Strict Local Martingales, Lost Mass, and Pathwise Intuition

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


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












A process $X_t$ is a local martingale if there exists an increasing sequence of stopping times $\{\tau_k,k=1,2,...\}$, with $\tau_k \to \infty$ almost surely, such that each 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.

## Answer by quasi (score 18)

https://quant.stackexchange.com/a/16133

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:

- 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.

- 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?

## Answer by xml (score 2)

https://quant.stackexchange.com/a/65899

A quantitative barrier to a local martingale being a true martingale is integrability. An example is as follows: $\int_0^t f(s) dB_s$ where $B_s$ is a Brownian motion and $f$ progressively measurable is a strict local martingale if $(E\int_0^t |f(s)|^2 ds)^{\alpha/2}=\infty$ for some $\alpha$ between $0$ and $1$. This is taken from Corollary 4 of `Strict local martingales: Examples' in 129 (2017) 65–68, by Xue-Mei Li.

## Answer by emcor (score 1)

https://quant.stackexchange.com/a/14234

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 $Q$.

## Answer by MiSa (score 0)

https://quant.stackexchange.com/a/25477

It seems that uniformly integrable martingales, as described by quasi, account for a specific class of strict local martingales.

A martingale on $[0,\infty)$ that is not uniformly integrable (like geometric Brownian motion) is a uniformly integrable martingale on $[0,t]$ for every $t\in [0,\infty)$. Consequently, mapping $[0,\infty]$ onto $[0,1]$ as done above creates a local martingale that is a true martingale on every interval $[0,s]$ for $s\in [0,1)$. The local martingale property appears only at $1$ (as the non-uniform integrability appeared at $\infty$).

Nevertheless, strict local martingales like $Y$, solution of the CEV model given as example by Kiwiakos, are strict on every interval $[0,s]$ for $s\in [0,1)$ ($E[Y_s]$ is strictly decreasing in $s$). This suggest that those processes are not associated with a non-uniformly integrable true martingale on $[0,\infty)$. Also, here stopping times will generally be path-dependent (and not deterministic).

I don't know if for such local martingales one can still relate local path behaviour to non-uniformly integrable martingales, however, viewing the local martingale property as appearing at a single "closure point" seems too narrow.

## Answer by Drew (score -2)

https://quant.stackexchange.com/a/16122

@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.