This is a homework question:

$y_t = \alpha y_{t-1} + \beta_0 x_t + \beta_1 x_{t-1} + \epsilon_t , 0<\alpha<1, (t=1,...,T)$ where $x_t$ is an I(1) process independent from $\epsilon_t$ for all $s$ and $t$, and $\epsilon_t , t=1,...,T $ are i.i.d with mean zero and variance $\sigma^2$


1) Suppose that $\beta_0 \not= -\beta_1$. Is it possible that $y_t$ is stationary?

2) Write down the error correction form of the model, showing the relationship between the new parameters and those in the original model. Argue that $y_t$ and $x_t$ are co-integrated and write down the co-integrating vector

Any help/pointers would be great.


For question 1), take the expectation of both sides, assume that the expectation of Y is the same in both periods (the definition of stationarity) and see what conditions the coefficients on X must satisfy.

For question 2), try rearranging the equation to be in terms of X and see what determines its variation.

P.S. I am still getting the hang of the mathematical notation on here, but I figure too much working out would be giving it away anyway

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  • $\begingroup$ Thanks for your reply. Here is my attempt at question 1) $y_t = \alpha y_{t-1} + \beta_0 x_t + \beta_1 x_{t-1} + \epsilon_t$ Taking expectations $E(y_t) = \alpha E(y_{t-1}) + \beta_0 E(x_t) + \beta_1 E(x_{t-1})$, expectation of error term is zero. Then assuming stationarity, $E(y_t) = E(y_{t-1})$ which using the above equation happens when $\alpha =1 , \beta_0 =0$ and $\beta_1 = 0$ $\endgroup$ – Kai_M Nov 11 '15 at 17:09
  • $\begingroup$ So $y_t$ could be stationary if these conditions are satisfied. $\endgroup$ – Kai_M Nov 11 '15 at 17:15
  • $\begingroup$ For part 2) It first asks to write down the error correction model and show the relationship between the new parameters and those in the original model: $y_t = \alpha y_{t-1} + \beta_0 x_t + \beta_1 x_{t-1} + \epsilon_t $ subtract $y_{t-1}$ from both sides. $\Delta y_t = - (1- \alpha) y_{t-1} + \beta_0 x_t + \beta_1 x_{t-1} + \epsilon_t $ subtract and then add back $\beta_0 x_{t-1}$ from the RHS and re-write as $\Delta y_t = \Delta \beta_0 x_t + \lambda(y_{t-1} - \delta x_{t-1}) + v_t $ where $\lambda = 1-\alpha$ and $\delta = \frac {\beta_0 + \beta_1}{1-\alpha}$ $\endgroup$ – Kai_M Nov 11 '15 at 17:27
  • $\begingroup$ For the next part of 2), when you say rearrange in terms of X, do you mean a) $\beta_0 x_t + \beta_1 x_{t-1} = y_t -\alpha y_{t-1} - \epsilon_t $ or b) $x_t = \frac {1}{\beta_0} (y_t -\alpha y_{t-1} - \beta_1 x_{t-1} - \epsilon_t) $ ? In either case im still having trouble progressing. $\endgroup$ – Kai_M Nov 11 '15 at 17:39
  • $\begingroup$ Any more ideas for part 2? $\endgroup$ – Kai_M Nov 13 '15 at 14:05

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