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I am looking at two very similar time series (correlation is about .997) - two commodity price series. Johansen coint. test indicates 1 coint. relationship for case 1 (no trend and no intercept) and no cointegration for other cases described by Johansen.

Lets denote the series A, B. If I use the order of series A, B, then in the estimated VECM, the coint. vector is (1, -1.003722)' The alpha coeff. for EC terms are 0.023158 (in equation for A) and 0.431335 (in eq. for B).

Now, if I estimate the VECM with reversed order, that is B, A (which should only influence the normalization of coint. vector) i get these results: Coint. vector ( 1, -0.996066)' The alpha coeff. for EC terms are -0.060448 (in equation for A) and -0.444057 (in eq. for B).

I wonder, why did the coefficients for EC terms changed their signs? I was under the impression that EC terms should have negative coefficients in order to maintain the coint. relationship.

Data: download CSV Use from obs. 62 onwards (a side question is that the cointegration results vary if you use all observations, or if you start from obs. 62)

Please, tell me if I understand it in the following example correctly.

Let's assume we observe: A=5,  B= 5.01;    

For order A, B:    
coint. vector: (1, -1.003722)';  
5 -1.003722*5.01 = -0.02864722;  
EC with loading in eq. for A:  0.023158 * -0.02864722 -> negative (A should go down?);   
EC with loading in eq. for B:  0.431335* -0.02864722 -> negative (B should go down?);  
for B, A:    
coint. vector: ( 1, -0.996066)';  
5.01-0.996066*5 = +0,02967;  
EC with loading in eq. for A:  -0.060448*+0,02967 -> negative (A should go down?);  
EC with loading in eq. for B:  -0.444057*0.02967 -> negative ( B should go down?);  
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    $\begingroup$ This could be software-related. Can you share the platform/code & data that you are using for this analysis, so that we can replicate these numbers? $\endgroup$ – tchakravarty Jan 23 '15 at 9:51
  • $\begingroup$ SW used was Eviews. Lag order 2 $\endgroup$ – vladimir Jan 23 '15 at 12:30
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Let the two cointegrating variables be $x$ and $y$.

The error correction term is $x-by$.

Consider equation 1 in the system (equation 1 or 2 does not matter, any one of them is enough to understand).

The error correction term with its loading is $a(x-by)$.

Now consider what happens when you switch the positions of $x$ and $y$ by renormalizing the cointegration vector. You use the following: $a(x-by)=-ba(y-\frac{1}{b}x)$ and get the new error correction term with its loading $a'(y-b'x)$ where $a'=-ba$ and $b'=\frac{1}{b}$.

That means you actually expect the opposite sign, and it's not a mistake. Your reported coefficient values seem to agree with this.

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  • $\begingroup$ OK, if we take coint. vector (1, -0.996), then let's assume value of A =5, B should be in equilibrium B=4.98. If B is actually 5.01, then we have error in the long-run: e=B-Bhat= 5.01-4.98 = +0.03 ; If we want this relationship to be maintained, shouldn't the loading coefficient be negative, i.e. if multiplied by the positive error of 0.03, it should be reflected in the movement of B downwards (to the equilibirum value)? $\endgroup$ – vladimir Jan 23 '15 at 10:45
  • $\begingroup$ I agree with you conceptually. On the other hand, doesn't my answer show how easily the sign gets reversed by just adopting an alternative representation of exactly the same object? If I have not made a mistake in the formulas above (have I?), it suggests that you look again into the estimation output and double-check whether it really does not make sense. Perhaps you will find out that the signs do match after all. Of course, I cannot guarantee you that since I cannot see the output :) It's just my guess. $\endgroup$ – Richard Hardy Jan 23 '15 at 11:11
  • $\begingroup$ So how is it with the loading coefficients? Can you comment on the example I provided? $\endgroup$ – vladimir Jan 29 '15 at 12:32
  • $\begingroup$ @vladimir Perhaps Examples 7-9 here are helpful: math.ku.dk/~sjo/papers/OverviewPreprint.pdf $\endgroup$ – hejseb Jan 14 '16 at 19:58
  • $\begingroup$ @hejseb, just checking, does my answer make sense? $\endgroup$ – Richard Hardy Jan 14 '16 at 20:04

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