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Finite sieve determines the probability of a zero count

Statement

Let Xn be nonnegative integer-valued random variables on finite probability spaces, and define (xr)=(x)r/r! for integers 0≤r≤x, and zero for r>x. Powers λr use natural integer exponents, with λ0=1 also when λ=0. If, for each fixed r≥0, E(Xn)r→λr for a fixed λ≥0, then P(Xn=0)→e−λ. No assertion about other values of Xn is needed.

Facts & Assumptions

Given: The objects and hypotheses of the Statement.

Goal: Prove the Statement.

[F1]

Expectation is a finite weighted sum, preserving pointwise inequalities (Expectation of a real random variable on a finite probability space).

[F2]

Factorials and falling factorials have their finite product definitions (The factorial n! and the falling factorial n^k̲, defined by recursion in ℕ).

[F3]

The exponential is the everywhere convergent series exp⁡(t)=∑r≥0tr/r! (The real exponential function and the number e by a power series).

[F4]

Convergence is interpreted along the sequence of finite probability spaces (Asymptotic quantifiers and uniform parameter ranges).

[F5]

A finite probability space (Ωn,wn) has nonnegative weights summing to one; every subset is an event whose probability is the sum of its weights. A real random variable is a function Xn:Ωn→R (Finite probability spaces, outcome weights, events, and event probabilities, Real random variables on finite probability spaces and their finite distributions).

[F6]

For a>0 and t∈R, real powers satisfy at=exp⁡(tlog⁡a), and the inverse logarithm satisfies log⁡(exp⁡(t))=t (Real powers for positive bases, with the zero-base positive-exponent convention, The natural logarithm as the inverse of the exponential function).

[F7]

Natural integer powers satisfy a0=1 and ar+1=ara for every real a and integer r≥0, including a=0 (Integer powers aᵐ).

Proof

technique · direct
1.1F2F7algebra

For integers x≥1 and m≥0, Pascal's identity, obtained by separating subsets according to whether they contain a specified element, telescopes to ∑r=0m(−1)r(xr)=(−1)m(x−1m); the binomial coefficient is zero beyond its range. The factorial formula agrees with subset counting because each r-subset admits r! orders. For x=0, the sum is one. Write 1{x=0} for the function equal to one when x=0 and zero otherwise. Thus, for every M≥0, ∑r=02M+1(−1)r(xr)≤1{x=0}≤∑r=02M(−1)r(xr).

2.1step 1.1F1F2F5F7algebra

On (Ωn,wn), the function ω↦1{Xn(ω)=0} has expectation ∑ω:Xn(ω)=0wn(ω)=P(Xn=0). Put qn=P(Xn=0) and Tn,k=∑r=0k(−1)rE(Xn)r/r!. Taking expectations in step 1.1 and distributing the finite weighted sum over the finite sieve sums gives Tn,2M+1≤qn≤Tn,2M for every M≥0.

3.1step 2.1F2F3F4F7givenchoosealgebra

Put sk=∑r=0k(−1)rλr/r! and L=exp⁡(−λ). The natural-power recursion gives (−λ)r=(−1)rλr by induction on r, starting at r=0 with both sides equal to one. Thus sk→L by the exponential series, including when λ=0. Fix ε>0 and choose M so that both ∣s2M+1−L∣<ε/3 and ∣s2M−L∣<ε/3. For each of the finitely many r∈{0,…,2M+1}, the given moment convergence supplies an Nr such that ∣E(Xn)r−λr∣<εr!/[3(2M+2)] for n≥Nr. For n≥N=max⁡0≤r≤2M+1Nr, the triangle inequality gives ∣Tn,k−sk∣≤∑r=0k∣E(Xn)r−λr∣/r!<ε/3 for both k=2M and k=2M+1. The bounds from step 2.1 now imply L−2ε/3<qn<L+2ε/3, so ∣qn−L∣<ε. This proves qn→exp⁡(−λ) directly from the convergence definition.

4.1step 3.1F3F6algebra∎

The series at one gives e=exp⁡(1)>0, and the inverse logarithm gives log⁡e=1; hence e−λ=exp⁡(−λlog⁡e)=exp⁡(−λ) by the real-power definition. Substituting this identity into step 3.1 yields the stated zero-count limit.

Depends on

Used by · 1

Dependency tree · 2 levels

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