Is there a process that is not square-integrable but still gives rise to a martingale when integrated w.r.t. Brownian motion?

Question

When an adapted process $X$ satisfies $\int_0^TX_t^2dt<\infty$ a.s. but not $E\int_0^TX_t^2dt<\infty$, the stochastic integral $\int_0^tX_sdB_s$, $0\le t\le T$, is only guaranteed to be a local martingale (where $B$ is the underlying Brownian motion). Is it still possible, nonetheless, that $\int_0^tX_sdB_s$ is a martingale?

Answer 1

Yes, that's possible.

Example Let $Y$ be a random variable which is independent from $(B_t)_{t \geq 0}$ and satisfies $$\mathbb{E}(|Y|) < \infty \quad \text{and} \quad \mathbb{E}(Y^2)=\infty$$ (e.g. an isotropic $\alpha$-stable random variable, $\alpha \in (1,2)$, or a random variable with density $f(x)=:=c/(1+|x|^3)$.) Consider the stochastic process $$X_t := Y \quad \text{for all $t \geq 0$}. $$Clearly, $$\int_0^T X_t^2 \, dt = Y^2 T < \infty$$ and $$\mathbb{E} \left( \int_0^T X_t^2 \, dt \right) =\mathbb{E}(Y^2) T =\infty,$$ i.e. using the criteria which you mentioned in your question we can only deduce that $M_t := \int_0^t X_s \, dB_s$ is a local martingale. However, because of the simple structure of $(X_t)_{t \geq 0}$, we have $$M_t = Y \cdot B_t;$$ using the independence of $Y$ and $(B_t)_{t \geq 0}$ it is not difficult to see that $(M_t)_{t \geq 0}$ is a true martingale with respect to the filtration $\mathcal{F}_t := \sigma(B_s, Y; s \leq t)$.