3 ms·
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 0.50
by raincole 10mo 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 10mo agoyep "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)
- raincole 10mo agoOh right, makes sense.
- jeffwass 10mo agoThe “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).