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Consider the following model:

$y_t=a+by_{t-1}+u_t\\ u_t=cu_{t-1}+v_t\\ v_t\sim N(0,\sigma^2)\quad IID $

I am trying to show that the OLS estimator of $y_t$ on $y_{t-1}$ and a constant for $b$ is not consistent. However, I am having some trouble with the calculations. Particularly, I am stuck on how to compute $E[y_{t-1}^2]$, and I am not sure if the OLS estimator $\widehat{b}$ would converge to the usual covariance over variance. I need help on how to proceed, maybe another strategy would be better? Thanks!

Edit 1: $b,c$ are such that both processes are invertible

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The expression for the OLS esimastor is

$$\hat b = \frac {\sum (y_ty_{t-1} - ay_{t-1})}{\sum y_{t-1}^2}$$

$$=\frac {\sum [(a+by_{t-1}+u_t)y_{t-1} - ay_{t-1}]}{\sum y_{t-1}^2}$$

$$=\frac {b\sum y_{t-1}^2+\sum u_ty_{t-1}}{\sum y_{t-1}^2}$$

$$=b + \frac {(1/T)\sum u_ty_{t-1}}{(1/T)\sum y_{t-1}^2}$$

As $T$ grows, the denominator converges to the variance of the process, but the main thing is that it converges to something other than infinity. Then what matters is what happens to the numerator:

$$(1/T)\sum u_ty_{t-1} = (1/T)\sum (cu_{t-1}+v_t)y_{t-1}$$

which does not converge to zero due to the correlation between $u_{t-1}$ and $y_{t-1}$.

So $\hat b$ is inconsistent.

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  • $\begingroup$ (+1) This is a very concise answer. $\endgroup$ – Digio Dec 26 '17 at 11:43

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