Difference of two independent Poisson Distributions

By B.A. Lee on June 3, 1998 :

From: B.A. Lee
On: 6/3/1998 at 12:37
I will give this a shot and see if anyone can help answer this question.

If X and Y are independent Poisson Distributions, we know that X + Y is also a Poisson Distribution with a mean which is the sum of the means of X and Y. What about X - Y? Is it also a Poisson Distribution? What is its mean then?

By Simon on June 4, 1998 :

The best way to approach this problem is by using PGFs (probability generating functions). However, I will assume that you don't yet know about them, because they're quite advanced. Incidentally, if you are interested, you might like to look up PGFs in a textbook - they're quite interesting and powerful.

The way you prove that X+Y is a Poisson r.v. (random variable) if X and Y are without resorting to technical things like PGFs, is to use the Law of Total Probability. In the case of a sum of Poisson r.v.s X and Y this is

$P (X + Y = n) = \sum_{m = 0}^{n} P (X + Y = n | X = m) \times P (X = m)$ . (1)
This is a standard result that will be explained in most books. We can use it to find the distribution of X+Y because we know how to calculate P(X+Y=n|X=m) = P(Y=n-m) and P(X=m). We can then plug these into our formula (1) and it turns out that we can write the sum in closed form (i.e. as a normal algebraic expression) as exactly the form of a Poisson probability. This is all very exciting and we think ``Hmm. Can we use the same method to find the distribution of the difference of X and Y?''

The answer is, we can try, but it doesn't quite work. We can use a similar formula to (1):

$P (X - Y = n) = \sum_{m = 0}^{n} P (X - Y = n | Y = m) \times P (Y = m)$ . (2)
Again, we can plug in the values of P(X-Y=n|Y=m) = P(X=m+n) and P(Y=m), but this time the sum is much more messy, and doesn't (at least to me) look like it could be written in a recognisable closed form (or possibly any closed form at all). In particular it is not a Poisson probability in general. Try it yourself, using formula (2).

So X-Y is not (in general at least) a Poisson r.v. if X and Y are.

The mean of the r.v. X-Y (where X is a Poisson r.v. with mean a, and Y is a Poisson r.v. with mean b, and X and Y are independent) is a-b, since we can use the formula E(X-Y) = E(X)-E(Y), which holds for any r.v.s.

I hope this has been helpful (even if it didn't give a particularly satisfying answer). The last problem is first year degree level stuff, so don't worry about it too much.

By Gordon Lee on June 5, 1998 :

I think I ought to explain a bit more what a probability generating function is. Because at its simplest, it is not very hard to understand.

Given a Random Variable X, we define its generating function G(s) as the following

$G (s) = \sum_{all integral k} P (X = k) s^{k} = E (s^{X})$
So G(s) is a function in terms of s.

Now this function is extremely useful in shortening calculations. Before we continue, I think we should note the following:

The p.g.f is completely determined by and also determines X
G'(1) is the expectation of X
G''(1) is the expectation of X(X-1).

So, how do we calculate the p.g.f of a Poisson random variable? Well, we just use the definition:

Let X be a Poisson Random Variable with parameter a. Its p.g.f, G(s) say is given by:

G(s) = (e^-a a⁰ /0!) s⁰ + (e^-a a¹ /1!) s¹ + (e^-a a² /2!) s²

+..... etc.

We can take the e^-a out of the series, since it is only a constant. Now note that the series we have left behind is just the expansion of e^as .

so G(s)=e^a(s-1) .

Now, as mentioned earlier G'(1) = E(X) = a , as expected.
Also, G''(1)= E(X(X-1)) = E(X² ) - E(X) = a²
so Var(X)= E(X² )- (E(X))² = a² + a - a² = a, as expected.

P.g.f's also allow us to add up independent random variables. Since if X and Y are independent, then we can find the p.g.f of X + Y by just multiplying the p.g.f's of the random variables together, since

E(s^X )E(s^Y )=E(s^X+Y ), if X and Y are independent.

The p.g.f for the random variable -Y where Y is a Poisson RV with parameter b is given by the following:

H (s) = \sum_{all integral k} P (- Y = k) s^{k} = \sum P (Y = - k) 1 / s^{- k}

Note that -k is +ve

= e^-b (1 + (b/s) + (1/2!)(b/s)² + (1/3!)(b/s)³ +....)
= e^{b(1/s - 1)} .

As we mentioned earlier, a p.g.f completely determines the random variable, so recalling that X - Y (= X + (-Y)) has p.g.f.

e^{a(s-1)+b(1/s-1)} ,

the 1/s in the above expression ensures that there's no way it is a Poisson random variable. So the answer to your question is basically this:

Although the mean of X - Y is just E(X)-E(Y), X - Y itself is NOT a Poisson distribution in general and cannot be modelled as such.

I am sorry that this is getting more technical, but it is the most elegant way of looking at these things. If you have ANY questions, please reply to us; we will be more than willing to answer any queries you have.

Hope that helps, We would appreciate a reply saying what you think of the answer.
Gordon