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Let’s say you have a one dimensional random walk, for example integers from 0 to infinity on the number line, and you start at some value $n$ with a probability $P(0)$ to take a step towards zero and a probability $P(\infty)$ to move towards infinity.

My intuition tells me that no matter how small $P(0)$ is, if you keep taking steps forever you will inevitably at some point hit the bound (zero). I also believe that if this is indeed the case, you will actually hit the bound infinitely many times.

- Conditional return time of simple random walk
- Random walk problem in the plane
- A prime number random walk
- Stationary distribution of random walk on a graph
- Asymmetric Random Walk / Prove that $T:= \inf\{n: X_n = b\}$ is a $\{\mathscr F_n\}_{n \in \mathbb N}$-stopping time
- biased random walk on line

The way I think about it is that even though the number of steps from zero changes at each step you take, you will at some point hit a sequence of consecutive steps towards zero such that you will hit it, no matter how unlikely (as long as $P(0) \neq 0$, of course).

Is my intuition correct? And is there a way to prove what is actually the case?

- Stationary distribution of random walk on a graph
- Returning Paths on Cubic Graphs
- Is there an intuitive way to see this property of random walks?
- Potential uses for viewing discrete wavelets constructed by filter banks as hierarchical random walks.
- Simple Symmetric Random Walk
- Showing that lim sup of sum of iid binary variables $X_i$ with $P = P = 1/2$ is a.s. infinite
- Identity for simple 1D random walk
- Modified gambler's ruin problem: quit when going bankruptcy or losing $k$ dollars in all

Suppose you start at some location $i>0$. Let $A$ be the event that we *ever* visit the location one step to the left of this (that is, to location $i-1$). So, $P[A]$ is the probability that we *ever* visit $0$, given we start at 1. By the repeated structure of the problem, $P[A]$ is also the probability that we *ever* visit $1$, given we start at 2. By the law of total probability:

$$ P[A] = \underbrace{P[A|\mbox{first step left}]}_{1}p_0 + P\underbrace{[A|\mbox{first step right}]}_{P[A]^2}(1-p_0) $$

Define $q=P[A]$. The above equation reduces to:

$$q =p_0 + q^2(1-p_0)$$

Solving the quadratic for $q$ gives two possible solutions: Either $q=1$ or $q = \frac{p_0}{1-p_0}$. Which one is the answer? Notice that $q$ is a probability, so it must be in the interval $[0,1]$. So, if $p_0\geq 1/2$, then $q=1$ is only possible choice. As Erick mentions in his comments above, if $p_0<1/2$, it can be shown that $q<1$ (although this is not obvious) and so $q=\frac{p_0}{1-p_0}$ in that case. So in Erick’s example, if $p_0=1/3$ then $q = \frac{1/3}{2/3} = 1/2$.

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