So I have to show that: Let $\{X_n\}$ be a Markov chain. Show that the property $P(X_{n+1}=i|X_n=j_n,X_0=j_0)=P(X_{n+1}=i|X_n=j_n)$ holds. Hint: use the Markov-property. The Markov Property being: $P(X_{n+1}=i|X_n=j_n,X_{n-1}=j_{n-1},...,X_1=j_1,X_0=j_0)=P(X_{n+1}=i|X_n=j_n)$
Isn't this kinda obvious? I mean the Markov property says that the last state is only dependent on the forelast state. What is here to prove? Isn't that too easy? What am I missing?
Fix $j_0, i, j_n$. By Markov property, for all $j_{n - 1}, \ldots, j_1 \in S$, where $S$ denotes the state space of this Markov chain, we have \begin{align} P[X_{n + 1} = i \mid X_n = j_n, X_{n - 1} = j_{n - 1}, \ldots, X_0 = j_0] = P[X_{n + 1} = i \mid X_n = j_n]. \end{align} Or equivalently, \begin{align} & P[X_{n + 1} = i, X_n = j_n, X_{n - 1} = j_{n - 1}, \ldots, X_0 = j_0] \\ = & P[X_{n + 1} = i \mid X_n = j_n]P[X_n = j_n, X_{n - 1} = j_{n - 1}, \ldots, X_0 = j_0]. \tag{1} \end{align} Now, summing $j_{n - 1}$ over $S$ on both sides of $(1)$ gives \begin{align} & P[X_{n + 1} = i, X_n = j_n, X_{n - 2} = j_{n - 2}, \ldots, X_0 = j_0] \\ = & P[X_{n + 1} = i \mid X_n = j_n]P[X_n = j_n, X_{n - 2} = j_{n - 2}, \ldots, X_0 = j_0]. \tag{2} \end{align} Next, summing $j_{n - 2}$ over $S$ on both sides of $(2)$ gives \begin{align} & P[X_{n + 1} = i, X_n = j_n, X_{n - 3} = j_{n - 3}, \ldots, X_0 = j_0] \\ = & P[X_{n + 1} = i \mid X_n = j_n]P[X_n = j_n, X_{n - 3} = j_{n - 3}, \ldots, X_0 = j_0]. \end{align} Continue in this way recursively to finally deduce that $$P[X_{n + 1} = i, X_n = j_n, X_0 = j_0] = P[X_{n + 1} = i \mid X_n = j_n] P[X_n = j_n, X_0 = j_0]$$ that is, \begin{align} P[X_{n + 1} = i \mid X_n = j_n, X_0 = j_0] = P[X_{n + 1} = i \mid X_n = j_n]. \end{align}