About the difference between residuals and disturbance error for AR and MA

Question

I'm confused with the error term and residuals for AR and MA models.

I thought residual Q is defined as $x_t-\hat{x_{t}}$, while error term is $\epsilon_t$, and residuals are estimated values for error terms?

For AR(1)

$$ x_t=\phi x_{t-1}+\epsilon_t$$

Hence, $Q=x_t-\hat{x_{t}}=x_t- \hat{\phi} x_{t-1}$

From the model, we have $\epsilon_t = x_t- \phi x_{t-1}$, so with estimated $\hat{\phi} $, we have $\hat{\epsilon_t} = x_t- \hat{\phi} x_{t-1}$

In this case , $Q=\hat{\epsilon_t}$.

For MA (1) $$ x_t=\epsilon_t- \theta \epsilon_{t-1}$$

Similarly, $Q=x_t-\hat{x_{t}}=x_t- \epsilon_t+\hat{\theta} \epsilon_{t-1} $

From the model, we have $\epsilon_t = x_t+\theta \epsilon_{t-1}$, so with estimated $\hat{\theta} $, we have $\hat{\epsilon_t} = x_t+ \hat{\theta} \epsilon_{t-1}$.

In this case, $Q\neq\hat{\epsilon_t}$.

What is the reason for this? residual calculation for MA part is wrong?

Answer 1

Let Q stand for the 1 period ahead prediction error; $I_t$ refers to the information set at time t (all that is known/knowable at time t). I think the tricky part here is in the case of the MA process to realize that E($\theta \epsilon_t|I_t)=\theta \epsilon_t$ and not $0$.

AR 1

$Q=x_{t+1}-E(X_{t+1}|I_t)=(\phi x_{t}+\epsilon_{t+1})-(E(\phi X_t+\epsilon_{t+1}|I_t))=(\phi x_{t}+\epsilon_{t+1})-(\phi x_t)=\epsilon_{t+1}$

MA 1

$Q=x_{t+1}-E(X_{t+1}|I_t)=(\epsilon_{t+1}-\theta \epsilon_t)-(E(\epsilon_{t+1}-\theta \epsilon_t|I_t))=(\epsilon_{t+1}-\theta \epsilon_t)-(0-\theta \epsilon_t)=\epsilon_{t+1}$