= Solution
Starting from $Y_{n-2}=0$ forces $X_{n-2}=0$. The first nonzero value is consequently $X_{n-1}=-1$. Remaining nonzero at the next step forces $X_n=1$, because $p_{-1,-1}=0$. Hence, on an event of positive probability,
$$
\boxed{\mathbb P(Y_{n+1}=0\mid Y_n=1,Y_{n-1}=1,Y_{n-2}=0)=p_{1,0}.}
$$
The two values are unequal by assumption. They expose memory of the hidden sign: immediately after leaving zero the sign is negative, whereas after two consecutive nonzero observations it is positive. Whenever the relevant observed histories coexist with positive probability, the <Markov property> for $Y$ fails, since its next-step law is not determined by its present value alone. In particular, the absolute-value projection does not preserve the <Markov property> for all initial laws; the <strong lumpability> condition would require $p_{-1,0}=p_{1,0}$.
There is a small qualification to a blanket conclusion about an unspecified initial law. The printed assumptions do not ensure that either conditioning event is possible, and a degenerate initial law may never reach the hidden states that cause the failure. For example, in the order $-1,0,1$, take
$$
P=\begin{pmatrix}0&1&0\\0&1&0\\0&0&1\end{pmatrix},\qquad X_0=0.
$$
All printed constraints hold, yet $Y_n=0$ always and is a <Markov chain>. Thus the correct general conclusion is \b[the transformed process is not necessarily Markov]; the two computed probabilities give the intended failure when their histories are available.
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