In the exercise 1.1.2 of "Markov chains - J.R.Norris"
Suppose that $(X_{n})$ is Markov $(\lambda,P)$. If $Y_{n}=X_{kn}$, show that $(Y_{n})$ is Markov $(\lambda, P^{k})$.
To prove the above: Let $(X_n)_{n\geq0}$ be a Markov chain $(\lambda, P)$ where $\mathbb{P}(X_{0}=i)=\lambda_i$ for any $i\in I$ (state space) . So Since $Y_0=X_0$, then \begin{equation*} \mathbb{P}(Y_o = i) = \mathbb{P}(X_0 = i) = \lambda_i \end{equation*} then $ \lambda = (\lambda_i) _ {i \ in I} $ for the initial distribution over the string $ (Y_n) _ {n \ in \mathbb{N}} $. On the other hand, let us note that \begin{equation*} \mathbb{P}(Y_{n}=i) = \mathbb{P}(X_{kn}=i) = (\lambda P^{kn})_{i} \end{equation*} then let's note that $\mathbb{P}(Y_{n}=i) = (\lambda P^{kn})_{i}$, so $P^{k}$ is the transition matrix of $ (Y_n)_{n\in\mathbb{N}}$, concluding that the Markov chain $(Y_n)_{n\in\mathbb{N}}$ is $(\lambda,P^{k})$.
But I am not sure about this procedure, as it could be reasoned by induction, the truth is that I still cannot sufficiently internalize the concepts of Markov chains. Does anyone know if this reasoning is correct? Is it common to consider $\lambda$ as a distribution for the random variable $X_0$? Can the other random variables be distributed differently?