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world:martingale
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Supermartingale convergence results
Theorem. Let \mcf=(\mcf_n)_{n\in\nset} be a filtration and let \seq{X_n}{n\in\nset} be an \mcf-adapted sequence of \lone-random variables such that
- M=\sup_{n\in\nset} \PE[(X_n)^-]<\infty
- for all n\geq 1, we have \PE[X_{n}|\mcf_{n-1}]\leq X_{n-1}.
that is, \seq{X_n}{n\in\nset} is a (\mcf_n)_{n\in\nset}-supermartingale, with negative part bounded in \lone. Then, almost surely, X_\infty=\lim_{n\to\infty} X_n exists and is in \lone.
Proof
Let a<b and define C_1=\indiacc{X_0<a} and for n\geq 2, C_n=\indiacc{C_{n-1}=1,X_{n-1} \leq b}+\indiacc{C_{n-1}=0,X_{n-1} < a} In words, the first time C_n flags 1 is when X_{n-1}<a. Then it flags 1 until X_{n-1} goes above b. Then it flags 0 until X_n goes below a. So consecutive sequences of C_n=1 are linked with upcrossings of [a,b] for (X_n) . Now, define Y_n=\sum_{k=1}^n C_k(X_k-X_{k-1}).
Define U_N[a,b] the number of upcrossings of [a,b] for (X_n)_{0\leq n \leq N } . Then, Y_N=\sum_{k=1}^N C_k(X_k-X_{k-1}) \geq (b-a) U_N[a,b]-(X_N-a)^- From the fact that \seq{X_n}{n\in\nset} is a \mcf-supermartingale and (C_n)_{n\in\nset} is \mcf-previsible, we deduce that \seq{Y_n}{n\in\nset} is also a \mcf-supermartingale so that \PE[Y_N]\leq 0. Then, (b-a) \PE[U_N[a,b]] \leq \PE[(X_N-a)^-]=\PE[(a-X_N)^+] \leq a^+ + \PE[(-X_N)^+]\leq a^+ + M Letting N goes to infinity, the monotone convergence theorem yields: (b-a) \PE[U_\infty[a,b]] \leq a^+ + M and thus, \PP(U_\infty[a,b]<\infty)=1 for all a<b. Now, \begin{align*} \PP(\liminf_n X_n<\limsup_n X_n) &\leq \sum_{a,b\in \mathbb{Q}, a<b} \PP\lr{\liminf_n X_n <a<b<\limsup_n X_n}\\ & \leq \sum_{a,b\in \mathbb{Q}, a<b} \PP(U_\infty[a,b]=\infty)=0 \end{align*} which shows that X_\infty=\lim_{n \to \infty} X_n exits almost surely.
Moreover, \PE[X_0] \geq \PE[X_n]=\PE[X^+_n]-\PE[X^-_n] so that \sup_n \PE[|X_n|]=\sup_n \lr{\PE[X^+_n]+\PE[X^-_n]} \leq \PE[X_0] +2 \sup_n \PE[X^-_n] \leq \PE[X_0] +2 M<\infty which implies by Fatou's lemma that \PE[|X_\infty|]=\PE[\liminf_{n}|X_n|] \leq \liminf_{n} \PE[|X_n|] \leq \sup_n \PE[|X_n|]<\infty. The proof is completed.
Corollary: Submartingale convergence results
As a consequence, the conclusion also holds if \seq{X_n}{n\in\nset} is a \mcf-submartingale, with positive part bounded in \lone : indeed, we only need to apply the previous result to -X_n.
Corollary Assume that \mcf=(\mcf_n)_{n\in\nset} is a filtration and let \seq{X_n}{n\in\nset} be an \mcf-adapted sequence of \lone random variables such that
- M=\sup_{n\in\nset} \PE[(X_n)^+]<\infty
- for all n\geq 1, we have \PE[X_{n}|\mcf_{n-1}]\geq X_{n-1}.
that is, \seq{X_n}{n\in\nset} is a \mcf-submartingale, with positive part bounded in \lone. Then, almost surely, X_\infty=\lim_{n\to\infty} X_n exists and is in \lone.
world/martingale.1616888974.txt.gz · Last modified: 2022/03/16 01:37 (external edit)
