Random walk: 50% chance of taking step forward, 50% chance of taking step back

Matteo

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Aug 24, 2018
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I have the following question.
We have a person who takes n steps with 50% chance of taking a step forward and 50% chance of taking a step backwards. What is the limit (for n -> inf) of the chance that the position X after n steps is lager or equal to √n ?
lim P(X >= √n)
 
Have you made no attempt on this yourself? I would think that, at least, you would calculate probabilities for various small values of n.

Or you should be able to realize that, since at each step there are two possibilities, this is a binomial distribution. After n steps there are n+ 1 possible positions, from -n to n by steps of 2, and the probability of being at step i (with i= 0 being -n, i= n+1 being n) is the binomial coefficient \(\displaystyle \begin{pmatrix}n \\ i \end{pmatrix}\).
 
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