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 p3(yy′y″) for the probability that [Yt=y,Yt+1=y′,Yt+2=y″], which is independent of time t. Assume that Z=phi(Y). For every z, z′ and z″ in phi(S), write q3(zz′z″) for the sum of p3(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 q3 can be factorized, in the sense that there exist functions r and s such that q3(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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