I read the proof of $L^p$ convergence theorem of martingales but I can't exactly figure out that whether we can find a square integrable martingale that converges almost surely but not in $L^2$.
a question about $L^p $ convergence theorem
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probability-theory
convergence
martingales
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0If you are assuming $\sup_t E[|M_t|^p]<\infty$, for some $p>1$, then the convergence is both in $L^p$ and a.s. The only tricky case is $L^1$ – 2012-11-29
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0No. I mean that is there a case where each ${X_n}^2$ is integrable but the whole thing (which although has an $a.s.$ convergence) does not converge in $L^2$ (in fact as you say we are looking for a martingale which $sup E|X_n|$ is not bounded but $E|X_n|$ is.) – 2012-11-29
1 Answers
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Let $\xi_j$ iid random variables, $\xi_j \sim \frac{1}{2} (\delta_{-1}+\delta_1)$. Then
$$X_n := \sum_{j=1}^n \xi_j \in L^2 \qquad \quad X_0 := 0$$
and $(X_n)_{n \in \mathbb{N}_0}$ is a martingale. Let $S:=\inf \{n \geq 0; X_n=-1\}$. By optional stopping we obtain that $Y_n := X_{S \wedge n}+1$ is a martingale. Since $Y_n \geq 0$ there exists
$$Y_\infty = \lim_n Y_n = X_S+1 = -1+1=0 \quad \text{a.s.}$$
But: $\mathbb{E}Y_n = \mathbb{E}Y_0=1$, $\mathbb{E}Y_\infty=0$. Hence $Y_n \not \to Y_\infty$ in $L^1$ (in particular $Y_n \not \to Y_\infty$ in $L^2$).