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--- world:martingale [2021/03/20 08:00]
rdouc [Proof]
+++ world:martingale [2022/03/16 07:40] (current)
@@ Line 1: / Line 1: @@
 {{page>:defs}}
-====== Statement: Martingale convergence results ======
+===== Supermartingale convergence results =====
 <WRAP center round tip 80%>
 **__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
   *   $M=\sup_{n\in\nset} \PE[(X_n)^-]<​\infty$
   *  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 finite.
+Then, almost surely,  $X_\infty=\lim_{n\to\infty} X_n$  exists and is in  $\lone$ .
 </WRAP>
@@ Line 15: / Line 16: @@
-===== Proof =====
+==== Proof ====
 Let  $a<b$  and define   $C_1=\indiacc{X_0<​a}$  and for  $n\geq 2$ ,
 $$
-C_n=\indiacc{C_{n-1}=1,X_{n-1} \leq b}+\indiacc{C_{n-1}=0,X_{n-1} \leq a}
+C_n=\indiacc{C_{n-1}=1,X_{n-1} \leq b}+\indiacc{C_{n-1}=0,X_{n-1} < a}
 $$
 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
@@ Line 26: / Line 27: @@
 $$
-Define $U_N[a,b](\omega) $the number of upcrossings of$ [a,b] $for$ (X_n)_{0\leq n \leq  } $. Then,
+Define  $U_N[a,b]$  the number of upcrossings of  $[a,b]$  for $(X_n)_{0\leq n \leq N } $. Then,
 $$
 Y_N=\sum_{k=1}^N C_k(X_k-X_{k-1}) \geq (b-a) U_N[a,b]-(X_N-a)^-
@@ Line 43: / Line 44: @@
   & \leq \sum_{a,b\in \mathbb{Q}, a<b}  \PP(U_\infty[a,b]=\infty)=0
 \end{align*}
-which completes the proof.
+which shows that  $X_\infty=\lim_{n \to \infty} X_n$  exits almost surely.
-===== Corollary =====
+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 =====
 {{anchor:submartingale:}}
@@ Line 50: / Line 57: @@
 <WRAP center round tip 80%>
-**__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
   *  $M=\sup_{n\in\nset} \PE[(X_n)^+]<​\infty$
   * 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 finite.
+Then, almost surely,  $X_\infty=\lim_{n\to\infty} X_n$  exists and is in  $\lone$ .
 </WRAP>

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