Extracting a smaller Markov chain from a larger Markov chain I am not very familiar with Markov chains, hence the probably ill titled questions. 
If we have 5 random variables $X, Y, Z, W$ and they form a Markov chain such that 
$$X \rightarrow Y \rightarrow Z \rightarrow W \rightarrow P$$
Is also the case that 
$$X \rightarrow Y \rightarrow P$$
or 
$$X \rightarrow Y \rightarrow W $$
Intuitively I would assume it is true because no information about $X$ can be gained as we move along the chain, thus it is okay to remove a link. 
Secondly if i recall correctly then if $Z = f(Y)$ then $X \rightarrow Y \rightarrow Z$ is a Markov chain. So in my example $f$ is just a composite function.
This is obviously a weak understanding at best, so any help is appreciated.
Edit:
My understanding is that 3 random variables $X, Y, Z$ form a markov chain if $X$ and $Z$ and conditionally independent given $Y$. So
$$X \rightarrow Y \rightarrow Z \iff p(x, y, z)=p(x) p(y | x) p(z | y)$$ 
 A: It suffices to show you can remove one link (then you can recursively remove as many as you like). Suppose: 
$$ X \rightarrow Y \rightarrow Z \rightarrow W$$
You want to show 
$$ X \rightarrow Y \rightarrow W$$
For simplicity assume all random variables are discrete and let $S_X, S_Y, S_Z, S_W$ be the finite or countably infinite sets of all possible values these variables can take (with positive probability). 
Then for all $w \in S_W, x \in S_X, y \in S_Y$ we have 
\begin{align}
&P[X=x, Y=y,W=w] \\
&\overset{(a)}{=} \sum_{z \in S_Z} P[X=x, Y=y, Z=z, W=w]\\
&\overset{(b)}{=}\sum_{z \in S_Z} P[X=x]P[Y=y|X=x]P[Z=z|Y=y]P[W=w|Z=z]\\
&= P[X=x]P[Y=y|X=x]\sum_{z \in S_Z} P[W=w|Z=z]P[Z=z|Y=y]\\
&\overset{(c)}{=}P[X=x]P[Y=y|X=x]\sum_{z \in S_Z} P[W=w|Z=z, Y=y]P[Z=z|Y=y]\\
&=P[X=x]P[Y=y|X=x]P[W=w|Y=y]
\end{align}
where (a) is from the law of total probability; (b) and (c) use the Markov chain assumption $X\rightarrow Y \rightarrow Z \rightarrow W$.
A: The question is not very clear. I am assuming that you have a Markov chain, $ \{ X_n : n \geq 0 \} $. You are interested in finding whether the chain $ \{ X_{2n} : n \geq 0 \} $ is also a Markov chain.
If that is the case, then the answer is yes. Actually $ \{ X_{kn} : n \geq 0 \} $ is a Markov chain for any integer $ k \geq 1 $. The transition probability matrix is given by $ \mathbf{P}^k $ where $ \mathbf{P}$ is the transition matrix of the original chain.
