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I know that an $I(1)$ process becomes stationary after differencing once. However, I somehow always equated that to its being a random walk because say having a unit root process like \begin{eqnarray} y_{t} = y_{t-1} + \epsilon_t \end{eqnarray} is clearly $I(1)$ and becomes converted to an $MA_\infty$ $y_t = \sum_{s=0}^t \epsilon_{t-s}$, which strikes me as a random walk as discussed here for example.

Now, is there a difference between a random walk a process $I(1)$?

Hirek
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1 Answers1

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If I understand it correctly, a random walk is a special case of an $I(1)$ process.

A random walk can be written as

$$y_t=y_{t-1}+\varepsilon_t$$

as you have done it. The increments $\varepsilon_t$ have to be independent of each other over time, e.g. $\varepsilon_t \sim i.i.N(0,\sigma^2$).

In case of a $I(1)$, the increments need not be independent; they have to be stationary, e.g. $\varepsilon_t$ follows ARMA($p,q$).

Hence, random walk is a special case of an $I(1)$ process. For example, a random walk is an ARIMA($0,1,0$); meanwhile, an $I(1$) process can be ARIMA($p,1,q$) with $p,q \geqslant 0$.

Richard Hardy
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