Let $B_t$ be Brownian motion, with $B_0=0$. Next define $M_t=-B_t$.

Have I understood it correctly that $M_t$ is not a Martingale?


$E[M_{t+1}|M_t]=-M_t$ and therefore not a Martingale?


Multiplying by a constant doesn't change the Martingale property: for $t>s$, $$ E[M_t|M_s]=E[-B_t|M_s]=E[-B_t|B_s]=-E[B_t|B_s]=-B_s=M_s. $$

  • 1
    $\begingroup$ Yes of course. Thank you for the quick response. $\endgroup$ – simme Oct 15 '14 at 0:18

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