0

Let $(W_t)$ be a Brownian motion. Under what minimal condition on the function $\sigma$ can we say that the solution of

$$dM_t = \sigma(M_t,t) dW_t$$

is a martingale with regard to its own filtration?

Same question if $(W_t)$ was a generic martingale, not necessarily a Brownian motion.

W. Volante
  • 2,364
  • Does this help? https://math.stackexchange.com/questions/38908/criteria-for-being-a-true-martingale – Jose Avilez May 27 '21 at 18:06
  • I still don't see, are the processes I mentioned always local martingales for any $\sigma$ ? – W. Volante May 27 '21 at 19:10
  • See also https://math.stackexchange.com/questions/232932/itō-integral-has-expectation-zero. – fes May 27 '21 at 19:26
  • @W.Volante as soon as $\sigma$ is adapted and satisfies $\int_0^t \sigma^2 ds < \infty$ almost surely, then $M$ is a local martingale. You may then use the links above to determine when $M$ is in fact a true martingale. – Jose Avilez May 27 '21 at 20:16
  • @JoseAvilez Is that obvious when the filtration is the filtration generated by $M$ instead of by $W$? I think $\sigma(M_t,t)$ will be adapted to $\mathcal F^M$ assuming some weak measurability conditions, but $W$ won't necessarily have to be. How would we conclude $M$ is a local martingale in that case? – user6247850 May 27 '21 at 20:25
  • @user6247850 It's unclear what $dM_t = \sigma (M_t, t) dW_t$ would mean if $\sigma$ and/or $W_t$ were not adapted to some underlying filtration. So for the purpose of this question, I would just assume that the filtration generated by $M$ is at least rich enough to house $W$ as well. – Jose Avilez May 27 '21 at 20:39
  • 2
    @JoseAvilez I agree that $\sigma$ and $W$ have to be adapted to SOME filtration, but saying it's the filtration generated by $M$ is placing a very non-trivial assumption on $\sigma$. I'm thinking of something like $dM_t = sgn(M_t)dW_t$ where we know there is a weak solution, but the filtrations generated by $M$ and $W$ are quite different. This example doesn't work because $\mathcal F^W \subset \mathcal F^M$, but I'm not sure exactly what condition on $\sigma$ would guarantee that. – user6247850 May 27 '21 at 20:54

0 Answers0