# Co-integration and Error correction [closed]

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 $\sigma^2$

Question

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.

• 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 $\beta_1 = 0$ – Kai_M Nov 11 '15 at 17:09
• So $y_t$ could be stationary if these conditions are satisfied. – Kai_M Nov 11 '15 at 17:15
• 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 $\delta = \frac {\beta_0 + \beta_1}{1-\alpha}$ – Kai_M Nov 11 '15 at 17:27
• 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. – Kai_M Nov 11 '15 at 17:39