Poisson Merging¶

Theorem¶

If $X\sim\operatorname{Pois}(\lambda_1)$ and $Y\sim\operatorname{Pois}(\lambda_2)$ are independent, then $X+Y\sim\operatorname{Pois}(\lambda_1+\lambda_2)$ $λ_{2})$ . Independence is needed: Poisson marginals alone do not make the sum Poisson.

The Poisson Distribution has MGF $\mathrm{e}^{\lambda(\mathrm{e}^t-1)}$ . Multiplying the MGFs of the independent counts adds their parameters, identifying the law of the total.

Interpretation¶

Independent Poisson sources combine into a Poisson total whose parameter is the sum of the source parameters. Counting two independent streams of events together gives a count of the same kind, with the means added.

Three timelines over one time unit: arrivals of a Poisson(3) count, arrivals of an independent Poisson(7) count, and both sets merged on one line as a Poisson(10) count.

Over one time unit, the blue arrivals form a $\operatorname{Pois}(3)$ count and the red arrivals an independent $\operatorname{Pois}(7)$ count. Placing both on one line gives the merged count $X+Y\sim\operatorname{Pois}(10)$ $Pois (10)$ ; in this sample there are $3+7=10$ arrivals. The merged line no longer records which source an arrival came from.

Two independent Poisson sources together give $\operatorname{Pois}(\lambda_1+\lambda_2)$ $λ_{2})$ .

The proof is a short MGF computation: multiplying the MGFs gives $\mathrm{e}^{\lambda_1(\mathrm{e}^t-1)}\mathrm{e}^{\lambda_2(\mathrm{e}^t-1)}=\mathrm{e}^{(\lambda_1+\lambda_2)(\mathrm{e}^t-1)}$ $e^{(λ_{1} + λ_{2}) (e^{t} - 1)}$ , and uniqueness identifies $\operatorname{Pois}(\lambda_1+\lambda_2)$ $λ_{2})$ . A direct check sums over the ways to split a total $n$ into $k$ and $n-k$ $k$ ; the binomial theorem then gives

\sum_{k=0}^n\frac{\lambda_1^k}{k!}\,\frac{\lambda_2^{n-k}}{(n-k)!}=\frac{(\lambda_1+\lambda_2)^n}{n!}.

Independence is essential. If $X\sim\operatorname{Pois}(\lambda)$ and $Y=X$ $X$ , both counts are Poisson, but $X+Y=2X$ takes only even values, so it is not Poisson. Merging and Poisson splitting describe the same situation from two sides: given the total $X+Y=n$ $n$ , each of the $n$ items comes from the first source with probability $\lambda_1/(\lambda_1+\lambda_2)$ $λ_{2})$ , independently, which is the binomial statement in the Key Results.

Key Results¶

For finitely many independent Poisson variables, $\displaystyle \sum_iX_i\sim\operatorname{Pois}(\sum_i\lambda_i)$ $Pois (i \sum λ_{i})$ .
For $X$ and $Y$ as in the theorem, $X\mid X+Y=n\sim\operatorname{Bin}(n,\lambda_1/(\lambda_1+\lambda_2))$ $λ_{2}))$ .

Examples¶

Independent arrival counts with means $3$ and $7$ combine to a $\operatorname{Pois}(10)$ count.
Two independent counts with means $2$ and $4$ have total-zero probability $\mathrm{e}^{-2}\mathrm{e}^{-4}=\mathrm{e}^{-6}$ $e^{- 6}$ .

Prerequisites¶

Sources¶

Nicky van Foreest · Probability Distributions · Week 1

Aliases: sum of independent Poisson random variables Symbols: lambda, sum

Last updated: 20 October 2026 03:02