Why is the proof of Markov's property so roundabout?

Question

We've defined a Markov chain as a sequence of random variables $(X_n)_{n\geq0}$ such that

$P(X_{n+1}=i_{n+1}|X_0=i_0,\dots,X_n=i_n)=P(X_{n+1}=i_{n+1}|X_n=i_n).$

That is, $X_{a}$ and $X_{b}$ are independent for any $|a-b|>1.$ (This is my interpretation of the claim, at least.)

We want to show Markov's property: If $(X_n)$ is Markov$(\lambda,P)$ and $X_{m}=i$, then $(X_{m+n})_{n\geq0}$ is Markov$(\delta_i,P)$ and independent of $X_0,\dots,X_m.$

The proof I've been given (see Theorem 1.1.2 in https://www.statslab.cam.ac.uk/~james/Markov/s11.pdf) goes on to consider an event $A$ determined by $X_0,\dots,X_m$, decompose that into "elementary" events and so on.

Why can't we just argue that $(X_{m+n})_{n\geq0}$ is a Markov chain as $X_a$ and $X_b$ independent for all $|a-b|>1$ implies $X_{m+a}$ and $X_{m+b}$ independent for all $(m+a)-(m+b)>1$, and it must have the initial distribution $\delta_i$ since we're conditioning on $X_{m+0}=i$?

A random walk is Markov. Think about why $X_{50}$ is dependent on $X_{48}$. — , Oct 20 '20 at 20:12

score 2 · Accepted Answer · answered Oct 20 '20 at 20:14

2

A random walk $X$ is a Markov process. Then $P(X_{50}=50)=\frac{1}{2^{50}}$. But $P(X_{50}=50|X_{48}=3)=0$.

Your interpretation is false.

answered Oct 20 '20 at 20:14

score 1 · Answer 2 · answered Oct 20 '20 at 21:12

The Markov property means that if we know the value of the previous state, we do not gain more knowledge about $X_n$ by knowing the full history. That is, the distribution of $X_n$ is only allowed to depend on the previous stage and on $n$. It is quite a lot more general than independence.

As the other answer points out, you can think of a random walk as a prototypical (time-homogenous) Markov process, where the previous stage completely controls the distribution of the next stage.

Why is the proof of Markov's property so roundabout?

2 Answers2