0
$\begingroup$

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.

$\endgroup$
  • 2
    $\begingroup$ $S_n\to +\infty$ with probability one by the strong law of large numbers. $\endgroup$ – spaceisdarkgreen Nov 4 '18 at 23:32
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.