Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Consider a process $\{X_n, n\geq 0\}$ with state space $S=\{0,1,2\}$ s.t. $$ P(X_{n+1}=j | X_n=i, X_{n-1}=i_{n-1}, \dots, X_0=i_0)=\begin{cases} P_{ij}^I \ \ \ n \ \mbox{ even},\\ P_{ij}^{II} \ \ \ n \ \mbox{odd} \end{cases} $$where $\sum_{j=0}^2 P_{ij}^I = \sum_{j=0}^2 P_{ij}^{II}=1$.

I don't see why this shouldn't be a Markov chain. It looks to me like a non-time homogeneous Markov chain.

Any ideas? Thanks a lot for any help! :)

share|improve this question
    
Yes, this is a time-inhomogenous Markov chain. Is this your question? –  Did Jan 20 '13 at 19:44
    
Yes :) Then there is no need to enlarge the state space to make it Markov? Does it satisfy the Chapman-Kolmogorov equations? –  Daniel Jan 20 '13 at 19:56
    
There is only dependence on the divisibility of $n$ by $2$, hence the enlargement is just $S\times \{0,1\}$. –  Ilya Jan 20 '13 at 20:12
    
The solution says the enlargement is $S=\{0,1,2,\hat{0},\hat{1}, \hat{2}\}$ where $X_n$ takes $0,1,2$ if even and $\hat{0}, \hat{1}, \hat{2}$ if odd. But why do we need to do that? Isn't $X$ a Markov chain anyways? –  Daniel Jan 20 '13 at 20:17
    
Once again, X is a Markov chain, an inhomogenous one. The enlargement trick you describe is only useful to make X a homogenous Markov chain. –  Did Jan 21 '13 at 7:30

1 Answer 1

Each inhomogenous Markov chain $(X_n)_{n\geqslant0}$ on the state space $S$ with transitions $(P^I_{ij})_{ij}$ and $(P^{II}_{ij})_{ij}$ alternatively, can be embedded in a homogenous Markov chain $(Z_n)_{n\geqslant0}$ with $Z_n=(X_n,Y_n)$, on the state space $S\times\{0,1\}$, with transitions $(Q_{(i,u),(j,v)})_{(i,u),(j,v)}$ defined by $$ Q_{(i,0),(j,1)}=P^I_{ij},\quad Q_{(i,1),(j,0)}=P^{II}_{ij},\quad Q_{(i,0),(j,0)}=Q_{(i,1),(j,1)}=0. $$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.