For me, a stochastic $(M_t)_{t\in [0,T]}$ is a martingale (w.r.t. $(\Omega ,(\mathcal F_t)_t,\mathbb P)$) if $M_t$ is $\mathcal F_t$ adapted and $$\mathbb E[M_t\mid \mathcal F_s]=M_s,\quad s\leq t.$$

A local martingale is a stochastic process $(M_t)_t$ s.t. there are stoping times $(\tau_n)$ almost increasing s.t. $\tau_n\to \infty$ a.s. and s.t. $(M_{t\wedge \tau_n})_{t\geq 0}$ is a martingale for all $n$.

Q1) So at the end, if $(M_t)_{t\in [0,T]}$ is a martingale for all $T>0$, then $(M_t)_{t\geq 0}$ is a local Martingale, right ?

Q2) If $(M_t)_{t\in [0,T]}$ is not a martingale, can it be a local martingale in the sense that there are stopping time $(\tau_n)_n$ that are a.s. increasing s.t. $\tau_n\to t$ and $(M_{t\wedge \tau_n})_{t\geq 0}$ or not really ?

  • 2
    $\begingroup$ You're correct: martingale implies local martingale (take $\tau_n=n$) and local martingale does not imply martingale (see examples here) $\endgroup$ Aug 7, 2019 at 16:01
  • $\begingroup$ @C.Windolf: So at the end, local martingale is really to extend martingale $(M_t)_{t\in [0,T]}$ as $(M_{t})_{t\geq 0})$ right ? And what about for Q2) ? $\endgroup$
    – user659895
    Aug 7, 2019 at 16:18
  • $\begingroup$ No -- the definition of martingale does not exclude the time domain being $[0,\infty)$ or even $\mathbb{R}$ or any other ordered set. See the general definition here $\endgroup$ Aug 7, 2019 at 16:23
  • $\begingroup$ As for Q2: your Q2 is whether there exist local martingales that are not martingales, or in other words, does the local martingale property imply the martingale property? I tried to address this by linking to examples of local martingales that are not full martingales in my first comment. $\endgroup$ Aug 7, 2019 at 16:23
  • $\begingroup$ Following your first link, local martingale looks to be with $T=\mathbb R$ only (that's why I ask the question Q2)). So I probably missed something, according to your first link, Q2) is something else than : local martingale implies martingale (that I know it's not correct). It's really : can $(M_t)_{t\in [0,T]}$ be not a Martingale but there is $(\tau_n)$ of stopping time s.t. $\tau_n$ a.s. increasing, $\tau_n\to t$ and $(M_{t,\tau_n})_t$ is a martingale for all $n$. $\endgroup$
    – user659895
    Aug 7, 2019 at 16:43

1 Answer 1


Before trying to understand the difference between martingales and local martingales on a technical level, it pays to have an intuitive understanding of the difference: that is what I will attempt to provide in the rest of this answer. But before doing that, let me quickly answer your two specific questions.

  1. As mentioned in the comments, it is easy to see that every martingale is also a local martingale.
  2. If you mimic the definition of a local martingale but on a bounded time interval instead of $[0,\infty)$ you only get martingales, not something more general - see here for a similar argument.

On an intuitive level, I think it pays to think first about what "local" means in the phrase "local martingale". There are some wonderful articles explaining what "localization" means in the theory of stochastic processes, but let me give my take on the concept as well. In many "geometrical" areas of math, localizing an object involves zooming in on a part of its domain to tease out limiting information about the object near a point. (The derivative is a quintessential example of a local operation.)

In the case of stochastic processes, localization is a way of "zooming in" on time "$t=\infty$". But instead of using a sequence of deterministic scales (like in the geometrical examples of localization you may be familiar with), we allow our scales to be stochastic (this should not come as a surprise...) which means that instead of "probing $t=\infty$" by using a deterministic sequence $t_n$ tending to $\infty$, we use a sequence of stopping times $\tau_n$ tending to $\infty$.

Why limit ourselves to stopping times and not just any sequence of random times? Well, it is easy to justify in the case of martingales that you are interested in: they are precisely the random times at which we can stop a martingale and still have a martingale. In other words, working with a martingale stopped at a stopping time is no more general than simply working with a martingale. Now we can "localize at $\infty$" by taking a limit of these stopped martingales, and the definition of a localized martingale pops out.

One other aspect of your question is the role played by the domain of the time variable. As pointed out in the comments, the time interval in a martingale is allowed to be completely general: it could be $[0,T]$ as in your question, or $[0,\infty)$, or even $[0,\infty]$. However, the definition of a local martingale requires the time domain to be $[0,\infty)$. The reason for this is that in order to zoom in on $\infty$, our process must be defined in a "neighborhood" of $\infty$, meaning that it must be defined for arbitrarily large times.

Thinking about $\infty$ as the boundary of the time domain $[0,\infty)$ brings the concept of local martingales in line with the idea of compactification. Compactification occurs whenever we want to promote limiting objects into actual concrete objects we can work with. An archetypical example here is the delta "function", which is not actually a function but rather a (weak) limit of functions spiking near the origin. Compactifying the space of functions in a suitable way to include the delta "function" leads to the space of distributions, a.k.a. generalized functions. Another (less flashy but perhaps more relevant) example of compactification is the one-point compactification, which when applied to the non-compact topological space $[0,\infty)$ yields the compact space $[0,\infty]$ (when given the topology that makes my earlier comment about "neighborhoods of $\infty$" precise).

The natural thing to say now is that the space of local martingales is the compactification of the space of martingales, but I actually don't know if this is true precisely - and formalizing it would be a little too far off topic for this post (but if you want to learn more about this here are some breadcrumbs: ucp convergence, completeness of local martingales, density in space of local martingales).

  • 2
    $\begingroup$ That was an embarrassing typo, thanks for the catch and apologies for the confusion. I have corrected it. $\endgroup$
    – pre-kidney
    Aug 8, 2019 at 6:30

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .