Consider the random walk $S_{n}$ for $n \geq 1$. Specifically, let $X_{1},X_{2},..$ be Independent with
$$ \mathbb{P}(X_{n}=1) = p,~~\mathbb{P}(X_{n}=-1) = 1- p =:q $$ and $S_{n} = \sum_{k=1}^{n}X_{k}$. Now I know that $|S_{n}|$ is a markov chain. But what about $\max(S_{n},0)$? Is this also a markov chain?
No it's not. Keep in mind your chain is now $Y_n:=\max(S_n,0)$, and so you have to rely on values of $Y_{n-1},\cdots,Y_0$ as opposed to knowing exactly what $S_{n-1},\cdots,S_0$ where.
To see this consider the ways in which $\max(S_n,0)=0$. For example imagine that $S_n$ is negative, so $Y_n=0$. You'd need to know the exact value of $S_n$ to infer whether $Y_{n+1}=\max(S_{n+1},0)=\max(S_n+X_{n+1},0)>0$, as you explicitly need to know if $S_n+X_{n+1}>0$. So $P(Y_{n+1}>0|Y_n=0,Y_{n-1},\cdots,Y_0)\neq P(Y_{n+1}>0|Y_{n}=0)$.