I wonder if two identical time series are cointegrated. Can anyone shed some light on this? Thanks
1 Answer
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If I understand your question correctly, you are asking if you take two identical time series (i.e. a time series and a direct copy of it), are these two cointegrated?
If that is your question, then it really reduces to a simpler form: is a time series stationary? As I understand it, cointegration is the phenomenon whereby non-stationary processes have linear combinations that are stationary. If you have two identical time series, then the set of all linear combinations of those time series is equivalent to the set of all scalar multiples of one of them. In this case, if a time series has a scalar multiple of it that is stationary, then the original time series was itself stationary.
I may be way off base, but this seems like the question you're asking, and it doesn't seem to be much more complicated than this. If you intended something more complicated (i.e. a time series and a lagged version of itself), then there may be some more depth here.
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$\begingroup$ +1, if series are stationary, then any lag of them is stationary also, so there is no depth here :) $\endgroup$– mpiktasCommented Apr 16, 2011 at 17:57
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$\begingroup$ I see what you are saying. But two time series are said to be co-integrated if there exists a linear combination of the two, so that the combination is stationary. The difference of the two identical time series are a all-zero time series. Is it stationary? $\endgroup$– DavidCommented Apr 17, 2011 at 3:07
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1$\begingroup$ @David, yes, trivially so. Let $X_n = Y_n - Z_n$ where $Y_n$ and $Z_n$ are your "two identical series". For any $k$ and any $n$, $\mathbb{P}( (X_1, X_2, \ldots, X_k) \in B ) = \mathbb{P}( (X_{1+n}, X_{2+n}, \ldots, X_{k+n}) \in B )$, i.e., $(X_t)$ is stationary. $\endgroup$– cardinalCommented Apr 17, 2011 at 3:40
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$\begingroup$ I think you are slightly confusing when you say If you have two identical time series, then the set of all linear combinations of those time series is equivalent to the set of all scalar multiples of one of them. In this case, if a time series has a scalar multiple of it that is stationary, then the original time series was itself stationary. Isn't the point here that even if the original times series is non-stationary, the difference between it and itself is stationary (indeed it is constant, since it is always zero)? So a time series is cointegrated with a copy of itself. $\endgroup$– HenryCommented Apr 17, 2011 at 11:38
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$\begingroup$ David (and Cardinal)'s points are well taken: the trivial case is obviously an extension. I was avoiding the case where the linear combination is to take the difference, as it didn't seem to be what the OP was asking. $\endgroup$ Commented Apr 17, 2011 at 19:36