# Strictly local martingales: what is the intuition behind them?

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.

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Very interesting question but please correct the typo: do you mean "supermartingale" or "submartingale"? thanks –  Richard Jul 31 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 at 11:50
Thanks @Richard, emcor also pointed out the typo. I made some more corrections and edits. –  Kiwiakos Jul 31 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: oxford-man.ox.ac.uk/~jruf/papers/nonEquiv.pdf –  nsw Jul 31 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 at 16:32

## 1 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):

$$X_t=X_0e^{W_t-\frac{t}{2}}$$

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 $Q$.

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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 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 at 15:33