If I use a RNG to generate a set of standard normal draws (Zi), anyone can get samples from a normal distribution with (μ,σ2) of their choosing via Xi=Zi.σ+μ. This doesn’t require them to do any random number generating themselves (i.e. they don’t need a RNG)

Is the same thing possible for Poisson distributions? i.e. If I generate a set of Poisson draws with λ=1 (or some other λ), is it possible for another person to get Poisson samples with whatever λ they want, without having to use an RNG?

(I realise you could use the normal approximation N(λ,√λ) if λ is large, but what if this isn’t the case?)

**Answer**

No, that is not possible.

For instance, assume we want to “transform” Poisson realizations with λ=1 to Poisson samples with λ′=5. The PMF at 0 for λ=1 is 1e≈0.368, so about 36.8% of the original samples will be 0. But the cumulative distribution function for λ′=5 is only 0.265 for x=3. That is, we would need somehow map an original observation of 0 to transformed observations 0,1,2,3,4 – and this in a way that satisfies the PMF for the new λ′. This is simply not possible without a RNG.

The same holds for “transformations” between any two discrete distributions (except of course for trivial cases).

**Attribution***Source : Link , Question Author : Mich55 , Answer Author : Stephan Kolassa*