Welcome to Randal Douc's wiki

A collaborative site on maths but not only!

User Tools

Site Tools


world:forward-downward-martingale

Differences

This shows you the differences between two versions of the page.

Link to this comparison view

Both sides previous revision Previous revision
Next revision
Previous revision
world:forward-downward-martingale [2026/02/03 15:58]
rdouc [Proof]
world:forward-downward-martingale [2026/02/07 15:03] (current)
rdouc
Line 1: Line 1:
 ====== Upcrossing Inequality and Martingale Convergence ====== ====== Upcrossing Inequality and Martingale Convergence ======
  
-**Intuition.** ​  +===== Doob's Upcrossing Inequality =====
-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. +
- +
-===== Theorem 1 (Doob's Upcrossing Inequality=====+
  
 Let $a<b$ and let $(X_k)_{k\ge0}$ be a real-valued process.  ​ Let $a<b$ and let $(X_k)_{k\ge0}$ be a real-valued process.  ​
Line 34: Line 29:
   * The process $Y_k$ accumulates only the increments of $X$ occurring during these upcrossing phases.   * The process $Y_k$ accumulates only the increments of $X$ occurring during these upcrossing phases.
  
-Let $U_n[a,b]$ denote the number of completed upcrossings from $a$ to $b$ between times $0$ and $n$.  ​+<WRAP center round tip 80%> 
 +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)^-.
 $$ $$
 +
 +</​WRAP>​
  
 ===== Proof ===== ===== Proof =====
Line 49: Line 47:
 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%>
 Let $(X_n)_{n\ge0}$ be a supermartingale such that Let $(X_n)_{n\ge0}$ be a supermartingale such that
 $$ $$
Line 60: Line 59:
 $$ $$
 exists almost surely. exists almost surely.
 +
 +</​WRAP>​
  
 ===== Proof ===== ===== Proof =====
Line 65: Line 66:
 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)^-].
Line 73: Line 74:
 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.  ​
Line 86: Line 87:
 The conditional expectations stabilize and converge to the conditional expectation with respect to the remaining common information. The conditional expectations stabilize and converge to the conditional expectation with respect to the remaining common information.
  
-===== Theorem ​(Lévy'​s Downward Convergence Theorem) =====+===== Theorem ​(Lévy'​s Downward Convergence Theorem) =====
  
-Let $(\mathcal ​G_{-n})_{n\ge1}$ be a decreasing sequence of $\sigma$-fields such that+<WRAP center round todo 80%> 
 +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}.
 $$ $$
 +
 +</​WRAP>​
  
 ===== Proof ===== ===== Proof =====
Line 108: Line 112:
 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 exactly ​the same reasons, ​ +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:  ​
-The number of upcrossing $U_n[a,b]$ between $-n$ and $-1$ can be bounded as follows:  ​+
 $$ $$
-Y_n \geq (b-a) U_n[a,b] - (X_n-a)^-+Y_{-1} ​\geq (b-a) U_{-n:-1}[a,b] - (X_{-1}-a)^-
 $$ $$
  
-Applying ​the same upcrossing construction to $(X_k)on the interval $[-n,-1]$, we obtain for any $a<b$+As before, ​the integrability of $Zensures that the expected number of upcrossings is finitewhich implies that $(X_k)converges almost surely as $k\to-\infty$.
-$$ +
-Y_n \ge (b-a)\,U_n[a,b] - (X_n-a)^-. +
-$$+
  
-As before, the integrability of $Z$ ensures that the expected number of upcrossings is finite, which implies that $(X_k)$ converges almost surely as $k\to-\infty$.+The limit must coincide with $\mathbb E[Z\mid\mathcal F_{-\infty}]$, which completes the proof∎ 
 + 
 +[[?​do=backlink|Linked pages]]
  
-The limit must coincide with $\mathbb E[Z\mid\mathcal G_{-\infty}]$,​ which completes the proof. ∎ 
  
world/forward-downward-martingale.1770130733.txt.gz · Last modified: 2026/02/03 15:58 by rdouc