# In Brownian motion, how does the absolute value of a particle's distance from the center scale with respect to number of iterations? [closed]

The mean distance is probably zero due to averaging out. But what about the mean of the absolute distance?

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Since distances cannot be negative, it's difficult to see how they could "average out" to zero! –  whuber Aug 11 '11 at 14:39
You would be looking at a half-normal distribution, whose expectation is $\sigma \sqrt{2/\pi}$, if the underlying normal random variable has variance $\sigma^2$. Furthermore, the variance in Brownian motion is commensurate with time.