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Trying to figure out how to solve problems on the 'form':

Find a real number $z$ and a square integrable, adapted process $\psi(s,w)$ such that

$$G(w) = z + \int \psi(s,w)\,dB_s(w)$$

for som process $G(w)$.

In the case I'm working on now I have $G(w) = (B^2_T(w)-T)\exp(B_T(w)-T)$.

So using the Martingale representation theorem I have that:

$$G(w) = E[G] + \int \psi(s,w)\,dB_s(w)$$

and I've already calculated $E[G]$ to be $T^2e^{-T/2}$. So it only remains to show what $\psi(s,w)$ is.

What I've done now is to apply the Itô formula on $G$, as he's done in other old exams, but I can't really understand what he's doing because his handwriting is terrible. But as I said he uses the Itô formula and uses the '$dB_s$'-term as the $\psi(s,w)$ but he's changing it and that step I can't really tell what he is doing. Does anyone know?

From the Itô formula I get $dG(w) = (B_s^2 + 2B_s - 2s)e^{B_s-s}dB_s(w) + (\ldots)dt$

Thanks in advance!

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    Are you sure that there is no typo in your definition of $G_T$...? (If there exists such a representation of $G$, then $(G_T)_T$ has to be a martingale, in particular $\mathbb{E}G_T$ should not depend on $T$.)2012-12-10
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    Sorry, now that you mentioned it I see that I forgot the '-T's. This should be correct now2012-12-10
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    'in particular $E[G_T]$ should not depend on T', yes this made me very confused aswell, but this is as he has solved it (http://www.math.kth.se/matstat/gru/tentor/5b1570/Solutions/SF2970_071027_loesning.pdf problem 5). I can't quite tell what he's done for the other part as I said though. But I guess it's for fixed T, I mean G is just a function of $\omega$.2012-12-10
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    Are you really sure about it? As far as I can see the "ds"-term does not disappear in this case...2012-12-10
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    The ds-term here: $dG(w) = (B_s^2 + 2B_s - 2s)e^{B_s-s}dB_s(w) + (\ldots)ds$?2012-12-10
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    Well, but that's weird what he is doing there... the integral $\int_0^T B_t \cdot Z_t \, dt$ still depends on $\omega$ - and he simply claims that it is equal to its expectation?! (And yes, this ds-term there...)2012-12-10

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