How do I construct the score process of a Markov model and verify that it is a Martingale?

Question

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 $-1 < \beta_0 < 1$ 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?

Answer 1

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.