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Generating Functions
There is a very powerful method for finding the probability distribution for the sum of two or more independent discrete random variables: the method of generating functions.
The idea is to associate with a discrete RV a polynomial-like function. We then have rules to translate back and forth between the probability table for a discrete RV and its generating function.
The rules are very simple: suppose we have the contingency table for the RV X, something like
x P (X = x) v 1 p 1 v 2 p 2
...... vn pn
The generating function corresponding to X is the function (of x)
p 1 xv^1 + p 2 xv^2 + · · · + pnxvn^.
Actually, we never use any function properties of the generating function; the x only plays the role of a “formal variable,” a placeholder for the calculations.
Example 1. Suppose X has the probability distribution
x P (X = x) 0 0. 1 0. 2 0.
Then the generating function for X is 0. 3 x^0 + 0. 2 x^1 + 0. 5 x^2 , which would more conventionally be written as 0 .3 + 0. 2 x + 0. 5 x^2.
Example 2. Going the other way, suppose we are given the generating function
0 .2 + 0. 3 x + 0. 1 x^2.^1 + 0. 4 x^3.^2.
Since this is the same as
- 2 x^0 + 0. 3 x^1 + 0. 1 x^2.^1 + 0. 4 x^3.^2 ,
we can reconstruct the probability table for X as:
x P (X = x) 0 0. 1 0. 2.1 0. 3.2 0.
So far, this doesn’t seem very interesting. All we seem to have done is to encode the same information in the table into a single polynomial-looking line. What makes the method powerful is:
Theorem 1. Suppose X, Y are independent discrete random variables with generating functions f (x), g(x), respectively. Then the generating function for X + Y is the product f (x)g(x). 1
2
Now, we called the generating functions polynomial-like, because a generating function
p 1 xv^1 + p 2 xv^2 + · · · + pnxvn
isn’t necessarily a polynomial–the vi don’t have to be positive integers (or zero). But polynomial- like functions are multiplied by the same rules you learned to multiply polynomials in high-school. Thus the problem of finding the probability distribution of the sum of two (or more) independent RV’s is reduced to a problem you’re been solving since ninth grade.
Example 3. Let us suppose X and Y are independent random variables on the same sample space having probability distributions
x P (X = x) 0 0. 1 0. 2 0.
y P (Y = y) 0 0. 1 0.
(The distribution for X is the same one we used in the first example.) This means, for example, that X has a 30% probability of taking the value 0, a 20% probability of taking the value 1, and a 50% probability of taking the value 2.
From these tables we immediately read off the generating functions:
- For X, 0. 3 x^0 + 0. 2 x^1 + 0. 5 x^2 , written for simplicity as
0 .3 + 0. 2 x + 0. 5 x^2.
Therefore the generating function for X + Y is
(0.3 + 0. 2 x + 0. 5 x^2 )(0.4 + 0. 6 x) = 0.12 + 0. 26 x + 0. 32 x^2 + 0. 3 x^3.
From this we read off the probability distribution for W = X + Y :
w P (W = w) 0 0. 1 0. 2 0. 3 0.
Example 4. Now suppose we want to find the probability distribution for a sum X 1 +X 2 +X 3 +X 4 of four independent RV’s which all have the same distribution, given by the table in the previous example. The resulting sum (which we shall call S) can take any integer values between 0 and 8 inclusive. The generating function for S will be the 4th power of the generating function of X,
(0.3 + 0. 2 x + 0. 5 x^2 )^4 = 0.0081 + 0. 0216 x + 0. 0756 x^2 + M a 0. 1176 x^3
- 2086 x^4 + 0. 196 x^5 + 0. 21 x^6 + 0. 1 x^7 + 0. 0625 x^8.
We can revert to the table form by reading off the probabilities from this generating function:
