I have stumbled upon an interesting problem with Poisson-distributed random variables the other day and I wasn't really able to come up with a solution - I wonder if anyone could help me out here. The question is as follows. Suppose you have two i.i.d. random variables $X_1$ and $X_2$ s.t. $X_1 \sim Poisson(\lambda)$ and $X_2 \sim Poisson(\lambda)$. Now let's draw a large number of pairs $(x_1, x_2)$ from $(X_1, X_2)$ and for each pair, do the following:

  • If $x_1$ > $x_2$, store $x_1$ in a list $L$.
  • If $x_1$ < $x_2$, do nothing, take the next pair.
  • If $x_1$ = $x_2$, store $x_1$ in $L$ with probability 0.5.

The question is: what is the distribution of numbers that we have placed in $L$?

Note: I made a few numeric simulations and for me the distribution of numbers in $L$ looks like another Poisson distribution with a mean higher than $\lambda$, but I can neither prove nor disprove it. Is there a closed-form solution for this? If not, is it possible to work out at least some kind of an iterative equation that I can use to obtain the distribution for any given $\lambda$?


1 Answer 1


I'd consider the alternative (slightly simpler) game: take the maximum of $(x_1,x_2)$ and place it in L.

Now, I claim that this second game is statistically equivalent to the original (they differ in the number of samples of course). Once you accept this, the problem is that of computing the probability function of the maximum of two Poisson variables.

In general, the maximum of two iid discrete variables is given by $P(y) = F_x(y)^2 - F_x(y-1)^2$ with $y=\max(x_1,x_2)$ (eg).

This gives me (errors aside) :

$$P(y) = e^{- 2 \lambda} \; \frac{\lambda^y}{y!} \left( \frac{\lambda^y}{y!} + 2 \sum_{k=0}^{y-1} \frac{\lambda^k}{k!} \right)$$

which resembles a Poisson, but not quite.

  • $\begingroup$ Great, thanks, this is exactly what I was looking for. $\endgroup$
    – Tamás
    Jun 8, 2011 at 21:20

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