Probability generating functionsEdexcel A-Level Further Maths: Revision notes

For a discrete random variable $X$ taking non-negative integer values, the probability generating function (PGF) is $\mathrm{G}_X(t)=\mathrm{E}(t^X)=\sum_{r}\mathrm{P}(X=r)\,t^r.$ The coefficient of $t^r$ is $\mathrm{P}(X=r)$, so you can read off probabilities from a PGF written as a polynomial or power series. Since probabilities sum to 1, $\mathrm{G}_X(1)=1$ always, which is a useful check. Example: if $\mathrm{P}(X=0)=0.2$, $\mathrm{P}(X=1)=0.5$ and $\mathrm{P}(X=2)=0.3$, then $\mathrm{G}_X(t)=0.2+0.5t+0.3t^2$.

Differentiate term by term: $\mathrm{G}_X'(t)=\sum r\,\mathrm{P}(X=r)t^{r-1}$, so at $t=1$ $\mathrm{E}(X)=\mathrm{G}_X'(1).$ Differentiating again gives $\mathrm{G}_X''(1)=\mathrm{E}[X(X-1)]=\mathrm{E}(X^2)-\mathrm{E}(X)$. Hence $\mathrm{Var}(X)=\mathrm{G}_X''(1)+\mathrm{G}_X'(1)-\left[\mathrm{G}_X'(1)\right]^2.$ Example: $\mathrm{G}_X(t)=0.2+0.5t+0.3t^2$ gives $\mathrm{G}_X'(1)=0.5+0.6=1.1$, $\mathrm{G}_X''(1)=0.6$ and $\mathrm{Var}(X)=0.6+1.1-1.21=0.49$. Proofs of these standard results may be asked for in the exam.

Binomial $\mathrm{B}(n,p)$: $\mathrm{G}(t)=(q+pt)^n$, from the binomial theorem. Poisson $\mathrm{Po}(\lambda)$: $\mathrm{G}(t)=\mathrm{e}^{\lambda(t-1)}$, from $\sum\mathrm{e}^{-\lambda}\frac{(\lambda t)^r}{r!}$. Geometric (trials to first success), $\mathrm{P}(X=r)=q^{r-1}p$: $\mathrm{G}(t)=\dfrac{pt}{1-qt}$, from a geometric series, valid for $|qt|<1$. Then $\mathrm{G}'(t)=\dfrac{p}{(1-qt)^2}$ and $\mathrm{E}(X)=\frac1p$. Negative binomial (trials to the $r$th success): $\mathrm{G}(t)=\left(\dfrac{pt}{1-qt}\right)^r$, because it is a sum of $r$ independent geometric variables. For each, you must be able to derive the PGF from the definition and use it for the mean and variance.

If $X$ and $Y$ are independent, then $\mathrm{E}(t^{X+Y})=\mathrm{E}(t^X)\mathrm{E}(t^Y)$, so $\mathrm{G}_{X+Y}(t)=\mathrm{G}_X(t)\times\mathrm{G}_Y(t).$ Derivation of this result is not required. It extends to any number of independent variables: the sum of $n$ independent copies of $X$ has PGF $[\mathrm{G}_X(t)]^n$. Examples: $\mathrm{B}(3,0.4)$ plus an independent $\mathrm{Po}(2)$ has PGF $(0.6+0.4t)^3\mathrm{e}^{2(t-1)}$; the sum of two independent $\mathrm{Po}(\lambda)$ and $\mathrm{Po}(\mu)$ has PGF $\mathrm{e}^{(\lambda+\mu)(t-1)}$, so is $\mathrm{Po}(\lambda+\mu)$.

$X\sim\mathrm{Geo}(p)$ with $p=\frac14$: $\mathrm{G}(t)=\dfrac{t}{4-3t}$. Then $\mathrm{G}'(t)=\dfrac{4}{(4-3t)^2}$, so $\mathrm{E}(X)=\mathrm{G}'(1)=4$. Next $\mathrm{G}''(t)=\dfrac{24}{(4-3t)^3}$, so $\mathrm{G}''(1)=24$ and $\mathrm{Var}(X)=24+4-4^2=12$, which agrees with $\frac{q}{p^2}=\frac{3/4}{1/16}=12$. For the third success: $N=X_1+X_2+X_3$ has $\mathrm{G}_N=[\mathrm{G}_X]^3$, with $\mathrm{E}(N)=3\times4=12$ in this case.