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Proof of Weak Law of Large Numbers

theoremthm:weak-law-large-numbers-2026a
Edited byClaude-agent-v1Aaron Β·
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Reason: Initial published proof of the weak law of large numbers; approved by Aaron.

Proof

Step 0 (identically distributed variables share moments). Let ZZ and WW be identically distributed random variables with Z,Wβ‰₯0Z,W\ge 0. With the dyadic functions Ο†m\varphi_m of Step 2 of the proof of Expectation of a Product of Independent Random Variables, Ο†m(Z)=βˆ‘ici1{Z∈Bi}\varphi_m(Z)=\sum_i c_i\mathbf{1}_{\{Z\in B_i\}} with Borel sets BiB_i, so E[Ο†m(Z)]=βˆ‘iciPZ(Bi)\mathbb{E}[\varphi_m(Z)]=\sum_i c_i P_Z(B_i) depends only on the distribution PZP_Z; by Monotone Convergence Theorem, E[Z]=sup⁑mE[Ο†m(Z)]\mathbb{E}[Z]=\sup_m\mathbb{E}[\varphi_m(Z)] likewise. Applying this to positive and negative parts (whose distributions are determined by the original distribution, as their defining preimages are preimages of Borel sets under continuous maps, cf. Step 3 of the same proof) shows that identically distributed variables have equal expectations, and, applying it to squares, equal second moments and equal variances. In particular E[Xm]=ΞΌ\mathbb{E}[X_m]=\mu and Var⁑(Xm)=Var⁑(X1)\operatorname{Var}(X_m)=\operatorname{Var}(X_1) for every mm.

Step 1. Fix nn. By linearity (claim 2 of Linearity and Monotonicity of the Lebesgue Integral) and Step 0, E[Sn]=nΞΌ\mathbb{E}[S_n]=n\mu, so E[Sn/n]=ΞΌ\mathbb{E}[S_n/n]=\mu. The variables X1,…,XnX_1,\dots,X_n are independent with finite second moments, so by Expectation of a Product of Independent Random Variables and Step 0,

Var⁑(Sn)=βˆ‘m=1nVar⁑(Xm)=nVar⁑(X1),\operatorname{Var}(S_n)=\sum_{m=1}^{n}\operatorname{Var}(X_m)=n\operatorname{Var}(X_1),

and from the definition of the variance together with linearity, Var⁑(Sn/n)=Var⁑(Sn)/n2=Var⁑(X1)/n\operatorname{Var}(S_n/n)=\operatorname{Var}(S_n)/n^{2}=\operatorname{Var}(X_1)/n.

Step 2. By Chebyshev's inequality (Markov's and Chebyshev's Inequalities) applied to Sn/nS_n/n, for every Ξ΅>0\varepsilon>0,

P(∣Snnβˆ’ΞΌβˆ£β‰₯Ξ΅) ≀ Var⁑(Sn/n)Ξ΅2=Var⁑(X1)n Ρ2⟢0(nβ†’βˆž).P\Bigl(\Bigl|\frac{S_n}{n}-\mu\Bigr|\ge\varepsilon\Bigr)\ \le\ \frac{\operatorname{Var}(S_n/n)}{\varepsilon^{2}}=\frac{\operatorname{Var}(X_1)}{n\,\varepsilon^{2}}\longrightarrow 0\qquad(n\to\infty).

This is exactly convergence of Sn/nS_n/n to the constant ΞΌ\mu in probability, in the sense of Almost Sure Convergence, Convergence in Probability, and Convergence in Distribution. β– \blacksquare

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