# Time series has unit root iff integrated of order 1

The wikipedia page for Unit Root says something like "..the stochastic process has a unit root or, alternatively, is integrated of order one..". Are these actually equivalent? Could someone point me to a reference of the proof?

## 2 Answers

Definition of Integration of order $$d$$ states that if $$X_t$$ is $$I(d)$$, then $$(1-L)^dX_t$$ is a stationary process, where $$L$$ is the lag operator. So, the characteristic equation (CE) of your equation has $$(1-L)^d$$ in its factorization (and no other $$(1-L)$$ factor in the remaining part), because the remaining part shouldn't have $$(1-L)$$ for stationarity.

For example, if your process is described by the CE: $$(L-0.5)(1-L)^2X_t$$, $$X_t$$ is non-stationary; but letting $$Y_t=(1-L)^2X_t$$, creates the process $$(L-0.5)Y_t$$, which doesn't have unit root, i.e. it is stationary.

As a result, if we have one unit root in the characteristic equation, the process is said to be $$I(1)$$, i.e. the statements are equivalent.

My understanding of a unit root is that this reflects non-stationarity which requires differencing for many methods (ARIMA for example and regression). A variable that is integrated of order one means that something needs to be differenced once to be stationary. So if you have a unit root you are likely integrated of order 1 [although it might be the case that this may requires two differencing to be stationary in which case you would be integrated of order 2 - I am not sure if a unit root ever requires that or not although some variables certainly do].