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I am having trouble solving the order of integration of a time series process. Consider the following processes:

\begin{align*} \epsilon_t &\sim i.i.d.(0,1) \\ x_t &= \alpha x_{t-1}+\epsilon_t \\ y_t &= y_{t-1}+x_t \\ z_t &= z_{t-1}+y_t \end{align*}

Assume $\alpha = 0.8$.

Then, what is the order of integration of $z$? As in $z$ is a $I(d)$-process, where I want to determine the order of integration $d$.

EulersNumber
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Hi: $y_{t}$ is I(1) so the first difference of $z_{t}$ is I(1). Therefore, you'd have to difference the difference of $z_{t}$ in order to obtain an I(0) process, so, by definition, it must be I(2). What I mean by "definition" is that the number of times one needs to difference a process in order to make it stationary, is its order of integration, $d$. I hope that helps.

EDIT: I realized after reading my answer that I did not explain why $y_t$ is I(1). So, the reason that $y_{t}$ is I(1) is the following:

$x_t$ is a stationary process ( i.e; an I(0) process ) because the absolute value of $\alpha$ is less than 1.0. This means that the first difference of $y_{t}$ is I(0). Since the first difference of the $y_{t}$ process is I(0), this implies that $y_{t}$ is an I(1) process.

mlofton
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  • Thanks, that definitely makes sense. So just to be sure that I understand things correctly: The first difference of $z_t$ is $y_t$, which is an $I(1)$ process. So then we take the difference of the difference of $z_t$ (difference of $y_t$), which is $x_t$. Well, $x_t$ is stationary because $|\alpha|<1$, so it is an $I(0)$ process. Hence $z_t$ must be an $I(2)$ process? – EulersNumber Dec 17 '21 at 07:40
  • In particular, because the second difference of $z_t$ is a stationary process. – EulersNumber Dec 17 '21 at 07:46
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    @EulersNumber: That's perfect. Note that, in practical cases, most series are I(0) or I(1). – mlofton Dec 17 '21 at 13:14