Martingale representation theorem

Question

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!

Answer 1

As far as I can see it's wrong what he is doing there. The claim is that

$$2 \int_0^T B_t \cdot \exp \left( B_t-\frac{t}{2} \right) \, dt = T^2$$

But this can't be true. Since $t \mapsto B_t(w) \cdot \exp \left( B_t-\frac{t}{2} \right)(w)$ is continuous almost surely we can apply the fundamental theorem of calculus and obtain

$$2 B_T \cdot \exp \left(B_T- \frac{T}{2} \right) = 2 T \quad \text{a.s.}$$

which would imply

$$B_T = T \cdot \exp \left(\frac{T}{2}-B_T\right) \geq 0$$

... and this is not correct.

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I applied Itô's formula to $f(t,x) := (x^2-t) \cdot \exp (x-t)$ and obtained

$$\underbrace{f(t,B_t)}_{G_t}-\underbrace{f(0,0)}_{0}= \int_0^t \exp(B_s-s) \cdot (B_s^2+2B_s-s) \, dB_s \\ + \frac{1}{2} \int_0^t \exp(B_s-s) \cdot \left(-\frac{1}{2} B_s^2+2B_s + \frac{s}{2} \right) \, ds$$