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 $d$.
Best Answer
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.