0
$\begingroup$

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?

$E[M_t]=0$

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

$\endgroup$
1
$\begingroup$

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

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.