guide:Bcd133fb9d: Difference between revisions
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\varphi(\E[M_n\mid\F_m])=\varphi(M_m)\leq \E[\varphi(M_n)\mid\F_m]a.s. | \varphi(\E[M_n\mid\F_m])=\varphi(M_m)\leq \E[\varphi(M_n)\mid\F_m]a.s. | ||
</math>}} | </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 | {{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. | <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. | ||
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</li> | </li> | ||
</ul>|}} | </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 | {{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"> | <math display="block"> | ||
\E[X_T]\leq \E[X_C]. | \E[X_T]\leq \E[X_C]. | ||
</math> | </math>|Exercise{{efn|The proof is the same as in [[guide:32e7bc2e71#The strong law of large numbers|The strong law of large numbers]]}}}} | ||
|Exercise{{efn|The proof is the same as in [[guide:32e7bc2e71#The strong law of large numbers|The strong law of large numbers]]}} | |||
{{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 | {{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 | ||
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</math> | </math> | ||
Moreover, this decomposition is a.s. unique. | Moreover, this decomposition is a.s. unique.|Let us define <math>A_0=0</math> and for <math>n\geq 1</math> | ||
|Let us define <math>A_0=0</math> and for <math>n\geq 1</math> | |||
<math display="block"> | <math display="block"> | ||
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</math> | </math> | ||
Therefore <math>L_n=M_n</math> and also <math>A_n=C_n</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 | {{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 | ||
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X_n=X_0+M_n-A_n, | X_n=X_0+M_n-A_n, | ||
</math> | </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>. | 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 | ||
|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"> | <math display="block"> | ||
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</math> | </math> | ||
and if we take <math>M_n=-L_n</math> and <math>A_n=C_n</math>, the claim follows.}} | 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>. | 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). | {{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). |
Latest revision as of 22:49, 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.
Show ProofLet 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
General references
Moshayedi, Nima (2020). "Lectures on Probability Theory". arXiv:2010.16280 [math.PR].
Notes
- The proof is the same as in The strong law of large numbers