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{{proofcard|Theorem|thm6|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 <math>(X_n)_{n\geq0}</math> converges a.s. to a r.v. <math>X_\infty\in L^1(\Omega,\F_\infty,(\F_n)_{n\geq0},\p)</math> if and only if there exists a r.v. <math>Z\in L^1(\Omega,\F,(\F_n)_{n\geq 0},\p)</math> such that <math>X_n=\E[Z\mid \F_n]</math> for all <math>n\geq0</math>, where <math>\F_\infty=\sigma\left(\bigcup_{n\geq 0}\F_n\right)</math>. | |||
{{alert-info | | |||
We shall see that one can always represent <math>X_n</math> as | |||
<math display="block"> | |||
X_n=\E[X_\infty\mid \F_n]. | |||
</math> | |||
}} | |||
|For the left right implication, we note that for all <math>m\geq n</math> and for all <math>A\in\F_n</math> we have | |||
<math display="block"> | |||
\E[X_n\one_A]=\E[X_m\one_A]. | |||
</math> | |||
Therefore we se that <math>X_m\xrightarrow{m\to\infty\atop L^1}X_\infty</math>, implies that | |||
<math display="block"> | |||
\lim_{m\to\infty}\E[X_m\one_A]=\E[X_\infty\one_A] | |||
</math> | |||
and therefore | |||
<math display="block"> | |||
X_n=\E[X_\infty\mid\F_n]. | |||
</math> | |||
For the left implication, we see that if <math>X_n=\E[Z\mid \F_n]</math>, then | |||
<math display="block"> | |||
\vert X_n\vert\leq \E[\vert Z\vert\mid \F_n] | |||
</math> | |||
and thus | |||
<math display="block"> | |||
\E[\vert X_n\vert]\leq \E[\vert Z\vert]. | |||
</math> | |||
This implies that | |||
<math display="block"> | |||
\sup_{n\geq 0}\E[\vert X_n\vert]\leq \E[\vert Z\vert] < \infty. | |||
</math> | |||
Hence we know now that <math>X_n\xrightarrow{n\to\infty_ a.s.}X_\infty</math>. It remains to show that | |||
<math display="block"> | |||
X_n\xrightarrow{n\to\infty\atop L^1}X_\infty. | |||
</math> | |||
First, we assume that <math>Z</math> is bounded, i.e. for all <math>\omega\in\Omega</math>, | |||
<math display="block"> | |||
\vert Z(\omega)\vert\leq M\in\R_+. | |||
</math> | |||
Hence <math>\vert X_n(\omega)\vert\leq M</math> and thus <math>L^1</math>-convergence follows from dominated convergence. For the general case, let <math>\varepsilon > 0</math> and <math>M\in\R_+</math>, such that | |||
<math display="block"> | |||
\E[\vert Z-Z\one_{\{ \vert Z\vert \leq M\}}\vert] < \varepsilon. | |||
</math> | |||
Thus, for all <math>n\geq 0</math> | |||
<math display="block"> | |||
\E[\vert X_n-\E[Z\one_{\{ \vert Z\vert \leq M\}}\mid \F_n]\vert]\leq \E[\E[\vert Z\vert \one_{\{\vert Z\vert > M\}}\mid \F_n]]=\E[\vert Z\vert \one_{\{ \vert Z\vert > M\}}] < \varepsilon. | |||
</math> | |||
Moreover, from the bounded case, it follows that | |||
<math display="block"> | |||
\left(\E[Z\one_{\{\vert Z\vert \leq M\}}\mid \F_n]\right)_{n\geq 0} | |||
</math> | |||
converges in <math>L^1</math>. Hence, there exists <math>n_0\in\N</math> such that for all <math>m,n\geq n_0</math>, we have | |||
<math display="block"> | |||
\E[\vert \E[Z\one_{\|\vert Z\vert \leq M\}}\mid\F_m]-\E[Z\one_{\{\vert Z\vert \leq M\}}\mid \F_n]\vert] < \varepsilon. | |||
</math> | |||
Now a simple application of the triangular inequality and the above estimates gives, for all <math>m,n\geq n_0</math> | |||
<math display="block"> | |||
\E[\vert X_m-X_n\vert]\leq 3\varepsilon. | |||
</math> | |||
Therefore <math>(X_n)_{n\geq 0}</math> is a Cauchy sequence in <math>L^1</math> and hence it converges in <math>L^1</math>.}} | |||
{{proofcard|Corollary|cor1|Let <math>(\Omega,\F,(\F_n)_{n\geq0},\p)</math> be a filtered probability space. Let <math>Z\in L^1(\Omega,\F,(\F_n)_{n\geq 0},\p)</math>. The unique martingale <math>X_n=\E[Z\mid \F_n]</math> converges a.s. and in <math>L^1</math> to <math>X_\infty=\E[Z\mid \F_\infty]</math>, where <math>\F_\infty=\sigma\left(\bigcup_{n\geq 0}\F_n\right)</math>. | |||
|First, we note that <math>X_\infty</math> is <math>\F_\infty</math>-measurable. Now choose <math>A\in\F_n</math>. Then | |||
<math display="block"> | |||
\lim_{n\to\infty}\E[Z\one_A]=\lim_{n\to\infty}\E[X_n\one_A]=\E[X_\infty\one_A], | |||
</math> | |||
and hence for all <math>A\in \bigcup_{n\geq 0}\F_n</math>, | |||
<math display="block"> | |||
\E[Z\one_A]=\E[X_\infty\one_A]. | |||
</math> | |||
The monotone class theorem implies that for every <math>A\in\F_\infty</math>, | |||
<math display="block"> | |||
\E[Z\one_A]=\E[X_\infty\one_A], | |||
</math> | |||
which implies that | |||
<math display="block"> | |||
X_\infty=\E[Z\mid \F_\infty]. | |||
</math>}} | |||
{{proofcard|Theorem (<math>L^p</math> martingale convergence theorem)|thm3|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. Assume that there exists <math>p\geq 1</math> such that | |||
<math display="block"> | |||
\sup_{n\geq 0}\E[\vert X_n\vert^p] < \infty. | |||
</math> | |||
Then | |||
<math display="block"> | |||
X_n\xrightarrow{n\to\infty\atop \text{$a.s.$ and $L^p$}}X_\infty. | |||
</math> | |||
Moreover, we have | |||
<math display="block"> | |||
\E[\vert X_\infty\vert^p]=\sup_{n\geq 0}\E[\vert X_n\vert^p] | |||
</math> | |||
and | |||
<math display="block"> | |||
\E[(X_\infty^*)^p]\leq \left( \frac{p}{p-1}\right)^p\E[\vert X_\infty\vert^p], | |||
</math> | |||
where | |||
<math display="block"> | |||
X_\infty^*=\sup_{n\geq 0}\vert X_n\vert. | |||
</math> | |||
{{alert-info | | |||
Let us summarize what we have seen so far. | |||
<ul style{{=}}"list-style-type:lower-roman"><li>If <math>(X_n)_{n\geq 0}</math> is bounded in <math>L^1</math>, we get <math>X_n\xrightarrow{n\to\infty\atop a.s.}X_\infty\in L^1</math>. | |||
</li> | |||
<li><math>X_n\xrightarrow{n\to\infty\atop \text{$a.s.$ and $L^1$}}X_\infty</math> if and only if <math>X_n=\E[X_\infty\mid \F_n]</math>. | |||
</li> | |||
<li>If <math>(X_n)_{n\geq 0}</math> is bounded in <math>L^p</math> with <math>p\geq 1</math>, then <math>X_n\xrightarrow{n\to\infty\atop\text{$a.s.$ and $L^p$}}X_\infty</math>. | |||
</li> | |||
</ul> | |||
}} | |||
|We first note that since <math>\sup_{n\geq 0}\E[\vert X_n\vert^p] < \infty</math>, we also have that | |||
<math display="block"> | |||
\sup_{n\geq 0}\E[\vert X_n\vert] < \infty. | |||
</math> | |||
Thus <math>X_n\xrightarrow{n\to\infty\atop a.s.}X_\infty</math>. From Doob's inequality, we get | |||
<math display="block"> | |||
\E[(X_\infty^*)^p]\leq \left(\frac{p}{p-1}\right)^p\sup_{n\geq 0}\E[\vert X_n\vert^p] < \infty | |||
</math> | |||
and therefore <math>X_\infty^*\in L^p(\Omega,\F,(\F_n)_{n\geq 0},\p)</math>. Moreover, for all <math>n\geq 0</math> we get <math>\vert X_n\vert\leq X_\infty^*</math> and | |||
<math display="block"> | |||
\vert X_n-X_\infty\vert^p\leq 2^p(X_\infty^*)p. | |||
</math> | |||
Using dominated convergence, we get | |||
<math display="block"> | |||
\E[\vert X_n-X_\infty\vert^p]\xrightarrow{n\to\infty}0 | |||
</math> | |||
and thus | |||
<math display="block"> | |||
X_n\xrightarrow{n\to\infty\atop L^p}X_\infty. | |||
</math> | |||
Finally, we note that <math>(\vert X_n\vert^p)_{n\geq 0}</math> is a positive submartingale. Hence we know that | |||
<math display="block"> | |||
(\E[\vert X_n\vert^p])_{n\geq 0} | |||
</math> | |||
is increasing, which implies that | |||
<math display="block"> | |||
\E[\vert X_\infty\vert^p]=\lim_{n\to\infty}\E[\vert X_n\vert^p]=\sup_{n\geq 0}\E[\vert X_n\vert^p]. | |||
</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 23:45, 8 May 2024
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 [math](X_n)_{n\geq0}[/math] converges a.s. to a r.v. [math]X_\infty\in L^1(\Omega,\F_\infty,(\F_n)_{n\geq0},\p)[/math] if and only if there exists a r.v. [math]Z\in L^1(\Omega,\F,(\F_n)_{n\geq 0},\p)[/math] such that [math]X_n=\E[Z\mid \F_n][/math] for all [math]n\geq0[/math], where [math]\F_\infty=\sigma\left(\bigcup_{n\geq 0}\F_n\right)[/math].
We shall see that one can always represent [math]X_n[/math] as
For the left right implication, we note that for all [math]m\geq n[/math] and for all [math]A\in\F_n[/math] we have
For the left implication, we see that if [math]X_n=\E[Z\mid \F_n][/math], then
Hence we know now that [math]X_n\xrightarrow{n\to\infty_ a.s.}X_\infty[/math]. It remains to show that
Let [math](\Omega,\F,(\F_n)_{n\geq0},\p)[/math] be a filtered probability space. Let [math]Z\in L^1(\Omega,\F,(\F_n)_{n\geq 0},\p)[/math]. The unique martingale [math]X_n=\E[Z\mid \F_n][/math] converges a.s. and in [math]L^1[/math] to [math]X_\infty=\E[Z\mid \F_\infty][/math], where [math]\F_\infty=\sigma\left(\bigcup_{n\geq 0}\F_n\right)[/math].
First, we note that [math]X_\infty[/math] is [math]\F_\infty[/math]-measurable. Now choose [math]A\in\F_n[/math]. Then
which implies 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. Assume that there exists [math]p\geq 1[/math] such that
and
Let us summarize what we have seen so far.
- If [math](X_n)_{n\geq 0}[/math] is bounded in [math]L^1[/math], we get [math]X_n\xrightarrow{n\to\infty\atop a.s.}X_\infty\in L^1[/math].
- [math]X_n\xrightarrow{n\to\infty\atop \text{$a.s.$ and $L^1$}}X_\infty[/math] if and only if [math]X_n=\E[X_\infty\mid \F_n][/math].
- If [math](X_n)_{n\geq 0}[/math] is bounded in [math]L^p[/math] with [math]p\geq 1[/math], then [math]X_n\xrightarrow{n\to\infty\atop\text{$a.s.$ and $L^p$}}X_\infty[/math].
We first note that since [math]\sup_{n\geq 0}\E[\vert X_n\vert^p] \lt \infty[/math], we also have that
Finally, we note that [math](\vert X_n\vert^p)_{n\geq 0}[/math] is a positive submartingale. Hence we know that
General references
Moshayedi, Nima (2020). "Lectures on Probability Theory". arXiv:2010.16280 [math.PR].