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  1. Learning objectives
  2. Core idea
  3. Key formulae
  4. Worked examples
  5. Explore the graph
  6. Common mistakes
  7. Exam tip
  8. Guided practice
  9. Practice questions
  10. Challenge
  11. Summary
  12. Quick quiz
  13. Same topic, different board
  14. Past paper questions
Learn9231Paper 4 · Further Probability & StatisticsProbability generating functionsProbability generating functions
Probability generating functions · Subtopic 1 of 1

Probability Generating Functions

GX(t)=E(tX)=∑P(X=x)txG_X(t)=E(t^X)=\sum P(X=x)t^x packs a whole distribution into one function. G(1)=1G(1)=1, E(X)=G′(1)E(X)=G'(1) and Var(X)=G′′(1)+G′(1)−[G′(1)]2\text{Var}(X)=G''(1)+G'(1)-[G'(1)]^2. For independent XX and YY, GX+Y=GXGYG_{X+Y}=G_XG_Y.

Cambridge 9231 Paper 4Edexcel 9FM0 Option≈ 45 minRevision 1 · updated 2026-10-27
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Section 1 of 14

Learning objectives

  • Write the PGF of a discrete distribution and read probabilities from it.
  • Find mean and variance from derivatives.
  • Find the PGF of a sum of independent variables.
Pause and think

The coefficient of t2t^2 in (0.5+0.5t)3(0.5+0.5t)^3 is 0.3750.375. What could that be?

Section 2 of 14

Core idea

Start hereA distribution in one formula

A probability table can be long. Multiply each probability by tt to the power of its value, add them up, and you get one function that stores everything: every probability, the mean and the variance, all recoverable with algebra and calculus.

The big idea

G(t)=E(tX)G(t)=E(t^X). Coefficients give probabilities, G′(1)G'(1) gives the mean, G′′(1)G''(1) gives E(X(X−1))E(X(X-1)), and independent sums multiply PGFs.

1. Moments from derivatives

Mean and variance
  1. G(t)=∑pxtxG(t)=\sum p_xt^x
    Definition.
  2. G′(t)=∑xpxtx−1⇒G′(1)=∑xpx=E(X)G'(t)=\sum xp_xt^{x-1}\Rightarrow G'(1)=\sum xp_x=E(X)
    Differentiate term by term; set t=1t=1.
  3. G′′(1)=∑x(x−1)px=E(X2)−E(X)G''(1)=\sum x(x-1)p_x=E(X^2)-E(X)
    Differentiate again.
  4. Var(X)=G′′(1)+G′(1)−[G′(1)]2\text{Var}(X)=G''(1)+G'(1)-[G'(1)]^2
    E(X2)−[E(X)]2E(X^2)-[E(X)]^2.
Worked exampleA small PGF

G(t)=14(1+2t+t2)G(t)=\frac14(1+2t+t^2). Find E(X)E(X) and Var(X)\text{Var}(X).

Goal
Mean and variance.
Strategy
Differentiate.
  1. G′(t)=14(2+2t), G′(1)=1G'(t)=\tfrac14(2+2t),\ G'(1)=1
    Mean.
  2. G′′(t)=12G''(t)=\tfrac12
    Second derivative.
  3. Var=12+1−1=12\text{Var}=\tfrac12+1-1=\tfrac12
    Formula.
AnswerE(X)=1E(X)=1, Var(X)=12\text{Var}(X)=\frac12.
✓ CheckThis is B(2,12)B(2,\frac12): np=1np=1, npq=12npq=\frac12 ✓.

2. Sums of independent variables

Why PGFs multiply
  1. GX+Y(t)=E(tX+Y)G_{X+Y}(t)=E(t^{X+Y})
    Definition.
  2. =E(tXtY)=E(t^Xt^Y)
    Index law.
  3. =E(tX)E(tY)=E(t^X)E(t^Y)
    Independence.
GX+Y=GXGYG_{X+Y}=G_XG_Y.
Worked examplePoisson sum

X∼Po(2)X\sim\text{Po}(2), Y∼Po(3)Y\sim\text{Po}(3) independent. Identify X+YX+Y.

Goal
Distribution.
Strategy
Multiply PGFs.
  1. e2(t−1)e3(t−1)e^{2(t-1)}e^{3(t-1)}
    Product.
  2. =e5(t−1)=e^{5(t-1)}
    Add exponents.
AnswerX+Y∼Po(5)X+Y\sim\text{Po}(5) (a PGF determines its distribution).
✓ CheckMeans add: 2+3=52+3=5 ✓.
What a PGF gives you

Everything comes from G.

Common misconceptionNot G″(1) alone G′′(1)=E(X2)−E(X)G''(1)=E(X^2)-E(X), so add G′(1)G'(1) back before subtracting the square.
TrickFind constants G(1)=1G(1)=1 always: use it to find an unknown coefficient.
Pause and think

What is G(0)G(0)?

Summary

  • G(t)=E(tX)G(t)=E(t^X), G(1)=1G(1)=1.
  • E(X)=G′(1)E(X)=G'(1).
  • Var=G′′(1)+G′(1)−G′(1)2\text{Var}=G''(1)+G'(1)-G'(1)^2.
  • Independent sums multiply.
Key words
PGF
E(tX)E(t^X) for a discrete variable taking non-negative integer values.
Factorial moment
E(X(X−1))E(X(X-1)), from G′′(1)G''(1).
HistoryDe Moivre and Laplace. De Moivre used generating functions for dice problems in 1730; Laplace developed them into a general tool in his 1812 "Théorie analytique des probabilités".
Fun factTwo dice. The PGF of the total of two dice is (t+t2+⋯+t66)2\left(\frac{t+t^2+\dots+t^6}6\right)^2. Expanding gives the famous 1,2,3,4,5,6,5,4,3,2,11,2,3,4,5,6,5,4,3,2,1 pattern.
Section 3 of 14

Key formulae

DefinitionG(t)=E(tX)=∑xP(X=x)tx,G(1)=1G(t)=E(t^X)=\sum_xP(X=x)t^x,\quad G(1)=1
MomentsE(X)=G′(1),Var(X)=G′′(1)+G′(1)−[G′(1)]2E(X)=G'(1),\quad\text{Var}(X)=G''(1)+G'(1)-[G'(1)]^2
StandardB(n,p):(q+pt)n;Po(λ):eλ(t−1);Geo(p):pt1−qtB(n,p):(q+pt)^n;\quad\text{Po}(\lambda):e^{\lambda(t-1)};\quad\text{Geo}(p):\frac{pt}{1-qt}
TrickSums multiply. tX+Y=tXtYt^{X+Y}=t^Xt^Y, and independence lets the expectation split, so GX+Y=GXGYG_{X+Y}=G_XG_Y.
Section 4 of 14

Worked examples

Example 1 · Moments0 / 3 steps
Use G(t)=eλ(t−1)G(t)=e^{\lambda(t-1)} to find the Poisson mean and variance.
    Example 2 · Sum0 / 2 steps
    X∼Po(2)X\sim\text{Po}(2) and Y∼Po(3)Y\sim\text{Po}(3) independent. Find GX+YG_{X+Y}.
      Section 5 of 14

      Explore the graph

      PGF of Po(λ)
      Loading the interactive graph…
      Try this
      1. Every PGF passes through (1,1)(1,1).
      2. The tangent there has gradient G′(1)=E(X)=λG'(1)=E(X)=\lambda.

      Gradient at t=1t=1 is the mean.

      In the real worldBranching processes. PGFs predict whether a family name, virus or chain reaction dies out: the extinction probability solves G(s)=sG(s)=s.
      Section 6 of 14

      Common mistakes

      IncorrectVar=G′′(1)\text{Var}=G''(1)
      CorrectG′′(1)+G′(1)−[G′(1)]2G''(1)+G'(1)-[G'(1)]^2
      G′′(1)=E(X(X−1))G''(1)=E(X(X-1)).
      IncorrectAdding PGFs for sums
      CorrectMultiply
      tX+Y=tXtYt^{X+Y}=t^Xt^Y.
      IncorrectEvaluating at t=0t=0
      CorrectAt t=1t=1
      G(0)=P(X=0)G(0)=P(X=0); moments come from t=1t=1.

      Whole distribution

      One function. Differentiate at t = 1 for snacks.

      Section 7 of 14

      Exam tip

      Exam tipCheck G(1) = 1. If an unknown constant is in the PGF, G(1)=1G(1)=1 usually finds it.
      Section 8 of 14

      Guided practice

      SA-ST-PG-G012 marksL1 FoundationNon-calculatorDefinition
      G(t)=k(1+2t+t2)G(t)=k(1+2t+t^2) is a PGF. Find kk.

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      Salam Academy Original Practice · Proposed mark scheme
      Section 9 of 14

      Practice questions

      SA-ST-PG-0011 markL1 FoundationNon-calculatorDefinition
      For G(t)=14(1+2t+t2)G(t)=\frac14(1+2t+t^2), find P(X=1)P(X=1).

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      Salam Academy Original Practice · Proposed mark scheme
      SA-ST-PG-0022 marksL2 StandardNon-calculatorMoments
      For G(t)=14(1+2t+t2)G(t)=\frac14(1+2t+t^2), find E(X)E(X).

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      Salam Academy Original Practice · Proposed mark scheme
      SA-ST-PG-0033 marksL3 ExamNon-calculatorMoments
      For G(t)=14(1+2t+t2)G(t)=\frac14(1+2t+t^2), find Var(X)\text{Var}(X).

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      Salam Academy Original Practice · Proposed mark scheme

      More questions on every skill in this subtopic are in the question bank under Practise.

      Section 10 of 14

      Challenge

      SA-ST-PG-0042 marksL3 ExamNon-calculatorSums
      X∼Po(2)X\sim\text{Po}(2) and Y∼Po(3)Y\sim\text{Po}(3) are independent. GX+Y(t)=ek(t−1)G_{X+Y}(t)=e^{k(t-1)}. Find kk.

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      Salam Academy Original Practice · Proposed mark scheme
      Section 11 of 14

      Summary

      • G(t)=E(tX)G(t)=E(t^X).
      • E(X)=G′(1)E(X)=G'(1).
      • Var=G′′(1)+G′(1)−G′(1)2\text{Var}=G''(1)+G'(1)-G'(1)^2.
      • Independent sums: multiply.
      Section 12 of 14

      Quick quiz

      Q1 · SA-ST-PG-Q011 markL1 FoundationNon-calculatorSums
      For independent XX, YY, the PGF of X+YX+Y is
      Choose one option

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      Salam Academy Original Practice · Proposed mark scheme
      Section 13 of 14

      Same topic, different board

      Specification
      9231 AS & A Level Further Mathematics
      Where
      Paper 4 · Probability generating functions
      Expected
      Define G(t) = E(t^X); PGFs of standard distributions; mean and variance from derivatives; PGF of a sum of independent variables.
      Calculator
      Permitted on all papers; exact answers are expected where the question asks.
      Typical style
      Typical: given a PGF, find P(X = 2), E(X) and Var(X); then find the PGF of X + Y.
      Open in
      9231 · Paper 4
      Section 14 of 14

      Past paper questions

      No past paper questions are tagged to this subtopic yet. We add papers every week.

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      Salam Academy original content · Content status: reviewed and published. Syllabus mapping: Cambridge International 9231, Current syllabus (official specification).