Suppose we have the following transition diagram for a homogeneous Markov chain $(Z_n)_{n \geq 0}$:
Now consider the sequence of random variables $(Z_n^2)_{n \geq 0}$.
How would I go about computing $\mathbb{P}(Z_2^2 = 0|Z_1^2 = 1)$?
If $Z_1^2 = 1$, then either $Z_1 = 1$ or $Z_1 = -1$, one having transition probability $1$ and the other having transition probability $\frac{1}{2}$, to $0$. I'm not sure how to combine these two facts to calculate what $\mathbb{P}(Z_2^2 = 0|Z_1^2 = 1)$ is, since it is not clear whether to start at $1$ or $-1$.
Also, I tried using the Law of Total Probability of conditional probabilities, but it would require computing quantities like $\mathbb{P}(Z_1 = 1|Z_1^2 = 1)$ and $\mathbb{P}(Z_1 = -1|Z_1^2 = 1)$.
Let $Y_n=Z_n^2$
Then $$\begin{align}P(Y_2 = y | Y_1 = 1)&= \frac{P(Y_2 = y , Y_1 = 1)}{P(Y_1 = 1)}\\ &=\frac{P(Y_2 = y \cap ( Z_1 = 1 \cup Z_1=-1))}{P( Z_1 = 1 \cup Z_1=-1))}\\ &=\frac{P(Y_2 = y | Z_1 = 1) P(Z_1 = 1) + P(Y_2 = y | Z_1 = -1) P(Z_1 = -1)}{P(Z_1 = 1)+P(Z_1 = -1)} \end{align}$$
Hence the $Y-$states transition probabilities alone are not enough.
Either you know the initial $Z-$states probabilities, and plug that into the formula above. Or either you compute the stationary $Z-$states probabilities (with that you get the $Y-$states probabilities) and with that you can compute the (steady) $Y-$states transition probabilities.