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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.

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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 $1 \ll \lambda_2 \ll \lambda_1$.

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  • $\begingroup$ This makes me wonder, if you are given $\lambda_2 - \lambda_1$ (individual values are not themselves known), a black box sampler for $Poisson(\lambda_1)$ and a $U[0,1]$ sampler, can we then sample from $Poisson(\lambda_2)$. Motivated by the case where both $\lambda_1$ and $\lambda_2$ are small then in the limit to Bernoulli distributions, I think we can now perform this sampling. $\endgroup$
    – rwolst
    Commented Jun 15, 2020 at 7:36
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    $\begingroup$ Yes, absolutely. You just can't do it without a source of randomness. For example, if $\lambda_2>\lambda_2$ you just the U[0,1] to generate Poisson($\lambda_2-\lambda_1$) and add it to what you have. $\endgroup$
    – Thomas Lumley
    Commented Jun 15, 2020 at 8:12
  • $\begingroup$ Makes sense, and if $\lambda_2 - \lambda_1 < 0$ this seems like it may need a more interesting trick as we cannot use the additive properties of independent Poisson. $\endgroup$
    – rwolst
    Commented Jun 16, 2020 at 7:47
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    $\begingroup$ It only works for discrete distributions. Entropy doesn't really mean the same thing for continuous distributions. You might be able to use an information argument to show you'd be able to get more Fisher information about $\lambda_1$ by transforming before estimating, and that sort of thing does carry through (the 'data processing theorem') $\endgroup$
    – Thomas Lumley
    Commented Jun 16, 2020 at 22:04
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    $\begingroup$ When $\lambda_1$ is small (and $\lambda_2$ is smaller) the two variables are effectively Bernoulli with different $p\approx 1-\exp(-\lambda)$. You can't turn binary with one $p$ into binary with a different $p$ deterministically, because all the 1s have to either map to 1s or to 0s. $\endgroup$
    – Thomas Lumley
    Commented Jun 17, 2020 at 22:15
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If you want it to be reversible and the result to have a Poisson distribution, it's not really possible.

Poisson distributions assign probability to the non-negative integers.

  1. What monotonic transformations take non-negative integers to non-negative integers?

    Unless you leave gaps in the middle of the sequence, I think all you have is shifts to the right, either of which doesn't leave you with another Poisson.

    So that rules out readily interpretable (monotonic) transformations. Assuming you want a 1-to-1 transformation (otherwise you cannot transform back), you're at best going to be left with "shuffling" the values in some way.

    But now note that the Poisson is unimodal (with an "edge case" of two adjacent modes when the parameter is an integer); the probabilities decrease either side of that. That severely limits your options (without even getting to the specific functional form)

  2. Now consider a specific Poisson -- say a Poisson(1). It has a specific set of probabilities. Any deterministic 1-to-1 transformation will simply move those probabilities somewhere else. In general no other Poisson can share more than a fraction of those probabilities; where do the others go? e.g. with a Poisson(1) you have two probabilities of 1/e -- you might perhaps be able to find another Poisson that has a probability of 1/e, but can you find one with two of them, both the largest possible probabilities? It turns out you can't.

  3. If you don't require a 1-to-1 transformation, so you can steal any number of probabilities from the far tail as needed, you might in some cases be able to have a transformation from a large parameter to a good approximation of a small one, but it might be hard to give a finite-time/space construction of one; I think this would not be a practical exercise in general.

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  • $\begingroup$ Not only shuffling, also summing is possible eg Poisson outcomes {1,2,3} -> 0 in different discrete distribution. $\endgroup$
    – innisfree
    Commented Jun 14, 2020 at 6:14
  • $\begingroup$ Probability integral transform and then the $Poisson(\lambda_2)$ inverse CDF, or does the discreetness wreck the probability integral transform? $\endgroup$
    – Dave
    Commented Jun 14, 2020 at 6:24
  • $\begingroup$ @innisfree You'll note I have "1-to-1" specified there (on the assumption that a reversible transform is required). This rules out adding probabilities. $\endgroup$
    – Glen_b
    Commented Jun 14, 2020 at 6:28
  • $\begingroup$ @Dave Try it :). Try to go from a Poisson(1/2) to a Poisson(5) and see what you end up with. $\endgroup$
    – Glen_b
    Commented Jun 14, 2020 at 6:29
  • $\begingroup$ It seems like the discreetness prevents the probability integral transform from giving a uniform distribution. Is this what you meant? Also, how did you pick $1/2$ and $5$? $\endgroup$
    – Dave
    Commented Jun 14, 2020 at 6:38

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