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?
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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.
gunes
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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].
user54285
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