Given a sequence of cadlag (right-continuous with left limits) martingales $X^n=(X^n_t)_{0\le t\le 1}$, we may use the well known criteria to determine whether it is weakly convergent, i.e. subtract a subsequence $(X^{n_k})_{k\ge 1}$ such that for all Skorokhod-continuous and bounded function $f$ one has


where the process $X$ is called the weak limit, that is again a martingale. Now, let us consider a different convergence. The sequence $(X^{n})_{n\ge 1}$ is said to be point-wise weakly convergent iff for any subsequence $(X^{n_k})_{k\ge 1}$ there exists a cadlag process $X$ (which is also a martingale) such that

$$(X_{t_1}^{n_k},\ldots, X_{t_m}^{n_k})~~\stackrel{Law}{\longrightarrow}~~(X_{t_1},\ldots, X_{t_m}) \mbox{ for all } 0\le t_1\le \cdots t_m\le 1.$$

My question is the following: Assume that the sequence of martingales have same marginal distributions, i.e. $Law(X_t^n)=\mu_t$ for all $n\ge 1$, where $(\mu_t)_{0\le t\le 1}$ is a sequence of distributions on $\mathbb R$, then could we show that the sequence $(X^{n})_{n\ge 1}$ is point-wise weakly convergent? I believe strongly that it is not true, but cannot find a counterexample. Does some give an example or prove this claim? Thanks a lot for the reply!

  • $\begingroup$ I am surprised with the name "weak convergence" for these two concepts, which to me seem just compactness properties. "Convergence" implies one and only one limit, in my book. Are you sure it is accepted usage ? $\endgroup$ Nov 12, 2015 at 14:34
  • $\begingroup$ @Jean Duchon: Thanks a lot the edit! $\endgroup$
    – CodeGolf
    Nov 12, 2015 at 14:34
  • $\begingroup$ Thanks for pointing that. It is indeed the problem about compactness. I say only finite-dimensional distribution convergence in order to avoid introduce the topology generated by point-wise convergence on Skorokhod space. $\endgroup$
    – CodeGolf
    Nov 12, 2015 at 14:37
  • $\begingroup$ What I am questioning is the name "convergence", not which topology to use on the set of probability measures on the Skorokhod space... To me it's compactness, not convergence, be it narrow (the first one) or pointwise. $\endgroup$ Nov 12, 2015 at 14:50

1 Answer 1


Since that this question does not draw much attention, I would like to present what I'm thinking about. First, thanks for the reply of Jean Duchon, I decide to make my question clearer.

Denote $D:=D([0,1],\mathbb R)$ the Skorokhod space of all cadlag functions $\omega=(\omega_t)_{0\le t\le 1}$ defined on $[0,1]$. Now let us endow $D$ with the topology induced by the point-wise convergence, i.e. $\omega^n$ converge to $\omega$ iff $\omega^n_t$ converges to $\omega_t$ for all $t\in [0,1]$.

For the sequence of martingales described above $(X^n)_{n\ge 1}$, I wonder whether it is compact w.r.t. to this topology, i.e. find some subsequence $(X^{n_k})_{k\ge 1}$ and another martingale $X$ such that

$$Law(X^{n_k}_{t_1},\ldots,X^{n_k}_{t_m})~~\longrightarrow~~Law(X^{n_k}_{t_1},\ldots,X^{n_k}_{t_m}) \mbox{ for all } 0\le t_1\le \cdots\le t_m\le 1.$$

How to prove or disapprove this claim? A natural idea is that we take first a arbitrary countable set $Q\subset [0,1]$ and define

$$D_{Q}~~:=~~\left\{\bar\omega=(\bar\omega_t)_{t\in Q}: \exists \omega\in D \mbox{ s.t. } \omega_t=\bar\omega_t \mbox{ for all } t\in Q\right\}.$$

If we denote by $\mathbb R^Q$ the product space of $\mathbb R$ and use again $\bar\omega$ to represent its elements, then we have the following result:

Given $\bar\omega\in\mathbb R^Q$, it belongs to $D_Q$ iff the map $t\mapsto \bar\omega_t$ is cadlag and the number of upcrossings of $\bar\omega$ through the interval $[0,1]$ is finite, see e.g. for the definition of the number of upcrossings in the blog of Almost Sure


Then fix such a countable set $1\in Q\subset [0,1]$, it is easy to show that there exists a subsequence $(X^{n_k})_{k\ge 1}$ and a martingale $X$ such that

$$Law(X^{n_k}_{t_1},\ldots,X^{n_k}_{t_m})~~\longrightarrow~~Law(X^{n_k}_{t_1},\ldots,X^{n_k}_{t_m}) \mbox{ for all } t_1\le \cdots\le t_m\in Q.$$

Now, with the supplementary condition that $Law(X^{n_k}_t)=\mu_t$ for all $k\ge 1$, we may deduce that $Law(X_t)=\mu_t$ for all $t\in [0,1]$. My question is whether we may show further that

$$Law(X^{n_k}_{t_1},\ldots,X^{n_k}_{t_m})~~\longrightarrow~~Law(X^{n_k}_{t_1},\ldots,X^{n_k}_{t_m}) \mbox{ for all } 0\le t_1\le \cdots\le t_m\le 1?$$

Thanks again Jean Duchon for his remarks!

  • 1
    $\begingroup$ @Jean Duchon: According to the Exercise 1.12 in the book "Exercises in Probability: A Guided Tour from Measure Theory to Random Processes", if the two random variables $Y_s=E[Y_t|Y_s]$ and $Y_t$ have the same law, they are equal almost surely. Could you show a bit more about the hint? $\endgroup$
    – CodeGolf
    Nov 12, 2015 at 15:45
  • 1
    $\begingroup$ No, the hint I had was wrong, and I have no idea whether the set of (laws of) martingales with given marginals is compact, in any topology, or not. $\endgroup$ Nov 14, 2015 at 13:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.