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4.5 Probability generating functions

Syllabus
9231–2028–2029
Topic
4.5
Level
A2

Learning objectives

4.5.1Probability generating functions• understand the concept of a probability generating function (PGF) and construct and use the PGF for given distributions Including the discrete uniform, binomial, geometric and Poisson distributions.4.5.2PGF mean and variance• use formulae for the mean and variance of a discrete random variable in terms of its PGF, and use these formulae to calculate the mean and variance of a given probability distribution4.5.3Sums of random variables• use the result that the PGF of the sum of independent random variables is the product of the PGFs of those random variables.

A PGF stores each probability as a power-series coefficient

For a non-negative integer-valued random variable $X$,G_X(s)=E[s^X]=\sum_{r=0}^{\infty}P(X=r)s^r.Thus $[s^r]G_X(s)=P(X=r)$, $G_X(0)=P(X=0)$ and $G_X(1)=1$.

distribution and support PGF
discrete uniform on 1,…,n1,\ldots,n s(1−sn)n(1−s)\dfrac{s(1-s^n)}{n(1-s)} for s≠1s\ne1, with G(1)=1G(1)=1
Bin⁡(n,p)\operatorname{Bin}(n,p), q=1−pq=1-p (q+ps)n(q+ps)^n
geometric P(X=r)=pqr−1P(X=r)=pq^{r-1}, r=1,2,…r=1,2,\ldots ps1−qs\dfrac{ps}{1-qs}
Po⁡(λ)\operatorname{Po}(\lambda) exp⁡(λ(s−1))\exp(\lambda(s-1))

For example, if $G(s)=0.2+0.5s+0.3s^2$, thenP(X=0)=0.2,\quad P(X=1)=0.5,\quad P(X=2)=0.3.Thecoefficientrulealsoletsaclosedformbeexpandedtorecoverprobabilities.The coefficient rule also lets a closed form be expanded to recover probabilities.

State the geometric support convention: starting at one produces the numerator ps, while a failures-before-success convention starts at zero. A PGF is not an MGF, and G(1) must equal one for a valid probability distribution.

PGF derivatives at one give factorial moments, then mean and variance

For $G(s)=E[s^X]$,E[X]=G'(1),\qquad E[X(X-1)]=G''(1).Since $X^2=X(X-1)+X$,\operatorname{Var}(X)=G''(1)+G'(1)-[G'(1)]^2.Differentiate before setting $s=1$.

For $X\sim\operatorname{Po}(\lambda)$,G(s)=e^{\lambda(s-1)},\quad G'(s)=\lambda e^{\lambda(s-1)},\quad G''(s)=\lambda^2e^{\lambda(s-1)}.Hence $G'(1)=\lambda$ and $G''(1)=\lambda^2$.

ThereforeThereforeE[X]=\lambda,andand\operatorname{Var}(X)=\lambda^2+\lambda-\lambda^2=\lambda.ThecancellationchecksthefamiliarPoissonequalityofmeanandvariance.The cancellation checks the familiar Poisson equality of mean and variance.

G double-prime at one is not E[X squared]; it omits one copy of E[X]. If an alleged variance is negative, recheck differentiation, substitution at one and the subtraction of the squared mean.

Independence turns the PGF of a sum into a product

If $X_1,\ldots,X_k$ are independent non-negative integer-valued variables and $S=\sum X_i$, thenG_S(s)=E\left[s^{\sum X_i}\right]=E\left[\prod s^{X_i}\right]=\prod G_{X_i}(s).Independencejustifiesthefactorisationoftheexpectation.Independence justifies the factorisation of the expectation.

If $X\sim\operatorname{Po}(\lambda)$ and $Y\sim\operatorname{Po}(\mu)$ independently,G_{X+Y}(s)=e^{\lambda(s-1)}e^{\mu(s-1)}=e^{(\lambda+\mu)(s-1)},sosoX+Y\sim\operatorname{Po}(\lambda+\mu).

If $X\sim\operatorname{Bin}(n_1,p)$ and $Y\sim\operatorname{Bin}(n_2,p)$ independently,G_{X+Y}(s)=(1-p+ps)^{n_1+n_2},hence $X+Y\sim\operatorname{Bin}(n_1+n_2,p)$. If the success probabilities differ, multiply the PGFs but do not label the result binomial without further justification.

Multiplying PGFs requires independence. Means still add for dependent variables with finite expectations, but the PGF factorisation and the standard-family conclusions above can fail; coefficients of the actual product may be read when no named family results.