TheoremBase

The mean of SnS_n is nm and its variance at most nσ² by additivity of variances over uncorrelated summands; Chebyshev's inequality gives the bound, which tends to 0 by the Archimedean property.

Proof

Fix a real ε>0\varepsilon>0, and for n∈Nn\in\mathbb{N} write Sn=∑k=1nXkS_{n}=\sum_{k=1}^{n}X_{k}.

Mean and variance of SnS_{n}. Let n∈Nn\in\mathbb{N}. Each XkX_{k} is integrable (Lᵖ Inclusions and Constants on a Finite Measure Space (i)), so by linearity (Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion (viii), with all coefficients equal to 11) and by Expectations and Variances of Independent Random Variables (iii), Sn∈L2(P)S_{n}\in\mathcal{L}^{2}(P) and

ESn=∑k=1nEXk=nμ,Var⁡Sn=∑k=1nVar⁡Xk≤nc\mathbb{E}S_{n}=\sum_{k=1}^{n}\mathbb{E}X_{k}=n\mu,\qquad\operatorname{Var}S_{n}=\sum_{k=1}^{n}\operatorname{Var}X_{k}\le nc

(the finite sums evaluated and compared by Finite Sums in a Commutative Ring and in an Ordered Field: Distributivity, Differences, Telescoping, Constant Terms, Comparison and the Triangle Inequality (iv) and Finite Sums in a Commutative Ring and in an Ordered Field: Distributivity, Differences, Telescoping, Constant Terms, Comparison and the Triangle Inequality (v)).

Proof of (i). Let n∈Nn\in\mathbb{N}. As n>0n>0, ∣Sn(ω)/n−μ∣≥ε|S_{n}(\omega)/n-\mu|\ge\varepsilon holds exactly when ∣Sn(ω)−nμ∣≥nε|S_{n}(\omega)-n\mu|\ge n\varepsilon, so Chebyshev's inequality (Markov's and Chebyshev's Inequalities (ii), with a=nεa=n\varepsilon) gives

P(∣Snn−μ∣≥ε)=P(∣Sn−ESn∣≥nε)≤Var⁡Snn2ε2≤cnε2.P\Bigl(\Bigl|\frac{S_{n}}{n}-\mu\Bigr|\ge\varepsilon\Bigr)=P\bigl(|S_{n}-\mathbb{E}S_{n}|\ge n\varepsilon\bigr)\le\frac{\operatorname{Var}S_{n}}{n^{2}\varepsilon^{2}}\le\frac{c}{n\varepsilon^{2}}.

Proof of (ii). Let ana_{n} be the probability bounded in (i), a real number with 0≤an≤10\le a_{n}\le1. Let η>0\eta>0 be real. By The Integers in the Real Numbers: Greatest Elements, the Integer Part, and Archimedean Consequences (vii) there is N∈NN\in\mathbb{N} with c/(ε2η)<nc/(\varepsilon^{2}\eta)<n for every n≥Nn\ge N, and for such nn, by (i),

∣an−0∣=an≤cnε2<η.|a_{n}-0|=a_{n}\le\frac{c}{n\varepsilon^{2}}<\eta.

Hence an→0a_{n}\to0 (Convergent Sequences of Real Numbers (i)).

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