Quadratic variation of continuous local martingales

Question

Dear all, I hope you can help me with the proof of the following result:

Fact If $X$ is a continuous local martingale, then $[X]_t < \infty $ a.s. for every $t \geq 0$, where $[X]$ denote the quadratic variation of the process $X$.

I have tried in vain different ways for approaching this problem: firstly using the definition of quadratic variation and secondly using the Doob-Meyer decomposition of $X$ but in both cases I got stuck in mountains of calculations.

Answer 1

Assuming your process is one dimensional, by representation theorem there exist a Brownian motion $B$ and a (predictible) process $Y_s$ verifying $\forall t>0$, $\int_0^tY_s^2ds<+\infty$ almost surely such that :

$X_t=E[X_0]+\int_0^tY_sdB_s$ for all $t>0$

Then $[X]_t=\int_0^tY_s^2ds$ which is almost surely finite by representation theorem.

Note that your statement is only almost surely true

Regards

Answer 2

I guess I have found another different proof of the fact I have asked. Could you check whether it is correct? Thanks

Lemma If $(M_t)_{t\geq 0}$ is a continuous local martingale, then $|M_t| < \infty$ a.s. for every $t\geq 0$.

Proof of the lemma: Suppose for sake of contradiction that $\exists s$ such that $P(|M_s|=\infty)>0$. If $(T_n)_n$ is a reducing sequence for $M$, then $\exists n$ such that $P(T_n > s)>0$. Thus $P(|M_{T_n \wedge t} |=\infty)>0$, but $M_{T_n \wedge t}$ is in $L^1$, since it is a martingale. $\Box$

Now, if we take $X_t$ continuous local martingale we have that also $X^2_t$ is a continuous local martingale and $|X^2_t| < \infty$ a.s. for every $t \geq 0$. (see comments)

The lemma applied to $X_t$ gives that $|X_t| < \infty$ a.s. for every $t \geq 0$ and thus $|X_t|^2 < \infty$ a.s. for every $t \geq 0$.

Finally, also $X^2_t-[X]_t$ is a continuous local martingale and hence we have $|X^2_t-[X]_t| < \infty$. From these facts follow that $[X]_t < \infty$ a.s. for every $t \geq 0$.