Consider $P(X<Y)$, $P(X=Y)$ and $P(X>Y)$ in $(0,1)$ that sum to $1$.

Is it always possible to find Poisson parameters $\lambda$ and $\mu$ such that independent $X\sim\text{Pois}(\lambda)$ and $Y\sim\text{Pois}(\mu)$ satisfy these conditions? (Equivalently, such that the corresponding Skellam distribution has the specified probability mass over the negative, zero and positive numbers.)

This was inspired by the discussion in this thread, where it was found that this is not always possibly with an additional condition for the sum $\lambda+\mu$ to equal a given number.


1 Answer 1


To start: If $P(X<Y)=P(Y>X)$ then $\lambda=\mu$ and $$P(X=Y)=\sum_i \frac{e^{-2\lambda}\lambda^{2i}}{x!^2}$$ which is 1 when $\lambda=\mu=0$ and decreases smoothly to zero as $\lambda=\mu\to\infty$ and so can be solved for any desired $P(X=Y)$, so we can get any $P(X<Y)=P(Y>X)$ we want (I tried unsuccessfully to get a counterexample this way).

Can we generalise? Suppose that instead we want a ratio $P(X<Y)=kP(Y>X)$ (for unspecified Ps). For any specified $\lambda$ we can find a $\mu$ to make $P(Y>X)$ as large as we like ($\mu\to\infty$) or as small as we like ($\mu\to 0$). So we have a continuous function $\lambda_k(\mu)$ and invoke the intermediate value theorem again.

Now, can we still get $P(X=Y)$ to be whatever we like? Obviously we will be able to for $k$ near enough to 1, so consider $k$ very large. We need $\lambda \gg \mu$. The potential problem is that if $\lambda \gg \mu$ we might need $\lambda$ large enough to force $P(X=Y)$ to be small, or $\mu$ small enough to force $P(X=Y)$ to be large.

The first is not a problem. We can take $P(X=Y)$ as large as we like (say $1-\epsilon$) by taking $\lambda$ and $\mu$ small. For very small $\epsilon$, then, $$P(X<Y)\approx P(X=0 \& Y=1)\approx e^{-\lambda}(1-e^{-\mu})\approx e^{-\lambda}$$ and $P(Y>X)\approx e^{-\mu}$. There's no problem in choosing $e^{-\lambda}=ke^{-\mu}$

And the second is not a problem. We can take $P(X=Y)$ as small as we like, and still have $P(X<Y)$ anywhere we like.

So for any $\mu$ we can obtain $P(X>Y)=kP(Y>X)$ by setting $\lambda=\lambda_k(\mu)$, and then choose $\mu$ to get $P(X=Y)$ as large or small as we like.

There might well be a way to make $\lambda_k(\mu)$ explicit for $\mu$ that isn't tiny, but I don't see it right now.

  • 1
    $\begingroup$ Are these functions monotonic? Would that imply there is a single solution $(\lambda,\mu)$ in each case? $\endgroup$
    – Henry
    Jun 3, 2020 at 9:19
  • $\begingroup$ There is a single solution, because for any $\lambda$, $P(X>Y)$ is strictly decreasing in $\mu$ $\endgroup$ Jun 3, 2020 at 9:36

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