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world:martingale [2021/03/20 09:04]
rdouc [Proof]
world:martingale [2022/03/16 07:40] (current)
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 ===== Supermartingale convergence results ===== ===== Supermartingale convergence results =====
 +
  
 <WRAP center round tip 80%> <WRAP center round tip 80%>
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 Let $a<b$ and define ​ $C_1=\indiacc{X_0<​a}$ and for $n\geq 2$,  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} ​\leq a}+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 ​ 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 ​
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 Moreover, $\PE[X_0] \geq \PE[X_n]=\PE[X^+_n]-\PE[X^-_n]$ so that 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]} ​ \geq \PE[X_0] +2 \sup_n \PE[X^-_n] \leq \PE[X_0] +2 M<​\infty  ​+\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 $\PE[|X_\infty|]=\PE[\liminf_{n}|X_\infty|] \leq \liminf_{n} \PE[|X_\infty|] \leq \sup_n ​  ​\PE[|X_n|]<​\infty$. The proof is completed.+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 ===== ===== Corollary: Submartingale convergence results =====
    
world/martingale.1616227488.txt.gz ยท Last modified: 2022/03/16 01:37 (external edit)