Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
3  
This looks like homework. It's customary here to use the "homework" tag in such cases, explain what you have tried and where you are stuck, and avoid phrasing your question as a command. –  Nate Eldredge May 8 '11 at 12:12
add comment

2 Answers

up vote 4 down vote accepted

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

share|improve this answer
    
Thanks a lot, TheBridge. I was struggling in vain with the definition of quadratic variation and mountains of calculations, but at the end the solution was easy and elegant, as you showed. Of course, my statement is true only almost surely. –  Ale Zok May 9 '11 at 7:17
    
Hi, I suggest that you edit your question to follow the recommandations of Nate Eldregde Regards –  TheBridge May 9 '11 at 14:00
add comment

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 $t \geq 0$.

share|improve this answer
    
if we take $X_t$ continuous local martingale we have that also $X^2_t$ is a continuous local martingale: sure about this? –  Did May 12 '11 at 20:27
    
Could you check whether it is correct? You could first apply Nate's recommendation. –  Did May 12 '11 at 20:28
    
You're right, $X^2_t$ is not necessarily a continuous local martingale. Actually the only thing I needed was $|X_t|^2 < \infty$ that simply follows the lemma applied to $X_t$. –  Ale Zok May 13 '11 at 7:25
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.