# Determining a square integrable martingale

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?

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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
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