| ▲ | raincole 9 hours ago | |||||||
Wait, are you saying that for a symmetrical random walk, the expected distance is of the order of sqrt(n), but even for a slightly biased random walk (like 0.5000001 chance to take right) it's of the order of n? Edit: well of course it is. I was thinking expected position (which should be 0) not distance | ||||||||
| ▲ | NooneAtAll3 9 hours ago | parent | next [-] | |||||||
yep "expected distance" is average abs(coordinate), so for biased walk (and big enough time) it's simply abs(bias)*time, and for unbiased it's deviation==sqrt(variance) | ||||||||
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| ▲ | jeffwass 9 hours ago | parent | prev [-] | |||||||
The “expected distance” is not what you think here. For a binomial distribution of probability p and (1-p), after N steps the expectation value of right steps is Np. The Variance is Np(1-p), so the standard deviation (or Root-Mean-Square) scales as Sqrt(N). | ||||||||