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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1If by "Brownian motion" you mean a random walk, then this may be relevant: http://math.stackexchange.com/questions/103142/expected-value-of-random-walk – user20637 Nov 02 '16 at 17:17
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1The 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 – kjetil b halvorsen Aug 16 '17 at 22:12
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"$B(t)$" 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 https://stats.stackexchange.com/questions/176702/how-to-calculate-the-expected-value-of-a-standard-normal-distribution/176814#176814. The $n^\text{th}$ moment of $B(t)$ therefore is found by multiplying those answers by $t^{n/2}$. – whuber Dec 09 '17 at 15:43
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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. – Alex R. Jun 04 '18 at 22:25