Markov Chain Derivation question

Question

This page defines the Markov Property as the following. Does anyone how form (b) becomes (c)? Is it due to $(X_{n-1} = i_{n-1}) \subset (X_{n-1} = i_{n-1}) \cap ... \cap (X_0 = i_0)$? Is it due to independence of the conditional events?

a) $$ \mathbb{P}(X_0 = i_0, X_1 = i_1, X_2 = i_2, ..., X_n = i+n) $$ b)

$$ = \mathbb{P}(X_0 = i_0) \mathbb{P}(X_1 = i_1 | X_0 = i_0) \mathbb{P}(X_2 = i_2 | X_1 = i_1, X_0 = i_0) ... \mathbb{P}(X_n = i_n | X_{n-1} =i_{n-1} , ..., X_1 = i_1, X_0 = i_0) $$

c) $$ = \mathbb{P}(X_0 = i_0) \mathbb{P}(X_1 = i_1 | X_0 = i_0) \mathbb{P}(X_2 = i_2 | X_1 = i_1) ... \mathbb{P}(X_n = i_n | X_{n-1} = i_{n-1}) $$

Answer 1

This follows from several uses of the Markov property: $$p_k:=\mathbb P\left(X_k=i_k\mid X_{k-1}=i_{k-1},X_{k-2}=i_{k-2},X_0=i_0\right)=\mathbb P\left(X_k=i_k\mid X_{k-1}=i_{k-1}\right)$$ and the probability into question in b) is $$\mathbb P(X_0=i_0)\prod_{k=1}^np_k=\mathbb P(X_0=i_0)\prod_{k=1}^n\mathbb P\left(X_k=i_k\mid X_{k-1}=i_{k-1}\right).$$

Answer 2

By the definition of a Markov chain, the probability of the $k$ th state depends only on the $k-1$ th state, it is independent of the $k-2, k-3, \cdots, 0$ th states.

This can be explained best with the help of a state transition diagram. For example, in this Markov chain, there are no arrows linking state $2$ to state $0$, without passing through state $1$. As the diagram clearly shows, the $k$ th state is linked to the $k-1$ th state only. Hence, $$\mathbb{P}(X_k = i_k|X_{k-1}=i_{k-1}, \ldots, X_0=i_0) = \mathbb{P}(X_k = i_k | X_{k-1} = i_{k-1})$$