4
$\begingroup$

I want to know the impact of misspecifying lag levels in AR models. Say the true model is AR(P), while in estimation AR(P-1) is estimated and used for forecast.

What will be the impact of estimating AR(P-1) on the estimated coefficients of $\phi_1 - \phi_{p-1}$, can they be consistently estimated. What about the forecast error.

$\endgroup$
  • $\begingroup$ What do you mean by impact? Your model will just be inaccurate if its misspecified. $\endgroup$ – EconJohn Dec 12 '17 at 4:37
  • $\begingroup$ Can I estimate $\phi_1 - \phi_{p-1}$ consistently or my estimates for all AR terms is inconsistent $\endgroup$ – Daniel Dec 12 '17 at 5:32
  • $\begingroup$ The exact paper you need is here. $\endgroup$ – 123 Dec 13 '17 at 12:47
2
$\begingroup$

Just think your true model is: $$y_t = \phi_0 + \phi_1 y_{t-1} + \phi_2 y_{t-2} + u_t \tag{1}$$

So, under your true model, $u_t$ is uncorrelated with all explanatory variables.

So, instead of using AR(2), use selected AR(1) model as: $$y_t = \phi_0 + \phi_1 y_{t-1} + \epsilon_t \tag{2}$$ So, now, $\epsilon_t = \phi_2 y_{t-2} + u_t$; therefore, $y_{t-1}$ will no longer will be uncorrelated with $\epsilon_t$ (due to correlation between $y_{t-1}$ & $y_{t-2}$). In simple words, this lead to the problem of endogeneity. And we know, OLS estimators are not consistent when residuals are correlated with any of the explanatory variables.

| improve this answer | |
$\endgroup$
  • 2
    $\begingroup$ Right. Another useful keyword is omitted variable bias. When the omitted variable is not orthogonal to all of the regressors (individually and their linear combinations), the estimates of the included regressors are biased both in finite samples and asymptotically, hence, inconsistent. $\endgroup$ – Richard Hardy Jan 17 '18 at 17:56
2
$\begingroup$

As Neeraj has explained correctly, omitting a variable will lead to inconsistent estimates of the included variables. They will be biased both in a finite sample and asymptotically because the included variables will be capturing the effect of the omitted variable. (The exception would be if the omitted variable is uncorrelated with the included regressors individually and their linear combinations; then there would be no bias, finite or asymptotic.)

Regarding the forecast error, it may or may not increase depending on how big the true effect of the variable is and how precisely you could estimate the coefficient had you not omitted the variable from the model.

  • If the effect is large and you can estimate the coefficient quite precisely, including the variable would help increase the forecast accuracy (reduce the forecast error), and conversely, omitting the variable would decrease forecast accuracy.
  • Meanwhile, if the effect is small and you cannot estimate the coefficient quite precisely, including the variable could induce too much unnecessary variability in the forecast and this way reduce the forecast accuracy.

This phenomenon is known as the bias-variance trade-off. Excluding a variable that truly belongs in the model introduces a bias, but it also reduces estimation variance. The total effect in terms of expected squared forecast error $$ \mathbb{E}\left((y-\hat y)^2\right)=\text{bias}^2+\text{variance}+\text{irreducible error} $$ is $$ \Delta\text{bias}^2+\Delta\text{variance} $$ where $\Delta$ denotes the change due to omitting the variable from the full model. If $$ \Delta\text{bias}^2+\Delta\text{variance}<0 $$ you will have an increase in forecast accuracy. If not, you will have a decrease. Hence, the answer is, it depends. Akaike's information criterion AIC gives a quick indication of which of the two cases your are in. The model with the lower AIC value is likely going to forecast better.

| improve this answer | |
$\endgroup$
0
$\begingroup$

If your model does not pass the Ramsey RESET for model specification then all of your results are untrustworthy. If this is the case, experiment with logarithmic transformations, etc.

| improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.