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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Statement

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

1. $X\sim\text{Poi}(\lambda)$ and
2. $Y\sim\text{Poi}(r)$

(And that these are statistically independent random variables

Then:

• Let $Z:\eq X+Y$

We claim

• $X+Y\eq: Z\sim\text{Poi}(\lambda+r)$

Proof

Let $Z'\sim\text{Poi}(\lambda+r)$

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

• Let $k\in\mathbb{N}_0$ be given, then:
  • $$\P{Z\eq k}:\eq\P{X+Y\eq k}\eq\sum^k_{i\eq 0}\overbrace{\P{X\eq i}\Pcond{Y\eq k-i}{X\eq i} }^{\eq\P{(X\eq i)\cap(Y\eq k-i)} }$$

    $$\eq\sum^k_{i\eq 0}\P{X\eq i}\P{Y\eq k-i}$$ - as $X$ and $Y$ are independent random variables

    $$\eq\sum^k_{i\eq 0}e^{-\lambda}\frac{\lambda^i}{i!}\cdot e^{-r}\frac{r^{k-i} }{(k-i)!}$$ - by definition of the Poisson distribution

    $$\eq e^{-(\lambda+r)}\sum^k_{i\eq 0}\lambda^i r^{k-i} \frac{1}{i!(k-i)!}$$

  • Recall that $${}^nC_r:\eq \frac{n!}{r!(k-r)!}$$
    • So $$\frac{ {}^kC_i}{k!}\eq\frac{1}{i!(k-i)!}$$ (by dividing by $n!$ in the definition on the line above)
  • Thus: $$\P{Z\eq k}\eq e^{-(\lambda+r)}\sum^k_{i\eq 0} \lambda^i r^{k-i} \frac{ {}^kC_i }{k!}$$

    $$\eq e^{-(\lambda+r)}\cdot\frac{1}{k!}\cdot\ \underbrace{\left(\sum^k_{i\eq 0} {}^kC_i\ \lambda_i r^{k-i} \right)}_{\eq(\lambda+r)^k}$$ - but note the sum is now just the binomial expansion of $(\lambda+r)^k$

  • Resulting in:
    • $$\P{Z\eq k}\eq e^{-(\lambda+r)} \frac{(\lambda+r)^k}{k!}$$ - which the definition of the probability of a Poisson random variable being equal to $k$ whose rate parameter is $\lambda+r$, specifically:
  • Now notice $$\P{Z'\eq k}:\eq e^{-(\lambda+r) } \frac{(\lambda+r)^k}{k!}$$
  • Thus we see that $\P{Z\eq k}\eq\P{Z'\eq k}$
• Since we showed this for an arbitrary $k\in\mathbb{N}_0$ we have shown it for all

This completes the proof