# Is a random sample of a Poisson distribution also Poisson distributed?

Car analogy:

Assume the traffic (number of cars per hour) on a road has a Poisson distribution, and the time between cars has the matching exponential distribution. If the chance of each car being red is independent from both time and the color of other cars, will the number of red cars per hour also have a Poisson distribution?

I strongly suspect so. Furthermore I expect E(red cars)=p(red) * E(cars), and because of the Poisson distribution σ(red cars) = p(red) * σ(cars). But how would I (dis)prove this?

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The answer to your first question is yes. If the sum of two independent variables is poisson then the individual variables are also poisson. See Raikov's Theorem

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@Srikant: Would you mind explaining which two independent variables are being summed? –  whuber Sep 15 '10 at 14:39
I'd assume red and non-red cars. –  MSalters Sep 15 '10 at 15:23
@whuber @msalters yes that is correct. My answer addresses the OP's first qn. No of cars per hour is poisson. No of cars per hr = No of red cars per hr + No of non-red cars per hr. Hence, No of red cars per hr is poisson. –  user28 Sep 15 '10 at 15:26
Hadn't realized it this explicitly, but I can indeed get at the results I need by treating my "all cars" not as an independent variable, but as the sum of two variables with related distributions. Then σ(cars) = σ(red cars) + σ(non-red cars) = σ(red cars) * (1/p(red)) –  MSalters Sep 15 '10 at 15:37
@Srikant: Cool! Thanks for the reference to Raikov's Theorem. (I might have spent a lot of time trying to think of a counterexample otherwise...) –  whuber Sep 15 '10 at 15:50

What you have described is a standard and important result from the theory of stochastic processes. Pick up any good book in the library on those, flip to the "Poisson Processes" chapter, and look for "decomposition" or sometimes "thinning".

Two of my favorites are Introduction to Stochastic Processes by Cinlar and Essentials of Stochastic Processes by Durrett.

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@G. Jay: Could you elaborate a little on what "standard and important result" this is and how it applies in this situation? –  whuber Sep 15 '10 at 14:40
Sure. I said "standard" because I have seen it usually covered in the first few chapters of books on stochastic processes. (Didn't mean to seem snooty; hope it didn't come off that way). I said "important" because a) Poisson processes form the foundation for renewal processes in general (where interarrival times are not necessarily exponentially dist'd but are independent with dist'n F), and b) decomposition of Poisson processes in particular gives, for instance, a handy way to simulate continuous time Markov chains with a computer. Durrett has a nice discussion of this point. –  G. Jay Kerns Sep 15 '10 at 15:45
By the way, as whuber noticed, the process determining which cars are "red" and which ones "not-red" should be independent of the Poisson process; otherwise, all bets are off. :-) The OP did mention independence, but it can be tricky to make things perfectly clear. –  G. Jay Kerns Sep 15 '10 at 15:48
@G. Jay: Thank you! –  whuber Sep 15 '10 at 15:56

I think you need additional assumptions to draw this conclusion.

Imagine this: the cars on the road are originally unpainted, but just before you observe them a demon decides what color they will become: red or some other color (of which there could be many). The demon does not have a clock (to ensure that whatever he does is independent of time), but he does have a fair coin. Ordinarily he flips the coin to determine whether the car will be red or some other color. However, his can of red paint can color ten cars and needs to be used as quickly as possible. So, if the coin decides red, the demon just goes ahead and paints the next ten cars red. Once the paint is exhausted, he goes back to flipping the coin. Other colors are painted in the same fashion.

Obviously the car count per unit time is unaffected by this procedure, but I hope it's just as obvious that the count of red cars per unit time is not a Poisson process, because it will be overdispersed: there will be too many high counts, due to the temporal clumping of red cars, and too many low counts, due to the temporal clumping of non-red cars.

It could be that this scenario violates your sense of "the chance of red being independent of the color of the other cars," but it's difficult to know exactly what this statement means, since it's open to various interpretations. In the present case, the probability that the next car to be observed is red, given that the previous car was non-red ("the other" colors), is independent of the color of the previous car.

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Sorry, I had thought it was clear that the color here should be taken as red or non-red. The chance of a car being red obviously doesn't depend on a previous car being blue or green, but does depend on the previous car being red. –  MSalters Sep 15 '10 at 15:29
@MSalters: thank you for the clarification. I'll leave my answer up for amusement value ;-). –  whuber Sep 15 '10 at 21:35
suppose we look at the number N of cars that pass in a fixed time [8 hours, perhaps]. N should have a poisson distribution, according to your assumptions. let $\lambda$ be the poisson parameter for N.
suppose all of the cars are colorless to begin with. as each car passes, we flip a coin for which P(heads) = $p$. [presumably $p$ will reflect the proportion of red cars in the population.] each time the coin lands 'heads' the car is colored red.
then if M denotes the number of red cars seen in the time period involved, M is also poisson with parameter $p\lambda$. [easy homework exercise, or see feller's book: intro to probability theory$\cdots$, vol 1, wiley.]