Sign up ×
Quantitative Finance Stack Exchange is a question and answer site for finance professionals and academics. It's 100% free, no registration required.

A process $X_t$ is a local martingale if for each increasing sequence of stopping times $\{\tau_k,k=1,2,...\}$ the stopped process is a martingale. All true martingales are local martingales, but the inverse is not true. A strict local martingale is a local martingale which is not a true martigale. In fact, a positive strict local martingale is a supermartingale -- i.e. the expectation decreases with horizon.

In quant finance strictly local martingales have appeared as models which exhibit volatility induced stationarity or models that describe financial bubbles.

All Ito processes desrcibed by a 'driftless SDE' are in fact local martingales, but not martingales (which is surprising to many). For example the familiar Geometric Brownian motion $$dX_t = X_t\ dW_t$$ is a local martingale and a true martingale, while the CEV model with exponent greater than one $$dY_t = Y_t^\alpha\ dW_t \text{ (for given }\alpha > 1\text{)}$$ is a local martingale but not a true martingale. In fact, starting from $Y_0$ the expectation $$E[Y_t|\mathcal{F}_0] < Y_0 \text{ (for all }t>0\text{)}$$

My question is on the intution behind their dynamics and paths:

  • What are the qualitative features that break the martingality for such a process as $\alpha$ crosses 1?
  • How do paths of strict local martingales look like (against true martingales)? They are not explosive, they are not hitting zero, and I cannot see anything particularly 'strange' or 'unusual' when I plot them.
  • How come that even when I simulate this process (with Euler) I get a negative drift, even though I am adding up a finite number of Gaussians with zero mean (although strictly local martingales are only a continuous time phenomenon)?

This blog post discusses these points, but I am looking for something more high level and (if possible) intuitive.

share|improve this question
Very interesting question but please correct the typo: do you mean "supermartingale" or "submartingale"? thanks – Richard Jul 31 '14 at 11:46
An application of strict local martingales is in the modelling of financial bubbles as Protter does see e.g. here – Richard Jul 31 '14 at 11:50
Thanks @Richard, emcor also pointed out the typo. I made some more corrections and edits. – Kiwiakos Jul 31 '14 at 12:33
@Kiwiakos, actually, this is directly related to my other question which you answered. Exchange rates that experience explosive hyperinflation could be an example of a strict local martingale (e.g., ARS on Jan 30, 2014). I don't necessarily agree that "they are not explosive." Indeed, the process is explosive during that short interval. You may Peter Carr's research on the subject helpful: – nsw Jul 31 '14 at 17:41
Thanks @nsw, this is a good point. I can agree that the strict local martingale looks 'explosive over a small interval', but my intuition cannot reconcile it with the fact the it then becomes a supermartingale, i.e. that the expectation becomes smaller the more 'explosive' it becomes. If it was the other way round, and the expectation also increased, then I would probably accept it easier. – Kiwiakos Aug 1 '14 at 16:32

3 Answers 3

I think to understand the martingale/local martingale distinction, it helps to bring in a third class of processes, the uniformly integrable martingale. I would argue that the local martingale and the non-uniformly integrable (true) martingale are actually fairly similar.

The key property that a uniformly integrable martingale has is the so-called closure property. Let $M_t, 0 \leq t < \infty$ be a uniformly integrable (UI) martingale. Then $M$ has a last element $M_\infty$, and the extended process $M_t, 0 \leq t \leq \infty$ is a martingale, and you can compute $E[M_\infty | \mathcal{F}_t ]$ to get all intermediate values of the martingale. We know that if a martingale is not uniformly integrable, then this is not the case.

One nice example is actually geometric brownian motion, which I'll call $X_t$. We know that almost surely as $t \rightarrow \infty, X_t \rightarrow 0$, although for all $t$ $E[X_t] = 1$. In the limit, as $t \rightarrow \infty$, the martingale loses its mass. This phenomenon is actually exactly the same thing that is happening with strict local martingales.

Consider some function that maps the unit interval to the positive reals, like $f(x) = \tan x$. Consider the process $\widetilde{X}_t := X_{f(t)}, 0 \leq t \leq 1$. You will now note that on the unit interval $[0,1]$, $\widetilde{X}$ is a strict local martingale. The sequence of stopping times can just be taken to be deterministic numbers increasing to $1$.

I hope this example shows how strict local martingales and non uniformly-integrable martingales are in effect the same thing. The crucial property is that their martingality does not extend to "the closure of the time interval". This happens because, somewhere, you have a set of random variables which is not uniformly integrable, and so in passing to the limit, you don't have continuity in $L^1$.

To answer a couple other questions:

  1. What do paths look like I think my discussion above shows that there is really no difference between the local path behavior of a strict local martingale, and the local path behavior of a true martingale. The breakdown occurs in the limit. In fact, this is where the name comes from: locally, a local martingale does look like a martingale.

  2. Simulation When you simulate, you should get zero drift. In the limit is where the mass will be lost. It's hard to say, could there be any underflow when $Y$ is small?

share|improve this answer

You are asking an interesting question.

Firstly, a Submartingale has increasing or equal expectation (not decreasing).

Secondly, the process $dX_t=X_tdW_t$ is a true martingale (not strictly local), since its solution (by Ito):


has $E(X_t)=X_0$ constant expectation ($e^{-\frac{t}{2}}E(e^{W_t})=1, W_t\sim N(0,t)$). The negative drift comes from $W_t$'s nonzero quadratic variation. Since $e^X$ is always positive, there needs to be a correcting factor $-t/2$ to ensure the martingale expectation property.

The dynamics are similar to the Black-Scholes Model:

$$dS_t=S_t(\mu dt+\sigma dW_t)$$

has known solution

$$S_t=S_0e^{(\mu-\frac{1}{2}\sigma^2) t +\sigma W_t}$$

This process is not a martingale because of its drift, it becomes so after applying a change of measure to $Q$.

share|improve this answer
You are right about 'sub': my mistake, I edited it to 'super'. But the GBM with drift is not a local martingale. – Kiwiakos Jul 31 '14 at 11:36
@Kiwiakos Thanks also, I corrected it. I asked a colleague it appears that local martingales are usually hard to find in practice, but for the theorems the local martingale property is sometimes more precise to use. There are some rather exotic examples on Wikipedia. – emcor Jul 31 '14 at 15:33

@emcor this is interesting. A log normal martingale has most paths drift downwards. And so appears to behave like a supermartingale. If you would like to check, simulate a BS model with drift $\mu < \frac{1}{2}\sigma^2$ and check. Almost all paths drift downwards. You should try to think of why this has to be, though enough numerical simulations will explain why. The key is the martingality, which is an expectational feature. Dupire has this result in his slides.

Any comments? FYI, any process without drift is essentially behaving like a Brownian I think.

share|improve this answer
As I wrote, $S_t$ with drift $\mu$ is not a martingale, so its either sub-martingale or super-martingale apparently (which does not surprise me). If the process has no drift such as zero expectation, it need not behave like a brownian e.g. the sinus curve is not a Brownian. – emcor Jan 7 at 23:28
Its a discounted martingale, thats not what Im pointing to. Discount it and try it, the deterministic thing isn't the point. – Drew Jan 8 at 0:37

Your Answer


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

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