PGF Boundary Derivatives and Count Moments

Why the Boundary at One Matters

For a nonnegative integer-valued random variable $N$, the probability-generating function (PGF) is $G(s)=E[s^N]$. Its value at $s=1$ is always finite: every possible count contributes its probability, so $G(1)=1$. But knowing that the probabilities sum to one says little about how quickly they decrease as the count grows. The boundary derivatives at one capture that behavior: they can be finite, or they can diverge, even though the PGF itself equals one.

The key is to approach one from within the PGF's natural interval $0\leq s\leq1$. For $s<1$, large counts are downweighted by powers of $s$, making differentiation well behaved. As $s$ increases to one, that downweighting disappears. The resulting derivative limits reveal whether the corresponding count moments exist.

Definition: Let $N$ take values in $\{0,1,2,\ldots\}$, with $P(N=k)=p_k$. Its PGF is $G(s)=\sum_{k=0}^{\infty}p_k s^k$ for $0\leq s\leq1$. For an integer $r\geq1$, define the falling factorial $(k)_r=k(k-1)\cdots(k-r+1)$ when $k\geq r$, and $(k)_r=0$ when $k<r$. The $r$th boundary derivative means the limit $\lim_{s\uparrow1}G^{(r)}(s)$, understood as an extended nonnegative value that may be $+\infty$.

The boundary derivative is a limit of derivatives evaluated at interior points; it is not initially an assumption that $G$ has a finite derivative at one. This distinction matters when the limit is infinite. For $0<s<1$, differentiating the power series inside its interval of convergence gives

G^{(r)}(s)=\sum_{k=r}^{\infty}(k)_r p_k s^{k-r}.

Each term is nonnegative and increases as $s$ increases to one. That monotonicity makes it possible to identify the limit without assuming that the moment is finite.

The Boundary Derivative Theorem

Theorem: For every integer $r\geq1$, $$ \lim_{s\uparrow1}G^{(r)}(s)=E[(N)_r]=\sum_{k=r}^{\infty}(k)_r p_k, $$ where either side is allowed to equal $+\infty$. In particular, the boundary derivative has a finite limit if and only if the $r$th factorial moment is finite.

The result holds whether the relevant series converges or diverges. The justification uses the monotone convergence theorem: it allows the increasing nonnegative terms to pass to their limits inside the sum, without requiring a finite dominating bound.

Differentiate inside the interval.
For $0<s<1$, the PGF power series can be differentiated term by term, giving $G^{(r)}(s)=\sum_{k=r}^{\infty}(k)_r p_k s^{k-r}$.

Increase toward the boundary.
For each $k\geq r$, $s^{k-r}$ increases to one as $s\uparrow1$. Thus the nonnegative summands increase to $(k)_r p_k$.

Apply monotone convergence.
The limit of the series is the sum of the termwise limits, which is exactly $E[(N)_r]$, finite or infinite.

More explicitly, take any sequence $s_m\uparrow1$. The monotone convergence theorem gives $\lim_{m\to\infty}\sum_{k=r}^{\infty}(k)_r p_k s_m^{k-r}=\sum_{k=r}^{\infty}(k)_r p_k$. Since the interior derivative is increasing in $s$, its limit as $s\uparrow1$ is the same, independent of the sequence. This proves the theorem. $\square$

If the $r$th factorial moment is finite, then the corresponding derivatives through order $r$ extend continuously from below to one, with their boundary values given by their factorial moments. In that case, these are also the successive left derivatives at the endpoint. When the limit of $G^{(r)}(s)$ is infinite, the derivative cannot extend continuously to a finite value there.

Worked Example: A Poisson Count Has Finite Boundary Derivatives

Let $N\sim\operatorname{Poisson}(\lambda)$, where $\lambda>0$. Its probabilities are $p_k=e^{-\lambda}\lambda^k/k!$, so its PGF is

G(s)=\sum_{k=0}^{\infty}e^{-\lambda}\frac{\lambda^k}{k!}s^k =e^{-\lambda}e^{\lambda s} =e^{\lambda(s-1)}.

Differentiating $r$ times gives $G^{(r)}(s)=\lambda^r e^{\lambda(s-1)}$. Taking the boundary limit,

\lim_{s\uparrow1}G^{(r)}(s)=\lambda^r.

By the theorem, $E[(N)_r]=\lambda^r$, which is finite at every fixed order. For example, $E[N]=\lambda$ and $E[N(N-1)]=\lambda^2$. The second ordinary moment is $E[N^2]=E[(N)_2]+E[N]=\lambda^2+\lambda$.

Finite PGF Values Do Not Guarantee Finite Moments

The PGF at one records total probability, not the size of the counts. A distribution can place sufficiently small probabilities on very large counts to make the probabilities sum to one while leaving a moment series divergent. The next example has a finite PGF at one but an infinite mean.

Worked Example: A Finite PGF With Infinite Mean

Let $P(N=k)=1/[k(k+1)]$ for $k\geq1$, and $P(N=0)=0$. These probabilities sum to one because

\sum_{k=1}^{m}\frac{1}{k(k+1)} =\sum_{k=1}^{m}\left(\frac{1}{k}-\frac{1}{k+1}\right) =1-\frac{1}{m+1} \longrightarrow 1.

Consequently, $G(1)=1$. For $0\leq s<1$, decompose $1/[k(k+1)]=1/k-1/(k+1)$. Using the power-series identity $\sum_{k=1}^{\infty}s^k/k=-\log(1-s)$, we obtain

G(s)=\sum_{k=1}^{\infty}\frac{s^k}{k(k+1)} =1+\left(\frac{1}{s}-1\right)\log(1-s), \qquad 0<s<1.

As $s\uparrow1$, $(1-s)\log(1-s)\to0$, so this expression also tends to one. But the first boundary derivative is

\lim_{s\uparrow1}G'(s) =E[N] =\sum_{k=1}^{\infty}\frac{k}{k(k+1)} =\sum_{k=1}^{\infty}\frac{1}{k+1} =+\infty.

Thus the PGF is finite at one while the mean is infinite. The theorem also says that $G'(s)$ increases without bound as $s\uparrow1$. The value $G(1)=1$ does not imply a finite slope at the boundary.

Factorial and Ordinary Moment Finiteness

For count variables, factorial and ordinary moments give different formulas, but they agree about whether a moment of a given order is finite. For example, $N^2=(N)_2+N$, so finite first and second factorial moments imply a finite second ordinary moment. More generally, the identity

N^r=\sum_{j=0}^{r}{r\brace j}(N)_j

expresses the $r$th ordinary power using Stirling numbers of the second kind ${r\brace j}$, which are nonnegative integers. Conversely, $(N)_r\leq N^r$. These facts show that finiteness of all factorial moments through order $r$ is equivalent to finiteness of the $r$th ordinary moment. In fact, for counts, finiteness of the $r$th factorial moment alone also suffices: it controls the tail of $N^r$, and the finitely many smaller count values cannot cause divergence.

Result: For any nonnegative integer-valued $N$ and integer $r\geq1$, $E[N^r]<\infty$ if and only if $E[(N)_r]<\infty$. Hence $E[N^r]$ is finite if and only if the $r$th boundary derivative of the PGF has a finite limit.

To see the direction not supplied by $(N)_r\leq N^r$, consider counts $k\geq2r$. Every factor in $(k)_r=k(k-1)\cdots(k-r+1)$ is at least $k/2$, so

(k)_r\geq (k/2)^r \quad\text{and therefore}\quad k^r\leq 2^r(k)_r.

For $k<2r$, $k^r\leq(2r)^r$. Combining these bounds for all $k$ gives $k^r\leq(2r)^r+2^r(k)_r$. Taking expectations proves that a finite $E[(N)_r]$ implies a finite $E[N^r]$. The reverse direction follows from $(N)_r\leq N^r$. Together with the boundary derivative theorem, this proves the result. $\square$

Worked Example: A Finite Mean but Infinite Second Moment

For $k\geq1$, set $P(N=k)=1/[\zeta(3)k^3]$, and set $P(N=0)=0$, where $\zeta(a)=\sum_{k=1}^{\infty}k^{-a}$ for $a>1$. The probabilities sum to one by the definition of $\zeta(3)$, so $G(1)=1$.

The mean is finite:

E[N]=\frac{1}{\zeta(3)}\sum_{k=1}^{\infty}\frac{k}{k^3} =\frac{\zeta(2)}{\zeta(3)}<\infty.

In contrast, the second factorial moment diverges:

E[(N)_2] =\frac{1}{\zeta(3)}\sum_{k=2}^{\infty}\frac{k(k-1)}{k^3} =\frac{1}{\zeta(3)}\sum_{k=2}^{\infty}\left(\frac{1}{k}-\frac{1}{k^2}\right) =+\infty.

Here the harmonic series diverges and $\sum_{k=2}^{\infty}k^{-2}$ converges. The boundary theorem therefore gives $\lim_{s\uparrow1}G''(s)=+\infty$, even though $\lim_{s\uparrow1}G'(s)=E[N]$ is finite. Since $N^2=(N)_2+N$, the second ordinary moment is infinite as well. This example shows why each derivative order must be checked separately.

Common Mistake: Confusing Total Probability With a Moment

A tempting but incorrect argument is: “The PGF converges at one and equals one, so differentiating it at one must give finite moments.” Convergence of $\sum_k p_k$ only controls total probability. The derivative series has different terms: $G'(s)$ weights $p_k$ by $k$, while $G''(s)$ weights them by $k(k-1)$. Those added factors can make the series diverge. The telescoping example has $G(1)=1$ but $E[N]=+\infty$, and the zeta example has a finite mean but an infinite second moment.

The correction is to differentiate at $s<1$, then take the limit from below and apply the boundary derivative theorem. A finite boundary derivative limit is exactly the condition for the corresponding factorial moment to be finite; the value $G(1)$ alone is not a substitute for that check.

Key takeaway: For a count $N$, $G(1)=1$ always, but $\lim_{s\uparrow1}G^{(r)}(s)=E[(N)_r]$ may be infinite. At order $r$, this boundary limit is finite exactly when the ordinary moment $E[N^r]$ is finite.

Check Your Understanding

Use the boundary derivative theorem and the examples above to answer these questions.

What does $\lim_{s\uparrow1}G'(s)$ equal, and can it be infinite when $G(1)=1$?
For a count $N$, what quantity is given by $\lim_{s\uparrow1}G^{(3)}(s)$?
Why does a finite first boundary derivative not guarantee that the second boundary derivative is finite?
If $E[(N)_r]<\infty$, explain why $E[N^r]<\infty$.
In the distribution with probabilities proportional to $k^{-3}$, which moment is finite and which is infinite: the mean or the second moment?