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E(X) of PGFs

How do I find E(X) of a PGF?

$E (X) = G_{X}' (1)$
- Differentiate the PGF and substitute in $t = 1$
This is because
- $G_{X} (t) = E (t^{X}) = \sum_{}^{} t^{x} P (X = x)$
- So $G_{X}' (t) = {\frac{d}{d t} [\sum_{}^{} t^{x} P (X = x)] = \sum}_{}^{} x t^{x - 1} P (X = x)$
- Substituting $t = 1$ gives $G_{X}' (1) = \sum_{}^{} x P (X = x) = E (X)$
You may need the chain, product or quotient rule.

Examiner Tips and Tricks

$E (X) = G'_{X} (1)$ is given in the Formulae Booklet

Worked Example

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

$G_{X} (t) = \frac{1}{81} t^{3} {(1 + 2 t)}^{4}$

Find $E (X)$ .

ex-of-pgfs

Var(X) of PGFs

How do I find Var(X) of a PGF?

The formula is $Var (X) = G_{X}'' (1) + G_{X}' (1) - {[G_{X}' (1)]}^{2}$
You may need the chain, product or quotient rule.
The first two terms in the formula are $E (X^{2})$
- $E (X^{2}) = G_{X}'' (1) + G_{X}' (1)$
The formula comes from
- $G_{X} (t) = E (t^{X}) = \sum_{}^{} t^{x} P (X = x)$
- So $G_{X}' (t) = {\frac{d}{d t} [\sum_{}^{} t^{x} P (X = x)] = \sum}_{}^{} x t^{x - 1} P (X = x)$
  - Recall $G_{X}' (1) = \sum x P (X = x) = E (X)$
- And $G_{X}'' (t) = {\frac{d}{d t} [\sum_{}^{} x t^{x - 1} P (X = x)] = \sum}_{}^{} x (x - 1) t^{x - 2} P (X = x)$
- Substituting $t = 1$ gives $G_{X}'' (1) = \sum_{}^{} x (x - 1) P (X = x) = E (X (X - 1)) = E (X^{2}) - E (X)$
- Make $E (X^{2})$ the subject
  - $E (X^{2}) = G_{X}'' (1) + E (X)$
- Then substitute it into $Var (X) = E (X^{2}) - {(E (X))}^{2}$
- And replace $E (X)$ with $G_{X}' (1)$

Examiner Tips and Tricks

$Var (X) = G_{X}'' (1) + G_{X}' (1) - {[G_{X}' (1)]}^{2}$ is given in the Formulae Booklet

Worked Example

The probability generating function for a discrete random variable, $X$ , is given by

$G_{X} (t) = 0.2 + 0.4 t + 0.3 t^{4} + 0.1 t^{5}$

Find $Var (X)$ .

varx-of-pgfs

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