I'm reading an article that says "a random walk generated by tossing a coin (one step forward if heads, one step backward if tails) would show a displacement from the starting point, depending on the square root of the number of tosses." so if i'm reading it correctly, after 100 tosses, you would expect to be 10 steps from the starting point. why wouldn't you expect to be simply at the starting point (on average), since you are equally likely to go forward or backwards?

From what I've read, you would be equally likely to be at the starting point. I always thought the square root example dealt with a 4 way option. Starting from a given point, one could move up, down, left or right. After 100 tosses, you would be highly likely to be within 10 spaces of the starting point (in any direction.)

The statement should be altered in something like this ... after 100 tosses, you would expect to be within 10 steps from the starting point in 68 (or xx) out of 100 scenarios. The standard deviation of the random walk is sqr(n) * stepsize. P.S. Unfortunately I donÂ¢t have FellerÂ¢s amazing textbook

agree with gbos, i think they mean displacement at every step inclusively, or Standard Deviation. rather than just displacement by the last step.

The ratio of heads-to-tails can remain at ~50% but the arithmetic difference between heads & tails can increase as you continue to do more flips, i.e. 5/5, 51/49, 510/490, 5100/4900. That can explain the variation from a zero baseline.

do you mean sqrt(Npq) instead of just sqrt(N)? in this example, N=100 p=.5 q=1-p=.5 so sqrt(100*.5*.5) instead of sqrt(100)? http://mathworld.wolfram.com/RandomWalk1-Dimensional.html

Like always when "Random Walk" comes up, many would benefit from reading an elementary text on probability theory. If you haven't, better don't try any further.