Let $X_t$ be a continuous square-integrable martingale. Then it is true (I think, please correct me if I am wrong) that:

$$\forall t \in [0,\infty), \quad \int_0^t 1_{[0,t]}(s) dX_s = X_t - X_0$$

So couldn't we use this as a definition of $\int_0^t 1 (s) dX_s$ (where $1$ is the constant function)? I.e.

$$\forall t \in [0,\infty), \quad \int_0^t 1(s) dX_s = \int_0^t 1_{[0,t]}(s)dX_s$$

I am confused because the integral on the right is supposed to be a square integrable martingale, but the one on the left is supposed to be just a local martingale, because the integrand on the left is not square-integrable, only locally $L^2-$bounded, whereas the integrand on the right is square-integrable.

Where is my mistake?

square-integrable = $L^2$-bounded: $\mathbb{E}[\int_0^{\infty} |f_s|^2 \mathrm{d}s]< \infty$

locally square-integrable/$L^2$-bounded: $\forall t \in [0,\infty) \quad \mathbb{E}[\int_0^t |f_s|^2 \mathrm{d}s] < \infty$

Seemingly we could do the same trick for any locally $L^2-$bounded process, and we would always get a square integrable martingale for any finite $t$ (just integrate $(f\cdot 1_{[0,t]})(s)$ instead of $f(s)$).

But Revuz and Yor says that the integrand needs to be square integrable, not just locally square integrable, for the stochastic integral to again be a continuous square-integrable martingale, and that if the integrand is only locally square integrable, then the resulting process is a continuous local martingale, but not a true martingale.

  • 2
    $\begingroup$ You seem to be confusing time intervals and state intervals. For example, the meaning of $$\int_0^t 1_{[0,t]} dX_t$$ is quite unclear since $1_{[0,t]}$ is defined on $\mathbb R_+$ while every $X_s$ is defined on $\Omega$. On the other hand $$\int_0^\infty 1_{[0,t]}(s)dX_s=\int_0^tdX_s=X_t-X_0.$$ $\endgroup$ – Did Jul 19 '16 at 8:43

Section 5.4 on p. 13 of this document, seems to confirm the notion that we can extend the class of allowable integrands to locally square-integrable processes (or at least when we restrict to predictable processes, instead of general progressively measurable processes). Not only that, but apparently we can also expand the class of integrators to include locally square-integrable martingales (including Brownian motion).

Actually re-reading the section of Revuz and Yor which I cited earlier, that appears to be what they are discussing before moving on to discuss local martingale integrals. My professor told me once (seemingly in error or more likely I wasn't paying attention and misheard them) that Brownian motion is a local martingale but not a true martingale, and now I have difficulty convincing myself that this is not true. It doesn't matter that Brownian motion isn't integrable, since it is integrable on $[0,t]$ for all $t$, that is enough.


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