guide:20cb216cb3

[math] \newcommand{\R}{\mathbb{R}} \newcommand{\A}{\mathcal{A}} \newcommand{\B}{\mathcal{B}} \newcommand{\N}{\mathbb{N}} \newcommand{\C}{\mathbb{C}} \newcommand{\Rbar}{\overline{\mathbb{R}}} \newcommand{\Bbar}{\overline{\mathcal{B}}} \newcommand{\Q}{\mathbb{Q}} \newcommand{\E}{\mathbb{E}} \newcommand{\p}{\mathbb{P}} \newcommand{\one}{\mathds{1}} \newcommand{\0}{\mathcal{O}} \newcommand{\mat}{\textnormal{Mat}} \newcommand{\sign}{\textnormal{sign}} \newcommand{\CP}{\mathcal{P}} \newcommand{\CT}{\mathcal{T}} \newcommand{\CY}{\mathcal{Y}} \newcommand{\F}{\mathcal{F}} \newcommand{\mathds}{\mathbb}[/math]

Theorem

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

[[math]] X_n=\E[X_\infty\mid \F_n]. [[/math]]

Show Proof

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]] \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]] \lim_{m\to\infty}\E[X_m\one_A]=\E[X_\infty\one_A] [[/math]]

and therefore

[[math]] 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]] \vert X_n\vert\leq \E[\vert Z\vert\mid \F_n] [[/math]]

and thus

[[math]] \E[\vert X_n\vert]\leq \E[\vert Z\vert]. [[/math]]

This implies that

[[math]] \sup_{n\geq 0}\E[\vert X_n\vert]\leq \E[\vert Z\vert] \lt \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]] 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]] \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 \gt 0[/math] and [math]M\in\R_+[/math], such that

[[math]] \E[\vert Z-Z\one_{\{ \vert Z\vert \leq M\}}\vert] \lt \varepsilon. [[/math]]

Thus, for all [math]n\geq 0[/math]

[[math]] \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 \gt M\}}\mid \F_n]]=\E[\vert Z\vert \one_{\{ \vert Z\vert \gt M\}}] \lt \varepsilon. [[/math]]

Moreover, from the bounded case, it follows that

[[math]] \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]] \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] \lt \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]] \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].

■

Corollary

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].

Show Proof

First, we note that [math]X_\infty[/math] is [math]\F_\infty[/math]-measurable. Now choose [math]A\in\F_n[/math]. Then

[[math]] \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]] \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]] \E[Z\one_A]=\E[X_\infty\one_A], [[/math]]

which implies that

[[math]] X_\infty=\E[Z\mid \F_\infty]. [[/math]]

■

Theorem ([math]L^p[/math] martingale convergence theorem)

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]] \sup_{n\geq 0}\E[\vert X_n\vert^p] \lt \infty. [[/math]]

Then

[[math]] X_n\xrightarrow{n\to\infty\atop \text{$a.s.$ and $L^p$}}X_\infty. [[/math]]

Moreover, we have

[[math]] \E[\vert X_\infty\vert^p]=\sup_{n\geq 0}\E[\vert X_n\vert^p] [[/math]]

and

[[math]] \E[(X_\infty^*)^p]\leq \left( \frac{p}{p-1}\right)^p\E[\vert X_\infty\vert^p], [[/math]]

where

[[math]] X_\infty^*=\sup_{n\geq 0}\vert X_n\vert. [[/math]]

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].

Show Proof

We first note that since [math]\sup_{n\geq 0}\E[\vert X_n\vert^p] \lt \infty[/math], we also have that

[[math]] \sup_{n\geq 0}\E[\vert X_n\vert] \lt \infty. [[/math]]

Thus [math]X_n\xrightarrow{n\to\infty\atop a.s.}X_\infty[/math]. From Doob's inequality, we get

[[math]] \E[(X_\infty^*)^p]\leq \left(\frac{p}{p-1}\right)^p\sup_{n\geq 0}\E[\vert X_n\vert^p] \lt \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]] \vert X_n-X_\infty\vert^p\leq 2^p(X_\infty^*)p. [[/math]]

Using dominated convergence, we get

[[math]] \E[\vert X_n-X_\infty\vert^p]\xrightarrow{n\to\infty}0 [[/math]]

and thus

[[math]] 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]] (\E[\vert X_n\vert^p])_{n\geq 0} [[/math]]

is increasing, which implies that

[[math]] \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

Moshayedi, Nima (2020). "Lectures on Probability Theory". arXiv:2010.16280 [math.PR].