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Let $(M_t)$ be a martingale w.r.t a filtration $(\mathcal F_t)$. The martingale representation theorem implies there exists a Brownian motion $(B_t)$ adapted to $(\mathcal F_t)$ such that $M_t$ is the solution of an SDE of the form

$$dM_t=\sigma(t,M_t)d B_t $$

Suppose that we found a process $(X_t)$ adapted to $(\mathcal F_t)$ such that

$$dM_t=\tilde \sigma(t,M_t)d X_t $$

Does that mean that $(X_t)$ is a Brownian motion? If so, what is the theorem behind that result?

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No. If $M_t$ is a martingale which is not a Brownian motion with respect to $(\mathcal{F}_t)$, then you can just pick $dX_t = dM_t$ and $\tilde{\sigma} = 1$.

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  • $\begingroup$ Makes sense. But in the case where $M_t$ and $X_t$ are distinct, would the answer be different ? $\endgroup$
    – W. Volante
    May 18, 2021 at 16:08
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    $\begingroup$ By scaling, the answer would still be no. More interestingly, if $dM_t = s(t,M_t) r(t, M_t) dB_t$ with $r, s \in L^2$, then you could always have $dX_t = r(t, M_t) dB_t$. $\endgroup$
    – Jose Avilez
    May 18, 2021 at 16:13

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