The sum of two random variables with Poisson distributions is a Poisson distribution itself

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There's still some work to do on this page, but the gist is very much present! Alec (talk) 22:46, 4 November 2017 (UTC)

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[ilmath]\newcommand{\P}[2][]{\mathbb{P}#1{\left[{#2}\right]} } \newcommand{\Pcond}[3][]{\mathbb{P}#1{\left[{#2}\!\ \middle\vert\!\ {#3}\right]} } \newcommand{\Plcond}[3][]{\Pcond[#1]{#2}{#3} } \newcommand{\Prcond}[3][]{\Pcond[#1]{#2}{#3} }[/ilmath]


Let [ilmath]\lambda,r\in\mathbb{R}_{>0} [/ilmath] be given. We start with the following two random variables

  1. [ilmath]X\sim\text{Poi}(\lambda)[/ilmath] and
  2. [ilmath]Y\sim\text{Poi}(r)[/ilmath]

(And that these are statistically independent random variables


  • Let [ilmath]Z:\eq X+Y[/ilmath]

We claim

  • [ilmath]X+Y\eq: Z\sim\text{Poi}(\lambda+r)[/ilmath]


Let [ilmath]Z'\sim\text{Poi}(\lambda+r)[/ilmath]

We will show that [ilmath]\forall k\in\mathbb{N}_0\big[\P{Z\eq k}\eq\P{Z'\eq k}\big][/ilmath]

  • Let [ilmath]k\in\mathbb{N}_0[/ilmath] be given, then:
    • [math]\P{Z\eq k}:\eq\P{X+Y\eq k}\eq\sum^k_{i\eq 0}\overbrace{\P{X\eq i}\Pcond{Y\eq k-i}{X\eq i} }^{\eq\P{(X\eq i)\cap(Y\eq k-i)} } [/math] [Note 1]
      [math]\eq\sum^k_{i\eq 0}\P{X\eq i}\P{Y\eq k-i} [/math] - as [ilmath]X[/ilmath] and [ilmath]Y[/ilmath] are independent random variables[Note 2]
      [math]\eq\sum^k_{i\eq 0}e^{-\lambda}\frac{\lambda^i}{i!}\cdot e^{-r}\frac{r^{k-i} }{(k-i)!} [/math] - by definition of the Poisson distribution
      [math]\eq e^{-(\lambda+r)}\sum^k_{i\eq 0}\lambda^i r^{k-i} \frac{1}{i!(k-i)!} [/math]
    • Recall that [math]{}^nC_r:\eq \frac{n!}{r!(k-r)!} [/math]
      • So [math]\frac{ {}^kC_i}{k!}\eq\frac{1}{i!(k-i)!} [/math] (by dividing by [ilmath]n![/ilmath] in the definition on the line above)
    • Thus: [math]\P{Z\eq k}\eq e^{-(\lambda+r)}\sum^k_{i\eq 0} \lambda^i r^{k-i} \frac{ {}^kC_i }{k!} [/math]
      [math]\eq e^{-(\lambda+r)}\cdot\frac{1}{k!}\cdot\ \underbrace{\left(\sum^k_{i\eq 0} {}^kC_i\ \lambda_i r^{k-i} \right)}_{\eq(\lambda+r)^k} [/math] - but note the sum is now just the binomial expansion of [ilmath](\lambda+r)^k[/ilmath]
    • Resulting in:
      • [math]\P{Z\eq k}\eq e^{-(\lambda+r)} \frac{(\lambda+r)^k}{k!} [/math] - which the definition of the probability of a Poisson random variable being equal to [ilmath]k[/ilmath] whose rate parameter is [ilmath]\lambda+r[/ilmath], specifically:
    • Now notice [math]\P{Z'\eq k}:\eq e^{-(\lambda+r) } \frac{(\lambda+r)^k}{k!} [/math]
    • Thus we see that [ilmath]\P{Z\eq k}\eq\P{Z'\eq k} [/ilmath]
  • Since we showed this for an arbitrary [ilmath]k\in\mathbb{N}_0[/ilmath] we have shown it for all

This completes the proof


  1. We could just have well have used
    • [ilmath]\P{Y\eq k}\Pcond{X\eq k-i}{Y\eq k} [/ilmath] in the sum, or
    • [ilmath]\P{Y\eq k-i}\Pcond{X\eq i}{Y\eq k-i} [/ilmath]
    and so forth (see conditional probability for details)
  2. Specifically that [ilmath]\Pcond{Y\eq k-i}{X\eq i}\eq\P{Y\eq k-i} [/ilmath] is used here. Which is fine as these events are independent events