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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!

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

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