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I did a Johansen test.
The max eigenvalue test resulted in 5 cointegrated variables with contegration rank being 1.
The trace test resulted in 5 cointegrated variables with contegration rank being 3.

Below is the estimated VECM with cointegration rank of 3:

cajorls(ca.jo(mydata, ecdet="none", type="trace", spec="transitory", K=29),r=3)
$rlm

Call:
lm(formula = substitute(form1), data = data.mat)

Coefficients:
              BIST100.d   Gold.d      Oil.d       TRY_USD.d   USD_EUR.d 
ect1          -4.843e-02   2.939e-05   1.634e-04  -7.808e-07   9.200e-07
ect2          -1.780e+02   1.322e-01   2.673e-01   5.668e-04   1.891e-03
ect3          -1.217e+01   2.622e-02   1.023e-02  -1.245e-04   4.935e-04
constant      -1.335e+04   1.941e+01  -1.246e+01   2.701e-01   1.689e-01
BIST100.dl1   -2.772e-02   2.683e-05  -1.036e-04  -4.862e-06  -2.745e-07
Gold.dl1       9.454e+01  -4.467e-03   6.449e-02   4.761e-03  -3.065e-04
......          ......      ......        ....       .....     .......
$beta
                ect1          ect2          ect3
BIST100.l1  1.000000e+00  0.000000e+00  4.336809e-19
Gold.l1    -1.136868e-13  1.000000e+00  0.000000e+00
Oil.l1     -3.197442e-14  2.775558e-17  1.000000e+00
TRY_USD.l1  1.152179e+05 -4.851482e+01 -5.183614e+01
USD_EUR.l1  8.984704e+04 -8.461897e+01 -3.174152e+02

I expect an error correction term of the format $y_1-\beta_2 y_2-\beta_3 y_3-\beta_4 y_4-\beta_5 y_5$. Or equivalently, all signs being "$-$" in the above ect1, ect2, ect3 terms. All are not "$-$", e.g., for the 1st ect1: -4.843e-02, +2.939e-05, +1.634e-04, -7.808e-07, +9.200e-07. There are "$+$" and "$-$" coefficients in the error correction terms above.

Does this show wrong unsuitable formulation in VECM? The cointegration rank is either 1 (max eigen) or 3 (trace). Both present "$+$" and "$-$" coefficients in error correction terms.

Below is the estimated VECM with cointegration rank of 1:

    cajorls(ca.jo(mydata, ecdet="none", type="eigen", spec="transitory", K=29),r=1)
$rlm

Call:
lm(formula = substitute(form1), data = data.mat)

Coefficients:
              BIST100.d   Gold.d      Oil.d       TRY_USD.d   USD_EUR.d 
ect1           6.592e-03  -1.138e-05   1.275e-05  -1.830e-07  -8.523e-08
constant      -1.035e+04   1.782e+01  -1.999e+01   2.857e-01   1.334e-01
BIST100.dl1   -5.707e-02   5.970e-05  -1.134e-05  -5.482e-06   6.044e-07
......          ......      ......        ....       .....     .......
$beta
                 ect1
BIST100.l1      1.000
Gold.l1     -5829.934
Oil.l1      -1100.681
TRY_USD.l1 455111.151
USD_EUR.l1 932542.892

Any idea? Can I go with the usual ritual vec2var(ca.jo(...), r=1 or r=3) or something must be handled properly?

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There is nothing wrong with the error correction term having both positive and negative loadings.

In fact, you could construct an example yourself.

  • Take $y_{1,t}=\sum_{\tau=1}^t \varepsilon_{1,\tau}, \ y_{2,t}=\sum_{\tau=1}^t \varepsilon_{2,\tau}, \ \dotsc, \ y_{k,t}=\sum_{\tau=1}^t \varepsilon_{1,\tau}$ where $\varepsilon_{i,\tau}$ are $i.i.d.$ across all $\{i,\tau\}$. Clearly, $y_{i,t}$s are integrated processes.
  • Define, for example, $y_{k+1,t}=y_{1,t}-y_{2,t}+y_{3,t}-y_{4,t}+y_{5,t}-\dotsc+(-1)^k y_{k,t}+\varepsilon_{k+1,t}$ where $\varepsilon_{k+1,t}$ is a stationary variable.
  • Then a linear combination $$y_{k+1,t}-y_{1,t}+y_{2,t}-y_{3,t}+y_{4,t}-y_{5,t}+\dotsc-(-1)^k y_{k,t}=\varepsilon_{k+1,t}$$ is stationary. So $(y_1,y_2\dotsc,y_k,y_{k+1})$ are cointegrated. The latter sum can serve as an error correction term. Note that it has altering signs. By changing the construction of $y_{k+1,t}$ (in the second bullet point) you can get whatever loadings you like (positive or negative) in the error correction term.
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  • $\begingroup$ Very practical and instructful answer; somewhat a top to bottom approach. Thx a lot Hardy. $\endgroup$ – Erdogan CEVHER Jul 10 '16 at 10:55
  • $\begingroup$ A constructive example helped me understand cointegration better than any other exposition. It is a pity not all textbooks start from there. I just loved it when I first saw something like it in a textbook by Davidson and MacKinnon, "Econometric Theory and Methods". By the way, I think of my exposition as bottom-up (rather than top-down) since it is constructive in essence. $\endgroup$ – Richard Hardy Jul 10 '16 at 10:57
  • $\begingroup$ Yeah, you are right, it is bottom-up [from (y1,t, ..., yk,t) to yk+1,t]. I said top-down with happily being understood the reasoning and focused on yk+1,t itself alone and looking reversively back to the coefficients from yk+1,t :) $\endgroup$ – Erdogan CEVHER Jul 10 '16 at 11:11

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