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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 | |||
<math display="block"> | |||
X_n^*:=\sup_{j\leq n}\vert X_j\vert. | |||
</math> | |||
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 | |||
<math display="block"> | |||
\p[X_n^*\geq \alpha]\leq \frac{\E[X_n^*]}{\alpha}, | |||
</math> | |||
with the obvious bound | |||
<math display="block"> | |||
\E[X_n^*]\leq \sum_{j=1}^n\E[\vert X_j\vert]. | |||
</math> | |||
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>. | |||
{{proofcard|Proposition|prop1|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 > 0,k\in\N</math>. Define | |||
<math display="block"> | |||
\begin{align*} | |||
A&:=\left\{\max_{0\leq n\leq k}X_n\geq \lambda\right\}\\ | |||
B&:=\left\{\min_{0\leq n\leq k}X_n\leq -\lambda\right\}. | |||
\end{align*} | |||
</math> | |||
Then the following hold. | |||
<ul style{{=}}"list-style-type:lower-roman"><li> | |||
<math display="block"> | |||
\lambda\p[A]\leq \E[X_k\one_A], | |||
</math> | |||
</li> | |||
<li> | |||
<math display="block"> | |||
\lambda\p[B]\leq \E[X_k\one_{B^C}]-\E[X_0]. | |||
</math> | |||
</li> | |||
</ul> | |||
{{alert-info | | |||
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 | |||
<math display="block"> | |||
\lambda\p[X^*_k\geq \alpha]\leq \E[\vert X_k\vert\one_A]\leq \E[\vert X_k\vert] | |||
</math> | |||
and hence | |||
<math display="block"> | |||
\p[X_k^*\geq \alpha]\leq \frac{\E[\vert X_k\vert]}{\alpha}. | |||
</math> | |||
}} | |||
|We need to show both points. | |||
<ul style{{=}}"list-style-type:lower-roman"><li>Let us introduce | |||
<math display="block"> | |||
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 display="block"> | |||
\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 display="block"> | |||
\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 display="block"> | |||
\E[X_T\one_A]\geq \lambda \E[\one_A]=\lambda \p[A]. | |||
</math> | |||
Therefore we get | |||
<math display="block"> | |||
\lambda\p[A]\leq \E[X_k\one_A]. | |||
</math> | |||
</li> | |||
<li>Let us define | |||
<math display="block"> | |||
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 display="block"> | |||
\E[X_S]\geq \E[X_0]. | |||
</math> | |||
Thus | |||
<math display="block"> | |||
\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 display="block"> | |||
\lambda\p[B]\leq \E[X_k\one_{B^C}]-\E[X_0]. | |||
</math> | |||
</li> | |||
</ul>}} | |||
{{proofcard|Proposition (Kolmogorov's inequality)|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, such that for all <math>n\geq 0</math> we have <math>\E[X_n^2] < \infty</math>. Then | |||
<math display="block"> | |||
\p\left[\max_{0\leq k\leq n}\vert X_k\vert\geq \lambda\right]\leq \frac{\E[X_k^2]}{\lambda^2}. | |||
</math> | |||
|We use the fact that <math>(X_n^2)_{n\geq 0}</math> is a positive submartingale. Therefore we get | |||
<math display="block"> | |||
\lambda^2\underbrace{\p\left[\max_{0\leq k\leq n}\vert X_k\vert^2\geq \lambda^2\right]}_{\p\left[\max_{0\leq k\leq k}\vert X_k\vert\geq \lambda\right]}\leq \E\left[X_k^2\one_{\left\{\max_{0\leq k\leq n}\vert X_k\vert^2\geq \lambda^2\right\}}\right]\leq \E[X_k^2] | |||
</math>}} | |||
{{proofcard|Theorem (Maximal inequality)|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. Then for all <math>\lambda\geq 0</math> and <math>n\in\N</math>, we get | |||
<math display="block"> | |||
\lambda\p\left[\max_{0\leq k\leq n}\vert X_k\vert\geq \lambda\right]\leq \E[X_0]+2\E[\vert X_n\vert]. | |||
</math> | |||
|Let <math>A</math> and <math>B</math> be defined as in Proposition 9.1. Then | |||
<math display="block"> | |||
\begin{align*} | |||
\lambda \p\left[\max_{0\leq k\leq n}\vert X_k\vert \geq \lambda\right]&=\lambda\p[A\cup B]\leq \E[X_k\one_A]-\E[X_0]+\E[X_k\one_{B^C}]\\ | |||
&\leq \E[\vert X_0\vert]+\E[\vert X_n\vert]+\E[\vert X_n\vert]=\E[\vert X_0\vert]+2\E[\vert X_n\vert]. | |||
\end{align*} | |||
</math>}} | |||
{{proofcard|Theorem (Doob's inequality)|thm2|Let <math>(\Omega,\F,(\F_n)_{n\geq0},\p)</math> be a filtered probability space. Let <math>p > 1</math> and <math>q > 1</math>, such that <math>\frac{1}{p}+\frac{1}{q}=1</math>. | |||
<ul style{{=}}"list-style-type:lower-roman"><li>If <math>(X_n)_{n\geq 0}</math> is a submartingale, then for all <math>n\geq 0</math> we have | |||
<math display="block"> | |||
\left\|\max_{0\leq k\leq n}X_k^+\right\|_p\leq q\left\|X_n^+\right\|_p. | |||
</math> | |||
</li> | |||
<li>If <math>(X_n)_{n\geq 0}</math> is a martingale, then for all <math>n\geq 0</math> we have | |||
<math display="block"> | |||
\left\|\max_{0\leq k\leq n}\vert X_k\vert\right\|_p\leq q\| X_n\|_p. | |||
</math> | |||
</li> | |||
</ul> | |||
{{alert-info | | |||
Recall that if <math>X\in L^1(\Omega,\F,(\F_n)_{n\geq 0},\p)</math>, then | |||
<math display="block"> | |||
\|X\|_p=\E[\vert X\vert^p]^{1/p}. | |||
</math> | |||
Moreover, if <math>p=q=2</math> and <math>(X_n)_{n\geq 0}</math> is a martingale, then for all <math>n\geq 0</math> we have | |||
<math display="block"> | |||
\E\left[\max_{0\leq k\leq n} X_k^2\right]\leq 4\E[X_n^2]. | |||
</math> | |||
In general, we have | |||
<math display="block"> | |||
\vert X_n\vert^2\leq \max_{0\leq k\leq n}\vert X_k\vert^p. | |||
</math> | |||
Therefore we get | |||
<math display="block"> | |||
\E[\vert X_n\vert^p]\leq \E\left[\max_{0\leq k\leq n}\vert X_k\vert^p\right]\leq ^{Doob}q^p\E[\vert X_n\vert^p]. | |||
</math> | |||
We shall also recall that for <math>X\in L^p(\Omega,\F,(\F_n)_{n\geq 0},\p)</math>, we can write | |||
<math display="block"> | |||
\E[\vert X\vert^p]=\E\left[\int_0^{\vert X\vert}p\lambda^{p-1}d\lambda\right]=\E\left[\int_0^\infty \one_{\{\vert X\vert\geq \lambda\}}p\lambda^{p-1}d\lambda\right]=\int_0^\infty p\lambda^{p-1}\p[\vert X\vert \geq \lambda]d\lambda | |||
</math> | |||
by using Fubini's theorem. | |||
}} | |||
|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 | |||
<math display="block"> | |||
\lambda\p\left[\max_{0\leq k\leq n}X_k^+\geq \lambda\right]\geq\E\left[X_n^+\one_{\{\max_{0\leq k\leq n}X_k^+\geq \lambda\}}\right]. | |||
</math> | |||
Now ler <math>Y_n:=\max_{0\leq k\leq n}X_k^+</math>. Then for any <math>k > 0</math>, we have | |||
<math display="block"> | |||
\begin{align*} | |||
\E[(Y_n\land k)^p]&=\int_0^\infty p\lambda^{p-1}\p[Y_n\land k\geq \lambda]d\lambda=\int_0^np\lambda^{p-1}\p[Y_n\geq\lambda]d\lambda\\ | |||
&\leq \int_0^n p\lambda^{p-1}\left(\frac{1}{\lambda}\E[X_n^+\one_{\{ Y_n\geq \lambda\}}]\right)d\lambda\\ | |||
&=\E\left[\int_0^n p\lambda^{p-1}X_n^+\one_{\{Y_n\geq\lambda\}}d\lambda\right]=\E\left[\int_0^{Y_n\land k}p\lambda^{p-2}X_n^+d\lambda\right]\\ | |||
&=\E\left[\frac{p}{p-1}(Y_n\land k)^{p-1}X_n^+\right]\\ | |||
&\leq q\E[(X_n^+)^p]^{1/p}\E[(Y_n\land k)^p]^{1/q}, | |||
\end{align*} | |||
</math> | |||
where we have used that <math>q=\frac{p}{p-1}</math> and Markov's inequality. Therefore we obtain | |||
<math display="block"> | |||
\E[(Y_n\land k)^p]\leq q\E[(X_n^+)^p]^{1/p}\E[(Y_n\land k)^p]^{1/q}. | |||
</math> | |||
Since <math>\E[(Y_n\land k)^p]\not=0</math>, we can divide by it to get | |||
<math display="block"> | |||
\E[(Y_n\land k)^p]^{1-1/q=1/p}\leq q\E[(X_n^+)^p]^{1/p} | |||
</math> | |||
and thus | |||
<math display="block"> | |||
\| (Y_k\land n)\|_p\leq q\| X_n^+\|_p. | |||
</math> | |||
Now for <math>k\to\infty</math>, monotone convergence implies that | |||
<math display="block"> | |||
\|Y_n\|_p\leq q\|X_n^+\|_p. | |||
</math>}} | |||
{{proofcard|Corollary|cor-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 and <math>p > 1</math>, <math>q > 1</math> such that <math>\frac{1}{p}+\frac{1}{q}=1</math>. Then | |||
<math display="block"> | |||
\left\| \sup_{n\geq 0}\vert X_n\vert\right\|_p\leq q\sup_{n\geq 0}\| X_n\|_p. | |||
</math> | |||
|Exercise{{efn|Use Doob's inequality.}}}} | |||
==General references== | |||
{{cite arXiv|last=Moshayedi|first=Nima|year=2020|title=Lectures on Probability Theory|eprint=2010.16280|class=math.PR}} | |||
==Notes== | |||
{{notelist}} |
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].