proof of Martingale criterion

Let $(\tau_{k})_{k\geq 1}$ be a localizing sequence of stopping times for $X$ . Then:

\displaystyle\Lambda_{\{\tau_{k}\geq n\}}

\displaystyle\uparrow\Lambda_{\Omega}\ \text{a.s.}\ k\to\infty,\forall n\in% \mathbb{N}

since $\{\tau_{k}\geq n\}\uparrow_{k}\bigcup_{1}^{\infty}\{\tau_{k}\geq n\}\forall\ n% \in\mathbb{N}$ .

\displaystyle\bigcup_{k=1}^{\infty}\{\tau_{k}\geq n\}

\displaystyle=\Omega\ \text{a.s., since}\ \tau_{k}\to\infty,\ \text{a.s.}

Now assume $EX_{n}^{-}<\infty,\forall n\geq n_{0}$ (the case $EX_{n}^{+}<\infty$ being analogous).

1) We have $EX_{n}^{-}<\infty\ \forall n\in\mathbb{N}$ .

We proceed by (backward) induction. For $n=n_{0}$ the statement holds.

$n\mapsto n-1$ :

\displaystyle(X^{\tau_{k}})^{-}

\displaystyle=(X_{\tau_{k}\wedge n}^{-})_{n\in\mathbb{N}}\ \text{submartingale}

We have:

	$\displaystyle\int_{\{\tau_{k}\geq n\}}X_{n-1}^{-}\,dP$	$\displaystyle=\int_{\{\tau_{k}\geq n\}}X_{\tau_{k}\wedge(n-1)}^{-}\,dP$
		$\displaystyle\leq\int_{\{\tau_{k}\geq n\}}X_{\tau_{k}\wedge n}^{-}\,dP=\int_{% \{\tau_{k}\geq n\}}X_{n}^{-}\,dP$
		$\displaystyle\leq\int X_{n}^{-}\,dP<\infty$

Where the first to second line is the submartingale property and the last line follows by induction hypothesis.

Using Fatou we get:

	$\displaystyle\int X_{n-1}^{-}\,dP$	$\displaystyle=\int\lim_{k\to\infty}X_{n-1}^{-}\Lambda_{\{\tau_{k}\geq n\}}\,dP$
		$\displaystyle\leq\liminf_{k\to\infty}\int X_{n-1}^{-}\Lambda_{\{\tau_{k}\geq n% \}}\,dP$
		$\displaystyle\leq\int X_{n}^{-}\,dP<\infty$

2) We have $X_{n}\in\mathscr{L}^{1}(n\in\mathbb{N})$ .

We have $X_{\tau_{k}\wedge n}^{+}\to X_{n}^{+}$ a.s., $k\to\infty,\forall n\in\mathbb{N}$ . With Fatou we get:

	$\displaystyle EX_{n}^{+}$	$\displaystyle\leq\liminf_{k\to\infty}EX_{\tau_{k}\wedge n}^{+}$
		$\displaystyle=EX_{0}+\liminf_{k\to\infty}EX_{\tau_{k}\wedge n}^{-}$
		$\displaystyle=EX_{0}+\liminf_{k\to\infty}E\left(\sum_{j=0}^{n-1}X_{j}^{-}% \Lambda_{\{\tau_{k}=j\}}+X_{n}^{-}\Lambda_{\{\tau_{k}\geq n\}}\right)$
		$\displaystyle\leq EX_{0}+\sum_{j=0}^{n}EX_{j}^{-}<\infty$

With 1) $X_{n}\in\mathscr{L}^{1}$ follows.

$X$ is a martingale, because $X_{n}^{\tau_{k}}\to X_{n}$ a.s. $k\to\infty$ and:

\displaystyle|X_{n}^{\tau_{k}}|

\displaystyle\leq\sum_{j=0}^{n}|X_{j}|\in\mathscr{L}^{1}\ \text{(}\ \mathscr{L% }^{1}\ \text{-bound)}

Thus $X_{n}^{\tau_{k}}\lx@stackrel{{\scriptstyle\begin{subarray}{c}L^{1}% \end{subarray}}}{{\longrightarrow}}X_{n},k\to\infty\ \forall n\in\mathbb{N}$ by bounded convergence theorem. Hence $X$ must be martingale and we are done. ∎

Title	proof of Martingale criterion
Canonical name	ProofOfMartingaleCriterion
Date of creation	2013-03-22 18:34:51
Last modified on	2013-03-22 18:34:51
Owner	karstenb (16623)
Last modified by	karstenb (16623)
Numerical id	5
Author	karstenb (16623)
Entry type	Proof
Classification	msc 60G07