# 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

First note that $G=X_TY_T\mathrm e^{-T/2}$ where the processes $X$ and $Y$ are defined for every $t\geqslant0$ by $$X_t=B_t^2-t,\qquad Y_t=\mathrm e^{B_t-t/2}.$$ The identity $M^T_t=E[X_TY_T\mid\mathcal F_t]$ defines a martingale $(M^T_t)_{0\leqslant t\leqslant T}$ such that $M^T_T=X_TY_T$ and $M^T_0=E[X_TY_T]$. Thus, $\mathrm dM^T_t=K^T_t\mathrm dB_t$ for some process $(K^T_t)_{0\leqslant t\leqslant T}$, and $$G=\mathrm e^{-T/2}M^T_0+\mathrm e^{-T/2}\int_0^TK^T_t\mathrm dB_t.$$ To sum up, this warm up paragraph shows that it suffices to identify $M_0^T$ and the process $K^T$.

You already know that $M_0^T=T^2\mathrm e^{-T/2}$. To identify $K^T$, fix some $t\lt T$, define $u=T-t$, and consider the processes $\bar B$, $\bar X$ and $\bar Y$ defined for every $s\geqslant0$ by $$\bar B_s=B_{t+s}-B_t,\qquad\bar X_s=\bar B_s^2-s,\qquad\bar Y_s=\mathrm e^{\bar B_s-s/2}.$$ Then, $$X_TY_T=((B_t+\bar B_u)^2-t-u)Y_t\bar Y_u=X_tY_t\bar Y_u+2B_tY_t\bar B_u\bar Y_u+Y_t\bar B_u^2\bar Y_u.$$ Since $\bar B$ is independent of $\mathcal F_t$ and $(\bar B,\bar Y)$ is distributed like $(B,Y)$, this yields $$M_t^T=E[X_TY_T\mid\mathcal F_t]=A_0(T-t)X_tY_t+2A_1(T-t)B_tY_t+A_2(T-t)Y_t,$$ where, for every integer $k\geqslant0$, $$A_k(u)=E[B_u^kY_u].$$ Since one already knows that $M^T$ is a martingale, one is only interested in the martingale part of the RHS of the last identity above giving $M_t^T$. Note that $\mathrm dX=2B\mathrm dB$ and $\mathrm dY=Y\mathrm dB$, hence the martingale part of $\mathrm d(BY)$ is $Y(B+1)\mathrm dB$ and the martingale part of $\mathrm d(XY)$ is $Y(X+2B)\mathrm dB$.

To compute the functions $A_k$ for $0\leqslant k\leqslant2$, note that, for every $x$, the identity $$Y^x_u=\exp((1+x)B_u-(1+x)^2u/2)=Y_u\exp(xB_u-xu-x^2u/2),$$ defines a martingale $Y^x$ such that $Y^x_0=1$ hence $E[Y^x_u]=1$ for every $u\geqslant0$. Expanding this in powers of $x$ yields $$E[Y_u]=1,\quad E[(B_u-u)Y_u]=0,\quad E[((B_u-u)^2-u)Y_u]=0,$$ that is, $$A_0(u)=1,\quad A_1(u)=uA_0(u)=u,\quad A_2(u)=2uA_1(u)-(u^2-u)A_0(u)=u^2+u.$$ Finally, for every $0\leqslant t\leqslant T$, $$K_t^T=(B_t^2-t+2(T-t+1)B_t+(T-t)^2+3(T-t))\cdot\mathrm e^{B_t-t/2}.$$

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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