From a Probability Law to a Generating Function
A probability law on the nonnegative integers is a sequence of numbers \(p_0,p_1,\ldots\), where \(p_k\) is the probability of the value \(k\), each \(p_k\ge0\), and \(\sum_{k=0}^{\infty}p_k=1\). A probability generating function packages this whole sequence into one function. This is useful because a function can be easier to manipulate than an infinite list of probabilities, while still retaining all the information in that list.
Let \(X\) be a nonnegative integer-valued random variable on a probability space \((\Omega,\mathcal{F},\mathbb{P})\), and write \(p_k=\mathbb{P}(X=k)\). For a complex number \(z\), use the convention \(z^0=1\), including when \(z=0\). A complex expectation \(\mathbb{E}[z^X]\) is defined when \(\mathbb{E}|z^X|<\infty\). The probability generating function, or PGF, is
at every \(z\) for which the series converges absolutely. In particular, this definition applies throughout the closed unit disk \(|z|\le1\). The coefficients are probabilities, not moments: the coefficient of \(z^k\) is exactly \(\mathbb{P}(X=k)\).
Convergence on the Unit Disk
For \(|z|\le1\), each term satisfies \(|p_kz^k|\le p_k\). Since the probabilities sum to \(1\), the series converges absolutely there. In fact, the convergence is uniform on the entire closed unit disk: after truncating at \(N\), the absolute size of the omitted terms is bounded by \(\sum_{k>N}p_k\), a quantity that tends to zero and does not depend on \(z\).
Proof. For every integer \(N\ge0\) and every \(|z|\le1\), the tail satisfies \(\left|\sum_{k>N}p_kz^k\right|\le\sum_{k>N}p_k\). The right-hand side tends to zero because \(\sum_{k=0}^{\infty}p_k=1\). This proves uniform convergence, and the same bound with \(N\) tending to infinity proves absolute convergence at each point. Every partial sum is a polynomial and hence continuous. A uniform limit of continuous functions is continuous, so \(G_X\) is continuous on the closed disk. On the open disk the series is a power series; the standard power-series theorem gives analyticity inside its radius of convergence. The radius is at least \(1\), as also follows from the coefficient bound \(\sum p_k=1\). \(\square\)
The radius may be larger than \(1\), or even infinite. In the usual power-series formula it is \(R=1/\limsup_{k\to\infty}p_k^{1/k}\), with reciprocal conventions when the limit superior is zero. This formula specifies absolute convergence for \(|z|<R\) and divergence for \(|z|>R\); it does not decide convergence at \(|z|=R\). The important guaranteed region for every PGF is the closed unit disk.
Worked Example: A Finite Distribution
Suppose \(X\) has probabilities \(\mathbb{P}(X=0)=\tfrac14\), \(\mathbb{P}(X=1)=\tfrac12\), and \(\mathbb{P}(X=3)=\tfrac14\), with probability zero at every other nonnegative integer. First check the probabilities: \(\tfrac14+\tfrac12+\tfrac14=1\). The PGF is the sum of the three nonzero terms:
For example, at \(z=-1\), the value is \(G_X(-1)=\tfrac14-\tfrac12-\tfrac14=-\tfrac12\). At \(z=1\), it is \(G_X(1)=\tfrac14+\tfrac12+\tfrac14=1\), as required by the total probability. The formula displays the distribution directly: the coefficients of \(1,z,z^3\) give the probabilities at \(0,1,3\), and the missing \(z^2\) term records probability zero at \(2\).
Check: Substituting \(z=1\) into the polynomial gives \(1\), and its coefficients sum to the checked total probability \(1\).
How PGF Values Encode the Distribution
One immediate check is \(G_X(1)=1\), since substituting \(1\) sums all probabilities. Also \(G_X(0)=p_0\), because all positive powers of zero vanish. More generally, the whole function on the interval \(0\le z<1\) determines every probability, not just these two facts. The next result gives a direct proof that does not require differentiating the PGF.
Proof. Set \(p_k=\mathbb{P}(X=k)\) and \(q_k=\mathbb{P}(Y=k)\). At \(x=0\), equality of the PGFs gives \(p_0=q_0\). Suppose for some \(m\ge1\) that \(p_k=q_k\) for every \(k<m\). For \(0<x<1\), subtract the equal lower-order terms from \(G_X(x)=G_Y(x)\) and divide by \(x^m\). This gives
The absolute value of the remaining sum is at most \(x\sum_{k>m}|p_k-q_k|\le2x\), because \(x^{k-m}\le x\) for \(k>m\), and \(\sum_{k>m}|p_k-q_k|\le\sum_{k>m}p_k+\sum_{k>m}q_k\le2\). As \(x\) decreases to zero this bound tends to zero. The displayed equality therefore implies \(p_m-q_m=0\). Starting from \(p_0=q_0\), induction proves \(p_k=q_k\) for every \(k\ge0\). Equal probabilities at every possible value are exactly equality of the distributions. \(\square\)
This is an encoding statement: a PGF is not merely a convenient summary that might discard information. If two PGFs agree throughout \(0\le x<1\), their probability laws agree. It does not say that one function value, such as \(G_X(\tfrac12)\), identifies a law; the theorem uses agreement over the whole interval.
Worked Example: Two Laws Distinguished by PGFs
Let \(X\) take the values \(0\) and \(2\), each with probability \(\tfrac12\). Let \(Y\) take \(0,1,2\) with probabilities \(\tfrac14,\tfrac12,\tfrac14\), respectively. Both lists sum to \(1\). Their PGFs are
At \(z=\tfrac12\), the first gives \(G_X(\tfrac12)=\tfrac12+\tfrac12\cdot\tfrac14=\tfrac58\). The second gives \(G_Y(\tfrac12)=\tfrac14+\tfrac12\cdot\tfrac12+\tfrac14\cdot\tfrac14=\tfrac14+\tfrac14+\tfrac1{16}=\tfrac9{16}\). Since \(\tfrac58=\tfrac{10}{16}\ne\tfrac9{16}\), these PGFs differ, consistent with the different probabilities assigned to the value \(1\).
Check: At \(z=1\), \(G_X(1)=\tfrac12+\tfrac12=1\) and \(G_Y(1)=\tfrac14+\tfrac12+\tfrac14=1\). Both functions pass the total-probability check, even though they encode different laws.
Worked Example: A Countably Supported Law
Define probabilities by \(p_0=\tfrac12\) and \(p_k=\tfrac{1}{2k(k+1)}\) for \(k\ge1\). To check that they form a probability law, use \(\tfrac1{k(k+1)}=\tfrac1k-\tfrac1{k+1}\). The partial sum through \(k=N\) is
As \(N\) tends to infinity this tends to \(\tfrac12\). Adding \(p_0=\tfrac12\) gives total probability \(1\). The PGF is therefore
For \(|z|\le1\), the absolute sum of the terms with \(k\ge1\) is bounded by \(\sum_{k=1}^{\infty}\tfrac1{2k(k+1)}=\tfrac12\). Thus this expression converges absolutely on the closed unit disk, including at \(z=-1\). At \(z=0\), it gives \(G_X(0)=\tfrac12\). At \(z=1\), the telescoping sum gives \(G_X(1)=\tfrac12+\tfrac12=1\).
Check: The coefficient of \(z^2\) is \(\tfrac1{2\cdot2\cdot3}=\tfrac1{12}\), which is exactly the assigned probability \(\mathbb{P}(X=2)\). The endpoint value at \(1\) also equals the verified total probability.
Where the Expectation Definition Stops
For \(|z|\le1\), the expectation and the absolutely convergent probability series always agree. Outside that disk, it is useful to distinguish two claims: the power series may or may not converge, and the complex random variable \(z^X\) has an expectation only if it is integrable. If the series converges absolutely, then \(\mathbb{E}|z^X|=\sum_{k\ge0}p_k|z|^k<\infty\), so the expectation is defined and equals the series. If a series converges only conditionally, it can be discussed as a series extension, but it is not the expectation \(\mathbb{E}[z^X]\).
Common mistake. It is incorrect to treat conditional convergence of \(\sum p_kz^k\) as sufficient for the expectation \(\mathbb{E}[z^X]\). For a concrete case, let \(p_k=C\,2^{-k}/(k+1)\) for \(k\ge0\), where \(C=(2\ln2)^{-1}\). The identity \(\sum_{k\ge0}2^{-k}/(k+1)=2\ln2\) shows that these probabilities sum to \(1\). At \(z=-2\), the formal series becomes \(C\sum_{k\ge0}(-1)^k/(k+1)\), which converges conditionally. But
Thus \((-2)^X\) is not integrable and its expectation is not defined, even though the formal series has a conditional sum. The correction is to define \(G_X(z)=\mathbb{E}[z^X]\) where the series converges absolutely; if one separately studies a conditionally convergent extension, define it by the series and do not identify it with an expectation.
Summary and Interpretation
A PGF begins with the probabilities of a nonnegative integer-valued random variable and forms the power series whose coefficients are those probabilities. The probability coefficients ensure absolute and uniform convergence on \(|z|\le1\), continuity on the closed disk, and analyticity in its interior. Its values across \(0\le z<1\) determine the entire distribution, as the coefficient-by-coefficient proof shows.
Keep three checks in view when using a PGF: the probabilities must sum to \(1\); \(G_X(1)=1\); and an expectation representation outside the unit disk requires absolute integrability, not merely convergence of a formal series. These conditions prevent confusing a valid encoding of a probability law with an unrelated or undefined complex expectation.
Check Your Understanding
Use the definition and convergence arguments above to answer the following.
- If \(X\) takes values \(0,2\) with probabilities \(\tfrac34,\tfrac14\), write \(G_X(z)\) and check \(G_X(1)=1\).
- Why does the PGF series converge absolutely for every \(z\) with \(|z|=1\), not only for \(|z|<1\)?
- If two PGFs agree for every real \(x\) with \(0\le x<1\), what induction argument shows their probabilities at each integer agree?
- What condition is required for \(\mathbb{E}[z^X]\) to be defined when \(z\) is complex?
- Explain why conditional convergence of \(\sum_k p_kz^k\) does not by itself justify calling its sum \(\mathbb{E}[z^X]\).