# First difference stationary process

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!

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

## 1 Answer

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.