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.

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