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Let $B_t$ be a standard Wiener motion, $I_t=\int_0^t|B_s|^2\!\text{ds}\ $and $S_t=\max_{0\leq s\leq t}B_s$. Let also $F:\mathbb{R}^2_+\times\mathbb{R}\times\mathbb{R}_+\rightarrow\mathbb{R}$ a $C^{1,2,1}$ function.

Show that if:$$F_t(t,a,x,s)+x^2F_a(t,a,x,s)+\frac{1}{2}F_{xx}(t,a,x,s)=0$$ for every $(t,a,x,s)\in\mathbb{R}^2_+\times\mathbb{R}\times\mathbb{R}_+$ and $F_s(t,a,x,s)=0$ for $x=s$,

then $F(t,I_t,B_t,S_t)$ is a continuous local martingale for $t\geq0$.

What I have done:

Since $I_t$ and $S_t$ are increasing processes $\Longrightarrow\ I_t, S_t$ are processes of bounded variation. Hence, we can apply Ito's formula. We have that:

$F(t,I_t,B_t,S_t)=F(0,I_0,B_0,S_0)+\int_0^tF_t(r,I_r,B_r,S_r)\text{dr}+\int_0^tF_a(r,I_r,B_r,S_r)\text{d}I_r+\\ +\int_0^tF_x(r,I_r,B_r,S_r)\text{d}B_r+\int_0^tF_s(r,I_r,B_r,S_r)\text{d}S_r+ \frac{1}{2}\int_0^tF_{xx}(r,I_r,B_r,S_r)\text{d}\left<B,B\right>_r=\\$

$=F(0,I_0,B_0,S_0)+\int_0^tF_t(r,I_r,B_r,S_r)\text{dr} +\int_0^tF_a(r,I_r,B_r,S_r)\text{d}I_r+\\ +\int_0^tF_x(r,I_r,B_r,S_r)\text{d}B_r+\int_0^tF_s(r,I_r,B_r,S_r)\text{d}S_r +\frac{1}{2}\int_0^tF_{xx}(r,I_r,B_r,S_r)\text{dr}$

In order to show that $F(t,I_t,B_t,S_t)$ is a continuous local martingale, we need to show that:

$\int_0^tF_t(r,I_r,B_r,S_r)\text{dr}+ \int_0^tF_a(r,I_r,B_r,S_r)\text{d}I_r+\int_0^tF_s(r,I_r,B_r,S_r)\text{d}S_r+\\ +\frac{1}{2}\int_0^tF_{xx}(r,I_r,B_r,S_r)\text{dr}=0$

What can we do with the terms $\int_0^tF_s(r,I_r,B_r,S_r)\text{d}S_r$ and $\int_0^tF_a(r,I_r,B_r,S_r)\text{d}I_r$? How will we relate them to the given relation?

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First hint: $dI_r = |B_r|^2 dr$. – Nate Eldredge Dec 12 '12 at 18:53
OK, since $I_r=\int_0^r|B_s|^2\text{d}s\ \Longrightarrow\ \text{d}I_r=|B_r|^2\text{d}r$ and this is the term $x^2F_a(t,a,x,s)$ that we have in the given relation. If $S_r=B_r\ \Longrightarrow\ \max_{0\leq s\leq r}B_s=B_r\ \Longrightarrow\ B_r\geq B_s$ for every $s\in[0,r]$ and this is my only conclusion! I need a second hint! :) – Nick Papadopoulos Dec 13 '12 at 8:17
Is this an exercise from a book...? – saz Dec 22 '12 at 9:49
It is an exercise from a class. A friend of mine has given me his notes from the class and I came across this exercise which I cannot find a way to solve it! – Nick Papadopoulos Dec 22 '12 at 9:59
We have that $M_t=c+\int_0^tH_s\text{d}B_s$ for some process $H$. Then $\text{d}M_t=H_t\text{d}B_t$. Isn't that right? – Nick Papadopoulos Dec 22 '12 at 21:01

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