0

Let $S$ be a countable state space and $(X_n)$ an $S$-valued discrete Markov chain. Then $$ \mathbb P [X_{n+1} = x_{n+1} |X_n = x_n, \ldots, X_0=x_0] = \mathbb P [X_{n+1} = x_{n+1} |X_n = x_n] $$ for all $x_0, \ldots, x_{n+1} \in S$. This is called the Markov property. Previously, I proved that

Theorem 1 $$ \mathbb P [X_{n+1} = x_{n+1} |X_n = x_n, \ldots, X_m=x_m] = \mathbb P [X_{n+1} = x_{n+1} |X_n = x_n] $$ for all $m \le n$ and $x_m, \ldots, x_{n+1} \in S$.

Now I would like to generalize it, i.e.,

Theorem 2 $$ \mathbb P [X_{n+1} = x_{n+1} |X_M = x_M, \ldots, X_m=x_m] = \mathbb P [X_{n+1} = x_{n+1} |X_M = x_M] $$ for all $m \le M \le n$ and $x_m, \ldots, x_{n+1} \in S$.

Could you confirm if my below understanding is correct?

Proof We proceed by backward induction on $M$. By Theorem 1, the statement holds for the base case $M=n$. Let it hold for $M$. Let's prove that it holds for $M-1$. We have $$ \begin{align*} & \mathbb P [X_{n+1} = x_{n+1} |X_{M-1} = x_{M-1}, \ldots, X_m=x_m] \\ =& \sum_{x\in S} \mathbb P [X_{n+1} = x_{n+1} | X_M = x,X_{M-1} = x_{M-1}, \ldots, X_m=x_m] \mathbb P [X_{M} = x_{M} |X_{M-1} = x_{M-1}, \ldots, X_m=x_m] \\ =& \sum_{x\in S} \mathbb P [X_{n+1} = x_{n+1} | X_M = x,X_{M-1} = x_{M-1}] \mathbb P [X_{M} = x_{M} |X_{M-1} = x_{M-1}] \\ =& \mathbb P [X_{n+1} = x_{n+1} |X_{M-1} = x_{M-1}]. \end{align*} $$

Above,

  • the first and the third equalities follows from the law of total conditional probability.
  • the second one follows from inductive hypothesis and Markov property.

This completes the proof.

Akira
  • 17,367
  • In Theorem 2 let us check first if the statement has any chance. Something should happen for all $m,n$ with $m\le n$ i suppose, and then for each two such fixed values for all possible $M$ between $m$ and $n$. If this is so, let me please fix $m=M=0$, and take some $n$ gigger $2023$ say. Than Theorem 2 claims that $$\Bbb P[X_{n+1}=x_{n+1}\ |\ X_0=x_0]=\Bbb P[X_{n+1}\ |\ X_n=x_n]\ ,$$and i have doubts. E.g. $X$ may start in $x_0$, i.e. $X_0=x_0$ happens with probability 1. Then i can imagine a process (random walk, step one) never passing from $x_n$ to $x_{n+1}$ in 1 step. But in $n$ steps... – dan_fulea Jan 04 '23 at 10:09
  • @dan_fulea Ah it's a typo. It should be $\mathbb P [X_{n+1} = x_{n+1} |X_M = x_M, \ldots, X_m=x_m] = \mathbb P [X_{n+1} = x_{n+1} |X_M = x_M]$. – Akira Jan 04 '23 at 10:12
  • 1
    Yes, it is ok now. – dan_fulea Jan 04 '23 at 10:23
  • @dan_fulea Thank you so much for your verification! – Akira Jan 04 '23 at 10:24

0 Answers0