Suppose $(X_n)_{n\geq0}$ is a Markov chain with transition probability matrix $P$ and initial distribution $\lambda$. Show that the process $Y_n = (X_{kn})_{n\geq0}$ with $k$ fixed is Markov with transition probability matrix $P^k$. What is the initial distribution?
I know all the general tricks for finding probabilities of nth steps through powers of the matrix and such, but I am not sure how to approach this one!
SMALL DISCOVERY: So of course the initial distribution $\lambda$ is a vector of length $n$ (if we have $n$ possible states) ie we can write
$\lambda = (\lambda_1, ..., \lambda_n)$ where $\lambda_i = \mathbb{P}(X_0 = i)$
So I guess as we have $X_0 = Y_0$ the initial distributions are both $\lambda$?
Clearly $Y_0=X_0$ so the initial distribution is still $\lambda$. As for the Markov property, for any nonnegative integers $0\leqslant t_1<t_2<\cdots<t_n$, $$ \begin{align*} \mathbb P(Y_{t_n+1}=j | Y_{t_n}=i_n, \ldots, Y_{t_1}=i_1) &= \mathbb P(X_{k(t_n+1)}=j |X_{kt_n} = i_n, \ldots, X_{kt_1}=i_1 )\\ &= \mathbb P(X_{k(t_n+1)} = j | X_{kt_n}=i_n)\\ &= \mathbb P(Y_{t_n+1} = j | Y_{t_n} = i_n). \end{align*} $$
Further, this probability is $$\mathbb P(X_{kt_n+k} = j | X_{kt_n}=i_n) = (P^k)_{ij}. $$