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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 sufficientand in . 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 is interested has in mind.

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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 and in fact, to be able to say anything about this problem, one should probably specify the kind of processes $Y$ and $Z$ one is interested in.