I'm working on this problem that I can't seem to figure out. The problem involves a 1-dimensional Brownian motion, $B_t$, where the subscript denotes the time, and it asks me to show that the expectation \begin{align} \mathbb{E}[|B_t-B_s|^4]=3|t-s|^2 \end{align} where $s<t$. I'm new to the subject of Brownian motion, so I'm not quite sure how to deal with this. I know that $B_t-B_s$ is a normally distributed random variable with mean zero and variance $t-s$. My intuition was therefore to compute \begin{align} \mathbb{E}[|B_t-B_s|^4] &= \int_{-\infty}^{\infty} |x|^4 ~f(x)~dx\\ &=\int_{-\infty}^{\infty} x^4~f(x)~dx \end{align} where $x$ denotes the value of the random variable $B_t-B_s$, and $f(x)$ denotes its distribution, $f(x)=\frac{1}{\sqrt{2\pi(t-s)}}e^{-\frac{x^2}{2(t-s)}}$. However, that's just the 4th moment of a normal distribution, which is $3(t-s)^4$, so I know that something is wrong with my intuition. Can someone point out what I'm doing wrong, or give a hint to what a better (or correct) approach would be?

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    $\begingroup$ If $X\sim N(0, t-s)$, where $t-s$ is the variance, and the fourth moment is $3\sigma^4$, then $E(X^4)=3\sigma^4=3\sqrt{t-s}^4=3(t-s)^2$, right? Sounds like your intuition is right, but mixing up standard deviation and variance. $\endgroup$ – hejseb Sep 20 '14 at 12:07
  • $\begingroup$ @hejseb That's totally true! Thanks for pointing that out! $\endgroup$ – user176740 Sep 20 '14 at 12:18

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