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Let $dX_t=u(t,w)dt+dB_t$, where $dB_t$ is a Brownian Motion, $u$ is bounded and measurable with respect to the filtration $F_t$ and $u$ be an ito process on $(\Omega, \{\mathcal{F}\},\mathcal{P})$. Find a martingale $M_t$ s.t. $M_0=1$ and $Y_t=X_tM_t$ is an $\mathcal{F}_t$-martingale.

It is easily derived that $X_t=X_0+\int^t_0 u(s,w)ds+B_t$.

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Could you also define $w$ ? –  Sasha Nov 14 '11 at 6:07
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Try Girsanov theorem –  TheBridge Nov 14 '11 at 8:41
    
$u(t,w)$ is some ito process and w is just some parameter –  Grim Reaper Nov 14 '11 at 14:36
    
I don't think $M_t=1$ will work, since $X_t$ is not a martingale –  Grim Reaper Nov 14 '11 at 14:46

2 Answers 2

Try $$ M=\exp\left(-\int_0^\cdot u_s\,\mathrm dB_s-\frac12\int_0^\cdot u_s^2\,\mathrm ds\right). $$ Then $M$ is an exponential martingale. Furthermore, $\mathrm dX=\mathrm dB+u\mathrm dt$ and $\mathrm dM=-uM\mathrm dB$ hence $\mathrm d\langle M,X\rangle=-uM\mathrm dt$ and $Y=MX$ is such that $$ \mathrm dY=X\mathrm dM+M\mathrm dX+\mathrm d\langle M,X\rangle=M(1-uX)\mathrm dB. $$ In particular, $Y$ is a martingale.

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Let $dM_t=\sigma (t,w)M_t dB_t$ then let $Y_t=X_t M_t$. we arrive at $dY_t=M_t(u(t,w)+\sigma(t,w))dt + (... )dB_t$, then we see that for $dY_t$ to be a martingale, we need to have zero drift. Hence $\sigma (t,w)=-u(t,w)$ thus $dM_t=-u (t,w)M_t dB_t$

So $M_t=e^{-\int_0^t u(s,w) dW_s}$

not too sure if this is the actual solution to the SDE.

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This $M_t$ is not a martingale. –  Did Jan 26 '13 at 14:14

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