Discrete time Martingales
Recall that the strong law of large numbers tells us, if [math](X_n)_{n\geq 1}[/math] is iid, [math]\E[\vert X_i\vert] \lt \infty[/math] and [math]\E[X_i]=\mu[/math], then
with [math]S_n=\sum_{i=1}^nX_i[/math]. We saw that the 0-1 law of Kolmogorov implied that in this case the limit, if it exists, is constant. It is of course of interest to have a framework in which the sequence of r.v.'s converges a.s. to another r.v. This can be achieved in the framework of martingales. In this chapter, we shall consider a probability space [math](\Omega,\F,\p)[/math] as well as an increasing family [math](\F_n)_{n\geq 0}[/math] of sub [math]\sigma[/math]-Algebras of [math]\F[/math], i.e. [math]\F_n\subset\F_{n+1}\subset \F[/math]. Such a sequence is called a filtration. The space [math](\Omega,\F,(\F_n)_{n\geq 0},\p)[/math] is called a filtered probability space. We shall also consider a sequence [math](X_n)_{n\geq 0}[/math] of r.v.'s. Such a sequence is generally called a stochastic process ([math]n[/math] is thought of as time). If for every [math]n\geq0[/math], [math]X_n[/math] is [math]\F_n[/math]-measurable, we say that [math](X_n)_{n\geq 0}[/math] is adapted (to the filtration [math](\F_n)_{n\geq0})[/math]. One can think of [math]\F_n[/math] as the information at time [math]n[/math] and the filtration [math](\F_n)_{n\geq0}[/math] as the flow of information in time.
Let us start with a stochastic process [math](X_n)_{n\geq 0}[/math]. We define
In general, if [math](\F_n)_{n\geq0}[/math] is a filtration, one denotes by
Let [math](\Omega,\F,(\F_n)_{n\geq0},\p)[/math] be a filtered probability space. A stochastic process [math](X_n)_{n\geq 0}[/math] is called a martingale, if
- [math]\E[\vert X_n\vert] \lt \infty[/math] for all [math]n\geq 0[/math].
- [math]X_n[/math] is [math]\F_n[/math]-measurable (adapted).
- [math]\E[X_n\mid \F_m]=X_m[/math] a.s. for all [math]m\leq n[/math].
The last point is equivalent to say
Let [math](X_n)_{n\geq0}[/math] be a sequence independent r.v.'s such that [math]\E[X_n]=0[/math] for all [math]n\geq 0[/math] (i.e. [math]X_n\in L^1(\Omega,\F,(\F_n)_{n\geq 0},\p)[/math]). Moreover, let [math]\F_n=\sigma(X_1,...,X_n)[/math] and [math]S_n=\sum_{i=1}^nX_i[/math] with [math]\F_0=\{\varnothing,\Omega\}[/math] and [math]S_0=0[/math]. Then [math](S_n)_{n\geq 0}[/math] is an [math]\F_n[/math]-martingale.
Show ProofWe need to check the assumptions for a martingale.
- The first point is clear by assumption on [math]X_1,...,X_n[/math] and linearity of the expectation.
[[math]] \E[\vert S_n\vert]\leq \sum_{i=1}\E[\vert X_i\vert] \lt \infty. [[/math]]
- It is clear that [math]S_n[/math] is [math]\F_n[/math]-measurable, since it is a function [math]X_1,...,X_n[/math], which are [math]\F_n[/math]-measurable.
- Observe that
[[math]] \begin{multline*} \E[S_{n+1}\mid \F_n]=\E[\underbrace{X_1+...+X_n}_{S_n}+X_{n+1}\mid\F_n]\\=\underbrace{\E[S_n\mid \F_n]}_{S_n}+\E[X_{n+1}\mid\F_n]=S_n+\E[X_{n+1}\mid\F_n]=S_n. \end{multline*} [[/math]]
Therefore, [math](S_n)_{n\geq0}[/math] is a martingale.
Let [math](\Omega,\F,(\F_n)_{n\geq 0},\p)[/math] be a filtered probability space and let [math]Y\in L^1(\Omega,\F,(\F_n)_{n\geq0},\p)[/math]. Define a sequence [math](X_n)_{n\geq0}[/math] by
Again, we show the assumptions for a martingale.
- Since [math]\vert X_n\vert\leq \E[\vert Y\vert\mid \F_n][/math], we get
[[math]] \E[\vert X_n\vert]\leq \E[\vert Y\vert] \lt \infty. [[/math]]
- [math]\E[Y\mid \F_n][/math] is [math]\F_n[/math]-measurable by definition.
- With the tower property, we get
[[math]] \E[X_{n+1}\mid \F_n]=\E[\underbrace{\E[Y\mid\F_{n+1}]}_{X_{n+1}}\mid\F_n]=\E[Y\mid\F_n]=X_n. [[/math]]
Therefore, [math](X_n)_{n\geq0}[/math] is a martingale.
Let [math](\Omega,\F,(\F_n)_{n\geq0},\p)[/math] be a filtered probability space. A martingale [math](X_n)_{n\geq 0}[/math] is said to be regular, if there exists a r.v. [math]Y\in L^1(\Omega,\F,(\F_n)_{n\geq0},\p)[/math] such that
Let [math](\Omega,\F,(\F_n)_{n\geq0},\p)[/math] be a filtered probability space. Let [math](X_n)_{n\geq 0}[/math] be a martingale. Then the map
is constant, i.e. for all [math]n\geq 0[/math]
By the definition of a martingale, we get
Let [math](\Omega,\F,(\F_n)_{n\geq0},\p)[/math] be a filtered probability space. A r.v. [math]T:\Omega\to\bar \N=\N\cup\{\infty\}[/math] is called a stopping time if for every [math]n\geq 0[/math]
Another, more general definition is used for continuous stochastic processes and may be given in terms of a filtration. Let [math](I,\leq )[/math] be an ordered index set (often [math]I=[0,\infty)[/math]) or a compact subset thereof, thought of as the set of possible [math]times[/math]), and let [math](\Omega,\F,(\F_n)_{n\geq 0},\p)[/math] be a filtered probability space. Then a r.v. [math]T:\Omega\to I[/math] is called a stopping time if [math]\{T\leq t\}\in\F_t[/math] for all [math]t\in I[/math]. Often, to avoid confusion, we call it a [math]\F_t[/math]-stopping time and explicitly specify the filtration. Speaking concretely, for [math]T[/math] to be a stopping time, it should be possible to decide whether or not [math]\{T\leq t\}[/math] has occurred on the basis of the knowledge of [math]\F_t[/math], i.e. [math]\{T\leq t\}[/math] is [math]\F_t[/math]-measurable.
Let [math](\Omega,\F,(\F_n)_{n\geq 0},\p)[/math] be a filtered probability space. Then
- Constant times are stopping times.
- The map
[[math]] T:(\Omega,\F)\to(\bar\N,\mathcal{P}(\bar\N)) [[/math]]is a stopping time if and only if [math]\{T\leq n\}\in\F_n[/math] for all [math]n\geq 0[/math].
- If [math]S[/math] and [math]T[/math] are stopping times, then [math]S\land T[/math], [math]S\lor T[/math] and [math]S+T[/math] are also stopping times.
- Let [math](T_n)_{n\geq 0}[/math] be a sequence of stopping times. Then [math]\sup_n T[/math], [math]\inf_n T_n[/math], [math]\liminf_n T_n[/math] and [math]\limsup_n T_n[/math] are also stopping times.
- Let [math](X_n)_{n\geq 0}[/math] be a sequence of adapted r.v.'s with values in some measure space [math](E,\mathcal{E})[/math] and let [math]H\in\mathcal{E}[/math]. Then, with the convention that [math]\inf\varnothing=\infty[/math],
[[math]] D_H=\inf\{ n\in\N\mid X_n\in H\} [[/math]]is a stopping time.
We need to show all points.
- This is clear.
- Note that
[[math]] \{T=n\}=\{T\leq n\}\setminus \{T\leq n-1\}\in\F_n [[/math]]and conversely,[[math]] \{T\leq n\}=\bigcup_{k=0}^n\{ T=k\}\in\F_n. [[/math]]
- Just observe that
[[math]] \{S\lor T\leq n\}=\{S\leq n\}\cap\{T\leq n\}\in\F_n [[/math]][[math]] \{S\land T\leq n\}=\{S\leq n\}\cup\{T\leq n\}\in\F_n [[/math]][[math]] \{S+T=n\}=\bigcup_{k=0}^n\underbrace{\{S=k\}}_{\in\F_k\subset\F_n}\cap\underbrace{\{T=n-k\}}_{\in\F_{n-k}\subset\F_n}\in\F_n [[/math]]
- First observe
[[math]] \{\sup_{k}T_k\leq n\}=\bigcap_k\{T_k\leq n\}\in\F_n \text{and} \{\inf_k T_k\leq n\}=\bigcup_k\{T_k\leq n\}\in\F_n. [[/math]]Now we can rewrite[[math]] \limsup_k T_k=\inf_k\sup_{m\geq k}T_m \text{and} \liminf_k T_k=\sup_k\inf_{m\leq k}T_m [[/math]]and use the relation above.
- For all [math]n\in\N[/math], we get
[[math]] \{D_H\leq n\}=\bigcup_{k=0}^n\underbrace{\{X_k\in H\}}_{\in\F_k\subset\F_n}\in\F_n [[/math]]
We say that a stopping time [math]T[/math] is bounded if there exists [math]C \gt 0[/math] such that for all [math]\omega\in\Omega[/math]
Let [math](\Omega,\F,(\F_n)_{n\geq 0},\p)[/math] be a filtered probability space. Let [math]T[/math] be a bounded stopping time and let [math](X_n)_{n\geq 0}[/math] be a martingale. Then we have
Assume that [math]T\leq N\in\N[/math]. Then
Let [math](\Omega,\F,(\F_n)_{n\geq0},\p)[/math] be a filtered probability space. Let [math]T[/math] be a stopping time for [math](\F_n)_{n\geq0}[/math]. We call the [math]\sigma[/math]-Algebra of events prior of [math]T[/math] and write [math]\F_T[/math] for the [math]\sigma[/math]-Algebra
We need to show that [math]\F_T[/math] is indeed a [math]\sigma[/math]-Algebra.
If [math]T[/math] is a stopping time, [math]\F_T[/math] is a [math]\sigma[/math]-Algebra.
Show ProofIt's clear that for a filtered probability space [math](\Omega,\F,(\F_n)_{n\geq0},\p)[/math], [math]\Omega\in\F_T[/math]. If [math]A\in\F_T[/math], then
If [math]T=n_0[/math] is constant, then [math]\F_T=\F_{n_0}[/math].
Let [math]S[/math] and [math]T[/math] be two stopping times.
- If [math]S\leq T[/math], then [math]\F_S\subset \F_T[/math].
- [math]\F_{S\land T}=\F_S\cap\F_T[/math].
- [math]\{S\leq T\}[/math], [math]\{S=T\}[/math] and [math]\{S \lt T\}[/math] are [math]\F_S\cap\F_T[/math]-measurable.
We need to show all three points.
- For [math]n\in\N[/math] and [math]A\in\F_S[/math] we get
[[math]] A\cap\{T\leq n\}= A\cap\underbrace{\{S\leq n\}\cap\{T\leq n\}}_{\{T\leq n\}}=(A\cap\{S\leq n\})\cap\underbrace{\{T\leq n\}}_{\in\F_n}\in\F_n. [[/math]]Therefore [math]A\in\F_T[/math].
- Since [math]S\land T\leq S[/math], we get by [math](i)[/math] that [math]\F_{S\land T}\subset\F_S[/math] and similarly that [math]\F_{S\land T}\subset\F_T[/math]. Let now [math]A\in \F_S\cap\F_T[/math]. Then
[[math]] A\cap\{S\land T\leq n\}=\left(\underbrace{A\cap \{ S\leq n\}}_{\in\F_n,(\text{since $A\in\F_S$})}\right)\cup \left( \underbrace{A\cap\{ T\leq n\}}_{\in\F_n,(\text{since $A\in\F_T$})}\right)\in\F_n. [[/math]]Therefore [math]A\in\F_{S\land T}[/math].
- Note that
[[math]] \{S\leq T\}\cap\{T=n\}=\{S\leq n\}\cap \{ T=n\}\in\F_n. [[/math]]Therefore [math]\{S\leq T\}\in\F_T[/math]. Note also that[[math]] \{S \lt T\}\cap\{T=n\}=\{S \lt n\}\cap \{T=n\}\in\F_n. [[/math]]Therefore [math]\{S \lt T\}\in\F_T[/math]. Finally, note that[[math]] \{S=T\}\cap\{T=n\}=\{S=n\}\cap\{T=n\}\in\F_n. [[/math]]Thus [math]\{S=T\}\in\F_T[/math]. Similarly one can show that these events are also [math]\F_S[/math]-measurable.
Let [math](\Omega,\F,(\F_n)_{n\geq0},\p)[/math] be a filtered probability space. Let [math](X_n)_{n\geq 0}[/math] be a stochastic process, which is adapted, i.e. [math]X_n[/math] is [math]\F_n[/math]-measurable for all [math]n\geq 0[/math]. Let [math]T[/math] be a finite stopping time, i.e. [math]T \lt \infty[/math] a.s., such that [math]X_T[/math] is well defined. Then [math]X_T[/math] is [math]\F_T[/math]-measurable.
Show ProofLet [math]\Lambda\in\B(\R)[/math] be a Borel measurable set. We want to show that
Let [math](\Omega,\F,(\F_n)_{n\geq0},\p)[/math] be a filtered probability space. Let [math](X_n)_{n\geq 0}[/math] be a martingale and let [math]S[/math] and [math]T[/math] be two bounded stopping times such that [math]S\leq T[/math] a.s. Then we have
Since we assume that [math]T\leq C\in\N[/math], we note that
Let [math](\Omega,\F,(\F_n)_{n\geq0},\p)[/math] be a filtered probability space. Let [math](X_n)_{n\geq 0}[/math] be a stochastic process such that for all [math]n\geq 0[/math]
Let [math]0\leq m \lt n \lt \infty[/math] and [math]\Lambda\in\F_m[/math]. Define for all [math]\omega\in\Omega[/math]
General references
Moshayedi, Nima (2020). "Lectures on Probability Theory". arXiv:2010.16280 [math.PR].