If a stochastic process has the same expectation value for all pisitive t, then is it a martingale? I don’t know how to show it whether that is right.


1 Answer 1


The Brownian motion $W_t$ is a martingale because for all $u>t$: $$ W_t = E(W_u | \mathcal{F}_t), $$

However, the process $X_t$
$$X_t = 1_{\{a<t<b\}} (2 . 1_{\{W_a < 0\}} -1) $$ has 0 expectation but is not a martingale.

  • $\begingroup$ Hmm... Why 21? If you give more information , I'll very thanks for you. $\endgroup$
    – Hobong
    Dec 10, 2018 at 15:33
  • $\begingroup$ I just added the dot between 2 and 1, this process takes value +1 or -1 between a and b, but it is not a martingale as it goes back to 0 for $t>b$ $\endgroup$
    – Sebapi
    Dec 10, 2018 at 15:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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