Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

In our homework assignment, we are supposed to prove:

If $ M $ is a countinuous local martingale and if for each $ T > 0, E[\sup_{t \leq T } |M_t|] < + \infty $ and $ H^T $ is a bounded predictable process, then $ H \cdot M $ is a true martingale.

In the hint it says that we should use the previous exercise, which was:

If $ M $ is a countinuous local martingale and $ H $ a predictable process such that
for each $ T > 0 $ we have

$ E[\sqrt{\int_0^T H_s^2 d \langle M \rangle_s }] < + \infty $,

then $ H \cdot M $ is a true martingale.

Well, I started like this:

$ | H^T | \leq C \ \Longrightarrow \ E[\sqrt{\int_0^T H_s^2 d \langle M \rangle_s }] \ \leq C E[\sqrt{ \langle M \rangle_T }] $

Moreover I know that $ E[\sup_{t \leq T } |M_t|] < + \infty, \forall T > 0, $ implies that $ M $ is a true martingale (but I don't know if we need this at all).

Anyway, I don't know how to proceed from here, or if the above was the right way to start at all.

Thanks for all your efforts! Regards, Si

share|cite|improve this question
Try to look at the Burkholder-Davis-Gundy inequality. – Kolmo Apr 30 '12 at 14:55
@Kolmo: Well, I used the BDG-inequality to prove the previous exercise. But what shall I do with it here? Could you give me a hint? – Mad Si Apr 30 '12 at 15:17
Well the inequality says that $c E\left[ [M]_T^{\frac{1}{2}}\right ] \leq E[\sup_{t\leq T}|M_t|]$ – Kolmo Apr 30 '12 at 16:00
@Kolmo: Ah, I see ;-) Thanks – Mad Si Apr 30 '12 at 16:18

Your Answer


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

Browse other questions tagged or ask your own question.