Good evening
Thank you for all the replies you sent me. in part, managed to control the problem of error that was giving applying log considering one of the suggestions that argued the possibility of specification is wrong.
But I heard some comments that perhaps this was not the most appropriate test.
Then I wanted to ask two items:
-What Test grelt could analyze coientegração with breaks
- Which test could use the order of variables in VEC modleo

I would like to point out that the use gretl, but still not use the scrips, only the functions that are already in command, then at each test that suggest me, I need you to tell me how to implement the gretl.
A thousand apologies for these issues and also not totally dominate the gretl

much appreciate the help of all.

Augusta Pelinski Raiher

2015-11-29 10:18 GMT-02:00 Pindar Os <pindar777@gmail.com>:

Hi there,

 

I’m still working on the restricted OLS regressions and in the meantime even uploaded an early version of a function package that uses a SQP of GNU R and a bootstrap routine to get coeffs and standard errors.

 

In order to produce a nice show case of the optimization problem I generate all combination of coefficients with a step size of n. If there are 3 coeffs that are >=0 and should sum up to 1 the matrix can be easily generated via loop. Since gretl has a special "for loop" one could use this loop version in order to fill the percentages in the matrix in one step. However, there is a bug. Have a look. I’m using the current x64 snapshot for windows.

 

Cheers and a nice first Sunday in Advent

Leon

 

<hansl>

scalar intN_1_a = 50 # step size of 2 percent

scalar intStep = 1/intN_1_a

eval intStep

matrix mCombis = NA

matrix mCombisALL = NA

 

set stopwatch

loop for i=0..intN_1_a --quiet

    loop for j=0..intN_1_a --quiet

        if i+j<=intN_1_a

            if i+j == 0

                mCombisALL = i ~ j ~ (intN_1_a-i-j)

            else   

                mCombisALL |= i ~ j ~ (intN_1_a-i-j)

            endif   

        endif  

    endloop 

endloop

eval $stopwatch/60

eval rows(mCombisALL)

# correct number of rows since

scalar intSum3 = (intN_1_a)*3 + (1+intN_1_a-3+1)*(1+intN_1_a-3)/2

# for three coeefs I found this closed formula, with more it’s another…

eval intSum3

 

 

set stopwatch

loop for (i=0; i<=1; i+=intStep) --quiet

    loop for (j=0; j<=1; j+=intStep) --quiet

        if i+j<=1

            if i+j == 0

                mCombis = i ~ j ~ (1-i-j)

            else   

                mCombis |= i ~ j ~ (1-i-j)

            endif   

        endif  

    endloop 

endloop

eval rows(mCombis)

eval $stopwatch

# correct number

scalar intSum3 = (intN_1_a)*3 + (1+intN_1_a-3+1)*(1+intN_1_a-3)/2

eval intSum3

# wrong, WHY? The reason has to do with the.xx999 values and the missing 1s.

 

<hansl>

 

 


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--
Augusta Pelinski Raiher
(Docente do curso de Economia e do Programa de Pós-Graduação em Ciências Sociais na Universidade Estadual de Ponta Grossa)