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

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{{ProbMacro}}
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==Statement==
Let {{M|\lambda,r\in\mathbb{R}_{>0} }} be given. We start with the following two [[random variables]]
# {{M|X\sim\text{Poi}(\lambda)}} and
# {{M|Y\sim\text{Poi}(r)}}
(And that these are [[statistically independent random variables]]


Then:
* Let {{M|Z:\eq X+Y}}
We claim
* {{M|X+Y\eq: Z\sim\text{Poi}(\lambda+r)}}
==Proof==
Let {{M|Z'\sim\text{Poi}(\lambda+r)}}

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

* Let {{M|k\in\mathbb{N}_0}} be given, then:
** {{MM|\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)} } }}  <ref group="Note" name="Cond1"/>
**: {{MM|\eq\sum^k_{i\eq 0}\P{X\eq i}\P{Y\eq k-i} }} - as {{M|X}} and {{M|Y}} are [[independent random variables]]<ref group="Note" name="Indep1"/>
**: {{MM|\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]]
**: {{MM|\eq e^{-(\lambda+r)}\sum^k_{i\eq 0}\lambda^i r^{k-i} \frac{1}{i!(k-i)!} }}
** Recall that [[nCr|{{MM|{}^nC_r:\eq \frac{n!}{r!(k-r)!} }}]]
*** So {{MM|\frac{ {}^kC_i}{k!}\eq\frac{1}{i!(k-i)!} }} (by dividing by {{M|n!}} in the definition on the line above)
** Thus: {{MM|\P{Z\eq k}\eq e^{-(\lambda+r)}\sum^k_{i\eq 0} \lambda^i r^{k-i} \frac{ {}^kC_i }{k!} }}
**: {{MM|\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 {{M|(\lambda+r)^k}}
** Resulting in:
*** {{MM|\P{Z\eq k}\eq e^{-(\lambda+r)} \frac{(\lambda+r)^k}{k!} }} - which the definition of the probability of a [[Poisson distribution|Poisson random variable]] being equal to {{M|k}} whose rate parameter is {{M|\lambda+r}}, specifically:
** Now notice {{MM|\P{Z'\eq k}:\eq e^{-(\lambda+r) } \frac{(\lambda+r)^k}{k!} }}
** Thus we see that {{M|\P{Z\eq k}\eq\P{Z'\eq k} }}
* Since we showed this for an arbitrary {{M|k\in\mathbb{N}_0}} we have shown it for all
This completes the proof


==Notes==
<references group="Note">
<ref name="Indep1">Specifically that {{M|\Pcond{Y\eq k-i}{X\eq i}\eq\P{Y\eq k-i} }} is used here. Which is fine as these {{link|events|probability}} are {{link|independent events|probability}}</ref>
<ref name="Cond1">We could just have well have used
* {{M|\P{Y\eq k}\Pcond{X\eq k-i}{Y\eq k} }} in the sum, or
* {{M|\P{Y\eq k-i}\Pcond{X\eq i}{Y\eq k-i} }}
and so forth (see [[conditional probability]] for details)</ref>
</references>
==References==
<references/>
{{Theorem Of|Elementary Probability|Probability|Statistics}}