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Retain each of $N$ objects independently with probability $r$, independently of the nonnegative integer-valued random variable $N$. The retained count $Y$ has the displayed conditional binomial distribution. The probability generating function of an independent random sum gives $G_Y(s)=G_N(1-r+rs)$. If $N$ has finite variance, the law of total expectation and law of total variance giveThe first term is the average conditional variance of the independent retention decisions; the second is the variance in their conditional expected value.

For $0\le s\le 1$, condition on $N=n$. Its independence from the $X_i$ preserves their independent common distribution, so part (ii) givesFor $n=0$, this is $1$, agreeing with $Y=0$. By the law of total expectation,This proves the probability generating function of an independent random sum. For nonnegative $s$ outside $[0,1]$, the same argument is valid whenever the right side is finite, by Tonelli's theorem for nonnegative sums. For signed or complex $s$, a sufficient absolute-convergence condition is $G_N(G_X(|s|))<\infty$; it holds automatically on $|s|\le 1$. Outside this domain, mere evaluation of an analytically extended expression is not a substitute for existence of the probability generating function.

Let $I_i$ indicate that student $i$ stays awake. These indicators have independent Bernoulli distributions with parameter $p$, and $W={\sum }_{i=1}^M{I}_i$. The Poisson distribution has probability generating functionThe probability generating function of an independent random sum now givesRecognizing the Poisson distribution,This is Poisson thinning.

Use the measurable partition $B_n=\{N=n\}$ for $n\ge 0$, omitting any null cells. For $0\le s\le 1$, the random variable $s^X$ is bounded, so the law of total expectation applies without moment assumptions. Conditional on $N=n$, the sum contains exactly $n$ terms. Independence of $N$ from the whole sequence preserves their joint law under this conditioning; independence of the $Y_k$ then givesFor $n=0$ the empty sum is zero and the empty product is one, in agreement with the formula. Therefore the probability generating function of an independent random sum is

The extra ancestor multiplies the probability generating function by $s$. The preceding probability generating function of an independent random sum formula therefore yieldsFor $0\le s<1$ the two roots are $1\pm \sqrt{1-s}$. A probability generating function lies in $[0,1]$ there, so the plus root is impossible. Also $Z\ge 1$ gives $G_Z(0)=0$. Consequently the total progeny with critical geometric offspring hasAt $s=1$ the value is $1$, consistent with almost-sure finiteness. For complex $|s|<1$, the same expression uses the analytic square root that equals $1$ at $s=0$.

For $0\le s\le 1$, condition on $N=k$. Independence of the descendant total progeny variables givesincluding the empty product $1$ when $k=0$. Taking expectation over $N$ proves the probability generating function of an independent random sum formulaThe same identity holds for complex $|s|<1$ by absolute convergence.

The probability generating function of an independent random sum states that, for a nonnegative integer-valued $N$ independent of independent and identically distributed nonnegative integer-valued summands, $G_{{\sum }_{i=1}^N{X}_i}(s)=G_N(G_{X_1}(s))$ wherever the expectations converge. Here $G_{X_1}(s)=1-r+rs$, so, writing $q=1-p$,It is valid at least for $|s|\le 1$, and its power series in fact converges for $|s|<(p+qr)/(qr)$.

Start a branching process with one ancestor and offspring distribution $\mathbb{P}(N=k)=2^{-k-1}$ for $k\ge 0$. Its probability generating function is $G_N(s)=1/(2-s)$ and its mean offspring count is $1$, making it a critical branching process. Splitting the family at the ancestor gives the total progeny recursion $Z=1+{\sum }_{i=1}^N{Z}_i$. The probability generating function of an independent random sum then gives $G_Z=s/(2-G_Z)$, whose root satisfying $G_Z(0)=0$ is $1-\sqrt{1-s}$. Its limit at $s\uparrow 1$ equals $1$, consistent with finite total progeny almost surely; its derivative diverges there, so the expectation of $Z$ is nevertheless infinite.