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world:martingale [2021/03/20 08:46]
rdouc [Proof]
world:martingale [2022/03/16 07:40] (current)
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 {{page>:​defs}} {{page>:​defs}}
  
-====== Statement: Martingale ​convergence results ======+===== Supermartingale ​convergence results ===== 
  
 <WRAP center round tip 80%> <WRAP center round tip 80%>
 **__Theorem__**. ​ **__Theorem__**. ​
-Let $\mcf=(\mcf_n)_{n\in\nset}$ be a filtration and let $\seq{X_n}{n\in\nset}$ be an $\mcf$-adapted sequence such that +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 
   *  $M=\sup_{n\in\nset} \PE[(X_n)^-]<​\infty$   *  $M=\sup_{n\in\nset} \PE[(X_n)^-]<​\infty$
   *  for all $n\geq 1$, we have $\PE[X_{n}|\mcf_{n-1}]\leq X_{n-1}$. ​     *  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 $\lone$. ​ 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 finite.+Then, almost surely, $X_\infty=\lim_{n\to\infty} X_n$ exists and is in $\lone$.
  
 </​WRAP>​ </​WRAP>​
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-===== Proof =====+==== Proof ====
  
 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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   & \leq \sum_{a,​b\in \mathbb{Q}, a<​b} ​ \PP(U_\infty[a,​b]=\infty)=0   & \leq \sum_{a,​b\in \mathbb{Q}, a<​b} ​ \PP(U_\infty[a,​b]=\infty)=0
 \end{align*} \end{align*}
-which shows that $X_\infty=\lim_{n \to \infty} X_n$ exits almost surely. ​Morevoer, $\PE[X_0] \geq \PE[X_n]=\PE[X^+_n]-\PE[X^-_n]$ so that+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]} ​ \geq \PE[X_0] +2 \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 $\PE[|X_\infty|]=\PE[\liminf_{n}|X_\infty|] \leq \liminf_{n} \PE[|X_\infty| \leq \sup_n ​  ​\PE[|X_n|]<​\infty$ +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 =====+===== Corollary: Submartingale convergence results ​=====
    
 {{anchor:​submartingale:​}} {{anchor:​submartingale:​}}
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-**__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 such that +**__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 
   * $M=\sup_{n\in\nset} \PE[(X_n)^+]<​\infty$   * $M=\sup_{n\in\nset} \PE[(X_n)^+]<​\infty$
   * for all $n\geq 1$, we have $\PE[X_{n}|\mcf_{n-1}]\geq X_{n-1}$.   * 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 $\lone$. ​ 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 finite.+Then, almost surely, $X_\infty=\lim_{n\to\infty} X_n$ exists and is in $\lone$.
  
 </​WRAP>​ </​WRAP>​
        
  
world/martingale.1616226371.txt.gz · Last modified: 2022/03/16 01:37 (external edit)