0
$\begingroup$

Given the model: \begin{aligned} Y_t &= \delta Y_{t-1}+u_t, \\ u_t &= \rho u_{t-1}+\epsilon_t, \end{aligned} where $\epsilon_t\sim i.i.d. (0,\sigma^2)$, $|\delta|,|\rho|<1$. Then how to find the prob. limit of the OLS estimator $\hat\delta$?

I have tried to use WLLN, since the $Y_t$ is not i.i.d., however the variance of $Y_t$ doesn't converge to 0. WLLN doesn't applies. So how to find the prob. limit of the OLS estimator $\hat\delta$?

Actually i calculated $\hat\delta$ directly using OLS estimator formula, and get \begin{equation} \hat\delta=\delta+\sum_{t=2}^T y_tu_{t-1}/\sum_{t=2}^Ty_{t-1}^2 \end{equation}. I wonder if i can expand $Y_t$ to infinite horizon to conclude $Y_t$ as covariance-stationary process or i can just expand $Y_t$ to finite horizon as $Y_t=\sum_{j=0}^{t-1}\delta^ju_{t-j}$. In order to use WLLN, I also used the above finite horizon expansion to calculate the $E(\sum_{t=2}^Ty_{t-1}^2/T-2)$ and find the probability limit is $\frac{(1+\rho \delta)\sigma^2}{(1-\rho \delta)(1-\delta^2)(1-\rho^2)}$. However, if i just consider the finite horizon expansion,then the fourth moment expection $E[(\sum_{t=2}^Ty_{t-1}^2/T-2)^2]$ is evidently doesn't converge to zero. So now i am quite confused. Is the OLS estimator $\hat\delta$ converge to some distribution in probability?

update: uploaded the full question enter image description here Thanks for everyone, i'd appreciate it if you can add your comment to this question.

$\endgroup$
2
  • $\begingroup$ I guess you mean $\vert\delta\vert,\vert\rho\vert<1$ and not 0. $\endgroup$
    – orsos
    Commented Dec 22, 2021 at 11:59
  • $\begingroup$ @orsos yes,sry it is a typo $\endgroup$
    – rookie
    Commented Dec 22, 2021 at 12:02

1 Answer 1

3
$\begingroup$

In this case, the OLS estimator for $\delta$ is inconsistent, i.e., $plim(\hat{\delta}) \neq \delta$. I think the easiest way to see this, is to subtract $\rho y_{t-1}$ in the first equation. You get: \begin{align} y_{t}-\rho y_{t-1}&=\delta y_{t-1}+u_t-\rho(\delta y_{t-2}+u_{t-1}) \end{align} Rearranging the terms yields: $$ y_t=(\delta+\rho)y_{t-1}-\delta \rho y_{t-2}+(u_t-\rho u_{t-1}) $$ Basically, $y_{t-2}$ is an omitted variable in your original model and you can apply the usual omitted variable bias methology. I haven't calculated it yet, but this prob limit should look like: $$ plim(\hat{\delta})=\delta-\delta\rho\frac{cov(y_{t-2},y_{t-1})}{var(y_{t-1})} $$ Whatever the second term will look like when it is simplified.

$\endgroup$
1
  • $\begingroup$ Thanks a lot! I updated some of my thought in the main question and you can have a glance. finally, thanks your help! $\endgroup$
    – rookie
    Commented Dec 23, 2021 at 5:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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