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.

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

1 Answer 1


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.

  • $\begingroup$ 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
    Dec 23, 2021 at 5:47

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