Probability generating functions
Encode a discrete distribution in a generating function, recover moments and combine independent random variables.
Exam scope and assessment notes
Mapped to Cambridge International AS & A Level Further Mathematics 9231. These original explanations and self-checks are independent study support, not Cambridge questions, marks or grades. The syllabus assumes prior knowledge of the relevant 9709 Mathematics content; check each component's prerequisites with your teacher. Full-paper assessment and human subject-specialist review remain outstanding.
Before you begin
- Cambridge expects learners to know the whole Cambridge International A Level Mathematics (9709) content. Paper 1 further pure mathematics draws on 9709 Papers 1 and 3. Confirm your exam year and component route with your school.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What makes a Further Mathematics solution convincing?
What you will learn
- Form and use a probability generating function
- Find mean and variance from derivatives at 1
- Multiply generating functions for independent sums
A PGF stores the probability sequence as coefficients
For a non-negative integer-valued X, G_X(t)=E(tˣ)=Σpₓtˣ. The coefficient of tˣ is P(X=x), and G_X(1)=1.
The mean is G′(1). The second factorial moment is G″(1)=E[X(X−1)], so Var(X)=G″(1)+G′(1)−(G′(1))².
Independent sums multiply PGFs
If X and Y are independent, G_{X+Y}(t)=G_X(t)G_Y(t). This turns a convolution of probability tables into multiplication of polynomials or series.
After multiplying, read the coefficient for the requested probability. Independence is essential; without it, the product rule need not hold.
Pause and explain
Which habit best checks your work on probability generating functions?
Worked example
Find the mean and variance of a Bernoulli variable X with P(X=1)=p and P(X=0)=1−p from its PGF.
Show the worked solution
- G(t)=1−p+pt.
- G′(1)=p, so E(X)=p; G″(1)=0.
- Var(X)=0+p−p²=p(1−p).
Answer Mean p; variance p(1−p).
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Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
A random variable has PGF G(t)=0.4+0.6t. Find its mean and variance.
Give me a hint
Differentiate once and twice, then use the variance identity.
Compare my reasoning
- G′(t)=0.6, so G′(1)=0.6.
- G″(t)=0, so G″(1)=0.
- Var(X)=0+0.6−0.6².
Mean 0.6; variance 0.24.
Look for these in your work
- First derivative evaluated
- Second derivative evaluated
- Variance identity applied
Independent Bernoulli variables X and Y have success probabilities 0.2 and 0.3. Find P(X+Y=1) using their PGFs.
Give me a hint
Multiply (0.8+0.2t)(0.7+0.3t) and read the coefficient of t.
Compare my reasoning
- The t coefficient is 0.8(0.3)+0.2(0.7).
- These are the disjoint cases (X=0,Y=1) and (X=1,Y=0).
- Add the two probabilities.
P(X+Y=1)=0.24+0.14=0.38.
Look for these in your work
- Independent PGFs multiplied
- Coefficient identified
- Two disjoint cases included
Explain why G_X(1)=1 for a valid PGF and what a coefficient of t³ represents.
Give me a hint
Evaluate the defining sum at t=1 and compare it with total probability.
Compare my reasoning
- G_X(1)=Σpₓ·1ˣ=Σpₓ.
- The probabilities of all possible values sum to 1.
- The coefficient of t³ is P(X=3).
G_X(1)=1; the coefficient of t³ is P(X=3).
Look for these in your work
- Definition substituted
- Total probability used
- Coefficient interpreted
Common mistakes
- Reading a probability from the function value instead of a coefficient
- Using Var(X)=G″(1) without converting the factorial moment
- Multiplying PGFs for dependent variables
What can you explain now?
What is E(X) in terms of a PGF G_X?
Compare with the explanation
G′_X(1)
Differentiate the power series and evaluate at 1.
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Make it stick
How do PGF derivatives encode mean and variance, and when can PGFs be multiplied?
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