pgfs-of-standard-distributions

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

Further Maths: Further Statistics 1

pgfs-of-standard-distributions

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?

Examiner Tips and Tricks

Worked Example

Responses