# 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

## closed as not a real question by whuber♦Aug 14 '12 at 12:49

It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, see the FAQ.

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.