Quadratic variation condition and square integrability

Question

Assume $M$ is local continuous martingale started from $0$. How would one go about showing that if it is a martingale bounded in $L^2$, then $E[\langle M\rangle_{\infty}]\lt \infty $?

Answer 1

Under your hypotheses the process $K_t:=M_t^2-\langle M\rangle_t$ is a continuous local martingale. Let $(T_n)$ be an increasing sequence of stopping times with $\lim_nT_n=\infty$ (a.s.) and each stopped process $K^{T_n}$ a martingale. By Doob's inequality, for each $t>0$ and each $n$, $$ \Bbb E[\langle M\rangle_{T_n\wedge t}]=\Bbb E[M^2_{T_n\wedge t}]\le\Bbb E[\sup_{0\le s\le t}M^2_s]\le 4\Bbb E[M^2_t]\le C<\infty, $$ for some constant $C$. By monotonicity, $\Bbb E[\langle M\rangle_\infty]\le C<\infty$.

Answer 2

This is not true. The condition $\mathbb{E}[\langle M,M \rangle_\infty] < \infty$ implies $M$ is a true martingale, but you can have continuous local martingales that are bounded in $L^2$ but not true martingales. For example, if $B_t$ is a Brownian motion in $\mathbb{R}^3$ started from $(1,0,0)$ then $M_t := \frac{1}{|B_t|}$ is a continuous local martingale bounded in $L^2$, but is not a true martingale. We can look at $M_t - 1$ instead to satisfy $M_0 = 0$.