Martingale Convergence Theorem

As the Martingale Convergence Theorem or Doob's Martingal Convergence Theorem (named after Joseph L. Doob ), certain statements about the convergence of martingales are called in probability theory . A martingale is a special stochastic process that can be seen as the formalization and generalization of a fair game of chance . Given additional requirements for the limitation of the process, its convergence can be deduced. The different versions of the theorem differ in terms of the type of restriction and the type of convergence. The crossover inequality is an essential aid in the proof . Analogous convergence theorems also exist for backward martingales .

requirements

On a probability space with a filtration and let a sequence of real random variables be given, which is adapted to the filtration and can be integrated. This means that the random variable is measurable with respect to and fulfilled for all . ${\displaystyle (\Omega ,{\mathcal {A}},P)}$ ${\displaystyle ({\mathcal {F}}_{n})_{n\in \mathbb {N} _{0}}}$ ${\displaystyle \textstyle {\mathcal {F}}_{\infty }=\sigma \left(\bigcup _{n=0}^{\infty }{\mathcal {F}}_{n}\right)}$ ${\displaystyle (X_{n})_{n\in \mathbb {N} _{0}}}$ ${\displaystyle n\in \mathbb {N} _{0}}$ ${\displaystyle X_{n}}$ ${\displaystyle {\mathcal {F}}_{n}}$ ${\displaystyle E(|X_{n}|)<\infty }$

The process is called martingale when the equation holds true for everyone . If it applies instead to everyone then the process is called a submartingale. In the case of everyone , the process is called the Supermartingale. Each martingale is a sub and a super martingale. A process is a super martingale if and only if it is a submartingale. ${\displaystyle (X_{n})_{n\in \mathbb {N} _{0}}}$ ${\displaystyle n\in \mathbb {N} _{0}}$ ${\displaystyle \mathbb {E} (X_{n+1}\mid {\mathcal {F}}_{n})=X_{n}}$ ${\displaystyle \mathbb {E} (X_{n+1}\mid {\mathcal {F}}_{n})\geq X_{n}}$ ${\displaystyle n\in \mathbb {N} _{0},}$ ${\displaystyle \mathbb {E} (X_{n+1}\mid {\mathcal {F}}_{n})\leq X_{n}}$ ${\displaystyle n\in \mathbb {N} _{0},}$ ${\displaystyle (X_{n})_{n\in \mathbb {N} _{0}}}$ ${\displaystyle (-X_{n})_{n\in \mathbb {N} _{0}}}$

Versions of the Martingale Convergence Theorem

Almost certain convergence

It is a submartingale and there is a constant with all that is, the expected value of the positive parts is limited. Then there is a measurable random variable with almost certain . ${\displaystyle (X_{n})_{n\in \mathbb {N} _{0}}}$ ${\displaystyle M>0}$ ${\displaystyle \mathbb {E} (X_{n}^{+})\leq M}$ ${\displaystyle n\in \mathbb {N} _{0},}$ ${\displaystyle X_{n}^{+}=\max(X_{n},0)}$ ${\displaystyle {\mathcal {F}}_{\infty }}$ ${\displaystyle X_{\infty }\in {\mathcal {L}}^{1}}$ ${\displaystyle X_{n}{\xrightarrow {n\to \infty }}X_{\infty }}$

proof

The so-called cross-over lemma is of decisive importance for the proof. This means that for two real numbers , the two stop times with and ${\displaystyle b>a}$ ${\displaystyle \sigma ,\tau }$ ${\displaystyle \sigma _{0}\equiv 0}$

${\displaystyle \tau _{k}:=\inf\{n\geq \sigma _{k-1}\mid X_{n}\leq a\}}$

${\displaystyle \sigma _{k}:=\inf\{n\geq \tau _{k}\mid X_{n}\geq b\}}$

and the random variable

${\displaystyle U_{n}^{a,b}:=\sup\{k\in \mathbb {N} \mid \sigma _{k}\leq n\}}$

the number of crossings the inequality

${\displaystyle \mathbb {E} [U_{n}^{a,b}]\leq {\frac {\mathbb {E} [(X_{n}-a)^{+}]-\mathbb {E} [(X_{0}-a)^{+}]}{b-a}}}$

Fulfills. From this, by means of the inequality from the assumed uniform restriction, it can be inferred that is also uniformly restricted. The monotonous Limes does exist, however, and it follows . But for any real numbers it holds ${\displaystyle \mathbb {E} [(X_{n}-a)^{+}]\leq |a|+\mathbb {E} [X_{n}^{+}]}$ ${\displaystyle \mathbb {E} [X_{n}^{+}]}$ ${\displaystyle \mathbb {E} [U_{n}^{a,b}]}$ ${\displaystyle U^{a,b}:=\lim _{n\to \infty }U_{n}^{a,b}}$ ${\displaystyle P(U^{a,b}<\infty )=1}$ ${\displaystyle b>a}$

${\displaystyle \left\{\liminf _{n\to \infty }X_{n}<a\right\}\cap \left\{\limsup _{n\to \infty }X_{n}>b\right\}\subset \left\{U^{a,b}=\infty \right\}}$

and it follows that the event

${\displaystyle \bigcup _{a,b\in \mathbb {Q} }\left\{\liminf _{n\to \infty }X_{n}<a\right\}\cap \left\{\limsup _{n\to \infty }X_{n}>b\right\}}$

almost certainly not occurring. So it will almost certainly converge against one . According to Fatou's lemma , on the one hand , similar one inferred. ${\displaystyle X_{n}}$ ${\displaystyle X_{\infty }}$ ${\displaystyle \mathbb {E} [X_{\infty }^{+}]\leq \sup\{\mathbb {E} [X_{n}^{+}]\mid n\geq 0\}<\infty }$ ${\displaystyle \mathbb {E} [X_{\infty }^{-}]<\infty }$

Convergence in p-th mean

Let there be a constant with for all that is, the sequence is limited in in space Then there is a -measurable random variable with almost certain and in . ${\displaystyle p>1}$ ${\displaystyle M>0}$ ${\displaystyle \mathbb {E} (|X_{n}|^{p})\leq M}$ ${\displaystyle n\in \mathbb {N} _{0},}$ ${\displaystyle (X_{n})_{n\in \mathbb {N} _{0}}}$ ${\displaystyle {\mathcal {L}}^{p}.}$ ${\displaystyle {\mathcal {F}}_{\infty }}$ ${\displaystyle X_{\infty }\in {\mathcal {L}}^{p}}$ ${\displaystyle X_{n}{\xrightarrow {n\to \infty }}X_{\infty }}$ ${\displaystyle {\mathcal {L}}^{p}}$

The statement is generally wrong for: A martingale bounded in in does not necessarily have to converge in. ${\displaystyle p=1}$ ${\displaystyle {\mathcal {L}}^{1}}$ ${\displaystyle {\mathcal {L}}^{1}}$

Convergence with equal integrability

If a submartingale is equally integrable , then there is a measurable random variable with almost certain and in . ${\displaystyle (X_{n})_{n\in \mathbb {N} _{0}}}$ ${\displaystyle {\mathcal {F}}_{\infty }}$ ${\displaystyle X_{\infty }\in {\mathcal {L}}^{1}}$ ${\displaystyle X_{n}{\xrightarrow {n\to \infty }}X_{\infty }}$ ${\displaystyle {\mathcal {L}}^{1}}$

Further applies and, in the event that there is a martingale, even . It is said that the martingale is closed by . ${\displaystyle X_{n}\leq \mathbb {E} (X_{\infty }\mid {\mathcal {F}}_{n})}$ ${\displaystyle (X_{n})_{n\in \mathbb {N} _{0}}}$ ${\displaystyle X_{n}=\mathbb {E} (X_{\infty }\mid {\mathcal {F}}_{n})}$ ${\displaystyle X_{\infty }}$

example

The simple symmetrical random walk with independent, identically distributed and is a martingale. Because of this , no path is convergent. ${\displaystyle X_{n}:=\sum _{j=1}^{n}Z_{j}}$ ${\displaystyle Z_{j}}$ ${\displaystyle P(Z_{j}=1)=P(Z_{j}=-1)={\frac {1}{2}}}$ ${\displaystyle |X_{n+1}-X_{n}|=1}$

For is given by a stop time and the stopped martingale with is also a martingale. Because of this , it fulfills the requirements of the Martingale Convergence Theorem for almost certain convergence. The only possible limit is , so it applies ${\displaystyle a\in \mathbb {N} =\{1,2,\dotsc \}}$ ${\displaystyle \tau :=\inf\{n>0:X_{n}=a\}}$ ${\displaystyle (M_{n})}$ ${\displaystyle M_{n}=X_{\min(n,\tau )}}$ ${\displaystyle M_{n}^{+}\leq a}$ ${\displaystyle a}$

${\displaystyle M_{n}{\xrightarrow {n\to \infty }}a}$ pretty sure.

In particular, it follows that . ${\displaystyle P(\tau <\infty )=1}$

Because of the Martingale is in limited. However, it does not converge in against , because in this case it should also converge against , in contradiction to for all . ${\displaystyle \mathbb {E} (|M_{n}|)=\mathbb {E} (2M_{n}^{+}-M_{n})=2\mathbb {E} (M_{n}^{+})-\mathbb {E} (M_{n})\leq 2a}$ ${\displaystyle (M_{n})}$ ${\displaystyle {\mathcal {L}}^{1}}$ ${\displaystyle {\mathcal {L}}^{1}}$ ${\displaystyle a}$ ${\displaystyle E(M_{n})}$ ${\displaystyle a>0}$ ${\displaystyle \mathbb {E} (M_{n})=\mathbb {E} (M_{0})=0}$ ${\displaystyle n}$

literature

• Achim Klenke: Probability Theory . 2nd Edition. Springer-Verlag, Berlin Heidelberg 2008, ISBN 978-3-540-76317-8 , section 11.2.