Why is it called a 'generating' function?

It is called a probability 'generating' function because the function, when written as a power series in $t$, 'generates' the probabilities of the random variable as the coefficients of the terms $t^0, t^1, t^2, \dots$.

Can I use PGFs for continuous random variables?

No, PGFs are specifically for discrete random variables that take non-negative integer values. The equivalent tool for continuous random variables is the Moment Generating Function (MGF), which is outside the scope of the A-Level Further Mathematics (9231) syllabus.

What's the most common mistake when using the variance formula?

The formula is $Var(X) = G_X''(1) + G_X'(1) - [G_X'(1)]^2$. The most common mistake is forgetting the middle '$+ G_X'(1)$' term. A good way to remember the structure is to know that $G_X''(1) = E(X(X-1))$. So the formula is equivalent to $Var(X) = E(X(X-1)) + E(X) - [E(X)]^2$, which simplifies to the familiar $E(X^2) - [E(X)]^2$.

Where can I learn Probability generating functions for Cambridge 9231?

This free lesson on MarkScheme covers syllabus point 4.5 (Probability generating functions) for Cambridge A-Level Further Mathematics. It includes visual revision, key points, and links to real past-paper questions with mark-scheme marking at markscheme.app/courses/9231/4-5-probability-generating-functions.

Is Probability generating functions (4.5) on 9231 free to study?

Yes. The Further Mathematics 9231 course on MarkScheme is 100% free — including premium visual lessons, exam tips, and past-paper marking. No paywall for Probability generating functions.

How do I revise Probability generating functions for 9231 exams?

Start with the visual steps on this page, note the key exam points, then attempt a 9231 past-paper question on Probability generating functions using MarkScheme's /mark tool. Mark strictly against the official scheme and re-read any marks you lost.

Probability generating functions 9231 — Free…

Probability generating functions

Welcome to the world of Probability Generating Functions, or PGFs. While we have many tools to describe probability distributions, PGFs offer a uniquely elegant and powerful approach. They package an entire discrete probability distribution into a single function, allowing us to 'generate' key properties like probabilities, mean, and variance through calculus. Their real power shines when dealing with sums of independent random variables, turning complex convolution problems into simple multiplication.

BY THE END, YOU CAN…

1Understand the definition of a PGF and derive it for a given discrete distribution.
2Use a PGF to find probabilities, the mean (expected value), and the variance of a random variable.
3Recall and use the PGFs for standard distributions including Bernoulli, Binomial, Geometric and Poisson.
4Apply the property that the PGF of a sum of independent random variables is the product of their individual PGFs.

In simple terms

A friendly intro before the formal notes — no formulas yet.

The Probability Vending Machine

A Probability Generating Function (PGF) is a special function that encodes all the probabilities of a discrete random variable into a single polynomial. By manipulating this function, we can easily extract key information like probabilities, the mean, and the variance without needing the full probability distribution.

THINK OF IT LIKE…

Imagine a vending machine for statistics. Instead of snacks, it dispenses information about a random variable. The PGF is this machine. You don't press a button for a specific probability. Instead, you perform mathematical operations: looking at the coefficient of $t^0$ gives you $P(X=0)$ , differentiating and setting $t=1$ gives you the mean, and a different formula involving derivatives gives you the variance. It's a compact, powerful tool that holds all the key information in one place.

1
Define the PGF for a discrete random variable $X$ as $G_X(t) = E(t^X) = \sum P(X=x)t^x$ .
2
Extract probabilities by finding the coefficient of $t^r$ in the expansion of $G_X(t)$ , or by using the formula $P(X=r) = \frac{G_X^{(r)}(0)}{r!}$ .
3
Calculate the mean and variance using derivatives evaluated at $t=1$ : $E(X) = G_X'(1)$ and $Var(X) = G_X''(1) + G_X'(1) - [G_X'(1)]^2$ .
4
For a sum of independent variables, $Z = X+Y$ , find the PGF of the sum by multiplying the individual PGFs: $G_Z(t) = G_X(t)G_Y(t)$ .

Explore the concept

Use the live diagram and synced steps — play it or tap a step card to walk through.

Key formulas

Tap any symbol to reveal exactly what it means and its units.

$G_X(t) = E(t^X) = \sum_{x=0}^{\infty} P(X=x)t^x = P(X=0) + P(X=1)t + P(X=2)t^2 + \dots$

Tap a symbol — great for exam definitions

$E(X) = G_X'(1)$

Tap a symbol — great for exam definitions

$Var(X) = G_X''(1) + G_X'(1) - [G_X'(1)]^2$

Tap a symbol — great for exam definitions

$G_Z(t) = G_X(t) \times G_Y(t)$

Tap a symbol — great for exam definitions

Full topic notes

Formal explanation with the rigour you need for the exam.

Defining the Probability Generating Function

For a discrete random variable $X$ that takes non-negative integer values $0, 1, 2, \dots$ , its Probability Generating Function, denoted $G_X(t)$ , is defined as a power series where the coefficients are the probabilities. The variable $t$ is a 'dummy' variable whose exponent 'carries' the value of the random variable.

$G_X(t) = E(t^X) = \sum_{x=0}^{\infty} P(X=x)t^x = P(X=0) + P(X=1)t + P(X=2)t^2 + \dots$

The PGF is defined for discrete random variables taking values in ${0, 1, 2, \dots}$ .
The coefficient of $t^r$ in the expansion of $G_X(t)$ is precisely $P(X=r)$ .
Substituting $t=1$ into the PGF gives $G_X(1) = \sum P(X=x) = 1$ , which is a useful check.

Finding Probabilities, Mean and Variance from a PGF

The PGF is not just a fancy way to write down probabilities; its true utility comes from the properties we can extract using calculus. By differentiating the PGF and evaluating at specific values of $t$ , we can find probabilities, the mean, and the variance.

E(X) = G_X'(1)

Var(X) = G_X''(1) + G_X'(1) - [G_X'(1)]^2

Probabilities: The coefficient of $t^r$ is $P(X=r)$ . Alternatively, $P(X=r) = \frac{G_X^{(r)}(0)}{r!}$ , where $G_X^{(r)}(0)$ is the $r$ -th derivative of $G_X(t)$ evaluated at $t=0$ . For $r=0, 1, 2$ , this gives $P(X=0)=G_X(0)$ , $P(X=1)=G_X'(0)$ , and $P(X=2)=G_X''(0)/2$ .
Mean (Expected Value): The mean is found by differentiating once and evaluating at $t=1$ .
Variance: The variance requires the first and second derivatives, both evaluated at $t=1$ .

EXAM TIP

Remember the key difference: derivatives at $t=0$ give you probabilities, while derivatives at $t=1$ give you moments (mean and variance). A very common mistake is to mix these up. Always double-check which value of $t$ you should be substituting.

Sums of Independent Random Variables

One of the most significant advantages of PGFs is in dealing with the sum of independent random variables. If we want to find the distribution of $Z = X+Y$ , where $X$ and $Y$ are independent, finding the PGF of $Z$ is remarkably simple.

If $X$ and $Y$ are independent discrete random variables, and $Z = X+Y$ , then the PGF of $Z$ is the product of the PGFs of $X$ and $Y$ : $G_Z(t) = G_X(t) \times G_Y(t)$

This property is extremely useful. By finding the PGF of the sum, we can often identify the resulting distribution if it matches a standard PGF. This avoids the more complex process of convolution.

Worked examples

See the formulas applied — reveal one step at a time, like the exam.

EXAMPLE 1

A discrete random variable $X$ has the probability distribution given by $P(X=1)=0.2$ , $P(X=2)=0.5$ , and $P(X=3)=0.3$ . Find the probability generating function of $X$ .

1
By definition, $G_X(t) = \sum P(X=x)t^x$ . We sum over the possible values of $X$ : 1, 2, and 3. $G_X(t) = P(X=1)t^1 + P(X=2)t^2 + P(X=3)t^3$ Substituting the given probabilities: $G_X(t) = 0.2t + 0.5t^2 + 0.3t^3$ This is the PGF for the random variable $X$ . As a check, note that $G_X(1) = 0.2(1) + 0.5(1)^2 + 0.3(1)^3 = 0.2+0.5+0.3 = 1$ .

✓ FULLY WORKED

EXAMPLE 2

The number of defects in a length of cloth is a random variable $X$ with PGF given by $G_X(t) = (0.8 + 0.2t)^5$ . Find: (i) The probability of there being exactly 2 defects. (ii) The mean number of defects. (iii) The variance of the number of defects.

1
(i) To find $P(X=2)$ , we need the coefficient of $t^2$ in the expansion of $G_X(t) = (0.8 + 0.2t)^5$ . This is a binomial expansion. The term in $t^2$ is given by $\binom{5}{2}(0.8)^{5-2}(0.2t)^2$ . Coefficient = $\binom{5}{2}(0.8)^3(0.2)^2 = 10 \times 0.512 \times 0.04 = 0.2048$ . So, $P(X=2) = 0.2048$ .

EXAMPLE 3

Let $X \sim Po(2)$ and $Y \sim Po(3)$ be two independent random variables. Let $Z = X+Y$ . Use probability generating functions to find the distribution of $Z$ .

1
First, we recall the PGF for a Poisson distribution. If $W \sim Po(\lambda)$ , its PGF is $G_W(t) = e^{\lambda(t-1)}$ .

How it all connects

The big idea sits in the middle — tap a linked idea to explore the link.

MAIN IDEAProbability generating functions

Tap a linked idea to see how it connects back to the main topic — that connection is what examiners reward.

Glossary

Try to recall each definition before you reveal it.

Quick check

Answer in your head first — then tap to check. No pressure.

Revision flashcards

Flip the card. Test yourself before the exam.

QUESTION · 1 / 12

What is the definition of the Probability Generating Function (PGF) for a discrete random variable $X$ ?

TAP OR SPACE TO FLIP

ANSWER

$G_X(t) = E(t^X) = \sum_{x=0}^{\infty} P(X=x)t^x$ . It's a power series in a dummy variable $t$ , where the coefficient of $t^x$ is the probability $P(X=x)$ .

TAP OR SPACE TO FLIP BACK

1 / 12 · ← → space

Key takeaways

Review these before you close the topic — retrieval beats re-reading.

✓
The PGF is defined for discrete random variables taking values in ${0, 1, 2, \dots}$ .
✓
The coefficient of $t^r$ in the expansion of $G_X(t)$ is precisely $P(X=r)$ .
✓
Substituting $t=1$ into the PGF gives $G_X(1) = \sum P(X=x) = 1$ , which is a useful check.

Practice — then mark it

The whole point: a real Cambridge question, marked mark-by-mark.

Real past paper[0]

Practice Questions

Do it on paper → mark itMARKED MARK-BY-MARK · B1 / M1 / A1 · OFFICIAL SCHEME

Extra simulations & links

PhET, GeoGebra and other curated tools — open in a new tab.

Cambridge International A Level Further Mathematics 9231 Syllabus↗

Frequently asked

✓

Checkpoint

One marked question is worth ten re-reads — close the loop before you move on.

Checkpoint

Reading it isn’t knowing it — prove it.

Before you move on: do Practice Questions on paper, snap a photo, and get examiner-style feedback on exactly where you win and lose marks.

Attempt & get marked →Takes about a minute · you’ll land back on this lesson

← PREVIOUS4.4 Non-parametric tests

Probability generating functions

BY THE END, YOU CAN…

1Understand the definition of a PGF and derive it for a given discrete distribution.
2Use a PGF to find probabilities, the mean (expected value), and the variance of a random variable.
3Recall and use the PGFs for standard distributions including Bernoulli, Binomial, Geometric and Poisson.
4Apply the property that the PGF of a sum of independent random variables is the product of their individual PGFs.

In simple terms

A friendly intro before the formal notes — no formulas yet.

The Probability Vending Machine

THINK OF IT LIKE…

1
Define the PGF for a discrete random variable $X$ as $G_X(t) = E(t^X) = \sum P(X=x)t^x$ .
2
Extract probabilities by finding the coefficient of $t^r$ in the expansion of $G_X(t)$ , or by using the formula $P(X=r) = \frac{G_X^{(r)}(0)}{r!}$ .
3
Calculate the mean and variance using derivatives evaluated at $t=1$ : $E(X) = G_X'(1)$ and $Var(X) = G_X''(1) + G_X'(1) - [G_X'(1)]^2$ .
4
For a sum of independent variables, $Z = X+Y$ , find the PGF of the sum by multiplying the individual PGFs: $G_Z(t) = G_X(t)G_Y(t)$ .

Explore the concept

Use the live diagram and synced steps — play it or tap a step card to walk through.

Key formulas

Tap any symbol to reveal exactly what it means and its units.

$G_X(t) = E(t^X) = \sum_{x=0}^{\infty} P(X=x)t^x = P(X=0) + P(X=1)t + P(X=2)t^2 + \dots$

Tap a symbol — great for exam definitions

$E(X) = G_X'(1)$

Tap a symbol — great for exam definitions

$Var(X) = G_X''(1) + G_X'(1) - [G_X'(1)]^2$

Tap a symbol — great for exam definitions

$G_Z(t) = G_X(t) \times G_Y(t)$

Tap a symbol — great for exam definitions

Full topic notes

Formal explanation with the rigour you need for the exam.

Defining the Probability Generating Function

$G_X(t) = E(t^X) = \sum_{x=0}^{\infty} P(X=x)t^x = P(X=0) + P(X=1)t + P(X=2)t^2 + \dots$

The PGF is defined for discrete random variables taking values in ${0, 1, 2, \dots}$ .
The coefficient of $t^r$ in the expansion of $G_X(t)$ is precisely $P(X=r)$ .
Substituting $t=1$ into the PGF gives $G_X(1) = \sum P(X=x) = 1$ , which is a useful check.

Finding Probabilities, Mean and Variance from a PGF

E(X) = G_X'(1)

Var(X) = G_X''(1) + G_X'(1) - [G_X'(1)]^2

Probabilities: The coefficient of $t^r$ is $P(X=r)$ . Alternatively, $P(X=r) = \frac{G_X^{(r)}(0)}{r!}$ , where $G_X^{(r)}(0)$ is the $r$ -th derivative of $G_X(t)$ evaluated at $t=0$ . For $r=0, 1, 2$ , this gives $P(X=0)=G_X(0)$ , $P(X=1)=G_X'(0)$ , and $P(X=2)=G_X''(0)/2$ .
Mean (Expected Value): The mean is found by differentiating once and evaluating at $t=1$ .
Variance: The variance requires the first and second derivatives, both evaluated at $t=1$ .

EXAM TIP

Sums of Independent Random Variables

If $X$ and $Y$ are independent discrete random variables, and $Z = X+Y$ , then the PGF of $Z$ is the product of the PGFs of $X$ and $Y$ : $G_Z(t) = G_X(t) \times G_Y(t)$

This property is extremely useful. By finding the PGF of the sum, we can often identify the resulting distribution if it matches a standard PGF. This avoids the more complex process of convolution.

Worked examples

See the formulas applied — reveal one step at a time, like the exam.

EXAMPLE 1

A discrete random variable $X$ has the probability distribution given by $P(X=1)=0.2$ , $P(X=2)=0.5$ , and $P(X=3)=0.3$ . Find the probability generating function of $X$ .

1
By definition, $G_X(t) = \sum P(X=x)t^x$ . We sum over the possible values of $X$ : 1, 2, and 3. $G_X(t) = P(X=1)t^1 + P(X=2)t^2 + P(X=3)t^3$ Substituting the given probabilities: $G_X(t) = 0.2t + 0.5t^2 + 0.3t^3$ This is the PGF for the random variable $X$ . As a check, note that $G_X(1) = 0.2(1) + 0.5(1)^2 + 0.3(1)^3 = 0.2+0.5+0.3 = 1$ .

✓ FULLY WORKED

EXAMPLE 2

1
(i) To find $P(X=2)$ , we need the coefficient of $t^2$ in the expansion of $G_X(t) = (0.8 + 0.2t)^5$ . This is a binomial expansion. The term in $t^2$ is given by $\binom{5}{2}(0.8)^{5-2}(0.2t)^2$ . Coefficient = $\binom{5}{2}(0.8)^3(0.2)^2 = 10 \times 0.512 \times 0.04 = 0.2048$ . So, $P(X=2) = 0.2048$ .

EXAMPLE 3

Let $X \sim Po(2)$ and $Y \sim Po(3)$ be two independent random variables. Let $Z = X+Y$ . Use probability generating functions to find the distribution of $Z$ .

1
First, we recall the PGF for a Poisson distribution. If $W \sim Po(\lambda)$ , its PGF is $G_W(t) = e^{\lambda(t-1)}$ .

How it all connects

The big idea sits in the middle — tap a linked idea to explore the link.

MAIN IDEAProbability generating functions

Tap a linked idea to see how it connects back to the main topic — that connection is what examiners reward.

Glossary

Try to recall each definition before you reveal it.

Quick check

Answer in your head first — then tap to check. No pressure.

Revision flashcards

Flip the card. Test yourself before the exam.

QUESTION · 1 / 12

What is the definition of the Probability Generating Function (PGF) for a discrete random variable $X$ ?

TAP OR SPACE TO FLIP

ANSWER

$G_X(t) = E(t^X) = \sum_{x=0}^{\infty} P(X=x)t^x$ . It's a power series in a dummy variable $t$ , where the coefficient of $t^x$ is the probability $P(X=x)$ .

TAP OR SPACE TO FLIP BACK

1 / 12 · ← → space

Key takeaways

Review these before you close the topic — retrieval beats re-reading.

✓
The PGF is defined for discrete random variables taking values in ${0, 1, 2, \dots}$ .
✓
The coefficient of $t^r$ in the expansion of $G_X(t)$ is precisely $P(X=r)$ .
✓
Substituting $t=1$ into the PGF gives $G_X(1) = \sum P(X=x) = 1$ , which is a useful check.

Practice — then mark it

The whole point: a real Cambridge question, marked mark-by-mark.

Real past paper[0]

Practice Questions

Do it on paper → mark itMARKED MARK-BY-MARK · B1 / M1 / A1 · OFFICIAL SCHEME

Extra simulations & links

PhET, GeoGebra and other curated tools — open in a new tab.

Cambridge International A Level Further Mathematics 9231 Syllabus↗

Frequently asked

✓

Checkpoint

One marked question is worth ten re-reads — close the loop before you move on.

Checkpoint

Reading it isn’t knowing it — prove it.

Before you move on: do Practice Questions on paper, snap a photo, and get examiner-style feedback on exactly where you win and lose marks.

Attempt & get marked →Takes about a minute · you’ll land back on this lesson

← PREVIOUS4.4 Non-parametric tests

Revise Probability generating functions for Cambridge 9231 (4.5)

In simple terms

Explore the concept

Key formulas

Full topic notes

Defining the Probability Generating Function

Finding Probabilities, Mean and Variance from a PGF

Sums of Independent Random Variables

Worked examples

How it all connects

Glossary

Quick check

Revision flashcards

Key takeaways

Practice — then mark it

Practice Questions

Extra simulations & links

Frequently asked

Checkpoint

Reading it isn’t knowing it — prove it.

In simple terms

Explore the concept

Key formulas

Full topic notes

Defining the Probability Generating Function

Finding Probabilities, Mean and Variance from a PGF

Sums of Independent Random Variables

Worked examples

How it all connects

Glossary

Quick check

Revision flashcards

Key takeaways

Practice — then mark it

Practice Questions

Extra simulations & links

Frequently asked

Checkpoint

Reading it isn’t knowing it — prove it.

Discuss Probability generating functions