Will a random walk on [0, inf) tend to infinity? Consider a random walk on [0, inf) where you start at 0. With probability p = 0.5, you increase by 1. With probability (1-p) = 0.5, you decrease by 1, but not below 0.
As time goes to infinity, will your position tend to infinity? If not, to what finite value does it converge? 
Edit: To be a bit more precise, what is the limit of the average position as time goes to infinity?
If your position tends to infinity for p = 0.5, for which other probabilities p is this true? (Clearly p > 0.5 will cause you to tend to infinity, so p < 0.5 is what I'm after)
What is the probability of being at position x after an arbitrary amount of steps?
I made a simple simulation to test the p = 0.5 case, and after 500 million iterations, it seems to tend to infinity, but I'd like a more solid explanation.
Thanks!
 A: As it has been said:


*

*if $p=\frac{1}{2}$ this is more or less the same thing as the absolute value of the standard symmetric random walk on $\mathbb{Z}$, wich is recurrent.

*if $p > \frac{1}{2}$, the law of large numbers immediately shows you that it tends to $+\infty$.

*if $p < \frac{1}{2}$, you can even check the detailed balance equations and find the invariant distribution: the Markov chain is positive recurrent.

A: The symmetric random walk $(X_k)$ on $\mathbb{Z}$ is recurrent. Therefore, with probability one, you will visit infinitely often $0$. The same is true for $(|X_k|)$, which is more ore less your random walk (the more or less depends on what happens exactly at the origin).
In other words, with probability one, the $\liminf$ is $0$. The $\limsup$ behaviour is the subject of the law of the iterated logarithm. 
A: Let me try again. Let $R_n$ be the random variable denoting the longest contiguous run of heads for $n$ independent $p$-biased coin tosses. 
It is well-known 1 that $E R_n\sim\log_{1/p}((1-p)n)$ plus small correction terms (the variance is $O(1)$). 
This means that $S_n=\sum_{i=1}^n X_i$ grows at least as $\log n$ for any fixed $p$, and proves almost-sure escape to $\infty$.
A: For any $p>0$ and any finite $N$, with probability one, eventually your random walk will exceed $N$.
