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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.

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  • $\begingroup$ assuming s < t? $\endgroup$ – BCLC Dec 28 '15 at 10:01
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    $\begingroup$ @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 -Reinstate Monica Dec 29 '15 at 8:47
  • $\begingroup$ @Glen_b because of the even powers? $\endgroup$ – BCLC Dec 29 '15 at 8:48
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    $\begingroup$ @BCLC Yes, plainly. $E[(-X)^4] = E[X^4]$ and $(-a)^2 = a^2$ $\endgroup$ – Glen_b -Reinstate Monica Dec 29 '15 at 8:53
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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.

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  • $\begingroup$ assuming s < t? $\endgroup$ – BCLC Dec 28 '15 at 10:01

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