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Let $B$ be a standard Brownian motion, and, $$ X_t=e^{\int_0^t f(B_s)ds}, $$ for some function $f$.
What are the condition on $f$ for $X_t$ to be of finite variation?

Let $Y_t=\int_0^t f(B_s)ds$, if $f$ is continuous then $Y_t$ is of finite variation.
Does it then imply that $X_t$ is of finite variation?
If $f$ is only bounded, I think nothing can be infered.

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If $f$ is continuous on $\mathbb{R}$, then for each path realization $\int_0^t f(B_s) ds$ is absolutely continuous with respect to the Lebesgue measure and therefore so is $X_t = e^{\int_0^t f(B_s) ds}$ which implies that $X_t$ is of finite variation. I don't know how to handle the bounded case offhand. –  Chris Janjigian Nov 13 '12 at 23:22
    
@Chris Very good answer. Thank you. –  Nicolas Essis-Breton Nov 14 '12 at 2:12
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