PGFs of Standard Distributions (Edexcel A Level Further Maths): Revision Note

Exam code: 9FM0

Author

Mark Curtis

Last updated

19 June 2025

PGFs of standard distributions

What are the PGFs of standard distributions?

You are given the following PGFs in the Formulae Booklet:

Distribution of $X$	$P (X = x)$	P.G.F.
Binomial $B (n, p)$	$(\begin{matrix} n \\ x \end{matrix}) p^{x} {(1 - p)}^{n - x}$	${(1 - p + p t)}^{n}$
Poisson $Po (λ)$	$e^{- λ} \frac{λ^{x}}{x!}$	$e^{λ (t - 1)}$
Geometric $Geo (p)$ on 1, 2, ...	$p {(1 - p)}^{x - 1}$	$\frac{p t}{1 - (1 - p) t}$
Negative binomial on $r$ , $r + 1$ , ...	$(\begin{matrix} x - 1 \\ r - 1 \end{matrix}) p^{r} {(1 - p)}^{x - r}$	${(\frac{p t}{1 - (1 - p) t})}^{r}$

How do I find the PGF of a binomial distribution?

If $X ~ B (n, p)$ then $G_{X} (t) = {(q + p t)}^{n}$ where $q = 1 - p$
- You can use this to to prove, by differentiation, that
  - $E (X) = n p$
  - $Var (X) = n p q$
To derive the PGF, create a probability distribution table
- Using the binomial probability function $(\begin{matrix} n \\ x \end{matrix}) p^{x} q^{n - x}$
- From $x = 0$ to $x = n$
- Showing all powers clearly

$x$	0	1	2	...	$n$
$P (X = x)$	$(\begin{matrix} n \\ 0 \end{matrix}) p^{0} q^{n}$	$(\begin{matrix} n \\ 1 \end{matrix}) p^{1} q^{n - 1}$	$(\begin{matrix} n \\ 2 \end{matrix}) p^{2} q^{n - 2}$	...	$(\begin{matrix} n \\ n \end{matrix}) p^{n} q^{0}$

Write down the PGF as a polynomial
- $G_{X} (t) = (\begin{matrix} n \\ 0 \end{matrix}) p^{0} q^{n} t^{0} + (\begin{matrix} n \\ 1 \end{matrix}) p^{1} q^{n - 1} t^{1} + (\begin{matrix} n \\ 2 \end{matrix}) p^{2} q^{n - 2} t^{2} + . . . + (\begin{matrix} n \\ n \end{matrix}) p^{n} q^{0} t^{n}$
Group the $p$ and $t$ together in brackets
- $G_{X} (t) = (\begin{matrix} n \\ 0 \end{matrix}) {(p t)}^{0} q^{n} + (\begin{matrix} n \\ 1 \end{matrix}) {(p t)}^{1} q^{n - 1} + (\begin{matrix} n \\ 2 \end{matrix}) {(p t)}^{2} q^{n - 2} + . . . + (\begin{matrix} n \\ n \end{matrix}) {(p t)}^{n} q^{0}$
Spot that this is the binomial expansion of ${(q + p t)}^{n}$
- Since ${(a + b)}^{n} = (\begin{matrix} n \\ 0 \end{matrix}) a^{0} b^{n} + (\begin{matrix} n \\ 1 \end{matrix}) a^{1} b^{n - 1} + (\begin{matrix} n \\ 2 \end{matrix}) a^{2} b^{n - 2} + . . . + (\begin{matrix} n \\ n \end{matrix}) a^{n} b^{0}$
- So $G_{X} (t) = {(q + p t)}^{n}$
The proof can also be done in summation (sigma) notation
- $G_{X} (t) = \sum_{x = 0}^{n} t^{x} P (X = x) = \sum_{x = 0}^{n} t^{x} (\begin{matrix} n \\ x \end{matrix}) p^{x} q^{n - x} = \sum_{x = 0}^{n} (\begin{matrix} n \\ x \end{matrix}) {(p t)}^{x} q^{n - x} = {(q + p t)}^{n}$

How do I find the PGF of a Poisson distribution?

If $X ~ Po (λ)$ then $G_{X} (t) = e^{λ (t - 1)}$
- You can use this to to prove, by differentiation, that
  - $E (X) = λ$
  - $Var (X) = λ$
To derive the PGF, create a probability distribution table
- Using the Poisson probability function $e^{- λ} \frac{λ^{x}}{x!}$
- From $x = 0$ x to infinity
- Showing all powers clearly
$x$
0
1
2
3
...
$P (X = x)$
$e^{- λ} \frac{λ^{0}}{0!}$
$e^{- λ} \frac{λ^{1}}{1!}$
$e^{- λ} \frac{λ^{2}}{2!}$
$e^{- λ} \frac{λ^{3}}{3!}$
...
Write down the PGF as a polynomial
- $G_{X} (t) = e^{- λ} \frac{λ^{0}}{0!} t^{0} + e^{- λ} \frac{λ^{1}}{1!} t^{1} + e^{- λ} \frac{λ^{2}}{2!} t^{2} + e^{- λ} \frac{λ^{3}}{3!} t^{3} + . . .$
Group the $λ$ and $t$ together in brackets
- $G_{X} (t) = e^{- λ} \frac{{(λ t)}^{0}}{0!} + e^{- λ} \frac{{(λ t)}^{1}}{1!} + e^{- λ} \frac{{(λ t)}^{2}}{2!} + e^{- λ} \frac{{(λ t)}^{3}}{3!} + . . .$
Factorise out $e^{- λ}$
- $G_{X} (t) = e^{- λ} [\frac{{(λ t)}^{0}}{0!} + \frac{{(λ t)}^{1}}{1!} + \frac{{(λ t)}^{2}}{2!} + \frac{{(λ t)}^{3}}{3!} + . . .]$
Spot that the Maclaurin series of $e^{λ t}$ is inside the brackets
- Since $e^{x} = 1 + x + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + . . .$
- So $G_{X} (t) = e^{- λ} e^{λ t}$
Add the powers then factorise out $λ$
- $G_{X} (t) = {e^{- λ}}^{+ λ t} = e^{λ (t - 1)}$
The proof can also be done in summation (sigma) notation
- $G_{X} (t) = \sum_{x = 0}^{\infty} t^{x} P (X = x) = \sum_{x = 0}^{\infty} t^{x} e^{- λ} \frac{λ^{x}}{x!} = e^{- λ} \sum_{x = 0}^{\infty} \frac{{(λ t)}^{x}}{x!} = {e^{- λ}}^{+ λ t} = e^{λ (t - 1)}$

How do I find the PGF of a geometric distribution?

If $X ~ Geo (p)$ then $G_{X} (t) = \frac{p t}{1 - q t}$ where $q = 1 - p$
- You can use this to to prove, by differentiation, that
  - $E (X) = \frac{1}{p}$
  - $Var (X) = \frac{q}{p^{2}}$
To derive the PGF, create a probability distribution table
- Using the Geometric probability function $p q^{x - 1}$
- From $x = 1$ to infinity ( $x \neq 0$ )
- Simplify the powers
- $x$
  1
  2
  3
  4
  ...
  $P (X = x)$
  $p$
  $p q$
  $p q^{2}$
  $p q^{3}$
  ...
Write down the PGF as a polynomial
- $G_{X} (t) = p t^{1} + p q t^{2} + p q^{2} t^{3} + p q^{3} t^{4} + . . .$
Spot that this is an infinite geometric series with first term $p t$ and common ratio $q t$
- Use $S_{\infty} = \frac{a}{1 - r}$
- So $G_{X} (t) = \frac{p t}{1 - q t}$

How do I find the PGF of a negative binomial distribution?

If $X ~ Negative B (r, p)$ then $G_{X} (t) = {(\frac{p t}{1 - q t})}^{r}$ where $q = 1 - p$
- You can use this to to prove, by differentiation, that
  - $E (X) = \frac{r}{p}$
  - $Var (X) = \frac{r q}{p^{2}}$
To derive the PGF, the proof is best done in summation (sigma) notation
- Use the negative binomial probability function $(\begin{matrix} x - 1 \\ r - 1 \end{matrix}) p^{r} q^{x - r}$
  - From $x = r$ to infinity
- $G_{X} (t) = \sum_{x = r}^{\infty} t^{x} P (X = x) = \sum_{x = r}^{\infty} t^{x} (\begin{matrix} x - 1 \\ r - 1 \end{matrix}) p^{r} q^{x - r} = \sum_{x = r}^{\infty} (\begin{matrix} x - 1 \\ r - 1 \end{matrix}) p^{r} q^{x - r} t^{x}$
To proceed, you will be given in the question the result that $\sum_{x = r}^{\infty} (\begin{matrix} x - 1 \\ r - 1 \end{matrix}) a^{x - r} = {(1 - a)}^{- r}$
- Write $t^{x}$ as $t^{x - r} t^{r}$ to group $q$ and $t$ together in brackets
  - $G_{X} (t) = \sum_{x = r}^{\infty} (\begin{matrix} x - 1 \\ r - 1 \end{matrix}) p^{r} q^{x - r} t^{x - r} t^{r} = \sum_{x = r}^{\infty} (\begin{matrix} x - 1 \\ r - 1 \end{matrix}) {(q t)}^{x - r} p^{r} t^{r}$
- Factorise $p^{r} t^{r}$ out (as it does not depend on $x$ )
  - $G_{X} (t) = p^{r} t^{r} \sum_{x = r}^{\infty} (\begin{matrix} x - 1 \\ r - 1 \end{matrix}) {(q t)}^{x - r}$
- Use the result given, where $a = q t$
  - $G_{X} (t) = p^{r} t^{r} {(1 - q t)}^{- r}$
- Write as one bracket to the power $r$
  - $G_{X} (t) = {(\frac{p t}{1 - q t})}^{r}$

Examiner Tips and Tricks

The words derive or from first principles mean you have to prove the PGF (not quote it)

Worked Example

Write down the probability generating function for the distribution $X ~ Geo (p)$ and use it to prove that $Var (X) = \frac{1 - p}{p^{2}}$ .

You've read 0 of your 5 free revision notes this week

Unlock more, it's free!

Join the 100,000+ Students that ❤️ Save My Exams

the (exam) results speak for themselves:

Did this page help you?

Previous:E(X) & Var(X) of PGFsNext:PGFs of Sums & Transformations

$x$	0	1	2	3	...
$P (X = x)$	$e^{- λ} \frac{λ^{0}}{0!}$	$e^{- λ} \frac{λ^{1}}{1!}$	$e^{- λ} \frac{λ^{2}}{2!}$	$e^{- λ} \frac{λ^{3}}{3!}$	...

PGFs of Standard Distributions (Edexcel A Level Further Maths): Revision Note

PGFs of standard distributions

What are the PGFs of standard distributions?

How do I find the PGF of a binomial distribution?

How do I find the PGF of a Poisson distribution?

How do I find the PGF of a geometric distribution?

How do I find the PGF of a negative binomial distribution?

You've read 0 of your 5 free revision notes this week

Unlock more, it's free!

Join the 100,000+ Students that ❤️ Save My Exams

Discrete Probability Distributions

Discrete Probability Distributions

E(X) & Var(X) of Discrete Random Variables

Linear Transformations of DRVs

Problem Solving with DRVs

Poisson & Binomial Distributions

The Poisson Distribution

Poisson Approximations of Binomials

Geometric & Negative Binomial Distributions

The Geometric Distribution

The Negative Binomial Distribution

Probability Generating Functions

Probability Generating Functions

Probability Generating Functions (PGFs)

E(X) & Var(X) of PGFs

PGFs of Standard Distributions

PGFs of Sums & Transformations

Central Limit Theorem

Central Limit Theorem

Central Limit Theorem

Hypothesis Testing

Poisson & Geometric Hypothesis Testing

Poisson Hypothesis Testing

Geometric Hypothesis Testing

Chi Squared Tests

Goodness of Fit

Chi Squared Tests for Standard Distributions

Chi Squared Tests for Contingency Tables

Quality of Tests

Type I & Type II Errors

Size & Power of Test

Author: Mark Curtis