0
$\begingroup$

I have a simply but still challenging question (at least to me). My question boils down to the following, if the first differences $\Delta y_t$ of an I(1) process is stationary, is then also the expectation of this $E[\Delta y_t]$ stationary? And if yes/no, why?

In greater detail: I have two I(1) time series processes $y_{2t}$ and $y_{1t}$ and I am meant to show that they can be cointegrated with vector (1 -1). What I obtain is $y_{2t}-y_{1t}=E_t{\Delta y_{1t+1}}$. Which means if $E_t{\Delta y_{1t+1}}$ would be stationary, then this would be true.

$\endgroup$
3
  • 1
    $\begingroup$ Hi: I don't think that you mean to use expectation here. The expectation of $\triangle{y_{t}}$ is a constant so to talk about stationarity with respect to a constant doesn't really make sense. I guess, technically speaking, one could say that the mean and variance of a constant are constants so it's stationary but I've never seen the concept of stationarity applied to constants. $\endgroup$
    – mlofton
    Commented Feb 11, 2022 at 18:05
  • $\begingroup$ I agree with mlofton. Also, the last equation does not make sense. $\endgroup$ Commented Feb 11, 2022 at 18:09
  • $\begingroup$ Thank you! This helps me $\endgroup$
    – Silke
    Commented Feb 14, 2022 at 8:43

1 Answer 1

1
$\begingroup$

My question boils down to the following, if the first differences $\Delta y_t$ of an I(1) process is stationary, is then also the expectation of this $E[\Delta y_t]$ stationary? And if yes/no, why?

If you deal with an $I(1)$ process you have that his first difference is stationary; this follow the definition of $I(1)$ process. Indeed the stationarity condition deal with the process not his expectation. Therefore your question is bad posed. Moreover it is true that if the process is stationary some conditions must be applied to his moments.

$\endgroup$
5
  • $\begingroup$ Moreover it is true that if the process is stationary some conditions must be applied to his moments. Does that have to do with weak vs. strong stationarity (where weak stationarity restricts the first two moments to be finite while strict stationarity does not)? $\endgroup$ Commented Feb 11, 2022 at 20:29
  • $\begingroup$ Yes this is true. Even if I had primarily in mind the quite general condition that under stationarity (strong) unconditional moments (if any) cannot depend on $t$. $\endgroup$
    – markowitz
    Commented Feb 11, 2022 at 20:36
  • $\begingroup$ Similarly, for weak stationary processes the same condition is true for mean and variance. $\endgroup$
    – markowitz
    Commented Feb 11, 2022 at 20:59
  • $\begingroup$ Thank you for your helpful comments! I am sorry if my question was not clear. But what I understand from your answers is that, as $\Delta y_t$ is stationary, we know that its expectation $E[\Delta y_t]=\mu$ is equal to a constant. Thus, we can say that $E[\Delta y_t]=\mu \sim I(0)$? $\endgroup$
    – Silke
    Commented Feb 14, 2022 at 8:27
  • $\begingroup$ We can say that $\Delta y_t$ is stationary and $E[\Delta y_t]$ is a constant. If you reread my answer you can understand why your last statement is confusing. $\endgroup$
    – markowitz
    Commented Feb 14, 2022 at 9:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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