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Let $X$ and $Y$ be independent. I know that if $X$ ~ $Poisson (a)$ and $Y$ ~ $Poisson (b)$ then $X + Y$ ~ $Poisson (a+b)$, but I don't fully understand why. I know how to develop the equation for it, but i'm lacking an intuitive sense of how this changes from a graphic perspective. Does anyone know of any visual source that can help explain this?

WaterDrop
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  • Not sure if this is what you're after, but this is the proof of the statement you referenced: https://math.stackexchange.com/questions/221078/poisson-distribution-of-sum-of-two-random-independent-variables-x-y – scoopfaze Jun 13 '20 at 19:06
  • I looking for more of a "visual" proof. Just something that can Poisson distribution for 2 independent sets and then shows the combined one. – WaterDrop Jun 13 '20 at 19:10
  • What does that mean? Poisson distribution of 2 sets? Poisson distribution is a measure! –  Jun 13 '20 at 19:18

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You can think of $X$ as the number of times an event occurs, in a period where on average it happens $a$ times, if the occurrence of such events is memoryless, i.e. no knowledge of events' histories affects the distribution of the number of events in any subsequent period. You can think of $Y$ as the number of times another such kind of event occurs, this time with average frequency $b$. So the interpretation is that defining "one event type or the other" events, if they're independent, retains the memorylessness property. This is the point @StashuKozlowski was making.

J.G.
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I'm not sure what you mean by 'visual', and the link @scoopfaze gave has a pretty comprehensive list of solutions. I'll try nevertheless to give a combinatorial, so to say, argument for the support of $X+Y$. Since both rvs have a non-negative support $X\geq 0$, their sum can take only non-negative integer values $n$.

At the same time, any positive integer $n$ is expressed as a sum of two non-negative integers (i.e. incl. $0$) in exactly $n+1$ number of ways: $n = k + (n-k), \ 0 \leq k \leq n$.

This determines the number of convolutions for each outcome of $X+Y$.

Alex
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