If L1 Convergent, Then Martingale is Closable
📂Probability TheoryIf L1 Convergent, Then Martingale is Closable
Theorem
Given a probability space (Ω,F,P) and a martingale {(Xn,Fn)}, if a stochastic process {(Xn,Fn)} converges to a random variable Y through L1, then {(Xn,Fn):n=1,⋯,∞} is a closable martingale.
Description
Even if Xn converges to Y through L1 and almost surely converges to X∞, it cannot be guaranteed that Y and X∞ have some relation.
Xn→L1Y∧Xn→a.s.X∞⇏Y=X∞
In formula terms, it can be restated as above, and during the proof, it is verified that in the case of martingales, Y=a.s.X∞ holds.
Proof
Part 1. E∣Xn∣→E∣Y∣
It is said that {Xn} converges to some random variable Y through L1, as follows.
n→∞lim∫Ω∣Xn−Y∣dP=n→∞limE∣Xn−Y∣=0
Without loss of generality, since ∣∣a∣−∣b∣∣≤∣a−b∣,
∣E∣Xn∣−E∣Y∣∣≤≤≤∣E(∣Xn∣−∣Y∣∣)E∣∣Xn∣−∣Y∣∣E∣Xn−Y∣
That is, when n→∞, E∣Xn∣→E∣Y∣ holds.
Part 2. Y=a.s.X∞
Submartingale Convergence Theorem: Given a probability space (Ω,F,P) and a submartingale {(Xn,Fn)}.
If it is assumed that n∈NsupEXn+<∞, then Xn almost surely converges to some random variable X∞:Ω→R, and EX∞<EX∞+<∞
Since it is E∣Xn∣→E∣Y∣, it becomes n∈NsupE∣Xn∣<∞. Of course, EXn+≤E∣Xn∣<∞ and since a martingale is a submartingale, according to the submartingale convergence theorem, the stochastic process {Xn} almost surely converges to some random variable X∞. Hence, Y=a.s.X∞ holds.
Part 3. E(X∞∣Fn)=Xn
Without loss of generality, since Z≥0∧EZ=0⟹Z≤0, it is sufficient to show that E∣E(X∞∣Fn)−Xn∣=0 holds. Considering m>n,
E∣E(X∞∣Fn)−Xn∣===≤=E∣E(X∞∣Fn)−E(Xn∣Fn)∣E∣E(X∞∣Fn)−E(Xm∣Fn)∣E∣E(X∞−Xm∣Fn)∣E[E(∣X∞−Xm∣∣Fn)]E∣X∞−Xm∣
Applying m→∞lim to both sides and according to Parts 1 and 2,
E∣E(X∞∣Fn)−Xn∣≤≤=m→∞limE∣E(X∞∣Fn)−Xn∣m→∞limE∣X∞−Xm∣0
Therefore, it is confirmed that {(Xn,Fn):n=1,⋯,∞} is a closable martingale.
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