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Probability Generating Functions

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) txG_X(t) = \mathrm{E}(t^X) = \sum_{x=0}^{\infty} \mathrm{P}(X = x)\, t^x

The coefficient of txt^x in the expansion of GX(t)G_X(t) is exactly P(X=x)\mathrm{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)=1G_X(1) = \sum \mathrm{P}(X = x) = 1: every PGF equals 1 at t=1t = 1. This is the standard check for unknown constants.
  • GX(0)=P(X=0)G_X(0) = \mathrm{P}(X = 0).
  • The PGF fully determines the distribution of XX.

2. Expectation and Variance from G(t)G(t)

Differentiate and evaluate at t=1t = 1:

E(X)=GX′(1),E(X(X−1))=GX′′(1)\mathrm{E}(X) = G_X'(1), \qquad \mathrm{E}(X(X-1)) = G_X''(1)

so

Var(X)=GX′′(1)+GX′(1)−(GX′(1))2\mathrm{Var}(X) = G_X''(1) + G_X'(1) - \bigl(G_X'(1)\bigr)^2

Worked example. GX(t)=12(1+t)2G_X(t) = \frac{1}{2}(1 + t)^2. Then

G′(t)=(1+t),G′(1)=2;G′′(t)=1,G′′(1)=1G'(t) = (1+t), \quad G'(1) = 2; \qquad G''(t) = 1, \quad G''(1) = 1 E(X)=2,Var(X)=1+2−22=1\mathrm{E}(X) = 2, \qquad \mathrm{Var}(X) = 1 + 2 - 2^2 = 1

(Expand to check: this is X∼B(2,12)X \sim \mathrm{B}(2, \tfrac12) — see the standard forms below.)

3. Standard PGFs to Memorise

DistributionGX(t)G_X(t)E(X)\mathrm{E}(X)Var(X)\mathrm{Var}(X)
B(n,p)\mathrm{B}(n, p)(q+pt)n(q + pt)^n, q=1−pq = 1-pnpnpnpqnpq
Po(λ)\mathrm{Po}(\lambda)eλ(t−1)e^{\lambda(t-1)}λ\lambdaλ\lambda
Geometric(p)(p) on {1,2,…}\{1, 2, \ldots\}pt1−qt\dfrac{pt}{1 - q t}1p\dfrac{1}{p}qp2\dfrac{q}{p^2}

Learn the binomial and Poisson forms derivationally: expand (q+pt)n(q+pt)^n by the binomial theorem and read off coefficients; expand eλ(t−1)e^{\lambda(t-1)} as a Maclaurin series and compare with the Poisson pmf.

4. Sums of Independent Random Variables

For independent XX and YY:

GX+Y(t)=GX(t) GY(t)G_{X+Y}(t) = G_X(t)\, G_Y(t)

This is the result the exam is built around.

Worked example. X∼B(4,0.3)X \sim \mathrm{B}(4, 0.3), Y∼B(6,0.3)Y \sim \mathrm{B}(6, 0.3), independent. Then

GX+Y(t)=(0.7+0.3t)4(0.7+0.3t)6=(0.7+0.3t)10G_{X+Y}(t) = (0.7 + 0.3t)^4 (0.7+0.3t)^6 = (0.7 + 0.3t)^{10}

which is the PGF of B(10,0.3)\mathrm{B}(10, 0.3): the sum of independent binomials with the same pp is binomial. The same argument for independent Poissons gives Po(λ1)+Po(λ2)=Po(λ1+λ2)\mathrm{Po}(\lambda_1) + \mathrm{Po}(\lambda_2) = \mathrm{Po}(\lambda_1 + \lambda_2).

A related form: if X1,…,XnX_1, \ldots, X_n are i.i.d. then GSn(t)=(GX(t))nG_{S_n}(t) = \bigl(G_X(t)\bigr)^n. For example, the sum of 12 independent geometric variables has PGF (pt1−qt)12\left(\dfrac{pt}{1-qt}\right)^{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)5G(t) = e^{2(t - 1)} \cdot (0.6 + 0.4t)^5 for independent X∼Po(2)X \sim \mathrm{Po}(2) and Y∼B(5,0.4)Y \sim \mathrm{B}(5, 0.4). Then X+YX + Y has PGF equal to the product, which has no simple closed form — but you can still compute E(X+Y)=2+2=3\mathrm{E}(X+Y) = 2 + 2 = 3 and Var(X+Y)=2+5(0.4)(0.6)=3.2\mathrm{Var}(X+Y) = 2 + 5(0.4)(0.6) = 3.2 from the product rule applied to derivatives at t=1t=1.

6. Common Pitfalls

  • Forgetting G′′(1)G''(1) is E(X(X−1))\mathrm{E}(X(X-1)), not Var(X)\mathrm{Var}(X): the variance formula has three terms.
  • Valid PGFs satisfy G(1)=1G(1) = 1: if an unknown constant kk appears, set G(1)=1G(1) = 1 first, then read off probabilities.
  • Sums of independent variables only. If XX and YY are dependent, the PGF product is wrong; state independence in your working.
  • Geometric support: some specifications define the geometric distribution on {0,1,2,…}\{0, 1, 2, \ldots\} (number of failures); its PGF is then p1−qt\dfrac{p}{1-qt}. 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+YX = c + Y and use GYG_Y.

7. Summary

TaskMethod
P(X=x)\mathrm{P}(X = x)Coefficient of txt^x
E(X)\mathrm{E}(X)G′(1)G'(1)
Var(X)\mathrm{Var}(X)G′′(1)+G′(1)−(G′(1))2G''(1) + G'(1) - (G'(1))^2
Sum of independent RVsMultiply the PGFs
Unknown constantUse G(1)=1G(1) = 1

Cross-References