Probability Generating Functions
Info: Board Coverage Edexcel FS1 (Section 5) | AQA Optional Topic | OCR (A) FS1
A probability generating function (PGF) packs the whole probability distribution of a discrete random variable taking non-negative integer values into one expression:
GX(t)=E(tX)=x=0∑∞P(X=x)txThe coefficient of tx in the expansion of GX(t) is exactly P(X=x). PGFs are examinable for three reasons: they give expectation and variance quickly, they classify standard distributions, and they handle sums of independent random variables with almost no work.
1. Basic Properties
- GX(1)=∑P(X=x)=1: every PGF equals 1 at t=1. This is the standard check for unknown constants.
- GX(0)=P(X=0).
- The PGF fully determines the distribution of X.
2. Expectation and Variance from G(t)
Differentiate and evaluate at t=1:
E(X)=GX′(1),E(X(X−1))=GX′′(1)so
Var(X)=GX′′(1)+GX′(1)−(GX′(1))2Worked example. GX(t)=21(1+t)2. Then
G′(t)=(1+t),G′(1)=2;G′′(t)=1,G′′(1)=1 E(X)=2,Var(X)=1+2−22=1(Expand to check: this is X∼B(2,21) — see the standard forms below.)
3. Standard PGFs to Memorise
| Distribution | GX(t) | E(X) | Var(X) |
|---|
| B(n,p) | (q+pt)n, q=1−p | np | npq |
| Po(λ) | eλ(t−1) | λ | λ |
| Geometric(p) on {1,2,…} | 1−qtpt | p1 | p2q |
Learn the binomial and Poisson forms derivationally: expand (q+pt)n by the binomial theorem and read off coefficients; expand eλ(t−1) as a Maclaurin series and compare with the Poisson pmf.
4. Sums of Independent Random Variables
For independent X and Y:
GX+Y(t)=GX(t)GY(t)This is the result the exam is built around.
Worked example. X∼B(4,0.3), Y∼B(6,0.3), independent. Then
GX+Y(t)=(0.7+0.3t)4(0.7+0.3t)6=(0.7+0.3t)10which is the PGF of B(10,0.3): the sum of independent binomials with the same p is binomial. The same argument for independent Poissons gives Po(λ1)+Po(λ2)=Po(λ1+λ2).
A related form: if X1,…,Xn are i.i.d. then GSn(t)=(GX(t))n. For example, the sum of 12 independent geometric variables has PGF (1−qtpt)12 — a negative binomial distribution.
5. Finding the Distribution from a PGF
Given an unfamiliar PGF, either match it to a standard form or expand enough coefficients to identify the probabilities, then name the distribution and state its parameters.
Worked example. G(t)=e2(t−1)⋅(0.6+0.4t)5 for independent X∼Po(2) and Y∼B(5,0.4). Then X+Y has PGF equal to the product, which has no simple closed form — but you can still compute E(X+Y)=2+2=3 and Var(X+Y)=2+5(0.4)(0.6)=3.2 from the product rule applied to derivatives at t=1.
6. Common Pitfalls
- Forgetting G′′(1) is E(X(X−1)), not Var(X): the variance formula has three terms.
- Valid PGFs satisfy G(1)=1: if an unknown constant k appears, set G(1)=1 first, then read off probabilities.
- Sums of independent variables only. If X and Y are dependent, the PGF product is wrong; state independence in your working.
- Geometric support: some specifications define the geometric distribution on {0,1,2,…} (number of failures); its PGF is then 1−qtp. Check which convention your board uses — the exam paper will match the formula booklet.
- Mixed distributions (constant + random parts) do not have PGFs on their own: write X=c+Y and use GY.
7. Summary
| Task | Method |
|---|
| P(X=x) | Coefficient of tx |
| E(X) | G′(1) |
| Var(X) | G′′(1)+G′(1)−(G′(1))2 |
| Sum of independent RVs | Multiply the PGFs |
| Unknown constant | Use G(1)=1 |
Cross-References