Generating a Poisson variable independent of "time".

Question

For $X \sim Poi(r,t)$ over a period of time $t$ with rate $r$, we would have: $$Pr(X = k) = \frac{(rt)^k}{k!}e^{-rt}$$ Is there any way I could simulate this distribution that is independent of $t$ (i.e. over a unit period of time), and apply a function on the resulting sample so that the overall random variable is distributed as $Poi(r,t)$?

Answer 1

Note: This is not a full answer, but is sufficiently long to not count as a comment.

If you had allowed random functions, you could do something with Poisson thinning (tossing away samples with some probability), but otherwise, no, I don't think you can do something for general rates. To see this, consider the Poisson RV Y distribution with a rate $r_0$ s.t. $0.71 > P(Y = 0) > 0.70 $. Now, consider the Poisson distribution with rate 1, which has probabilities $P(X = 0) = P(X = 1) = \frac{1}{e} ~ 0.36 $. Now, if you are asking whether there exists a deterministic function f s.t. $f(X) = Y$, then the answer is no, since there is no subset of integers that will have their probabilities sum up to an interval between 0.71 and 0.70.

Of course, you can do more if you are willing to use multiple samples of $X$ and generate smaller probabilities (to help you add up to whatever you want), you can get better and better approximations, but the best way I can think of a mapping would be to tolerate some error and plug it into a linear program to find the best mapping.