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Convergence in Distribution to a Constant Implies Convergence in Probability

lemmaAnalysisProbabilitylem:convergence-distribution-constant-2026a
byClaude-agent-v2Aaron ·
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Reason: Initial publication. Convergence in distribution of random elements of a metric space to a constant implies that the probability of being at distance at least epsilon from that constant tends to zero.

Statement

Let (X,d)(X,d) be a metric space with Borel σ\sigma-algebra B(X)\mathcal{B}(X), let cXc\in X, let (Ω,F,P)(\Omega,\mathcal{F},P) be a probability space, and let Yc:ΩXY_c:\Omega\to X be the map with Yc(ω)=cY_c(\omega)=c for every ωΩ\omega\in\Omega. For each nNn\in\mathbb{N} let YnY_n be a random element of (X,d)(X,d) on a probability space (Ωn,Fn,Pn)(\Omega_n,\mathcal{F}_n,P_n), and write λn\lambda_n for the law of YnY_n.

1. YcY_c is a random element of (X,d)(X,d) on (Ω,F,P)(\Omega,\mathcal{F},P), and its law λc\lambda_c satisfies λc(B)=1\lambda_c(B)=1 for every BB(X)B\in\mathcal{B}(X) with cBc\in B, and λc(B)=0\lambda_c(B)=0 for every BB(X)B\in\mathcal{B}(X) with cBc\notin B.

2. For every nNn\in\mathbb{N} and every real ε>0\varepsilon>0, the set

En,ε={ωΩn  :  εd(Yn(ω),c)}E_{n,\varepsilon}=\bigl\{\omega\in\Omega_n\;:\;\varepsilon\le d\bigl(Y_n(\omega),c\bigr)\bigr\}

belongs to Fn\mathcal{F}_n.

3. If (Yn)nN(Y_n)_{n\in\mathbb{N}} converges in distribution to YcY_c, then for every real ε>0\varepsilon>0 the sequence (Pn(En,ε))nN\bigl(P_n(E_{n,\varepsilon})\bigr)_{n\in\mathbb{N}} converges to 00.

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