Where a random walk can be

Topic: Markov chains · more questions on this topic
‹ A walk with a floor

A random walk on the integers starts at 0 and steps +1 with probability p and -1 with probability q=1-p. Find \mathbb{P}(X_n=k). Why is \mathbb{P}(X_n=0)=0 when n is odd?

Worked answer (click to show)

Let K\sim\operatorname{Bin}(n,p) be the number of steps up. Then X_n=K-(n-K)=2K-n, so X_n=k exactly when K=(n+k)/2:

\mathbb{P}(X_n=k)=\binom{n}{(n+k)/2}p^{(n+k)/2}q^{(n-k)/2},

for |k|\le n with n+k even, and 0 otherwise. X_n=2K-n has the same parity as n, so an odd number of steps can never end at 0.