How do I construct the score process of a Markov model and verify that it is a Martingale? The following is a specific question that is useful for demonstrating a general idea.

Consider the following autoregressive model:
$$
X_{t+1} = \alpha_0 + \beta_0 (X_t - \alpha_0) + W_{t+1},
$$
where $-1 &lt; \beta_0 &lt; 1$ and $W_{t+1}$ is distributed as a normal with mean zero and variance one. 

How should I go about constructing the bivariate score process associated with the parameters $\alpha_0$ and $\beta_0$? How can I verify that it is a martingale?

Progress:

I would begin by constructing the log-likelihood process as follows (conditioning on $X_0$):
$$
\ell_t(\theta \mid \textbf X) = -\frac t2 \ln(2 \pi) - \frac 12 \sum_{j=1}^t
   (X_j - \alpha_0 - \beta_0(X_{j-1} - \alpha_0))^2.
$$
Then, the score process can be constructed as
$$
s_t(\theta \mid \textbf X) =
\begin{bmatrix}
 (1 - \beta_0) \sum_{j=1}^t (X_j - \alpha_0 - \beta_0(X_{j-1} - \alpha_0)) \\
 \sum_{j=1}^t (X_j - \alpha_0 - \beta_0(X_{j-1} - \alpha_0)) (X_{j-1} - \alpha_0)
\end{bmatrix}.
$$
Is this correct? How do I proceed?
The derivation of the score process is correct. To verify that the process is a Martingale, recall the definition. It becomes clear that if we substitute $W_{t+1}$ back into the equation
$$
s_t(\theta \mid \textbf X) =
\begin{bmatrix}
 (1 - \beta_0) \sum_{j=1}^t W_j \\
 \sum_{j=1}^t W_j (X_{j-1} - \alpha_0)
\end{bmatrix}.
$$
Because $W_{t+1}$ are Normal with mean 0 and variance 1 (I'm assuming they're iid), then
$$
E[s_{t+1} \mid s_t ] = s_t + E \left [
\begin{matrix}
(1 - \beta_0) W_{t+1} \\
W_{t+1} (X_t - \alpha_0)
\end{matrix} \middle |
s_t
\right ]
= s_t
$$ 
and we are finished.