I have tried to solve this exercises... Let ${S_{n}}$ be a randon walk (in $Z$), where $S_{0} = 0$ with $p$ to move to rigth, and $q = 1-p$ to move to left.
I want to know if the following processes are homogeneous Markov chains.
$\textbf{1. $X_{n} = |S_{n}$|}$ I'm confused because $P(X_{2} = 2 | X_{1}=1)= p+q = 1$, because:
$P(S_{2} = -2|S_{1}=-1) =q$ and $P(S_{2} = 2|S_{1}=1) = p$
but also $P(X_{2} = 0) = 1 $ for the same reason, so I don't know what to do here.
$\textbf{2. $Z_{n} = S_{n}-S_{n-1}$} $
I think there is a Markov Chain, but it's not homogeneous because: Here the space of states is $ \{-1, 1 \} $ so
$P(Z_{2} = 1| Z_{1}=-1) = P(S_{3}=-1|S_{2}=-2) + P(S_{3}=1|S_{2}=0) = p+p=2p$
and
$P(Z_{1} = 1| Z_{0}=-1) = P(S_{2}=0|S_{1}=-1) = p$
Then it is not homogeneous.
$\textbf{3. $Y_{n} = M_{n} - S_{n} $|}$ where $M_{n} = Max \{ S_k| k \epsilon \{0,...,n\} \}$
$P(Y_{2} = 1| Y_{1}=0) = q$ and
$P(Y_{3} = 1| Y_{2}=0) = 2q$
So it is not homogeneous.
Can anybody tell me...I am fine? And can anybody help me with exercise 1 please
I think we have to careful about the exact statement we are trying to show here. I assume that we are taking the definition of homogenous Markov chain to be that $\{X_n\}_{n \geq 1}$ satisfies $P(X_n \mid X_1, \dots, X_{n-1}) = P(X_n \mid X_{n-1})$. Using that definition, we have $S_n = \sum_{i=1}^n X_i$, where $X_i \sim U(\{-1,1\})$ and the $X_i$ are all independent. Then, we naturally have $$P(S_n \mid S_1, \dots, S_{n-1}) = P(\sum_{i=1}^n X_i \mid X_1, X_1 + X_2, \dots, \sum_{i=1}^{n-1} X_i) = P(\sum_{i=1}^n X_i \mid \sum_{i=1}^{n-1} X_i)$$
With that in mind, part (1) follows similarly $$P(|S_n| \mid |S_1|, |S_2|, \dots, |S_{n-1}|) = P(|\sum_{i=1}^n X_i| \mid |X_1|, |X_1 + X_2|, \dots, |\sum_{i=1}^{n-1} X_i|) $$ $$= P(|\sum_{i=1}^n X_i| \mid |\sum_{i=1}^{n-1} X_i|) = P(|S_n| \mid |S_{n-1}|)$$
Part (2) follows from the fact that $S_n - S_{n-1} = X_n$ and the fact that the $X_i$ are independent from each other (i.e. they form a trivial Markov chain with no memory).
Hope that helps!