I want to show that: if for all $\lambda \in \mathbb{R}$ the process $\left(\exp\left(\lambda X_t-\frac{\lambda ^2}{2}t\right)\right)_{t\geq0}$ is a $\mathcal{F}^X$ local martingale, then the $\mathbb{R}$-valued process $X$ is a $\mathcal{F}^X$-Brownian motion

I think it is enough to show that $X_t$ is a continuous local martingale and that for all $t$: $[X]_t=t$, because then by Lévy's Characterization $X$ has to be a Brownian motion.

I tried using Ito's Lemma but couldn't get there.

  • $\begingroup$ A start: Assuming that $X$ is a continuous semimartingale, use $M_t$ to denote the local martingale $\exp(\lambda X_t-\lambda^2t/2)$. Now apply Ito to $\log M$, and use the resulting development to relate $[X]$ and $[M]$. $\endgroup$ – John Dawkins Apr 30 '16 at 19:07
  • $\begingroup$ Do you assume that $(X_t)_{t \geq 0}$ has continuous sample paths? $\endgroup$ – saz Apr 30 '16 at 19:57
  • $\begingroup$ Thank you for the hint, but this is the exact step i got stuck at. I got: $\lambda^2 [X]_t = \frac{1}{2} \int_0^t exp(-2(\lambda X_s + \frac{\lambda^2 s}{2}))d[M]_s $ @saz: Yes, i think so $\endgroup$ – Adsertor Justitia May 1 '16 at 12:39
  • $\begingroup$ Now use this and the Ito expansion of $\log M$ to show that the finite variation part of $X$ must vanish, meaning that $X$ is a local martingale. $\endgroup$ – John Dawkins May 1 '16 at 15:49
  • $\begingroup$ Thank you, this did help a lot! $\endgroup$ – Adsertor Justitia May 1 '16 at 22:21

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