Failure of the Martingale Property For Stochastic Integration

If X is a cadlag martingale and {\xi} is a uniformly bounded predictable process, then is the integral

\displaystyle  Y=\int\xi\,dX (1)

a martingale? If {\xi} 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 {L^p}-integrable martingale, any {p > 1}, then so is Y. Furthermore, integrability of the maximum {\sup_{s\le t}\lvert X_s\rvert} 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 {\xi} to be locally bounded. In the general situation for cadlag martingales X and bounded predictable {\xi}, 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 {\{X_t\}_{t\in[0,1]}} be any of the examples of cadlag martingales with non-integrable maximum given in the previous post. Denote the running maximum by

\displaystyle  X^*_t=\sup_{s\le t}X_s.

For the given examples, {X^*} is a continuous process such that {X^*_1} is non-integrable. We consider letting the integrand in (1) be a bounded measurable function of {X^*}, {\xi_t=u(X^*_t)}. Then,

\displaystyle  \setlength\arraycolsep{2pt} \begin{array}{rl} \displaystyle \int_0^tu(X^*_s)\,dX_s&\displaystyle=U(X^*_t)-U(X_0)-u(X^*_t)(X_t-X_0),\smallskip\\ \displaystyle U(x)&\displaystyle=\int u(x)\,dx. \end{array} (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

\displaystyle  u(x)=\sum_{n=-\infty}^\infty 1_{\{n\le x < n+1\}}(-1)^n.

Then, {\lvert u(x)\rvert=1} and {\lvert U(x)-U(y)\rvert\le1} for all x and y. Using (2),

\displaystyle  Y_1=U(X^*_1)-U(X_0)-u(X^*_1)(X^*_1-X_1).

Taking absolute values,

\displaystyle  \lvert Y_1\rvert\ge \lvert X^*_1\rvert-\lvert X_1\rvert - 1

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

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