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 Incoming links: Poisson characterization by independent binomial splitting

Past exam of the mathematics course of the University of Cambridge / 2002 / ia / Paper 2 / 10F / c / Solution Created 2026-09-24 Updated 2026-10-14

With equal type probabilities,

F

and

S

have the same marginal probability generating function, namely

G_{N} ((1 + s) /2)

. Their assumed independence, together with

N = F + S

, gives

G_{N} (s) = G_{F} (s) G_{S} (s) = G_{N} (\frac{1 + s}{2})^{2} .

H (t) = G_{N} (1 - t)

for

0 \leq t \leq 1

. The identity becomes

H (t) = H (t /2)^{2}

and, after

k

H (t) = H (t / 2^{k})^{2^{k}} .

The finite expected value gives

G_{N} (1 - h) = 1 - μ h + o (h)

h ↓ 0

. To justify this without any analyticity assumption at the endpoint, note that

(1 - (1 - h)^{N}) / h \to N

and is bounded above by

N

; dominated convergence gives the derivative

G_{N}^{'} (1 -) = μ

. Substituting the expansion into the iterated identity yields

H (t) = (1 - \frac{μ t}{2 ^{k}} + o (2^{- k}))^{2^{k}} ⟶ e^{- μ t} .

The left side is independent of

k

, so it equals the limit. Thus

G_{N} (s) = e^{μ (s - 1)}

on

[0, 1]

and uniqueness of the power-series coefficients proves $N \sim Poisson (μ)$ . Only finite expected value was needed; the given finite variance is stronger. This proves the Poisson characterization by independent binomial splitting rather than assuming the original count distribution.

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