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Simple probability doubt

  1. Dec 17, 2011 #1
    Hi all,

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

    However if [itex]V_{i} = (*,0,0)[/itex] and [itex]V_{i-1} = (1, 0, *)[/itex], then [itex]Pr(V_{i+1}|V_{i},V_{i-1}) = 0[/itex], while if [itex]V_{i-1} = (0, 0, *)[/itex] then [itex]Pr(V_{i+1}|V_{i},V_{i-1})[/itex] 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 [itex]Pr(V_{i} = (*, 0, 0) | V_{i-1} = (1,0,*) ) = 0[/itex]. Yet it seems not very reasonable to me to neglect the case when it comes to [itex]Pr(V_{i+1}|V_{i},V_{i-1})[/itex], as the variables can take in any possible value.

    Did I miss something?
     
    Last edited: Dec 17, 2011
  2. jcsd
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