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Probability generating function of an independent random sum 2026-10-18

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 sum

S = \sum_{j = 1}^{N} Y_{j},

with the empty sum equal to zero. Given

N = n

, the probability generating function of an independent sum yields the conditional expectation

E [s^{S} ∣ N = n] = G_{Y} (s)^{n}

. By the law of total expectation,

G_{S} (s) = E [G_{Y} (s)^{N}] = G_{N} (G_{Y} (s)), ∣ s ∣ \leq 1.

The composition is valid because

∣ G_{Y} (s) ∣ \leq 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

λ

, substitution into

G_{N} (u) = exp (λ (u - 1))

gives a compound Poisson distribution with

G_{S} (s) = exp (λ (G_{Y} (s) - 1)) .

If the jumps have finite second moment, conditioning also gives mean

E [S] = λ E [Y]

Var (S) = λ 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 Poisson (2)

and jumps taking values one and two with equal probability. Keep the independence assumptions above. Then

G_{Y} (s) = \frac{s + s ^{2}}{2}, G_{S} (s) = exp (2 [\frac{s + s ^{2}}{2} - 1]) = exp (s + s^{2} - 2) .

Here

E [Y] = 3/2

,

Var (Y) = 1/4

, and

E [Y^{2}] = 5/2

. The law of total expectation and law of total variance give

E [S] = 2 \cdot \frac{3}{2} = 3, Var (S) = 2 \cdot \frac{1}{4} + 2 (\frac{3}{2})^{2} = 5.

Thus the aggregate is not itself a Poisson distribution, whose mean and variance would coincide.

To recover the probability mass function

p_{n} = P (S = n)

without repeated differentiation, use

G_{S}^{'} (s) = (1 + 2 s) G_{S} (s)

and compare coefficients:

p_{0} = e^{- 2}, n p_{n} = p_{n - 1} + 2 p_{n - 2} (n \geq 1), p_{n} = 0 (n < 0) .

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

p_{n} = e^{- 2} \sum_{b = 0}^{⌊ n /2 ⌋} \frac{1}{( n - 2 b )! b !} .

Figure 1.
A random number of one-or-two jumps gives a compound Poisson count
. The upper diagram separates the Poisson distribution count, the independent and identically distributed jumps, and their random sum. Below are calculated coefficients of $exp (s + s^{2} - 2)$ for $0 \leq n \leq 20$ , with the omitted tail probability reported explicitly. They are exact model probabilities evaluated numerically, not observed data; the displayed masses have not been renormalized. Details.

The plotted probability masses stop at 20, although the support of a probability distribution is unbounded. The omitted tail probability is

P (S > 20) = 1 - e^{- 2} \sum_{n = 0}^{20} \sum_{b = 0}^{⌊ n /2 ⌋} \frac{1}{( n - 2 b )! b !} \approx 1.95613 \times 1 0^{- 7} .

This 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 ∣ \leq 1

domain.

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