Showing that a process is not a Markov chain

Question

Exercise: Let $\xi_1, \xi_2, \ldots$ be i.i.d. $\in\{-1,1\}$, taking each value with probability $1 / 2$. Let $S_0=0, S_n=\xi_1+\cdots > \xi_n$ and $X_n= \max \left\{ S_m: 0 \leq m \leq n \right\}$. Show that $X_n$ is not a Markov chain.

Durrett's Solution: $X_n$ is not a Markov chain since $X_{n+1}=X_n+1$ with probability $1 / 2$ when $X_n=S_n$ and with probability $0$ when $X_n>S_n$

I am a bit distressed about these problems in Durrett, as someone coming from a couple courses in measure theory, the solutions do not feel rigorous to me, and I am not convinced that this is a valid proof.

In order for $X_n$ to not be a Markov Chain, it must be that the law conditioned on the past equals the law conditioned on the present. Therefore, we must have some dependency on the past for this to not be a Markov Chain.

Let $\mathcal{F}_n$ be the natural filtration of the $\{\xi_n: n>0\}$, which represents the information we know at each timestep. Then I somehow need to show that we cannot write $P(X_{n+1}\in B| \mathcal{F_n}) = p(X_n,B)$ for any $p$ How does Durrett's solution show this?

Answer 1

If it is a Markov chain then $P(X_{n+1}=i_{n+1}|X_1=i_1,X_2=i_2,\cdots, X_n=i_n)$ depends only on $i_n$ and $i_{n+1}$ (and not on $i_1,i_2,\cdots,i_n$). You can easily find examples where $i_1+i_2+\cdots+i_n=i_n$ and $i_1+i_2+\cdots+i_n<i_n$ (fixing $i_n$ and $i_{n+1}$). It follows that $P(X_{n+1}=i_{n+1}|X_1=i_1,X_2=i_2,\cdots, X_n=i_n)$ takes the values $\frac 1 2 $ as well as $0$, leading to a contradiction.

Answer 2

Another way to express what Durrett is saying is to observe that $$ X_{n+1} = X_n+1_{\{X_n=S_n\}}\cdot 1_{\{\xi_{n+1}=1\}}, $$ bearing in mind that $\xi_{n+1}$is independent of the pair $(X_n,S_n)$. (Indeed, independent of all of $\mathcal F_n$.) This makes it clear that the conditional distribution of $X_{n+1}$, given $\mathcal F_n$, is different from the conditional distribution of $X_{n+1}$, given $X_n$.

It may be worth noting that the identity displayed above implies that $(X_n,S_n)$ is a Markov chain.