2
$\begingroup$

I have the following two processes: \begin{align} x_t &= x_{t-1} + u_t \tag{1} \\ x_t &= {\beta}_0 + {\beta}_1t + u_t \tag{2} \end{align}

Differencing once leads to:

\begin{align} \Delta x_t &= u_t \tag{1} \\ \Delta x_t &= {\beta}_1 + u_t + u_{t-1} \tag{2} \end{align}

From my notes, I understand that process (1) is stationary as $u_t$ is a white noise process. However, my lecture notes also state that process (2) is not stationary.

Why is process (2) not stationary?

$\endgroup$
2
$\begingroup$

Assuming $E[u_t]=0$ and $\beta_1 \not = 0$ then in the second question $$E[x_t]-E[x_{t-1}] = E[\Delta x_t] = \beta_1 \not = 0$$ which imples the process is not stationary.

$\endgroup$
  • $\begingroup$ What if B_1 is a constant? The first condition for stationarity is that the process must be constant in mean, so if B_1 is a constant, the condition is satisfied $\endgroup$ – A K May 3 '15 at 14:53
  • $\begingroup$ If $\beta_1$ is a non-zero constant then $E[x_t]$ is increasing with $t$, so the original process is not constant in mean. $\endgroup$ – Henry May 3 '15 at 16:07
  • $\begingroup$ @Henry first process is stationary after differencing, second contains unit root in MA process... $\endgroup$ – Analyst May 4 '15 at 9:38
1
$\begingroup$

First process is stationary in differences but non-stationary in levels.

Second process is linear time trend and is non-stationary in levels. It is non-stationary also in differences since it contains non-invertible root in the MA(1) process. But you can easily see that deviation from the mean trend is white noise just by subtracting constant and trend value from the y(t).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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