1, because the probabilities sum to 1.

PGF of the geometric distribution (trials to first success)?

\dfrac{pt}{1-qt}, valid for |qt|<1

PGF of the negative binomial (trials to the rth success)?

\left(\dfrac{pt}{1-qt}\right)^r

Mean of a geometric distribution, from its PGF?

G'(t)=\frac{p}{(1-qt)^2}, so E(X)=\frac1p.

Why is the negative binomial PGF a power of the geometric PGF?

It is the sum of r independent geometric variables, and PGFs of independent sums multiply.

Probability generating functionsEdexcel A-Level Further Maths: Flashcards

Card 1 of 130 of 13 known

Question

Define the probability generating function of $X$.

Tap or press Space to reveal

Tap card or press Space to flip

See all 13 cards

Define the probability generating function of $X$ .: $\mathrm{G}_X(t)=\mathrm{E}(t^X)=\sum\mathrm{P}(X=r)t^r$
What is $\mathrm{G}_X(1)$ ?: $1$ , because the probabilities sum to 1.
How do you find $\mathrm{E}(X)$ from the PGF?: $\mathrm{E}(X)=\mathrm{G}_X'(1)$
How do you find $\mathrm{Var}(X)$ from the PGF?: $\mathrm{G}_X''(1)+\mathrm{G}_X'(1)-[\mathrm{G}_X'(1)]^2$
What does $\mathrm{G}_X''(1)$ equal?: $\mathrm{E}[X(X-1)]=\mathrm{E}(X^2)-\mathrm{E}(X)$
PGF of $\mathrm{B}(n,p)$ ?: $(q+pt)^n$
PGF of $\mathrm{Po}(\lambda)$ ?: $\mathrm{e}^{\lambda(t-1)}$
PGF of the geometric distribution (trials to first success)?: $\dfrac{pt}{1-qt}$ , valid for $|qt|<1$
PGF of the negative binomial (trials to the $r$ th success)?: $\left(\dfrac{pt}{1-qt}\right)^r$
PGF of $X+Y$ for independent $X$ and $Y$ ?: $\mathrm{G}_X(t)\times\mathrm{G}_Y(t)$
How do you read $\mathrm{P}(X=r)$ from a PGF?: It is the coefficient of $t^r$ .
Mean of a geometric distribution, from its PGF?: $\mathrm{G}'(t)=\frac{p}{(1-qt)^2}$ , so $\mathrm{E}(X)=\frac1p$ .
Why is the negative binomial PGF a power of the geometric PGF?: It is the sum of $r$ independent geometric variables, and PGFs of independent sums multiply.

Exam questions on Probability generating functions

The discrete random variable $X$ has $\mathrm{P}(X=0)=0.2$ , $\mathrm{P}(X=1)=0.5$ and $\mathrm{P}(X=2)=0.3$ . The probability generating function of $X$ is $\mathrm{G}_X(t)$ .
Use $\mathrm{G}_X(t)$ to find $\mathrm{Var}(X)$ .2 marks
The random variable $X$ has a geometric distribution with parameter $p=\frac14$ , so $X$ is the number of trials up to and including the first success, with $\mathrm{P}(X=r)=q^{r-1}p$ for $r=1,2,3,\ldots$ and $q=1-p$ .
Given that $\mathrm{G}_X'(t)=\dfrac{4}{(4-3t)^2}$ , find $\mathrm{G}_X''(1)$ and hence show that $\mathrm{Var}(X)=12$ .2 marks
The random variable $X$ has distribution $\mathrm{B}(3,0.4)$ and the random variable $Y$ has distribution $\mathrm{Po}(2)$ . The variables $X$ and $Y$ are independent and $Z=X+Y$ .
Show, from the definition of a probability generating function, that $\mathrm{G}_X(t)=(0.6+0.4t)^3$ .3 marks

See the full worksheet

Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).