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I have a sequence of continuous time random variables $X_n(t)$ where $t \in [0,1]$. Suppose that there is a filtration $F_t$ such that for each $n$, $X_n$ is a martingale with respect to this filtration. Note that the filtration does not depend on $n$. Also assume that $$\sup_n E[\sup_{0 \leq t \leq 1 } |X_n(t)|] < \infty.$$

Finally suppose for every $t \in [0,1]$ $X_n(t)$ converge uniformly to a limit $X(t)$ and $X_n(t)$ 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.

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  • $\begingroup$ 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? $\endgroup$
    – Gerald Edgar
    Commented Dec 13, 2011 at 1:51
  • $\begingroup$ @Gerald: Yes, but he didn't say anything about convergence of X(T) for stopping times T. $\endgroup$
    – George Lowther
    Commented Dec 13, 2011 at 2:53
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    $\begingroup$ 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$$\mathbb{E}[1_A(X(t)-X(s))]=\lim_n\mathbb{E}[1_A(X_n(t)-X_n(s))]=0.$$ $\endgroup$
    – George Lowther
    Commented Dec 13, 2011 at 2:55
  • $\begingroup$ @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. $\endgroup$
    – Ben
    Commented Dec 13, 2011 at 3:18

1 Answer 1

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Posting the answer already given in comments:

All you really need is that $X_n(t) \to X(t)$ in $L^1$ for each $t$. Conditional expectation with respect to any $\sigma$-field is continuous with respect to $L^1$ convergence; this is an elementary property of conditional expectation which follows from the inequality $\left|E[X \mid \mathcal{F}]\right| \le E [|X| \mid \mathcal{F}]$ and taking unconditional expectation of both sides.

Given this, since for each $s \le t$ and each $n$ we have $X_n(s) = E[X_n(t) \mid \mathcal{F}_s]$, you may pass to the $L^1$ limit on both sides to get $X(s) = E[X(t) \mid \mathcal{F}_s]$. So $X$ is a martingale.

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  • $\begingroup$ Since a martingale is required to be adapted, we'll need completeness of the filtration for the conclusion from the $L^1$-convergence. $\endgroup$
    – 0xbadf00d
    Commented Jun 3, 2018 at 23:43

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