Since the simple affine transformation does not preserve Poisson distribution, I’m wondering if there is any trick to apply a (deterministic) transformation to a Poisson random variable with mean $\lambda_1$ such that it remains Poisson but with mean $\lambda_2$?

One idea I had is to do the Anscombe transformation to get an approximate normally distributed random variable, and then apply a linear transformation to get the desired mean, followed by the inverse Anscombe. Of course, this is only approximate and I’m not sure if it’s even valid.

**Answer**

Not in general.

It’s not possible to do it exactly if $\lambda_2>\lambda_1$, since a Poisson variable with mean $\lambda_2$ has higher entropy than one with mean $\lambda_1$, so it takes more information to specify it, even if you are willing to have a crazy non-monotone transformation.

For $\lambda_2<\lambda_1$, it is at least not always possible. Suppose $\lambda$ is small, so that the variable basically has only values 0 and 1, and the probability of 0 is $\exp(-\lambda)$. You can’t transform between two distributions like this.

I can’t see any easy way to rule out that it’s possible in some cases with $1 \ll \lambda_2 \ll \lambda_1$.

**Attribution***Source : Link , Question Author : lucusk , Answer Author : J.G.*