Show that if for $n \ge 1$, $(X_0, \ldots, X_n)$ has probability function $f_n$, then $X$ is a Markov Chain with transition matrix $\mathbb P$

Question

Let $S$ a set of states and $\mathbb P=\{p_{i,j}\}_{i,j \in S}$ be a transition matrix.

I've proved that $f_n(i_0,\ldots,i_n) := f_0(i_0) p_{i_0 i_1} \dots p_{i_{n-1} i_n}, \ \ (i_0, \ldots, i_n) \in S^{n+1}$ is a probability function on $S^{n+1}$ given that $f_0(i_0), \ \ i_0 \in S$ is a probability function on $S$.

Let $X := (X_n)_{n\ge 0}$ be a sequence of discrete stochastic variables with set of states $S$. Show that if for $n \ge 1$, $(X_0, \ldots, X_n)$ has probability function $f_n$, then $X$ is a Markov Chain with transition matrix $\mathbb P$.

Idea:

It is enough to show that for every $n \ge 1: P(X_n = i_n \mid X_{n-1} = i_{n-1}, \ldots, X_0 = i_0) = p_{i_{n-1,n}}$ ?

But $f_n(i_0,\ldots,i_n) := f_0(i_0) p_{i_0 i_1} \dots p_{i_{n-1} i_n}, \ \ (i_0, \ldots, i_n) \in S^{n+1}$ does not reflect this property ?

Answer 1

If $P(X_{n-1} = i_{n-1}, \ldots, X_0 = i_0) \neq 0$, \begin{align*} P(X_n = i_n \mid X_{n-1} = i_{n-1}, \ldots, X_0 = i_0)&=\frac{P(X_n = i_n , X_{n-1} = i_{n-1}, \ldots, X_0 = i_0)}{P(X_{n-1} = i_{n-1}, \ldots, X_0 = i_0)}\\ &= \frac{P(X_n = i_n , X_{n-1} = i_{n-1}, \ldots, X_0 = i_0)}{\sum_{i_n}P(X_n = i_n , X_{n-1} = i_{n-1}, \ldots, X_0 = i_0)}\\ & = \frac{f_n(i_0,\cdots, i_n)}{\sum_{i_n}f_n(i_0,\cdots, i_n)} \\ &= p_{i_{n-1}i_n} \text{ (we used that $(p_{ij})$ is a transition matrix )} \end{align*}