PGFs Of Bernoulli, Binomial, Geometric, And Poisson Variables

From a Probability Mass Function to a PGF

For a nonnegative integer-valued random variable, a probability generating function packages all its point probabilities into a single function. Standard distributions often have compact PGFs, but those formulas depend on the support and parameterization being used. In particular, geometric variables are defined with two common support conventions, and the resulting formulas differ by a factor of \(s\).

Let \(X\) be a nonnegative integer-valued random variable on a probability space \((\Omega,\mathcal{F},\mathbb{P})\), with probability mass function \(p_k=\mathbb{P}(X=k)\), \(k=0,1,2,\ldots\). For \(0\le s\le1\), its probability generating function (PGF) is

At \(s=0\), use the convention \(0^0=1\) in this series, so \(G_X(0)=p_0\). The series converges for every \(s\) in \([0,1]\), since \(0\le p_k s^k\le p_k\) and \(\sum_{k=0}^{\infty}p_k=1\). A basic check is therefore

Why the check works. At \(s=1\), every term \(s^k\) equals one, including the \(k=0\) term. Thus evaluating a proposed PGF at one must recover the total probability. This check detects many errors, although passing it alone does not prove that a formula is correct.

Four Standard PGFs

Proof. For Bernoulli \(X\), the only possible values are zero and one, with probabilities \(1-p\) and \(p\). Substitution into the defining sum gives \(G_X(s)=(1-p)s^0+ps^1=1-p+ps\). This also covers \(p=0\) and \(p=1\), which give the degenerate variables zero and one, respectively.

For a binomial variable, the mass function is \(\mathbb{P}(X=k)=\binom{n}{k}p^k(1-p)^{n-k}\) for \(k=0,\ldots,n\). Hence \(G_X(s)=\sum_{k=0}^{n}\binom{n}{k}(ps)^k(1-p)^{n-k}=(1-p+ps)^n\), by the finite binomial expansion. If \(n=0\), the sum has just its \(k=0\) term and the formula is \(1\); if \(p=0\) or \(p=1\), the same expansion gives the corresponding degenerate distribution.

For the stated geometric convention, \(\mathbb{P}(X=k)=p(1-p)^k\) for \(k=0,1,\ldots\). Put \(q=1-p\). Since \(p>0\), \(0\le q<1\); for \(0\le s\le1\), \(0\le qs<1\). The finite geometric sum is \(\sum_{k=0}^{m}(qs)^k=(1-(qs)^{m+1})/(1-qs)\). As \(m\) tends to infinity, \((qs)^{m+1}\) tends to zero, so \(G_X(s)=\sum_{k=0}^{\infty}p(qs)^k=p/(1-qs)=p/(1-(1-p)s)\). This includes \(p=1\), when \(q=0\) and \(X=0\) with probability one.

For a Poisson variable, \(\mathbb{P}(X=k)=e^{-\lambda}\lambda^k/k!\) for \(k=0,1,\ldots\). Substitution gives \(G_X(s)=e^{-\lambda}\sum_{k=0}^{\infty}(\lambda s)^k/k!\). By the exponential power series, the sum is \(e^{\lambda s}\), so \(G_X(s)=e^{-\lambda}e^{\lambda s}=e^{\lambda(s-1)}\). When \(\lambda=0\), the distribution is concentrated at zero and this formula gives \(1\), as required. \(\square\)

Each formula can now be interpreted as a compact version of its mass function. The finite-support cases produce polynomials. The geometric and Poisson cases produce infinite series, summed using the geometric-series and exponential-series identities. These derivations use the actual probability masses, not an assumption that PGFs of different distributions combine in a particular way.

Reading the Formulas Without Changing the Convention

The four derivations share the same procedure: write the support and mass function, substitute into \(\sum_k\mathbb{P}(X=k)s^k\), and then simplify the resulting sum. The support matters because it determines the powers of \(s\). For a Bernoulli or binomial variable, only finitely many terms occur. For a geometric or Poisson variable, infinitely many terms occur, so a convergent series identity is needed.

The geometric parameterization deserves particular care. This tutorial defines \(X\) as the number of failures before the first success, so the first possible value is zero and \(\mathbb{P}(X=0)=p\). If instead \(Y\) counts the trial on which the first success occurs, then \(Y\) takes values \(1,2,\ldots\) and \(\mathbb{P}(Y=k)=p(1-p)^{k-1}\). Substitution gives \(G_Y(s)=\sum_{k=1}^{\infty}p(1-p)^{k-1}s^k=ps/[1-(1-p)s]\). The extra factor \(s\) records the shift in support.

Common mistake. For failures before the first success, writing \(ps/[1-(1-p)s]\) is incorrect: it assigns zero probability to \(X=0\), even though \(\mathbb{P}(X=0)=p\). For example, with \(p=\tfrac13\), that incorrect expression evaluates at zero to \(0\), while the mass function gives \(G_X(0)=\mathbb{P}(X=0)=\tfrac13\). The correct expression is \((\tfrac13)/(1-\tfrac23s)\), which evaluates at zero to \(\tfrac13\) and at one to \(1\). Always check both the support’s first value and \(G_X(1)=1\).

PGFs are useful because they preserve the probability-mass information in a form that can be manipulated algebraically. At this stage, the essential skill is deriving the correct expression and checking its normalization and support convention. A formula that has the right appearance but comes from a shifted or misread mass function will encode a different random variable.