2
$\begingroup$

Let $B(t)$ and $B(s)$ are brownian-motion I want to show $$E(B(t)-B(s))^4=3(t-s)^2$$

thanks for help.

$\endgroup$
4
  • $\begingroup$ assuming s < t? $\endgroup$
    – BCLC
    Dec 28, 2015 at 10:01
  • 1
    $\begingroup$ @BCLC There's no need to assume $t>s$ in the question. However, it can be assumed for convenience in an answer without loss of generality $\endgroup$
    – Glen_b
    Dec 29, 2015 at 8:47
  • $\begingroup$ @Glen_b because of the even powers? $\endgroup$
    – BCLC
    Dec 29, 2015 at 8:48
  • 1
    $\begingroup$ @BCLC Yes, plainly. $E[(-X)^4] = E[X^4]$ and $(-a)^2 = a^2$ $\endgroup$
    – Glen_b
    Dec 29, 2015 at 8:53

1 Answer 1

2
$\begingroup$

If $B(t)$, $t\ge 0$ is a Brownian motion, then $B(t)-B(s)$ has $N(0,t-s)$ distribution. From there, you just need to figure out how to compute the 4th moment of a Gaussian with given variance, which I trust you can do.

$\endgroup$
1
  • $\begingroup$ assuming s < t? $\endgroup$
    – BCLC
    Dec 28, 2015 at 10:01

Your Answer

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

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