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.

I'm preparing for an exam in my course Martingales & Stochastic Integrals. Currently I'm having a look at some old exams, and there's a question on one of them that I'm not able to figure out. The teacher has not provided any solutions so now I'm asking you.

Assignment: Determine the continuous square integrable martingale $M = ( M_t , t \geq 0 )$ and the associated filtration, given $ M_0 = 0 $ and the quadratic variation $\langle M\rangle_t =\log(1+t^2) $ (the natural logarithm of course).

The only thing I've come up with is that $ M_t^2 -\langle M\rangle_t $ is supposed to be a martingale, hence $$ E(M_t^2 - \langle M\rangle _t) = E(M_0^2 -\langle M\rangle _0), $$ which by definition yields $$ E(M_t^2 -\langle M\rangle _t) = 0. $$ Thus $$ M_t^2 = \log(1+t^2) $$ $$ M_t = \sqrt{\log(1+t^2)}. $$

Could this really be correct? Does anyone know the solution or at least have the link to some free lecture notes where I can learn to solve this?

share|improve this question
    
How does the "Thus" follow? I mean, you haven't really used any property of $M_t$ other than that it's a martingale. So I could put whatever I want (instead of $\log(1+t^2)$) for the quadratic variation and the answer would be the same. So the solution is not complete. –  Alex R. Jun 5 '12 at 23:02
1  
when you have a deterministic QV the obvious thing to try is a $\int f(t)dW_t$ which is a martingale with QV $\int^s f(s)^2 ds$ which leads to $f^2(t) = \frac {2t}{1+t^2}$. I don't understand the "the" because it's not unique. –  mike Jun 5 '12 at 23:26

Your Answer

 
discard

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

Browse other questions tagged or ask your own question.