# Simple probability doubt

1. Dec 17, 2011

### hikaru1221

Hi all,

This is not a homework question, but I think it's reasonable to post it here.
Say, for a Markov process, $Pr(Z_{i+1}|Z_{i},...,Z_0) = Pr(Z_{i+1}|Z_{i},Z_{i-1})$, people prove that if we consider $V_{i} = (Z_{i+1},Z_{i},Z_{i-1})$, it would be a 1st-order Markov process, i.e. $Pr(V_{i+1}|V_{i},...,V_0) = Pr(V_{i+1}|V_{i})$.

However if $V_{i} = (*,0,0)$ and $V_{i-1} = (1, 0, *)$, then $Pr(V_{i+1}|V_{i},V_{i-1}) = 0$, while if $V_{i-1} = (0, 0, *)$ then $Pr(V_{i+1}|V_{i},V_{i-1})$ may be non-zero. That is, it's not really reducible to the 1st-order. I understand that in a sense, the former case is rather invalid, as $Pr(V_{i} = (*, 0, 0) | V_{i-1} = (1,0,*) ) = 0$. Yet it seems not very reasonable to me to neglect the case when it comes to $Pr(V_{i+1}|V_{i},V_{i-1})$, as the variables can take in any possible value.

Did I miss something?

Last edited: Dec 17, 2011