Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Suppose I have a sequence of continuous time random variables $X_n(t)$ where $t \in [0,1]$, adapted to a filtration $F_t$, that are martingales with respect to this filtration and that $\sup_n E[\sup_{0 \leq t \leq 1 } |X_n(t)|] < \infty$. Note that the filtration does not depend on $n$. I assume quite strong convergence of the $X_n(t)$ to a limit $X(t)$, namely that they converge uniformly on $[0,1]$ and for each $t$, $X_n(t)$ converges in $L^1$. I would like to know what are the weakest conditions I can assume about the $X_n(t)$ so that $X(t)$ will also be a martingale. If I assume the $X_n(t)$ are continuous or even left continuous then I believe that the limit will also be a martingale. If I just want to assume right continuity, I think I need that the fixed time distributions of $X(t)$ are non-singular. What I'd really like to know is if some kind of continuity of the $X_n$'s is a requirement to push the martingale through to the limit.

share|improve this question
    
Isn't saying $X(t)$ is a martingale for $(F_t)$ equivalent to saying $E[X(T)] = E[X(0)]$ for all stopping times $T$? If that is so, then won't your uniform convergence preserve this? –  Gerald Edgar Dec 13 '11 at 1:51
    
@Gerald: Yes, but he didn't say anything about convergence of X(T) for stopping times T. –  George Lowther Dec 13 '11 at 2:53
2  
But, the fact that $X_n(t)$ converges to $X(t)$ in $L^1$ for each time $t$ is enough. For any $0\le s < t\le1$ and $A\in F_s$, you have$$\mathbb{E}[1_A(X(t)-X(s))]=\lim_n\mathbb{E}[1_A(X_n(t)-X_n(s))]=0.$$ –  George Lowther Dec 13 '11 at 2:55
    
@George. Yes, of course. I got confused about the definition of a martingale. Your calculation shows that for fixed $t,s$, $E[X(t)|F_s] = X(s)$ a.s., and I thought I wanted something like the statement, a.s. for every fixed $t,s$, $E[X(t)|F_s] = X(s)$ which in retrospect doesn't make sense since the equality is only defined up to measure zero sets. Thanks. –  Ben Dec 13 '11 at 3:18

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.