Failure of the Martingale Property For Stochastic Integration

If X is a cadlag martingale and is a uniformly bounded predictable process, then is the integral

(1)

a martingale? If is elementary this is one of most basic properties of martingales. If X is a square integrable martingale, then so is Y. More generally, if X is an -integrable martingale, any , then so is Y. Furthermore, integrability of the maximum is enough to guarantee that Y is a martingale. Also, it is a fundamental result of stochastic integration that Y is at least a local martingale and, for this to be true, it is only necessary for X to be a local martingale and to be locally bounded. In the general situation for cadlag martingales X and bounded predictable , it need not be the case that Y is a martingale. In this post I will construct an example showing that Y can fail to be a martingale.

The integral (1) can only fail to be a martingale when the absolute maximum of X is non-integrable. So, start by choosing be any of the examples of cadlag martingales with non-integrable maximum given in the previous post. Denote the running maximum by

For the given examples, is a continuous process such that is non-integrable. We consider letting the integrand in (1) be a bounded measurable function of , . Then,

(2)

This is a fundamental identity in the theory of Azema-Yor processes, and a quick proof was given in an earlier post for the case where u is continuously differentiable. Using bounded convergence, (2) also holds whenever u is a limit of a uniformly bounded sequence of continuously differentiable functions. I will let u be the square wave function

Then, and for all x and y. Using (2),

Taking absolute values,

is not integrable. So, Y is not a martingale.