Probability generating function ($G_X(s)=\mathbb{E}[s^X]$)

For a nonnegative integer-valued random variable $X$, write $p_k=\mathbb{P}(X=k)$. Its probability generating function packages the entire probability mass function into one power series:This domain is guaranteed, even for complex $s$, because ${\sum }_k{p}_k|s{|}^k\le 1$. At $s=0$ the constant monomial is one, including when $X=0$: the convention $0^0=1$ here gives $G_X(0)=p_0$. Also $G_X(1)=1$. The power series may converge beyond this domain, but that needs an additional argument. Encoding a probability distribution this way turns convolution of independent random variables into multiplication.^[ref]

Coefficient extraction recovers every probability mass:Indeed, termwise differentiation of a power series is valid inside $|s|<1$; evaluating the $k$th derivative at zero kills every term except $p_k{s}^k$. Thus two probability generating functions with the same coefficients describe the same probability distribution, without any finite-moment assumption.

At one, approach along the real line from below. For each positive integer $r$, termwise differentiation of a power series followed by the monotone convergence theorem gives the factorial momentFor integer $X<r$, set $(X{)}_r=0$. The limit may be infinite; it is a finite boundary value only when the corresponding factorial moment is finite. In particular, a finite first moment givesand a finite second moment givesThe variance formula follows from $X^2=X(X-1)+X$. These are boundary limits of derivatives; they do not assert that the probability generating function extends smoothly through one.

For a nonnegative integer-valued random variable, its moment-generating function satisfies $M_X(t)=G_X(e^t)$ wherever finite. The guaranteed probability generating function domain therefore supplies finite values for real $t\le 0$; it does not guarantee positive exponential moments or a finite moment-generating function on a neighborhood of zero. Bounded nonnegative integer-valued random variables have polynomial probability generating functions, but their convenient behavior is not a general existence theorem.

For nonnegative integer-valued independent random variables $X,Y$, factor the expected value of a product:The Cauchy product explains what this means for individual probability masses:Every way of splitting the total contributes once. This is convolution of independent random variables; multiplying probability generating functions performs the bookkeeping automatically.

For an original example, take independent $X\sim \mathrm{Bernoulli}(0.3)$ and $Y\sim \mathrm{Binomial}(2,0.5)$, using the Bernoulli distribution and binomial distribution. ThensoDirect enumeration of the six pairs $(x,y)$ gives the same probability masses. For example, total one collects $(0,1)$ and $(1,0)$, with contributions $0.7\cdot 0.5+0.3\cdot 0.25=0.425$. The coefficients sum to one. The expected value is $G_{X+Y}^{\prime }(1)=1.3$; since $G_{X+Y}^{\prime \prime }(1)=1.1$, the variance is $1.1+1.3-1.3^2=0.71$, agreeing with $0.3\cdot 0.7+2\cdot 0.5\cdot 0.5$.

Uncorrelated random variables are insufficient for the product identity. For a concrete counterexample, let $U$ be uniform on $\{-1,0,1\}$ and set $A=U+1$, $B=U^2$. Then $\mathrm{Cov}(A,B)=\mathbb{E}[U^3]=0$, but $\mathbb{P}(A+B=0)=0$, whereas $G_A(0)G_B(0)=1/9$. Zero covariance controls one mixed moment, not the full joint probability mass function.

Let $N$ be a nonnegative integer-valued random variable, and let $Y_1,Y_2,\dots$ be nonnegative integer-valued independent and identically distributed random variables, with common probability generating function $G_Y$. Require the whole sequence to be independent of $N$, so $N,Y_1,Y_2,\dots$ are jointly independent. Set the random sumwith the empty sum equal to zero. Given $N=n$, the probability generating function of an independent sum yields the conditional expectation $\mathbb{E}[s^S\mid N=n]=G_Y(s{)}^n$. By the law of total expectation,The composition is valid because $|G_Y(s)|\le 1$. Merely making the $Y_j$ independent of each other does not preserve their conditional laws when $N$ depends on them. This is the probability generating function version of transform composition for an independent random sum.

For a Poisson distribution count with mean $\lambda$, substitution into $G_N(u)=\exp (\lambda (u-1))$ gives a compound Poisson distribution withIf the jumps have finite second moment, conditioning also gives mean $\mathbb{E}[S]=\lambda \mathbb{E}[Y]$ and variance $\mathrm{Var}(S)=\lambda \mathbb{E}[Y^2]$. The extra spread comes from randomness in the count as well as randomness in the jumps.^[ref]

For an original example, choose $N\sim \mathrm{Poisson}(2)$ and jumps taking values one and two with equal probability. Keep the independence assumptions above. ThenHere $\mathbb{E}[Y]=3/2$, $\mathrm{Var}(Y)=1/4$, and $\mathbb{E}[Y^2]=5/2$. The law of total expectation and law of total variance giveThus the aggregate is not itself a Poisson distribution, whose mean and variance would coincide.

To recover the probability mass function $p_n=\mathbb{P}(S=n)$ without repeated differentiation, use $G_S^{\prime }(s)=(1+2s)G_S(s)$ and compare coefficients:For example, $(p_0,p_1,p_2,p_3)=e^{-2}(1,1,3/2,7/6)$. As an independent check, expanding $e^{s+s^2-2}=e^{-2}{e}^s{e}^{s^2}$ gives

The plotted probability masses stop at 20, although the support of a probability distribution is unbounded. The omitted tail probability isThis model has bounded jumps and an unbounded count. Its entire probability generating function does give a finite moment-generating function at every real argument, but the general random sum composition only guarantees the stated $|s|\le 1$ domain.

Encode a fair die with face labels $a_1,\dots ,a_s$ by $A(z)={\sum }_i{z}^{a_i}$. Its probability generating function is $A(z)/s$. Two independent dice with polynomials $A,B$ have the same sum probability distribution as another pair $C,D$ if $AB=CD$ and all four dice have $s$ faces. Redistributing factors can construct unexpected examples, but each resulting polynomial must have nonnegative integer coefficients, no constant term if labels must be positive, and value $s$ at $z=1$. For example, reallocating $1-z+z^2$ from one copy of $z+\cdots +z^6$ to the other gives face labels $(1,2,2,3,3,4)$ and $(1,3,4,5,6,8)$.

For a positive integer-valued random variable of finite support, this transform is a finite Dirichlet series. Its values encode ordinary moments, $q_X(-r)=\mathbb{E}[X^r]$, and its derivatives at zero encode logarithmic moments: $(-1{)}^r{q}_X^{(r)}(0)=\mathbb{E}[(\log X{)}^r]$. Positive support is essential to these definitions; the transform is distinct from the ordinary probability generating function $\mathbb{E}[s^X]$.