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Suppose I have a brownian motion $B(t)$,

how to calculate the Expected value of $B(t)$ to the power of any integer value $n$?

Intuition told me should be all 0. But how to make this calculation?

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    $\begingroup$ If by "Brownian motion" you mean a random walk, then this may be relevant: math.stackexchange.com/questions/103142/… $\endgroup$ – user20637 Nov 2 '16 at 17:17
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    $\begingroup$ The marginal distribution for the Brownian motion (as usually defined) at any given (pre)specified time $t$ is a normal distribution ... Write down that normal distribution and you have the answer $\endgroup$ – kjetil b halvorsen Aug 16 '17 at 22:12
  • $\begingroup$ "$B(t)$" is just an alternative notation for a random variable having a Normal distribution with mean $0$ and variance $t$ (which is just a standard Normal distribution that has been scaled by $t^{1/2}$). The expectation of a power is called a moment. You can find the moments of a unit-variance distribution all worked out at stats.stackexchange.com/questions/176702/…. The $n^\text{th}$ moment of $B(t)$ therefore is found by multiplying those answers by $t^{n/2}$. $\endgroup$ – whuber Dec 9 '17 at 15:43
  • $\begingroup$ Certainly not all powers are 0, otherwise $B(t)=0$! It will however be zero for all odd powers since the normal distribution is symmetric about 0. $\endgroup$ – Alex R. Jun 4 '18 at 22:25

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