\sum_{\substack{k_1,\ldots,k_r\geq0\\k_1+\cdots+k_r=m}} \prod_{i=1}^{r} e^{-\lambda_i}\frac{\lambda_i^{k_i}}{k_i!}\\ &=e^{-\Lambda} \sum_{\substack{k_1,\ldots,k_r\geq0\\k_1+\cdots+k_r=m}} \prod_{i=1}^{r}\frac{\lambda_i^{k_i}}{k_i!}\\ &=e^{-\Lambda}\frac{1}{m!} \sum_{\substack{k_1,\ldots,k_r\geq0\\k_1+\cdots+k_r=m}} \frac{m!}{k_1!\cdots k_r!} \prod_{i=1}^{r}\lambda_i^{k_i}\\ &=e^{-\Lambda}\frac{\Lambda^m}{m!}. \end{aligned}

The last equality is the multinomial theorem: expanding $(\lambda_1+\cdots+\lambda_r)^m$ groups terms according to the counts $k_1,\ldots,k_r$, with coefficient $m!/(k_1!\cdots k_r!)$. Thus the probability at every $m\geq0$ is the Poisson mass with rate $\Lambda$. The sum cannot take negative or noninteger values, so this identifies its entire distribution. If all rates are zero, each variable is zero with probability one and the same conclusion holds with $\Lambda=0$. $\square$

Worked Example: Three Different Rates

Let $X_1,X_2,X_3$ be mutually independent with respective rates $0.5,\;1.5,\;2$. Find the probability that their total is at most 1.

The combined rate is

\Lambda=0.5+1.5+2=4.

By the finite-sum theorem, $S=X_1+X_2+X_3\sim\operatorname{Poisson}(4)$. Since $S\leq1$ means $S=0$ or $S=1$,

\begin{aligned} P(S\leq1) &=P(S=0)+P(S=1)\\ &=e^{-4}\frac{4^0}{0!}+e^{-4}\frac{4^1}{1!}\\ &=5e^{-4}\approx0.0916. \end{aligned}

The same probability can be checked from the component counts. A total of zero requires all three counts to be zero, with probability $e^{-0.5}e^{-1.5}e^{-2}=e^{-4}$. A total of one requires exactly one component to equal one and the others to equal zero. The three possibilities have probabilities $0.5e^{-4}$, $1.5e^{-4}$, and $2e^{-4}$, which sum to $4e^{-4}$. Adding the zero-count probability gives $5e^{-4}$, as above.

Worked Example: Probability of No Events Across Several Sources

Three independent monitoring systems record event counts $A\sim\operatorname{Poisson}(0.4)$, $B\sim\operatorname{Poisson}(1.1)$, and $C\sim\operatorname{Poisson}(0.5)$. Find the probability that at least one event is recorded in total.

The combined count $T=A+B+C$ is Poisson with rate $0.4+1.1+0.5=2$. The event $T\geq1$ is the complement of $T=0$, so

\begin{aligned} P(T\geq1) &=1-P(T=0)\\ &=1-e^{-2}\frac{2^0}{0!}\\ &=1-e^{-2}\approx0.8647. \end{aligned}

Independence also verifies the zero-count probability directly: all three systems must record zero, giving $e^{-0.4}e^{-1.1}e^{-0.5}=e^{-2}$. The combined-rate calculation agrees.

What the Result Does—and Does Not—Say

The rate-sum rule lets us replace several independent Poisson counts by one Poisson count whenever the quantity of interest is their total. For example, it makes probabilities such as $P(\sum_i X_i=m)$, $P(\sum_iX_i\leq m)$, and $P(\sum_iX_i\geq m)$ straightforward to calculate from a single rate. For a cumulative probability, sum the corresponding Poisson masses after combining the rates.

Two conditions matter. First, the variables must be independent; the convolution step used $P(X=k,Y=m-k)=P(X=k)P(Y=m-k)$. Second, each summand must have a Poisson distribution. Having the same expected value as a Poisson variable is not enough to establish the sum law.

Common mistake: adding the rates without checking independence. The formula for the convolution depends on independence. Without it, the joint probabilities of the different splits need not be products, and the conclusion does not follow.
Common mistake: adding probabilities at the same count. The total equals $m$ through every split $k+(m-k)=m$, not only through the split $m+0$. The convolution sums all those mutually exclusive possibilities.
Common mistake: assuming equal rates are required. The rates can differ. The finite-collection result uses their sum, not their average, and does not require identical distributions.

Key takeaway: For mutually independent Poisson variables, the sum is Poisson and its rate is the sum of the individual rates. The result follows by discrete convolution and the binomial theorem for two variables, or the multinomial theorem for a finite collection.

Check Your Understanding

Use the convolution argument and the rate-sum rule to answer the following.

If independent $X\sim\operatorname{Poisson}(1.2)$ and $Y\sim\operatorname{Poisson}(2.8)$, what is the distribution of $X+Y$?
For independent Poisson variables with rates 2 and 1, write the convolution sum for the probability that their total is 3.
Three independent Poisson variables have rates 0.5, 1, and 1.5. What is the probability that all three record zero?
Which step in the two-variable proof would fail if $X$ and $Y$ were dependent?
Explain why the formula for a total of $m$ must include every nonnegative split of $m$, rather than just one split.

Addition of Independent Poisson Variables

What You'll Learn