# E[ | X - Y | ] where X and Y are independent Poisson random variable

What is the expected value of the absolute difference of two independent Poisson variables?

$$E[ |X - Y| ]$$

Seems like an easy question but I haven't found an easy solution.

I've split the double sum into the correct regions but not sure what to do with the partial sums remaining.

I have:

$$\sum_0^\infty p(x) \sum_0^\infty |X - Y| p(y)$$

...since p(x,y) = p(x)p(y)

$$= \sum_0^\infty p(x) [\sum_0^x (X - Y) p(y) + \sum_x^\infty (Y - X) p(y)]$$

Should get something like | E[X] - E[Y] | + some variance or covariance term, the latter of which will be 0 since X and Y are independent.

• In what context did this question arise? If homework or similar exercise, please see the FAQ for why this is not an appropriate forum for this question. Otherwise, I'm still not sure this is the site you are looking for. Aug 5 '12 at 10:50
• I agree that you should give more context. It's not clear whether to expect there to be a simple solution or not. The "variance or covariance term" suggestion seems wrong. Integrating or summing the absolute values of simple functions can produce much more complicated results. Anyway, for equal parameters, Mathematica sums over the region where $X\gt Y$ (half of the answer) and gets $\exp(-2 \lambda)(BesselI[1,2\lambda] + \lambda BesselI[2,2 \lambda])$. Aug 5 '12 at 11:02
• No its not a homework question (I left school many years ago!). Its something I came across while working through some derivations in a Mathematical Finance book recently. Maybe I was on the wrong track, but it just seems E[|X - Y|] should be expressible in terms of E[X] and E[Y]. Aug 5 '12 at 11:16
• In general, you can't express the expected absolute value that way. Consider $f(x) = x^5+2x-1$ on $[0,1]$. The average value of $f$ is easy to compute, but to compute the average value of $|f|$ you need to be able to express the root of $f$ on that interval. That can be an example of an absolute value of the difference between random variables where one is the constant $0$. Aug 5 '12 at 11:35
• The difference of two Poisson distributions is called a Skellam distribution and can be expressed in terms of modified Bessel functions. I don't see why the expectation of the absolute value should have a simple expression. Aug 5 '12 at 13:33