guide:B79b5d9111: Difference between revisions
No edit summary |
mNo edit summary |
||
(One intermediate revision by one other user not shown) | |||
Line 1: | Line 1: | ||
<div class="d-none"><math> | |||
\newcommand{\R}{\mathbb{R}} | |||
\newcommand{\A}{\mathcal{A}} | |||
\newcommand{\B}{\mathcal{B}} | |||
\newcommand{\N}{\mathbb{N}} | |||
\newcommand{\C}{\mathbb{C}} | |||
\newcommand{\Rbar}{\overline{\mathbb{R}}} | |||
\newcommand{\Bbar}{\overline{\mathcal{B}}} | |||
\newcommand{\Q}{\mathbb{Q}} | |||
\newcommand{\E}{\mathbb{E}} | |||
\newcommand{\p}{\mathbb{P}} | |||
\newcommand{\one}{\mathds{1}} | |||
\newcommand{\0}{\mathcal{O}} | |||
\newcommand{\mat}{\textnormal{Mat}} | |||
\newcommand{\sign}{\textnormal{sign}} | |||
\newcommand{\CP}{\mathcal{P}} | |||
\newcommand{\CT}{\mathcal{T}} | |||
\newcommand{\CY}{\mathcal{Y}} | |||
\newcommand{\F}{\mathcal{F}} | |||
\newcommand{\mathds}{\mathbb}</math></div> | |||
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] < \infty</math> and <math>\E[X_i]=\mu</math>, then | |||
<math display="block"> | |||
\frac{1}{n}S_n\xrightarrow{n\to\infty\atop a.s.}\mu, | |||
</math> | |||
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. | |||
{{alert-info | | |||
Let us start with a stochastic process <math>(X_n)_{n\geq 0}</math>. We define | |||
<math display="block"> | |||
\F_n=\sigma(X_0,...,X_n)=\sigma(X_k\mid0\leq k\leq n). | |||
</math> | |||
By construction, <math>\mathcal{F}_n\subset\F_{n+1}</math> and <math>(X_n)_{n\geq0}</math> is <math>(\F_n)_{n\geq0}</math>-adapted. In this case <math>(\F_n)_{n\geq 0}</math> is called the ''natural filtration'' of <math>(X_n)_{n\geq 0}</math>. | |||
}} | |||
{{alert-info | | |||
In general, if <math>(\F_n)_{n\geq0}</math> is a filtration, one denotes by | |||
<math display="block"> | |||
\F_\infty=\bigvee_{n\geq0}\F_n=\sigma\left(\bigcup_{n\geq 0}\F_n\right). | |||
</math> | |||
the tail <math>\sigma</math>-Algebra. | |||
}} | |||
{{definitioncard|Martingale| | |||
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 | |||
<ul style{{=}}"list-style-type:lower-roman"><li><math>\E[\vert X_n\vert] < \infty</math> for all <math>n\geq 0</math>. | |||
</li> | |||
<li><math>X_n</math> is <math>\F_n</math>-measurable (adapted). | |||
</li> | |||
<li><math>\E[X_n\mid \F_m]=X_m</math> a.s. for all <math>m\leq n</math>. | |||
</li> | |||
</ul> | |||
The last point is equivalent to say | |||
<math display="block"> | |||
\E[X_{n+1}\mid\F_n]=X_na.s., | |||
</math> | |||
which can be obtained by using the tower property and induction. | |||
}} | |||
{{proofcard|Example|Exc3|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.|We need to check the assumptions for a martingale. | |||
<ul style{{=}}"list-style-type:lower-roman"><li>The first point is clear by assumption on <math>X_1,...,X_n</math> and linearity of the expectation. | |||
<math display="block"> | |||
\E[\vert S_n\vert]\leq \sum_{i=1}\E[\vert X_i\vert] < \infty. | |||
</math> | |||
</li> | |||
<li>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. | |||
</li> | |||
<li>Observe that | |||
<math display="block"> | |||
\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> | |||
</li> | |||
</ul> | |||
Therefore, <math>(S_n)_{n\geq0}</math> is a martingale.|}} | |||
{{proofcard|Example|ex1|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 | |||
<math display="block"> | |||
X_n:=\E[Y\mid\F_n]. | |||
</math> | |||
Then <math>(X_n)_{n\geq 0}</math> is an <math>\F_n</math>-martingale.|Again, we show the assumptions for a martingale. | |||
<ul style{{=}}"list-style-type:lower-roman"><li>Since <math>\vert X_n\vert\leq \E[\vert Y\vert\mid \F_n]</math>, we get | |||
<math display="block"> | |||
\E[\vert X_n\vert]\leq \E[\vert Y\vert] < \infty. | |||
</math> | |||
</li> | |||
<li><math>\E[Y\mid \F_n]</math> is <math>\F_n</math>-measurable by definition.</li> | |||
<li>With the tower property, we get | |||
<math display="block"> | |||
\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> | |||
</li> | |||
</ul> | |||
Therefore, <math>(X_n)_{n\geq0}</math> is a martingale.}} | |||
{{definitioncard|Regularity of Martingales|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 | |||
<math display="block"> | |||
X_n=\E[Y\mid\F_n] | |||
</math> | |||
for all <math>n\geq 0</math>. | |||
}} | |||
{{proofcard|Proposition|prop-1|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 | |||
<math display="block"> | |||
n\mapsto \E[X_n] | |||
</math> | |||
is constant, i.e. for all <math>n\geq 0</math> | |||
<math display="block"> | |||
\E[X_n]=\E[X_0]. | |||
</math>|By the definition of a martingale, we get | |||
<math display="block"> | |||
\E[X_n]=\E[\E[X_n\mid \F_0]]=X_0. | |||
</math>}} | |||
{{definitioncard|Discrete Stopping time| | |||
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> | |||
<math display="block"> | |||
\{T\leq n\}\in\F_n. | |||
</math> | |||
}} | |||
{{alert-info | | |||
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. | |||
}} | |||
{{proofcard|Proposition|prop-2|Let <math>(\Omega,\F,(\F_n)_{n\geq 0},\p)</math> be a filtered probability space. Then | |||
<ul style{{=}}"list-style-type:lower-roman"><li>Constant times are stopping times. | |||
</li> | |||
<li>The map | |||
<math display="block"> | |||
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>. | |||
</li> | |||
<li>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. | |||
</li> | |||
<li>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. | |||
</li> | |||
<li>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 display="block"> | |||
D_H=\inf\{ n\in\N\mid X_n\in H\} | |||
</math> | |||
is a stopping time. | |||
</li> | |||
</ul>|We need to show all points. | |||
<ul style{{=}}"list-style-type:lower-roman"><li>This is clear. | |||
</li> | |||
<li>Note that | |||
<math display="block"> | |||
\{T=n\}=\{T\leq n\}\setminus \{T\leq n-1\}\in\F_n | |||
</math> | |||
and conversely, | |||
<math display="block"> | |||
\{T\leq n\}=\bigcup_{k=0}^n\{ T=k\}\in\F_n. | |||
</math> | |||
</li> | |||
<li> Just observe that | |||
<math display="block"> | |||
\{S\lor T\leq n\}=\{S\leq n\}\cap\{T\leq n\}\in\F_n | |||
</math> | |||
<math display="block"> | |||
\{S\land T\leq n\}=\{S\leq n\}\cup\{T\leq n\}\in\F_n | |||
</math> | |||
<math display="block"> | |||
\{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> | |||
</li> | |||
<li>First observe | |||
<math display="block"> | |||
\{\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 display="block"> | |||
\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. | |||
</li> | |||
<li>For all <math>n\in\N</math>, we get | |||
<math display="block"> | |||
\{D_H\leq n\}=\bigcup_{k=0}^n\underbrace{\{X_k\in H\}}_{\in\F_k\subset\F_n}\in\F_n | |||
</math> | |||
</li> | |||
</ul>}} | |||
{{alert-info | | |||
We say that a stopping time <math>T</math> is bounded if there exists <math>C > 0</math> such that for all <math>\omega\in\Omega</math> | |||
<math display="block"> | |||
T(\omega)\leq C | |||
</math> | |||
Without loss of generality, we can always assume that <math>C\in\N</math>. In this case we shall denote by <math>X_T</math>, the r.v. given by | |||
<math display="block"> | |||
X_T(\omega)=X_{T(\omega)}(\omega)=\sum_{n=0}^\infty X_n(\omega)\one_{\{T(\omega)=n\}}. | |||
</math> | |||
Note that the sum on the right hand side perfectly defined since <math>T(\omega)</math> is bounded. | |||
}} | |||
{{proofcard|Theorem|thm-1|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 | |||
<math display="block"> | |||
\E[X_T]=\E[X_0]. | |||
</math>|Assume that <math>T\leq N\in\N</math>. Then | |||
<math display="block"> | |||
\begin{align*} | |||
\E[X_T]&=\E\left[\sum_{n=0}^\infty X_n\one_{\{T=n\}}\right]=\E\left[\sum_{n=0}^N X_n\one_{\{T=n\}}\right]=\sum_{n=0}^N\E[X_n\one_{\{T=n\}}]=\sum_{n=0}^N\E[\E[X_N\mid\F_n]\one_{\{T=n\}}]\\ | |||
&=\sum_{n=0}^N\E[\E[X_n\one_{\{T=n\}}\mid\F_n]]=\sum_{n=0}^N\E[X_n\one_{\{T=n\}}]=\E\left[X_n\sum_{n=0}^N\one_{\{T=n\}}\right]=\E[X_n]=\E[X_0] | |||
\end{align*} | |||
</math>}} | |||
{{definitioncard|Stopping time <math>\sigma</math>-Algebra| | |||
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 | |||
<math display="block"> | |||
\F_T=\{A\in\F\mid A\cap\{T\leq n\}\in\F_n,\forall n\geq0\}. | |||
</math> | |||
}} | |||
{{alert-info | | |||
We need to show that <math>\F_T</math> is indeed a <math>\sigma</math>-Algebra. | |||
}} | |||
{{proofcard|Proposition|prop-3|If <math>T</math> is a stopping time, <math>\F_T</math> is a <math>\sigma</math>-Algebra.|It'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 | |||
<math display="block"> | |||
A^C\cap \{T\leq n\}=\underbrace{\{T\leq n\}}_{\in\F_n}\setminus (\underbrace{A\cap \{T\leq n\}}_{\in\F_n})\in\F_n | |||
</math> | |||
and hence <math>A^C\in\F_n</math>. If <math>(A_i)_{i\geq0}\in\F_T</math>, then | |||
<math display="block"> | |||
\bigcup_{i\geq 0}A_i\cap \{T\leq n\}=\bigcup_{i=1}^\infty A_i\cap \underbrace{\{T\leq n\}}_{\in\F_n}. | |||
</math> | |||
Hence <math>\bigcup_{i\geq 0}A_i\in\F_T</math>. Therefore <math>\F_T</math> is a <math>\sigma</math>-Algebra.}} | |||
{{alert-info | | |||
If <math>T=n_0</math> is constant, then <math>\F_T=\F_{n_0}</math>. | |||
}} | |||
{{proofcard|Proposition|prop-4|Let <math>S</math> and <math>T</math> be two stopping times. | |||
<ul style{{=}}"list-style-type:lower-roman"><li>If <math>S\leq T</math>, then <math>\F_S\subset \F_T</math>. | |||
</li> | |||
<li><math>\F_{S\land T}=\F_S\cap\F_T</math>. | |||
</li> | |||
<li><math>\{S\leq T\}</math>, <math>\{S=T\}</math> and <math>\{S < T\}</math> are <math>\F_S\cap\F_T</math>-measurable. | |||
</li> | |||
</ul> | |||
|We need to show all three points. | |||
<ul style{{=}}"list-style-type:lower-roman"><li>For <math>n\in\N</math> and <math>A\in\F_S</math> we get | |||
<math display="block"> | |||
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>. | |||
</li> | |||
<li>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 display="block"> | |||
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>. | |||
</li> | |||
<li>Note that | |||
<math display="block"> | |||
\{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 display="block"> | |||
\{S < T\}\cap\{T=n\}=\{S < n\}\cap \{T=n\}\in\F_n. | |||
</math> | |||
Therefore <math>\{S < T\}\in\F_T</math>. Finally, note that | |||
<math display="block"> | |||
\{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. | |||
</li> | |||
</ul>}} | |||
{{proofcard|Proposition|prop-5|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 < \infty</math> a.s., such that <math>X_T</math> is well defined. Then <math>X_T</math> is <math>\F_T</math>-measurable.|Let <math>\Lambda\in\B(\R)</math> be a Borel measurable set. We want to show that | |||
<math display="block"> | |||
\{X_T\in\Lambda\}\in\F_T, | |||
</math> | |||
that is, for all <math>n\geq 0</math> | |||
<math display="block"> | |||
\{X_T\in\Lambda\}\cap\{T\leq n\}\in\F_n. | |||
</math> | |||
Observe that | |||
<math display="block"> | |||
\{X_T\in\Lambda\}\cap\{T\leq n\}=\bigcup_{k=1}^n\{X_T\in \Lambda\}\cap\{T\leq k\}=\bigcup_{k=1}^n\underbrace{\{X_k\in\Lambda\}}_{\in\F_k\subset\F_n}\cap\underbrace{\{T=k\}}_{\in\F_k\subset\F_n}, | |||
</math> | |||
which implies that <math>\{X_T\in\Lambda\}\cap\{T\leq n\}\in\F_n</math> and the claim follows.}} | |||
{{proofcard|Theorem|thm-2|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 | |||
<math display="block"> | |||
\E[X_T\mid \F_S]=X_Sa.s. | |||
</math>|Since we assume that <math>T\leq C\in\N</math>, we note that | |||
<math display="block"> | |||
\vert X_T\vert\leq \sum_{i=0}^C\vert X_i\vert\in L^1(\Omega,\F,(\F_n)_{n\geq 0},\p). | |||
</math> | |||
Let now <math>A\in\F_S</math>. We need to show that | |||
<math display="block"> | |||
\E[X_T\one_A]=\E[X_S\one_A]. | |||
</math> | |||
Let us define the random time | |||
<math display="block"> | |||
R(\omega)=S(\omega)\one_A(\omega)+T(\omega)\one_{A^C}(\omega). | |||
</math> | |||
We thus note that <math>R</math> is a stopping time. Indeed, we have | |||
<math display="block"> | |||
\{R\leq n\}=(\underbrace{A\cap \{S\leq n\}}_{\in\F_n})\cup (\underbrace{A^C\cap \{T\leq n\}}_{\in\F_n}). | |||
</math> | |||
Consequently, since <math>S,T</math> and <math>R</math> are bounded, we have | |||
<math display="block"> | |||
\E[X_S]=\E[X_T]=\E[X_R]=\E[X_0]. | |||
</math> | |||
Therefore we get | |||
<math display="block"> | |||
\E[X_R]=\E[X_S\one_A+X_T\one_{A^C}]\text{and}\E[X_T]=\E[X_T\one_A+X_T\one_{A^C}] | |||
</math> | |||
and thus | |||
<math display="block"> | |||
\E[X_S\one_A]=\E[X_T\one_A]. | |||
</math> | |||
Moreover, since <math>X_S</math> is <math>\F_S</math>-measurable, we conclude that | |||
<math display="block"> | |||
\E[X_T\mid \F_S]=X_Sa.s. | |||
</math>}} | |||
{{proofcard|Proposition|prop-6|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> | |||
<math display="block"> | |||
\E[\vert X_n\vert] < \infty | |||
</math> | |||
and with <math>X_n</math> being <math>\F_n</math>-measurable. If for all bounded stopping times <math>T</math>, we have | |||
<math display="block"> | |||
\E[X_T]=\E[X_0], | |||
</math> | |||
then <math>(X_n)_{n\geq 0}</math> is a martingale.|Let <math>0\leq m < n < \infty</math> and <math>\Lambda\in\F_m</math>. Define for all <math>\omega\in\Omega</math> | |||
<math display="block"> | |||
T(\omega)=m\one_{\Lambda^C}(\omega)+n\one_\Lambda(\omega). | |||
</math> | |||
Then <math>T</math> is a stopping time. Therefore | |||
<math display="block"> | |||
\E[X_0]=\E[X_T]=\E[X_m\one_{\Lambda^C}+X_n\one_\Lambda]=\E[X_m]. | |||
</math> | |||
Hence we get | |||
<math display="block"> | |||
\E[X_m\one_\Lambda]=\E[X_n\one_\Lambda] | |||
</math> | |||
and thus | |||
<math display="block"> | |||
\E[X_n\mid \F_m]=X_ma.s. | |||
</math>}} | |||
==General references== | |||
{{cite arXiv|last=Moshayedi|first=Nima|year=2020|title=Lectures on Probability Theory|eprint=2010.16280|class=math.PR}} |
Latest revision as of 22:14, 8 May 2024
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].