I'm working on a problem in which the random variable is the difference between two random variables. We know from math that the mean will be the difference of the means, and that the variance will be the sum of the variances. I know that for a Poisson distributions the sum of two independent variables will be another Poisson distribution with mean the sum of the means, and this will be identical to the variance.
Therefore I can treat the difference between Poisson distributed random variables the same way. However, a Poisson distribution has only positive values, and I know that the difference between the random variables will have some value that turns out to be negative. What am I missing?
Best Answer
Here are two Poisson distributions:
Here is the associated Skellam Distribution (which is the solution to your problem):
Indeed, the distribution is defined for negative values. Moreover, the mean of the Skellam distribution is the difference of the means of the two Poisson distributions (even if that value is negative) and the variance of the Skellam distribution is the (positive) sum of the variances of the Poisson distributions.