# 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 $EX_{n}^{-}<\infty\ \forall n\in\mathbb{N}$.

$n\mapsto n-1$:

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

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.

3)

$X$ 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 ProofOfMartingaleCriterion 2013-03-22 18:34:51 2013-03-22 18:34:51 karstenb (16623) karstenb (16623) 5 karstenb (16623) Proof msc 60G07