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world:forward-downward-martingale [2026/02/03 16:06]
rdouc [Proof]
world:forward-downward-martingale [2026/02/07 15:03] (current)
rdouc
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 ====== Upcrossing Inequality and Martingale Convergence ====== ====== Upcrossing Inequality and Martingale Convergence ======
- 
-**Intuition.**  ​ 
-A process cannot oscillate infinitely many times between two fixed levels $a<b$ without paying a cost.  ​ 
-Each complete upcrossing from $a$ to $b$ forces an increase of at least $b-a$.  ​ 
-The upcrossing inequality makes this idea precise and leads to almost sure convergence results for martingales and supermartingales. 
  
 ===== Doob's Upcrossing Inequality ===== ===== Doob's Upcrossing Inequality =====
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 <WRAP center round tip 80%> <WRAP center round tip 80%>
-Let $U_n[a,b]$ denote the number of completed upcrossings from $a$ to $b$ between times $0$ and $n$.  ​+Let $U_{0:n}[a,b]$ denote the number of completed upcrossings from $a$ to $b$ between times $0$ and $n$.  ​
 Then the following inequality holds: Then the following inequality holds:
 $$ $$
-Y_n \ge (b-a)\,U_n[a,b] - (X_n-a)^-.+Y_n \ge (b-a)\,U_{0:n}[a,b] - (X_n-a)^-.
 $$ $$
  
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 Summing the contributions of all completed upcrossings and accounting for the final correction yields the stated inequality. ∎ Summing the contributions of all completed upcrossings and accounting for the final correction yields the stated inequality. ∎
  
-===== Theorem ​(Doob'​s Forward Convergence Theorem) =====+===== Theorem ​(Doob'​s Forward Convergence Theorem) =====
  
 <WRAP center round todo 80%> <WRAP center round todo 80%>
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 By the upcrossing inequality, for any $a<b$, By the upcrossing inequality, for any $a<b$,
 $$ $$
-(b-a)\,​\mathbb E[U_n[a,b]]+(b-a)\,​\mathbb E[U_{0:n}[a,b]]
 \le \le
 \mathbb E[Y_n] + \mathbb E[(X_n-a)^-]. \mathbb E[Y_n] + \mathbb E[(X_n-a)^-].
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 The $L^1$ boundedness assumption implies that $\sup_n \mathbb E[(X_n-a)^-]<​\infty$. The $L^1$ boundedness assumption implies that $\sup_n \mathbb E[(X_n-a)^-]<​\infty$.
  
-Therefore,+By the monotone convergence theorem
 $$ $$
-\sup_n ​\mathbb E[U_n[a,​b]]<​\infty,​+\mathbb E[U_{0:​\infty}[a,​b]]<​\infty,​
 $$ $$
 which implies that the total number of upcrossings is almost surely finite.  ​ which implies that the total number of upcrossings is almost surely finite.  ​
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 <WRAP center round todo 80%> <WRAP center round todo 80%>
-Let $(\mathcal ​G_{-n})_{n\ge1}$ be a decreasing sequence of $\sigma$-fields such that+Let $(\mathcal ​F_{-n})_{n\ge1}$ be a decreasing sequence of $\sigma$-fields such that
 $$ $$
-\mathcal ​G_{-\infty}+\mathcal ​F_{-\infty}
 = =
-\bigcap_{k\ge1}\mathcal ​G_{-k} +\bigcap_{k\ge1}\mathcal ​F_{-k} 
-\subseteq \cdots \subseteq \mathcal ​G_{-(n+1)} \subseteq \mathcal ​G_{-n} \subseteq \cdots \subseteq \mathcal ​G_{-1}.+\subseteq \cdots \subseteq \mathcal ​F_{-(n+1)} \subseteq \mathcal ​F_{-n} \subseteq \cdots \subseteq \mathcal ​F_{-1}.
 $$ $$
  
 For any random variable $Z$ satisfying $\mathbb E[|Z|]<​\infty$,​ we have For any random variable $Z$ satisfying $\mathbb E[|Z|]<​\infty$,​ we have
 $$ $$
-\lim_{n\to\infty}\mathbb E[Z\mid\mathcal ​G_{-n}]+\lim_{n\to\infty}\mathbb E[Z\mid\mathcal ​F_{-n}]
 = =
-\mathbb E[Z\mid\mathcal ​G_{-\infty}]+\mathbb E[Z\mid\mathcal ​F_{-\infty}]
 \quad\text{almost surely}. \quad\text{almost surely}.
 $$ $$
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 Define the process Define the process
 $$ $$
-X_k := \mathbb E[Z\mid\mathcal ​G_k], \qquad -n\le k\le -1.+X_k := \mathbb E[Z\mid\mathcal ​F_k], \qquad -n\le k\le -1.
 $$ $$
 This is a martingale bounded in $L^1$. This is a martingale bounded in $L^1$.
  
-Similarly as before, we define $C_1=\mathsf{1}_{X_{-n} < a}$ . For all $-n  < k \leq -1$, we set $C_k=\mathsf{1}_{C_{k-1}=1} \mathsf{1}_{X_{k-1} \leq b}+ \mathsf{1}_{C_{k-1}=0} \mathsf{1}_{X_{k-1} <a}$. Define also $Y_k=\sum_{\ell=-n}^k C_\ell (X_\ell - X_{\ell-1})$. Then, for the same reasons, the number of upcrossing $U_n[a,b]$ between $-n$ and $-1$ can be bounded as follows:  ​+Similarly as before, we define $C_1=\mathsf{1}_{X_{-n} < a}$ . For all $-n  < k \leq -1$, we set $C_k=\mathsf{1}_{C_{k-1}=1} \mathsf{1}_{X_{k-1} \leq b}+ \mathsf{1}_{C_{k-1}=0} \mathsf{1}_{X_{k-1} <a}$. Define also $Y_k=\sum_{\ell=-n}^k C_\ell (X_\ell - X_{\ell-1})$. Then, for the same reasons, the number of upcrossing $U_{-n:-1}[a,b]$ between $-n$ and $-1$ can be bounded as follows:  ​
 $$ $$
-Y_{-1} \geq (b-a) U_n[a,b] - (X_{-1}-a)^-+Y_{-1} \geq (b-a) U_{-n:-1}[a,b] - (X_{-1}-a)^-
 $$ $$
  
 As before, the integrability of $Z$ ensures that the expected number of upcrossings is finite, which implies that $(X_k)$ converges almost surely as $k\to-\infty$. As before, the integrability of $Z$ ensures that the expected number of upcrossings is finite, which implies that $(X_k)$ converges almost surely as $k\to-\infty$.
  
-The limit must coincide with $\mathbb E[Z\mid\mathcal ​G_{-\infty}]$,​ which completes the proof. ∎+The limit must coincide with $\mathbb E[Z\mid\mathcal ​F_{-\infty}]$,​ which completes the proof. ∎ 
 + 
 +[[?​do=backlink|Linked pages]] 
  
world/forward-downward-martingale.1770131190.txt.gz · Last modified: 2026/02/03 16:06 by rdouc