# Transform “Standard Poisson” to any Poisson

If I use a RNG to generate a set of standard normal draws $$(Zi)(Z_i)$$, anyone can get samples from a normal distribution with $$(μ,σ2)(\mu, \sigma^2)$$ of their choosing via $$Xi=Zi.σ+μX_i=Z_i.\sigma+\mu$$. 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 $$λ=1\lambda=1$$ (or some other $$λ\lambda$$), is it possible for another person to get Poisson samples with whatever $$λ\lambda$$ they want, without having to use an RNG?

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

For instance, assume we want to “transform” Poisson realizations with $$λ=1\lambda=1$$ to Poisson samples with $$λ′=5\lambda'=5$$. The PMF at $$00$$ for $$λ=1\lambda=1$$ is $$1e≈0.368\frac{1}{e}\approx 0.368$$, so about 36.8% of the original samples will be $$00$$. But the cumulative distribution function for $$λ′=5\lambda'=5$$ is only $$0.2650.265$$ for $$x=3x=3$$. That is, we would need somehow map an original observation of $$00$$ to transformed observations $$0,1,2,3,40,1,2,3,4$$ – and this in a way that satisfies the PMF for the new $$λ′\lambda'$$. This is simply not possible without a RNG.