I have a question regarding an I(1) process.

Suppose we have the following equation:

$$ Y_0 = \alpha_1Y_{-1}+\alpha_2Y_{-2}+\alpha_3Y_{-3}+...+\alpha_4Y_{-4} $$ $$ with \sum\limits_{i=1}^4 \alpha_{i} = 1 $$

How can I prove mathematically that Y is already an I(1) process, given that the sum of the lagged coefficients are equal to one?

Thanks in advance!

  • 1
    $\begingroup$ One equation does not prove anything about the properties of the process. $\endgroup$ – Dilip Sarwate Mar 17 at 16:11

As @Dilip says, your notation is wrong, it should be general as follows, having $t$ in it, together with its ch. eqn: $$Y_t=\sum_{i=1}^4\alpha_iY_{t-i}\rightarrow 1=\sum_{i=1}^4\alpha_iL^i$$

If the ch. eqn. has $(1-L)^d$ as a factor, than it is $I(d)$. Checking if at least we have $(1-L)$ in its factorization is easy, because if it is, $L=1$ should be a root of the characteristic equation. Substituting into the above equation yieds: $$1=\sum_{i=1}^4\alpha_i$$ which is already true. So, $L=1$ is a root of the characteristic equation, and $Y_t$ is at least $I(1)$. We can't say anything about larger orders.


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.