Symmetric random walk that does not hit $0$

Question

Suppose a random walk $\{X_n\}_{n\in \mathbb Z^{\ge 0}}$ on the integer number line, $ \mathbb {Z}$ , which starts at $0$ and at each step moves $+1$ or $−1$ with equal probability. The increments between any two consecutive steps are independent. I want to find the probability that $X$ does not hit $0$ during time $\mathbb Z^{\ge 1}$.

$M_n = \max(X_0,...,X_n)$, for $n \in \mathbb N^{\ge 0}$.

My question: It seems like that $P[X_1\le 0 , X_2 \le 0, ..., X_{n-1}\le 0] = P[M_{n-1} = 0]$ only explains the lower half of the graph. By symmetry, don't we need to multiply $P[M_{n-1} = 0]$ by $2$? But, multiplying by $2$ does not work even for $n=1$. Why do we not need to multiply $P[M_{n-1} = 0]$ by $2$?

Answer 1

The solution shows how to find the probability given that $X_1=-1$. The probability given $X_1=1$ is the same. So from here it is just conditional probability. Let $E=\{X_1\ne 0,...,X_n\ne 0\}$. Then:

$p_n=P(E)=P(E|X_1=-1)\times P(X_1=-1)+P(E|X_1=1)\times P(X_1=1)=$

$=P(E|X_1=-1)\times (\frac{1}{2}+\frac{1}{2})=P(E|X_1=-1)$

Answer 2

The probability that the walk does not return to 0 is

$P[$the first step is $-1]$ $\times$ $P[X_1\le 0 , X_2 \le 0, ..., X_{n-1}\le 0]$ $+$ $P[$the first step is $+1]$ $\times$ $P[X_1\ge 0 , X_2 \ge 0, ..., X_{n-1}\ge 0]$.

However, note that $P[$the first step is $-1]$ $=$ $P[$the first step is $+1]$ $= 1/2$, and $P[X_1\le 0 , X_2 \le 0, ..., X_{n-1}\le 0]$ $=$ $P[X_1\ge 0 , X_2 \ge 0, ..., X_{n-1}\ge 0]$, giving the result as in the textbook that you quoted.