Let $X_n$ denote the two-state Markov chain with transition probability matrix P= $ \begin{bmatrix} \alpha & 1-\alpha\\ 1-\beta & \beta \end{bmatrix} $ given states 0 and 1. Let $Z_n=(X_{n-1},X_n)$ be a Markov chain over the four states (0,0), (0,1), (1,0), and (1,1). Determine the transition probability matrix.
Here are my thoughts so far: If $Z_n=(X_{n-1},X_n)$, then $Z_{n-1}=(X_{n-2},X_{n-1})$. I believe I'm looking for $P[Z_n=(x_1,y_1)|Z_{n-1}=(x_2,y_2)]=\frac{P[X_{n-1}=x_1, X_n=y_1, X_{n-2}=x_2, X_{n-1}=y_2]}{P[X_{n-2}=x_2, X_{n-1}=y_2]}$=$P(X_n=y_1|X_{n-1}=x_1, X_{n-2}=x_2)$=$P(X_n=y_1|X_{n-1}=x_1)$ by the Markov property.
For example, using the points (0,0)$\rightarrow$(0,0) would be $P[Z_n=(0,0)|Z_{n-1}=(0,0)]=\frac{P[X_{n-1}=0, X_n=0, X_{n-2}=0, X_{n-1}=0]}{P[X_{n-2}=0, X_{n-1}=0]}$=$P(X_n=0|X_{n-1}=0, X_{n-2}=0)$=$P(X_n=0|X_{n-1}=0)$ which corresponds to $p_{00}=\alpha$.
Following this process I obtained P=$\begin{bmatrix} \alpha &0 &\alpha &0\\ 1-\alpha &0 &1-\alpha &0\\ 0 &1-\beta &0 &1-\beta\\ 0 &\beta &0 &\beta\\ \end{bmatrix}$.
$0$'s denote a probability that is not possible (i.e. $X_{n-1}$ cannot be both 0 and 1).
I've seen the answer expressed as the matrix P= $\begin{bmatrix} \alpha &1-\alpha &0 &0\\ 0 &0 &1-\beta &\beta\\ \alpha &1-\alpha &0 &0\\ 0 &0 &1-\beta &\beta\\ \end{bmatrix}$
Based on the logic in when a probability equals zero I really think my transition probability matrix is correct rather than the alternative choice. Any thoughts?
Your matrix is just the transpose of the answer. It occurs probably because your definition is different from the answer's. For instance, for the transition probability matrix $P$, the row sums equal to 1. It means that $$ \sum_{j} p_{ij} = 1 $$ because it is summing over all possible outcomes of $i \to j$. It does not matter in computing the probability for $(0,0) \to (0,0)$, but it does matter for $(1,0) \to (0,0)$. Examine your calculation and make sure that it is defined as $$ p_{ij} = \mathbb{P}(Z_{n+1}=j|Z_n=i) $$ Yours may be the other way around and result in the transpose.