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Let $Y$ be a discrete stationary stochastic process. Suppose that $Y$ is not $n$-step Markov for any positive integer $n$. Let $Z$ be a 1-block factor of $Y$. For what condition on $Y$ or the factor map, will $Z$ be Markov?

The problem has been well-studied when $Y$ is assumed to be Markov. Then $Z$ is a hidden Markov chain. However, I am not familiar with any literature addressing the question when $Y$ is not Markov. Any references would be greatly appreciated. Thank you.

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Stephen, could you perhaps explain what you mean by "1-block factor of Y"? I'll delete this post once you see the message. thanks. –  John Jiang Mar 31 '10 at 7:18
    
I believe a 1-block factor is an equivalence class on the values of Y, or a function from the values of Y to a new set of labels which is not expected to be injective. –  Douglas Zare Mar 31 '10 at 7:45
    
A 1-block factor is a function from the symbols in Y to a new set of symbols which may or may not be injective. However, in this case, since Y is not Markov, if the map is injective, the image will not be Markov. –  Stephen Shea Mar 31 '10 at 11:37
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1 Answer

Some necessary conditions for $Z$ to be Markov are easy to understand and to write down.

For every $y$, $y'$ and $y''$ in the state space $S$ of $Y$, write $p_3(yy'y'')$ for the probability that $[Y_t=y,Y_{t+1}=y',Y_{t+2}=y'']$, which is independent of time $t$. Assume that $Z=\phi(Y)$. For every $z$, $z'$ and $z''$ in $\phi(S)$, write $q_3(zz'z'')$ for the sum of $p_3(yy'y'')$ over every $y$, $y'$ and $y''$ such that $z=\phi(y)$, $z'=\phi(y')$ and $z''=\phi(y'')$. Then a necessary condition is that $q_3$ can be factorized, in the sense that there exist functions $r$ and $s$ such that $q_3(zz'z'')=r(zz')s(z'z'')$, for every $z$, $z'$ and $z''$ in $\phi(S)$.

Of course, this condition is far from sufficient. In fact, to be able to say anything even moderately interesting about this problem, one should probably specify the kind of processes $Y$ and $Z$ one has in mind.

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