TheoremBase

Proof

Claim 1 (almost sure implies in probability). Fix ε>0\varepsilon>0 and let Ck=⋃m≥k{∣Xm−X∣≥ε}C_k=\bigcup_{m\ge k}\{|X_m-X|\ge\varepsilon\}, events by Sigma-Algebra and Measurable Space since each ∣Xm−X∣|X_m-X| is a random variable (Step 0(a) of the proof of Linearity and Monotonicity of the Lebesgue Integral). The CkC_k decrease, and every ω∈⋂kCk\omega\in\bigcap_k C_k satisfies ∣Xm(ω)−X(ω)∣≥ε|X_m(\omega)-X(\omega)|\ge\varepsilon for infinitely many mm, so Xm(ω)↛X(ω)X_m(\omega)\not\to X(\omega); hence ⋂kCk\bigcap_k C_k is contained in the complement of the convergence event of Almost Sure Convergence, Convergence in Probability, and Convergence in Distribution, which has probability 00 by hypothesis. By continuity from above of probability measures (Preliminaries of the proof of Borel-Cantelli Lemmas), P(Ck)→0P(C_k)\to 0; and monotonicity gives P(∣Xk−X∣≥ε)≤P(Ck)→0P(|X_k-X|\ge\varepsilon)\le P(C_k)\to 0.

Claim 2 (in probability implies in distribution). Fix t∈Rt\in\mathbb{R} at which FXF_X is continuous, and let ε>0\varepsilon>0. If Xm≤tX_m\le t then either X≤t+εX\le t+\varepsilon or ∣Xm−X∣>ε|X_m-X|>\varepsilon; by finite subadditivity of PP,

FXm(t) ≤ FX(t+ε)+P(∣Xm−X∣≥ε).F_{X_m}(t)\ \le\ F_X(t+\varepsilon)+P\bigl(|X_m-X|\ge\varepsilon\bigr).

Symmetrically, if X≤t−εX\le t-\varepsilon then either Xm≤tX_m\le t or ∣Xm−X∣>ε|X_m-X|>\varepsilon, so

FX(t−ε) ≤ FXm(t)+P(∣Xm−X∣≥ε).F_X(t-\varepsilon)\ \le\ F_{X_m}(t)+P\bigl(|X_m-X|\ge\varepsilon\bigr).

Letting m→∞m\to\infty with ε\varepsilon fixed, every subsequential behavior of FXm(t)F_{X_m}(t) is confined to the interval [FX(t−ε),FX(t+ε)][F_X(t-\varepsilon),F_X(t+\varepsilon)] in the sense that

FX(t−ε) ≤ lim inf⁡mFXm(t) ≤ lim sup⁡mFXm(t) ≤ FX(t+ε),F_X(t-\varepsilon)\ \le\ \liminf_m F_{X_m}(t)\ \le\ \limsup_m F_{X_m}(t)\ \le\ F_X(t+\varepsilon),

with lim inf⁡\liminf and lim sup⁡\limsup of bounded real sequences as in the proof of Dominated Convergence Theorem. Letting ε→0\varepsilon\to 0 and using continuity of FXF_X at tt, both bounds tend to FX(t)F_X(t), so FXm(t)→FX(t)F_{X_m}(t)\to F_X(t), as required by Almost Sure Convergence, Convergence in Probability, and Convergence in Distribution.

Claim 3 (convergence in distribution to a constant). Let FcF_c denote the cumulative distribution function of the constant cc, so Fc(t)=0F_c(t)=0 for t<ct<c and Fc(t)=1F_c(t)=1 for t≥ct\ge c, and FcF_c is continuous at every t≠ct\ne c. Fix ε>0\varepsilon>0. Then

P(∣Xm−c∣≥ε) ≤ P(Xm≤c−ε)+P(Xm>c+ε2) = FXm(c−ε)+1−FXm(c+ε2),P\bigl(|X_m-c|\ge\varepsilon\bigr)\ \le\ P\bigl(X_m\le c-\varepsilon\bigr)+P\bigl(X_m>c+\tfrac{\varepsilon}{2}\bigr)\ =\ F_{X_m}(c-\varepsilon)+1-F_{X_m}\bigl(c+\tfrac{\varepsilon}{2}\bigr),

using {Xm≥c+ε}⊆{Xm>c+ε/2}\{X_m\ge c+\varepsilon\}\subseteq\{X_m>c+\varepsilon/2\} and finite subadditivity. The points c−εc-\varepsilon and c+ε/2c+\varepsilon/2 are continuity points of FcF_c, so by hypothesis FXm(c−ε)→Fc(c−ε)=0F_{X_m}(c-\varepsilon)\to F_c(c-\varepsilon)=0 and FXm(c+ε/2)→Fc(c+ε/2)=1F_{X_m}(c+\varepsilon/2)\to F_c(c+\varepsilon/2)=1; hence the right side tends to 00, proving convergence in probability to cc. ■\blacksquare

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