Absolute value of a symmetric 'simple' random walk

Question

I have the symmetric 'simple' random walk $\{S_{n}: n=0,1,2,...\}$ defined by $S_{n+1}=S_{n}+A_{n}, n=0,1,2,...$ where $A_{0}, A_{1}, A_{2},...$ are independent and identically distributed copies of a random variable A with $\mathbb{P}(A=-1)=\mathbb{P}(A=1)=\frac{1}{2}$.

Denote by $X_{n}=|S_{n}|$ the absolute value of $S_{n}$. Show that $X_{n}$ is a Markov chain.

My question is what exactly is different to $X_{n}$ than $S_{n}$? Aren't all values of $S_{n}$ positive already?

Answer 1

No, the values of $S_{n}$ can be negative as each $A_{n}$ can be -1.

You could have a sequence of $A_{j}$ like {-1, -1, -1} which would clearly make your $S_{n}$ negative.

I think your confusion came from the fact that the probabilities are positive, while the values of $A_{j}$ are not necessarily positive.