PGFs of Sums & Transformations | Edexcel A Level Further Maths: Further Statistics 1 Revision Notes 2017

PGFs of X+Y

If $Z = X + Y$ then $G_{Z} (t) = G_{X} (t) \times G_{Y} (t)$
- Multiply the PGFs together
This only works if $X$ and $Y$ are independent distributions

Let $X$ be a discrete random variable with PGF $G_{X} (t)$
If $Z = X_{1} + X_{2} + . . . + X_{n}$ where each $X_{i}$ is its own independent $X$ distribution
- Then $G_{Z} (t) = G_{X_{1}} (t) G_{X_{2}} (t) . . . G_{X_{n}} (t) = {[G_{X} (t)]}^{n}$
For example
- $X$ could be the winnings on a game played once
- $X_{1}, X_{2} . . ., X_{5}$ represent five different winnings when played five times
  - assuming each game is independent of the last
- The winnings after five games has the PGF ${[G_{X} (t)]}^{5}$

If $X ~ Po (λ)$ and $Y ~ Po (μ)$ are independent then
- $G_{X} (t) = e^{λ (t - 1)}$ and $G_{Y} (t) = e^{μ (t - 1)}$
So $Z = X + Y$ has the PGF $G_{Z} (t) = G_{X} (t) G_{Y} (t) = e^{λ (t - 1)} e^{μ (t - 1)}$
- Add the powers in the exponentials
- Factorise out $(t - 1)$
- So $G_{Z} (t) = e^{(λ + μ) (t - 1)}$
This is exactly the PGF for the distribution of $Po (λ + μ)$
- So the sum of two independent Poisson distributions is a Poisson distribution

A fair three-sided spinner, labelled 2, 4 and 8, is spun.
A biased coin, labelled 4 and 6, is flipped, where there is a 25% chance of landing on a 4.

Use probability generating functions to find the probability that the sum of the scores is 8.

pgfs-of-sums-1 pgfs-of-sums-2

Let $Y = a X + b$ be a linear transformation of $X$
Then $G_{Y} (t) = t^{b} G_{X} (t^{a})$
- Replace $t$ with $t^{a}$
- Multiply the PGF by $t^{b}$
For example
- Take $G_{X} (t) = {(0.9 + 0.1 t)}^{10}$
- Let $Y = 8 X + 3$
- So $G_{Y} (t) = t^{3} {(0.9 + 0.1 t^{8})}^{10}$

The formula comes from
- $G_{Y} (t) = E (t^{Y}) = E (t^{a X + b}) = E (t^{a X} t^{b}) = t^{b} E ({(t^{a})}^{X}) = t^{b} G_{X} (t^{a})$

Beware, you need to learn $G_{Y} (t) = t^{b} G_{X} (t^{a})$ as it is not given in the Formulae Booklet

The discrete random variable $X$ has a probability generating function given by

$G_{X} (t) = \frac{t}{5 - 4 t}$

Find and simplify the probability generating function of $Y$ , where $Y = 2 (X + 5)$ .

pgfs-of-ax-and-b