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 $r$
  - $G_{X} (t) = {(\frac{p t}{1 - q t})}^{r}$

Exam Tip

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 $Var (X) = \frac{1 - p}{p^{2}}$ .

pgfs-of-standard-distributions-1

PGFs of Standard Distributions (Edexcel A Level Further Maths: Further Statistics 1)

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?

Exam Tip

Worked example

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Author: Mark

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 $r$ , $r + 1$ , ...	$(\begin{matrix} x - 1 \\ r - 1 \end{matrix}) p^{r} {(1 - p)}^{x - r}$	${(\frac{p t}{1 - (1 - p) t})}^{r}$

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

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