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I have two random variables (X and Y), each of which is poisson distributed. Can I assume the ratio of both variables (X/Y) to be poisson distributed as well (for an empirical analysis)? Note that Y has very few 0s, and if Y=0, X=0 as well (magically). I have seen other papers posing that 0/0=0 for such cases.

Mathematically that is not clean, but empirically how much of a problem is that?

Note a similar question has been asked in this forum, however the explanation seems insufficient for my case.

Thank you for your precious help and best, m

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Even assuming that 0/0 =0 would not be sufficient to assume that X/Y is Poisson. The Poisson distribution only takes on integer values, so the ratio clearly cannot be Poisson. What exactly do you mean by an empirical analysis? –  jsk Apr 28 at 8:36

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I have two random variables (X and Y), each of which is poisson distributed.

How do you know?

What is the bivariate distribution like?

Can I assume the ratio of both variables (X/Y) to be poisson distributed as well Note that Y has very few 0s, and if Y=0, X=0 as well (magically).

(Can you explain how Y=0 guarantees X=0 while still maintaining this marginal Poissoness? That sounds like a very interesting bivariate distribution.)

First problem: Even a single (0,0) is undefined.

Second problem: Even if you explicitly excluded the (0,0) part of the bivariate density, or explicitly defined it to be say 0, the result is still clearly not Poisson, because a Poisson distribution is defined on the integers, while the resulting ratio is defined on the non-negative rationals.

The black step function below is the empirical cdf for the ratio of two independent Poissons (with mean large enough that 0/0 doesn't occur even in 100,000 observations), while the blue step function is a Poisson with the same mean as the simulated ratio:

enter image description here

Third Problem: Note that you can't even say "well, maybe it's scaled Poisson", because the gaps in a scaled Poisson are still of a minimum size (there are no other values between them), whereas any two rationals always have more rationals between them. The two things are very different.

Here's a different ratio of Poissons (still independent) with moment-matched (to mean & variance) scaled Poisson:

enter image description here

This is far from a worst case!**

You might do better with a (possibly zero-inflated) continuous distribution (the "ratio" is actually discrete, but continuous distributions may still offer a better approximation than a Poisson. A decent first approximation (in some circumstances) -- where 0's are extremely rare, for starters, and assuming you do something suitable with 0's -- might be a Normal based off a Taylor approximation.

**Dependent Poissons can do some very interesting things ... and given what you said, you can't have independence.

Here's an example of what one particular form of high dependence (the only one I checked) can do:

enter image description here

Good luck approximating that with either a Poisson or a scaled Poisson. Or any continuous distribution. [You might get somewhere near to it with a mixture of several continuous distributions and a discrete distribution, but I am not certain what purpose would be served by that.]

I have seen other papers posing that 0/0=0 for such cases.
Mathematically that is not clean, but empirically how much of a problem is that?

You can define your transformation $Z=f(X,Y)$ how you like, but it's no longer really a ratio. It may or may not be much of an issue depending on your audience and your data and what you're doing with it ... but even if that's all fine, it doesn't make the result Poisson. It will very rarely be the case that the ordinary Poisson is even a half-decent approximation.

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