Markov Chains - Can we set the Initial distribution as $X_{1}$ instead of $X_{0}$ to calculate a non-conditional probability? Had a small question when I was going through Ross – Probability Models. Given a discrete-time Markov chain
$\{X_{n}: n=1,2,3 \ldots\}$ with a state space $\mathcal{S} = \{0,1\}$, and an initial distribution $P(X_{0} = 0) = \alpha = 1 - P(X_{0} = 1)$, we can write the unconditional distribution of $X_{n}$ as:
$P(X_{n} = k) = \sum_{i\in \mathcal{s}} P(X_{n} = k | X_{0} = i)P(X_{0} = i)$
Assuming we calculate the distribution of $X_{1}$ from a given transition probability matrix and the given distribution of $X_{0}$, can we write:
$P(X_{n+1} = k) = \sum_{i\in \mathcal{s}} P(X_{n+1} = k | X_{1} = i)P(X_{1} = i)$.
Here, all I do is change the initial distribution from $X_{0}$ to $X_{1}$. Is this allowed?
 A: The short answer is yes—what you've stumbled upon is a case of variable elimination. You might as well change the $n+1$s to $n$, just to reduce clutter:
$$
P(X_{n} = k) = \sum_{i\in \mathcal{S}} P(X_{n} = k | X_{1} = i)P(X_{1} = i)
\text{.}
$$

Short proof:
$$
\begin{align}
P(X_{n} = k) &= \sum_{i\in \mathcal{S}} P(X_{n} = k | X_{1} = i)P(X_{1} = i) & \text{Your starting point} \\
P(X_{n} = k) &= \sum_{i\in \mathcal{S}} P(X_{n} = k | X_{1} = i) \sum_{j \in \mathcal{S}} P(X_{1} = i \mid X_{0} = j) P(X_0 = j) & \text{Definition of } P(X_1) \text{ from Ross}\\
P(X_{n} = k) &= \sum_{j \in \mathcal{S}} \sum_{i\in \mathcal{S}} P(X_{n} = k | X_{1} = i)  P(X_{1} = i \mid X_{0} = j) P(X_0 = j) & \text{Push constant inside second summation}\\
P(X_{n} = k) &= \sum_{j \in \mathcal{S}} \sum_{i\in \mathcal{S}} P(X_{n} = k | X_{1} = i, X_{0}=j)  P(X_{1} = i \mid X_{0} = j) P(X_0 = j) & \text{Markov assumption}\\
P(X_{n} = k) &= \sum_{j \in \mathcal{S}} P(X_{n} = k \mid X_{0} = j) P(X_0 = j) & \text{Definition of marginal probability}\\
\end{align}
$$
Some discussion:
You've provided the following formula from the Ross book. The formula is correct, but it's one of those 'Why would you ever write it this way?' things.
$$P(X_{n} = k) = \sum_{i\in \mathcal{S}} P(X_{n} = k | X_{0} = i)P(X_{0} = i)$$
All of the fun of the Markov chain is in this term: $P(X_{n} = k | X_{0} = i)$, which we tend to decompose with the chain rule and the Markov conditional independence assumption: $P(x_n | x_0) = \sum_{x_{n-1}} \cdots \sum_{x_1} P(x_n | x_{n-1}) \ldots P(x_1 | x_0)$. Usually we'll write it explicitly. But Ross's equation is still correct.
A: Your conclusion holds even if you drop the assumption that $X_n$ is a Markov chain. In general, if $Y$ and $Z$ are any two random variables both having state space $\{0, 1\}$, then by the second version of the law of total probability given by Wikipedia,
$$
P(Z = k) = P(Z = k | Y = 0)P(Y=0) + P(Z = k | Y = 1)P(Y=1)
$$
