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{{definitioncard|Submartingale and Supermartingale| | |||
Let <math>(\Omega,\F,(\F_n)_{n\geq0},\p)</math> be a filtered probability space. A stochastic process <math>(X_n)_{n\geq0}</math> is called a submartingale (resp. supermartingale) 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)_{n\geq 0}</math> is <math>\F_n</math>-adapted. | |||
</li> | |||
<li><math>\E[X_n\mid \F_m]\geq X_m</math> a.s. for all <math>m\leq n</math> (resp. <math>\E[X_n\mid \F_m]\leq X_m</math> a.s. for all <math>m\leq n</math>) | |||
</li> | |||
</ul>}} | |||
{{alert-info | | |||
A stochastic process <math>(X_n)_{n\geq 0}</math> is a martingale if and only if it is a submartingale and a supermartingale. A martingale is in particular a submartingale and a supermartingale. If <math>(X_n)_{n\geq0}</math> is a submartingale, then the map <math>n\mapsto \E[X_n]</math> is increasing. If <math>(X_n)_{n\geq 0}</math> is a supermartingale, then the map <math>n\mapsto \E[X_n]</math> is decreasing. | |||
}} | |||
'''Example''' | |||
Let <math>(\Omega,\F,(\F_n)_{n\geq0},\p)</math> be a filtered probability space. Let <math>S_n=\sum_{j=1}^{n}Y_j</math>, where <math>(Y_n)_{n\geq1}</math> is a sequence of iid r.v.'s. Moreover, let <math>S_0=0</math>, <math>\F_0=\{\varnothing,\Omega\}</math> and <math>\F_n=\sigma(Y_1,...,Y_n)</math>. Then we get | |||
<math display="block"> | |||
\E[S_{n+1}\mid\F_n]=S_n+\E[Y_{n+1}]. | |||
</math> | |||
If <math>\E[Y_{n+1}] > 0</math>, then <math>\E[S_{n+1}\mid\F_n]\geq S_n</math> and thus <math>(S_n)_{n\geq 0}</math> is a submartingale. On the other hand, if <math>\E[Y_{n+1}] < 0</math>, then <math>\E[S_{n+1}\mid\F_n]\leq S_n</math> and thus <math>(S_n)_{n\geq 0}</math> is a supermartingale. | |||
{{proofcard|Proposition|prop-1|Let <math>(\Omega,\F,(\F_n)_{n\geq0},\p)</math> be a filtered probability space. If <math>(M_n)_{n\geq 0}</math> is a martingale and <math>\varphi</math> is a convex function such that <math>\varphi(M_n)\in L^1(\Omega,\F,(\F_n)_{n\geq 0},\p)</math> for all <math>n\geq0</math>, then | |||
<math display="block"> | |||
(\varphi(M_n))_{n\geq 0} | |||
</math> | |||
is a submartingale. | |||
|The first two conditions for a martingale are clearly satisfied. Now for <math>m\leq n</math>, we get | |||
<math display="block"> | |||
\E[M_n\mid \F_m]=M_ma.s., | |||
</math> | |||
since <math>(M_n)_{n\geq 0}</math> is assumed to be a martingale. Hence, with Jensen's inequality, we get | |||
<math display="block"> | |||
\varphi(\E[M_n\mid\F_m])=\varphi(M_m)\leq \E[\varphi(M_n)\mid\F_m]a.s. | |||
</math>}} | |||
{{proofcard|Corollary|cor-1|Let <math>(\Omega,\F,(\F_n)_{n\geq0},\p)</math> be a filtered probability space. If <math>(M_n)_{n\geq 0}</math> is a martingale, then | |||
<ul style{{=}}"list-style-type:lower-roman"><li><math>(\vert M_n\vert)_{n\geq0}</math> and <math>(M^+_n)_{n\geq 0}</math> are submartingales. | |||
</li> | |||
<li>if for all <math>n\geq 0</math>, <math>\E[M_n^2] < \infty</math>, then <math>(M_n^2)_{n\geq 0}</math> is a submartingale. | |||
</li> | |||
</ul>|}} | |||
{{proofcard|Theorem|thm-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 submartingale and let <math>T</math> be a stopping time bounded by <math>C\in\N</math>. Then | |||
<math display="block"> | |||
\E[X_T]\leq \E[X_C]. | |||
</math> | |||
|Exercise{{efn|The proof is the same as in Theorem 7.7.}}}} | |||
{{proofcard|Theorem (Doob's decomposition)|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 submartingale. Then there exists a martingale <math>M=(M_n)_{n\geq 0}</math> with <math>M_0=0</math> and a sequence <math>A=(A_n)_{n\geq 0}</math>, such that <math>A_{n+1}\geq A_n</math> a.s. with <math>A_0=0</math> a.s., which is called an increasing process, and with <math>A_{n+1}</math> being <math>\F_n</math>-measurable, which we will call predictable, such that | |||
<math display="block"> | |||
X_n=X_0+M_0+A_n. | |||
</math> | |||
Moreover, this decomposition is a.s. unique. | |||
|Let us define <math>A_0=0</math> and for <math>n\geq 1</math> | |||
<math display="block"> | |||
A_n=\sum_{k=1}^n\E[X_k-X_{k-1}\mid\F_{k-1}]. | |||
</math> | |||
Since <math>(X_n)_{n\geq 0}</math> is a submartingale, we get | |||
<math display="block"> | |||
\E[X_k-X_{k-1}\mid\F_{k-1}]\geq 0 | |||
</math> | |||
and hence <math>A_{n+1}-A_n\geq 0</math>. Therefore <math>(A_n)_{n\geq 0}</math> is an increasing process. Moreover, from the definition of the conditional expectation, <math>A_n</math> is <math>\F_{n-1}</math>-measurable for <math>n\geq 1</math>. Thus <math>A_n</math> is predictable as well. We also note that | |||
<math display="block"> | |||
\E[X_n\mid \F_{n-1}]-X_{n-1}=\E[X_n-X_{n-1}\mid \F_{n-1}]=A_n-A_{n-1}. | |||
</math> | |||
Hence we get | |||
<math display="block"> | |||
\underbrace{\E[X_n\mid\F_{n-1}]}_{\E[X_n-A_n\mid \F_{n-1}]}-A_n=X_{n-1}-A_{n-1}. | |||
</math> | |||
If we set <math>M_n=X_n-A_n-X_0</math>, it follows that <math>M=(M_n)_{n\geq 0}</math> is a martingale with <math>M_0=0</math>. This proves the existence part. For uniqueness, we note that if we have two such decompositions | |||
<math display="block"> | |||
X_n=X_0+M_n+A_n=X_0+L_n+C_n, | |||
</math> | |||
where <math>L_n</math> denotes the martingale part and <math>C_n</math> the increasing process part, it follows that | |||
<math display="block"> | |||
L_n-M_n=A_n-C_n. | |||
</math> | |||
Now since <math>A_n-C_n</math> is <math>\F_{n-1}</math>-measurable, we get that <math>L_n-M_n</math> is also <math>\F_{n-1}</math>-measurable. Thus | |||
<math display="block"> | |||
L_n-M_n=\E[L_n-M_n\mid \F_{n-1}]=L_{n-1}-M_{n-1}, | |||
</math> | |||
because of the martingale property. By induction, we have a chain of equalities | |||
<math display="block"> | |||
L_n-M_n=L_{n-1}-M_{n-1}=\dotsm =L_0-M_0=0. | |||
</math> | |||
Therefore <math>L_n=M_n</math> and also <math>A_n=C_n</math>.}} | |||
{{proofcard|Corollary|cor-2|Let <math>(\Omega,\F,(\F_n)_{n\geq0},\p)</math> be a filtered probability space. Let <math>X=(X_n)_{n\geq 0}</math> be a supermartingale. Then there exists a.s. a unique decomposition | |||
<math display="block"> | |||
X_n=X_0+M_n-A_n, | |||
</math> | |||
where <math>M=(M_n)_{n\geq 0}</math> is a martingale with <math>M_0=0</math> and <math>A=(A_n)_{n\geq 0}</math> is a increasing process with <math>A_0=0</math>. | |||
|Let <math>Y_n=-X_n</math> for all <math>n\geq 0</math>. Then the stochastic process obtained by <math>(Y_n)_{n\geq0}</math> is a submartingale. Theorem 8.4. tells us that there exists a unique decomposition | |||
<math display="block"> | |||
Y_n=Y_0+L_n+C_n, | |||
</math> | |||
where <math>L_n</math> denotes the martingale part and <math>C_n</math> the increasing process part. Hence we get | |||
<math display="block"> | |||
X_n=X_0-L_n-C_n | |||
</math> | |||
and if we take <math>M_n=-L_n</math> and <math>A_n=C_n</math>, the claim follows.}} | |||
Now consider a stopped process. Let <math>(\Omega,\F,(\F_n)_{n\geq0},\p)</math> be a filtered probability space. Let <math>T</math> be a stopping time and let <math>(X_n)_{n\geq 0}</math> be a stochastic process. We denote by <math>X^T=(X^T_n)_{n\geq 0}</math> the process <math>(X_{n\land T})_{n\geq 0}</math>. | |||
{{proofcard|Proposition|prop-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 (resp. sub- or supermartingale) and let <math>T</math> be a stopping time. Then <math>(X_{n\land T})_{n\geq 0}</math> is also a martingale (resp. sub- or supermartingale). | |||
|Note that | |||
<math display="block"> | |||
\{T\geq n+1\}=\{T\leq n\}^C\in\F_n. | |||
</math> | |||
Hence we have | |||
<math display="block"> | |||
\E[X_{n+1\land T}-X_{n\land T}\mid \F_n]=\E[(X_{n+1\land T}-X_{n\land T})\one_{\{T\geq n+1\}}\mid\F_n]=\one_{\{ T\geq n+1\}}\E[X_{n+1}-X_n\mid \F_n]. | |||
</math> | |||
If <math>(X_n)_{n\geq 0}</math> is a martingale, we deduce that | |||
<math display="block"> | |||
\E[X_{n+1\land T}-X_{n\land T}\mid\F_n]=0. | |||
</math> | |||
Moreover, <math>X_{n\land T}</math> is <math>\F_n</math>-measurable. Therefore | |||
<math display="block"> | |||
\E[X_{n+1\land T}\mid \F_n]=X_{n\land T}. | |||
</math> | |||
The same holds for sub-and super martingales.}} | |||
{{proofcard|Theorem|thm-3|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 submartingale (resp. supermartingale) and let <math>S</math> and <math>T</math> be two bounded stopping times, such that <math>S\leq T</math> a.s. Then | |||
<math display="block"> | |||
\E[X_T\mid\F_S]\geq X_Sa.s.(resp. \E[X_T\mid \F_S]\leq X_Sa.s.) | |||
</math> | |||
|Let us assume that <math>(X_n)_{n\geq 0}</math> is a supermartingale. Let <math>A\in\F_S</math> such that <math>S\leq T\leq C\in\N</math>. We already know that <math>(X_{n\land T})_{n\geq 0}</math> is a supermartingale. Therefore we get | |||
<math display="block"> | |||
\begin{align*} | |||
\E[X_T\one_A]&=\sum_{j=0}^C\E[X_{\underbrace{C\land T}_{T}}\one_A\one_{\{S=j\}}]=\sum_{j=0}^C\E[X_{C\land T}\one_{\underbrace{A\cap\{S=j\}}_{\in\F_j}}]\\ | |||
&\leq \sum_{j=0}^C\E[X_{j\land T}\one_{A\cap \{S=j\}}]=\sum_{j=0}^C\E[X_j\one_A\one_{\{S=j\}}]\\ | |||
&=\E\left[\sum_{j=0}^CX_j\one_{\{S=j\}}\one_A\right]=\E[X_S\one_A]=\E[X_T\mid\F_S]\leq X_S. | |||
\end{align*} | |||
</math>}} | |||
{{proofcard|Corollary|cor-3|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 submartingale (resp. supermartingale) and let <math>T</math> be a bounded stopping time. Then | |||
<math display="block"> | |||
\E[X_T]\geq \E[X_0](resp.\E[X_T]\leq \E[X_0]). | |||
</math> | |||
Moreover, if <math>S\leq T</math>, for <math>S</math> and <math>T</math> two bounded stopping times, we have | |||
<math display="block"> | |||
\E[X_T]\geq \E[X_S](resp.\E[X_T]\leq \E[X_S]). | |||
</math>|}} | |||
\begin{exer} | |||
Let <math>(\Omega,\F,(\F_n)_{n\geq0},\p)</math> be a filtered probability space. Let <math>X=(X_n)_{n\geq 0}</math> be a supermartingale and let <math>T</math> be a stopping time. Then | |||
<math display="block"> | |||
X_T\in L^1(\Omega,\F,(\F_n)_{n\geq 0},\p) | |||
</math> | |||
and | |||
<math display="block"> | |||
\E[X_T]\leq \E[X_0] | |||
</math> | |||
in each case of the following situations. | |||
<ul style{{=}}"list-style-type:lower-roman"><li><math>T</math> is bounded. | |||
</li> | |||
<li><math>X</math> is bounded and <math>T</math> is finite. | |||
</li> | |||
<li><math>\E[T] < \infty</math> and for some <math>k\geq 0</math>, we have | |||
<math display="block"> | |||
\vert X_n(\omega)-X_{n-1}(\omega)\vert\leq k, | |||
</math> | |||
for all <math>\omega\in\Omega</math>. | |||
</li> | |||
</ul> | |||
\end{exer} | |||
==General references== | |||
{{cite arXiv|last=Moshayedi|first=Nima|year=2020|title=Lectures on Probability Theory|eprint=2010.16280|class=math.PR}} | |||
==Notes== | |||
{{notelist}} |
Revision as of 01:53, 8 May 2024
Let [math](\Omega,\F,(\F_n)_{n\geq0},\p)[/math] be a filtered probability space. A stochastic process [math](X_n)_{n\geq0}[/math] is called a submartingale (resp. supermartingale) if
- [math]\E[\vert X_n\vert] \lt \infty[/math] for all [math]n\geq 0[/math]
- [math](X_n)_{n\geq 0}[/math] is [math]\F_n[/math]-adapted.
- [math]\E[X_n\mid \F_m]\geq X_m[/math] a.s. for all [math]m\leq n[/math] (resp. [math]\E[X_n\mid \F_m]\leq X_m[/math] a.s. for all [math]m\leq n[/math])
A stochastic process [math](X_n)_{n\geq 0}[/math] is a martingale if and only if it is a submartingale and a supermartingale. A martingale is in particular a submartingale and a supermartingale. If [math](X_n)_{n\geq0}[/math] is a submartingale, then the map [math]n\mapsto \E[X_n][/math] is increasing. If [math](X_n)_{n\geq 0}[/math] is a supermartingale, then the map [math]n\mapsto \E[X_n][/math] is decreasing.
Example
Let [math](\Omega,\F,(\F_n)_{n\geq0},\p)[/math] be a filtered probability space. Let [math]S_n=\sum_{j=1}^{n}Y_j[/math], where [math](Y_n)_{n\geq1}[/math] is a sequence of iid r.v.'s. Moreover, let [math]S_0=0[/math], [math]\F_0=\{\varnothing,\Omega\}[/math] and [math]\F_n=\sigma(Y_1,...,Y_n)[/math]. Then we get
If [math]\E[Y_{n+1}] \gt 0[/math], then [math]\E[S_{n+1}\mid\F_n]\geq S_n[/math] and thus [math](S_n)_{n\geq 0}[/math] is a submartingale. On the other hand, if [math]\E[Y_{n+1}] \lt 0[/math], then [math]\E[S_{n+1}\mid\F_n]\leq S_n[/math] and thus [math](S_n)_{n\geq 0}[/math] is a supermartingale.
Let [math](\Omega,\F,(\F_n)_{n\geq0},\p)[/math] be a filtered probability space. If [math](M_n)_{n\geq 0}[/math] is a martingale and [math]\varphi[/math] is a convex function such that [math]\varphi(M_n)\in L^1(\Omega,\F,(\F_n)_{n\geq 0},\p)[/math] for all [math]n\geq0[/math], then
The first two conditions for a martingale are clearly satisfied. Now for [math]m\leq n[/math], we get
Let [math](\Omega,\F,(\F_n)_{n\geq0},\p)[/math] be a filtered probability space. If [math](M_n)_{n\geq 0}[/math] is a martingale, then
- [math](\vert M_n\vert)_{n\geq0}[/math] and [math](M^+_n)_{n\geq 0}[/math] are submartingales.
- if for all [math]n\geq 0[/math], [math]\E[M_n^2] \lt \infty[/math], then [math](M_n^2)_{n\geq 0}[/math] is a submartingale.
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 submartingale and let [math]T[/math] be a stopping time bounded by [math]C\in\N[/math]. Then
Exercise[a]
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 submartingale. Then there exists a martingale [math]M=(M_n)_{n\geq 0}[/math] with [math]M_0=0[/math] and a sequence [math]A=(A_n)_{n\geq 0}[/math], such that [math]A_{n+1}\geq A_n[/math] a.s. with [math]A_0=0[/math] a.s., which is called an increasing process, and with [math]A_{n+1}[/math] being [math]\F_n[/math]-measurable, which we will call predictable, such that
Moreover, this decomposition is a.s. unique.
Let us define [math]A_0=0[/math] and for [math]n\geq 1[/math]
Let [math](\Omega,\F,(\F_n)_{n\geq0},\p)[/math] be a filtered probability space. Let [math]X=(X_n)_{n\geq 0}[/math] be a supermartingale. Then there exists a.s. a unique decomposition
Let [math]Y_n=-X_n[/math] for all [math]n\geq 0[/math]. Then the stochastic process obtained by [math](Y_n)_{n\geq0}[/math] is a submartingale. Theorem 8.4. tells us that there exists a unique decomposition
Now consider a stopped process. Let [math](\Omega,\F,(\F_n)_{n\geq0},\p)[/math] be a filtered probability space. Let [math]T[/math] be a stopping time and let [math](X_n)_{n\geq 0}[/math] be a stochastic process. We denote by [math]X^T=(X^T_n)_{n\geq 0}[/math] the process [math](X_{n\land T})_{n\geq 0}[/math].
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 (resp. sub- or supermartingale) and let [math]T[/math] be a stopping time. Then [math](X_{n\land T})_{n\geq 0}[/math] is also a martingale (resp. sub- or supermartingale).
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 submartingale (resp. supermartingale) and let [math]S[/math] and [math]T[/math] be two bounded stopping times, such that [math]S\leq T[/math] a.s. Then
Let us assume that [math](X_n)_{n\geq 0}[/math] is a supermartingale. Let [math]A\in\F_S[/math] such that [math]S\leq T\leq C\in\N[/math]. We already know that [math](X_{n\land T})_{n\geq 0}[/math] is a supermartingale. Therefore we get
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 submartingale (resp. supermartingale) and let [math]T[/math] be a bounded stopping time. Then
\begin{exer} Let [math](\Omega,\F,(\F_n)_{n\geq0},\p)[/math] be a filtered probability space. Let [math]X=(X_n)_{n\geq 0}[/math] be a supermartingale and let [math]T[/math] be a stopping time. Then
and
in each case of the following situations.
- [math]T[/math] is bounded.
- [math]X[/math] is bounded and [math]T[/math] is finite.
- [math]\E[T] \lt \infty[/math] and for some [math]k\geq 0[/math], we have
[[math]] \vert X_n(\omega)-X_{n-1}(\omega)\vert\leq k, [[/math]]for all [math]\omega\in\Omega[/math].
\end{exer}
General references
Moshayedi, Nima (2020). "Lectures on Probability Theory". arXiv:2010.16280 [math.PR].