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In the martingale convergent theorem:

$X_n$ is a martingale with $\sup_n\mathbb E[|X_n|]<\infty$.

Then there exist a $X_\infty\in L^1$ such that $X_n\to X_\infty\text{ a.s.}$

I want to find a example that $\mathbb E[X_\infty\mid\mathcal F_n]=X_n$ for all $n$ not hold.

Thanks in advance.

saz
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Counterexample 1: Let $\xi_j$ be independent identically distributed random variables such that $$\mathbb{P}(\xi_j = 0) = \mathbb{P}(\xi_j =2)=\frac{1}{2}.$$ Then $$X_n := \prod_{j=1}^n \xi_j$$ is a non-negative martingale satisfying $\mathbb{E}(X_n)=1$. On the other hand, it is not difficult to see that $X_n \to 0$ almost surely. As $$\mathbb{E}(X_{\infty} \mid \mathcal{F}_n) = 0 \neq X_n$$ we are done.

Counterexample 2: Let $\xi_j$ be independent identically distributed random variables such that $\mathbb{P}(\xi_j = 1)= \mathbb{P}(\xi_j = -1) = 1/2$. Obviously, $X_n := \sum_{j=1}^n \xi_j$ defines a martingale. Define a stopping time $\tau$ by

$$\tau := \inf\{n \geq 0; X_n = -1\}$$

and set $Y_n := 1+ X_{\tau \wedge n}$. By the optional stopping theorem, $(Y_n)_{n \in \mathbb{N}}$ is a martingale. Since $Y_n$ is non-negative we find $$\sup_n \mathbb{E}(|Y_n|) = \sup_n \mathbb{E}(Y_n) = \mathbb{E}Y_0 = 1.$$ It follows from the martingale converence theorem that $$Y_{\infty} = \lim_{n \to \infty} (X_{\tau \wedge n}+1) = X_{\tau}+1 = 0.$$ Again, this implies that $$\mathbb{E}(Y_{\infty} \mid \mathcal{F}_n) \neq Y_n.$$

saz
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  • For example 1, how do you show almost sure convergence? I can show convergence in probability ($P(X_n \neq 0 ) = (1/2)^n \to 0$), how do I make this almost sure?

    Is it that we know $X_n$ has to converge almost surely by Doob's forward convergence theorem, so the a.s limit must be the probabilisitic liimt? (sorry if I'm over-complicating this)

     – user428487 Jun 24 '18 at 10:25
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    @user428487 Since $\sum_{j \geq 1} \mathbb{P}(\xi_j=0) = \infty$ it follows from the Borel-Cantelli lemma that there exists for almost all $\omega \in \Omega$ at least one $j=j(\omega)$ such that $\xi_j(\omega)=0$ (in fact, Borel Cantelli gives that there are infinitely many such $j$). By the very definition of $X_n$ this implies $X_n(\omega)=0$ for all $n \geq j(\omega)$. This shows, in particular, that $X_n \to 0$ almost surely. – saz Jun 24 '18 at 10:48