# On simple random walk with $\mathbb{P}\left(X=+1\right)=p>\frac{1}{2}$

I'm reading Ross' "Stochastic Processes" book, in the part of random walks, and I have troubles to understand an argument.

Consider $$\{S_n\}_{n\ge 0}$$ the simple random walk, with $$\mathbb{P}\left(X=+1\right)=p>\frac{1}{2}$$. And we denote $$\alpha=\mathbb{P}\left(\text{return to zero}|X_1=+1\right)$$ Ross says that $$\mathbb{P}\left(\text{return to zero}|X_1=-1\right)=1$$. I'm trying to see why it is so obvious. I tried to count the paths of "length" $$n$$. Maybe is an argument so easy that I can not see it.

The help will be well received.

Sheldon Ross; "Stochastic Processes". 2nd Edition, page:332.

• $S_n\to +\infty$ with probability one by the strong law of large numbers. – spaceisdarkgreen Nov 4 '18 at 23:32

## 1 Answer

Note that:

P(return to zero|X1 = -1) = P(Sn touches 1 | x0 = 0) = 1 - P(Sn <= 0, for all n)

let A = { w | Sn(w) <= 0, for any n}, An = {w | Sn(w) <=0}

A = joint of n of An

then P(A) <= P(An) for any n

but P(An) -> 0, because Sn is a Binomial(double sided)(n, p), p > 0.5.