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单词 ProofOfMartingaleConvergenceTheorem
释义

proof of martingale convergence theorem


Let (Xn)n be a supermartingalePlanetmathPlanetmath such that 𝔼|Xn|M, and let a<b. We define a random variableMathworldPlanetmath counting how many times the process crosses the stripe between a and b:

Un:=max{0,r{1,,n}0s1<t1<s2<t2<<sr<trni{1,,r}:XsiaXtib}.

Obviously Un+1Un therefore U=limnUn exists almost surely. Next we will construct a new process that mirrors the movement of Xn but only if the original process is underway of going from below a to over b, and is constant otherwise. To do this let C1:=χ[X0<a], Ck:=χ[Ck-1=0Xk-1<a]+χ[Ck-1=1Xk-1b] for k2, and define Y0:=0, Yn:=k=1nCk(Xk-Xk-1). Then Yn is also a supermartingale, and the inequalityMathworldPlanetmath Yn(b-a)Un-|Xn-a| holds, which gives 0𝔼(Yn)(b-a)𝔼(Un)-𝔼|Xn-a|. After rearrangement we get

𝔼(Un)𝔼|Xn-a|b-a𝔼|Xn|+|a|b-aM+|a|b-a.

Therefore by the monotone convergence theoremMathworldPlanetmath

𝔼(U)=limn𝔼(Un)M+|a|b-a<,

which means (U=)=0. Since a and b were arbitrary X=limnXn exists almost surely. Now the Fatou lemmaMathworldPlanetmath gives

𝔼|X|=𝔼(limn|Xn|)lim infn𝔼|Xn|M<.

Thus Xn is in fact convergentMathworldPlanetmathPlanetmath almost surely, and XL1.

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