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There is a martingale representation theorem

If $M$ is a continuous $L^2$-martingale, there is a Brownian motion $B$ and a cadlag adapted function $\sigma$ such that $$ M_t = M_0 + \int_0^t \sigma(B_s)dB_s, \text{ a.s., for all } t \ge 0. $$

Is there an equivalent for continuous process of finite variation?
Something like

If $V$ is a continuous process of finite variation, there is a Brownian motion $B$ and a cadlag adapted function $\mu$ such that $$ V_t = V_0 + \int_0^t \mu(B_s)ds, \text{ a.s., for all } t \ge 0. $$

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1  
The integral $\int_0^t\mu(B_s)\,\mathrm dB_s$ can't be of finite variation, can it? –  Stefan Hansen Dec 4 '12 at 6:13
    
@StefanHansen Good point. I meant $ds$ in the integral not $dB_s$. I corrected it. –  Nicolas Essis-Breton Dec 4 '12 at 12:33
4  
1. you seem to be absolutely continuous, how about a cantor function 2. you can have a process like that on a space that does not support a brownian motion, which cannot happen in the contiuous martingale case –  mike Dec 4 '12 at 17:16

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