- #1

CAF123

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## Homework Statement

A drunk lurches from one lamp post to the next on his way home. At each lamp post he pauses and is equally likely move towards or away from home. Suppose the posts are separated by a distance ##a## and find the mean and standard deviation of his displacement ##d## from the starting point, after ##N## steps.

## Homework Equations

Binomial distribution may be incorporated

## The Attempt at a Solution

I assume that each step the drunk makes is of length ##a##. He starts at some origin and then takes a total of N steps to reach a displacement d. Let ##r## be the number of steps to the right and let ##\ell## be the number of steps to the left. Then ##r + \ell = N##. At a given post, the probability that the drunk moves right is 1/2 = probability that the drunk moves left. The probability of getting ##r## right steps is then binomially distributed;$$P(\text{r right steps}) = {N \choose r} p^r (1-p)^{N-r},$$ and since ##p = 1/2##, this is the same as $${N \choose r} \frac{1}{2^N}$$ Similarly, $$P(\text{l left steps}) = {N \choose N-r} \frac{1}{2^N}$$ which is the same distribution.

I do not know how to proceed from here. I was also thinking that the expected value of displacement is given by the sum of all possible displacements multiplied by their corresponding probabilities, however, I am not sure how to obtain this in practice.

Many thanks.