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| | |We need to show both points. | ||
We need to show both points. | |||
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| | |It is enough to prove <math>(ii)</math>. Since <math>(X_n)_{n\geq 0}</math> is a submartingale, we know that <math>(X_n^+)_{n\geq 0}</math> is a submartingale. Hence | ||
It is enough to prove <math>(ii)</math>. Since <math>(X_n)_{n\geq 0}</math> is a submartingale, we know that <math>(X_n^+)_{n\geq 0}</math> is a submartingale. Hence | |||
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Latest revision as of 23:38, 8 May 2024
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 stochastic process, such that [math]X_n[/math] is [math]\F_n[/math]-measurable for all [math]n\geq 0[/math]. We denote
Note that [math](X_n^*)_{n\geq 0}[/math] is increasing and [math]\F_n[/math]-adapted. Therefore if [math]X_n\in L^1(\Omega,\F,(\F_n)_{n\geq 0},\p)[/math] for all [math]n\geq 0[/math], then [math](X_n^*)_{n\geq 0}[/math] is a submartingale.
Maximal inequality and Doob's inequality
Recall Markov's inequality in terms of [math](X_n^*)_{n\geq0}[/math], which is given by
with the obvious bound
We shall see for instance that when [math](X_n)_{n\geq 0}[/math] is a martingale, one can replace [math]\E[X_n^*][/math] by [math]\E[\vert X_n\vert][/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 submartingale and let [math]\lambda \gt 0,k\in\N[/math]. Define
Then the following hold.
-
[[math]] \lambda\p[A]\leq \E[X_k\one_A], [[/math]]
-
[[math]] \lambda\p[B]\leq \E[X_k\one_{B^C}]-\E[X_0]. [[/math]]
If [math](X_n)_{n\geq 0}[/math] is a martingale, then [math](\vert X_n\vert)_{n\geq 0}[/math] is a submartingale. Moreover, from [math](i)[/math] we get
We need to show both points.
- Let us introduce
[[math]] T=\inf\{n\in\N\mid X_n\leq \lambda\}\land k. [[/math]]Then [math]T[/math] is a stopping time, which is bounded by [math]k[/math]. We thus have[[math]] \E[X_T]\leq \E[X_k]. [[/math]]We note that [math]X_T=X_k[/math] if [math]T=k[/math], which happens for [math]\omega\in A^C[/math]. Hence we get[[math]] \E[X_T]=\E[X_T\one_A+X_T\one_{A^C}]=\E[X_T\one_A]+\E[X_k\one_{A^C}]\leq \underbrace{\E[X_k]}_{\E[X_k(\one_A+\one_{A^C})]}. [[/math]]Now we note that[[math]] \E[X_T\one_A]\geq \lambda \E[\one_A]=\lambda \p[A]. [[/math]]Therefore we get[[math]] \lambda\p[A]\leq \E[X_k\one_A]. [[/math]]
- Let us define
[[math]] S=\inf\{n\in\N\mid X_n\leq -\lambda\}\land k. [[/math]]Again [math]S[/math] is a stopping time, which is bounded by [math]k[/math]. We hence have[[math]] \E[X_S]\geq \E[X_0]. [[/math]]Thus[[math]] \E[X_0]\leq \E[X_S\one_B]+\E[X_S\one_{B^C}]\leq -\lambda \p[B]+\E[X_k\one_{B^C}]. [[/math]]Therefore we get[[math]] \lambda\p[B]\leq \E[X_k\one_{B^C}]-\E[X_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, such that for all [math]n\geq 0[/math] we have [math]\E[X_n^2] \lt \infty[/math]. Then
We use the fact that [math](X_n^2)_{n\geq 0}[/math] is a positive submartingale. 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. Then for all [math]\lambda\geq 0[/math] and [math]n\in\N[/math], we get
Let [math]A[/math] and [math]B[/math] be defined as in Proposition 9.1. Then
Let [math](\Omega,\F,(\F_n)_{n\geq0},\p)[/math] be a filtered probability space. Let [math]p \gt 1[/math] and [math]q \gt 1[/math], such that [math]\frac{1}{p}+\frac{1}{q}=1[/math].
- If [math](X_n)_{n\geq 0}[/math] is a submartingale, then for all [math]n\geq 0[/math] we have
[[math]] \left\|\max_{0\leq k\leq n}X_k^+\right\|_p\leq q\left\|X_n^+\right\|_p. [[/math]]
- If [math](X_n)_{n\geq 0}[/math] is a martingale, then for all [math]n\geq 0[/math] we have
[[math]] \left\|\max_{0\leq k\leq n}\vert X_k\vert\right\|_p\leq q\| X_n\|_p. [[/math]]
Recall that if [math]X\in L^1(\Omega,\F,(\F_n)_{n\geq 0},\p)[/math], then
It is enough to prove [math](ii)[/math]. Since [math](X_n)_{n\geq 0}[/math] is a submartingale, we know that [math](X_n^+)_{n\geq 0}[/math] is a submartingale. Hence
Now ler [math]Y_n:=\max_{0\leq k\leq n}X_k^+[/math]. Then for any [math]k \gt 0[/math], we have
where we have used that [math]q=\frac{p}{p-1}[/math] and Markov's inequality. Therefore we obtain
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 [math]p \gt 1[/math], [math]q \gt 1[/math] such that [math]\frac{1}{p}+\frac{1}{q}=1[/math]. Then
Exercise[a]
General references
Moshayedi, Nima (2020). "Lectures on Probability Theory". arXiv:2010.16280 [math.PR].