# Martingale representation theorem

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 $dG(w) = (B_s^2 + 2B_s - 2s)e^{B_s-s}dB_s(w) + (\ldots)dt$

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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 $T$.) – saz Dec 10 '12 at 13:02 Sorry, now that you mentioned it I see that I forgot the '-T's. This should be correct now – Good guy Mike Dec 10 '12 at 13:06 '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 (math.kth.se/matstat/gru/tentor/5b1570/Solutions/… 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 $\omega$. – Good guy Mike Dec 10 '12 at 13:32 Are you really sure about it? As far as I can see the "ds"-term does not disappear in this case... – saz Dec 10 '12 at 13:34 The ds-term here: $dG(w) = (B_s^2 + 2B_s - 2s)e^{B_s-s}dB_s(w) + (\ldots)ds$? – Good guy Mike Dec 10 '12 at 13:44
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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.

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

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

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 I don't follow your first part unfortunately. The second part is what I got aswell though, I just omitted the 'ds'-part because as I first interpreted his answer he was just looking at the dB part from the Itôs formula. In a similar question he did the following: G is in this case: $G(w) = B^2_T(w) exp(B_T(w))$ from which you get that $E[G] = e^{T/2}(T + T^2)$, then you should do this representation of $G(w)$ as before. His solution to this is as follows: – Good guy Mike Dec 10 '12 at 15:18 "Use the Martingale Representation Theorem to get $G(w) = z + \int \psi(s,w)dB_s$ where z = E[G]. Now to get $\psi$, we apply the Itô formula to $B^2_t e^{B_t}$: to identify (as is done in several previous exercises) $\psi$. We get $\psi(s,w) = 2e^{B_t}B_t + B_t^2 e^{B_t}$." So that's why I thought we only cared about the dB-term in the Ito-formula. – Good guy Mike Dec 10 '12 at 15:18 If you compare the 5th line below (where he wrote $X_T=\ldots$) and the last one ($G_T=\ldots$), you see that he claims $2 \int_0^T B_t \cdot \exp \left(B_t - \frac{t}{2} \right) \, dt = T^2$. And I wanted to show that this cannot be true. (By the representation theorem there exists some $\psi$ such that $G=z+\int \psi \, dB$, but I think his one is not the correct one). – saz Dec 10 '12 at 16:55 Ok, now I see what you've done. He's taken the expectation in between though. So $E[2 \int_0^T B_t \cdot \exp \left( B_t-\frac{t}{2} \right) \, dt] = 2 \int_0^T E[B_t \cdot \exp \left( B_t-\frac{t}{2} \right)] \, dt = 2 \int_0^T E^Q[B^Q_t + t] \, dt = 2 \int_0^T t \, dt = T^2$. So this is true. The question is now why he did that. – Good guy Mike Dec 10 '12 at 17:26 The point is: He has a decomposition $X_T = Z+\int \ldots \, dB_s$ where $Z$ is some random variable ($Z$ is not a constant!). And now he takes the expectation of $Z$ and claims $X_T = \mathbb{E}Z+ \int \ldots \, dB_s$. But that's not true in general! – saz Dec 10 '12 at 17:35