I(0) and I(1) Factors

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CharlieEVIEWS
Posts: 202
Joined: Tue Jul 17, 2012 9:47 am

I(0) and I(1) Factors

Postby CharlieEVIEWS » Wed Sep 18, 2013 4:24 pm

Dear all,

Here's a subroutine I wrote to recursively (!j) estimate Stock and Watson (2002) I(0) factors and Bai (2004) I(1) factors based on a Gauss code by Igor Masten where N>T.

Note: X is our matrix of time series created from g1_standstat1 - a group of standardized stationary variables.

Code: Select all

'Stock and Watson Code:

%start = "1980q1"
   %endminusone = "2002q4-1"
   !noiterations = 20
for !j = 1 to !noiterations
smpl {%start} {%endminusone}+!j
stom(g1_standstat1,X)
scalar T = @rows(X)
scalar N = @columns(X)
'maximum number of factors:
scalar k_max = 4
'Because N>T:
  sym XX = X*@transpose(X)
           scalar il = T-1-k_max+1
           scalar iu = T-1
'Note: VA corresponds to VEctors, VE corresponds to VAlues
matrix va = @eigenvectors(XX)
vector ve = @eigenvalues(XX)
for !i = il to iu
vector factor!i = ((T-1)^(0.5))*@columnextract(va,!i)
matrix(T,k_max) factors
colplace(factors,factor!i,!i-il+1)
next
matrix loadings = @transpose(factors)*X/(T-1)
'For a normalization, multiply/divide by the element [kmax,1] of the loadings matrix.
scalar norm = loadings(k_max,1)
factors = factors*norm
loadings = loadings/norm
'Re-extract normalized factors from this matrix as series for future subroutines, in reverse order:
'i.e. the factor associated with the largest eigenvalue is F_1, second is F_2,...,F_k_max
mtos(factors,factorgroup1)
for !i = 1 to k_max
scalar temp = k_max-!i+1
rename ser0{temp} f_1_!i_!j
next
delete temp
delete factorgroup1


This is our subroutine for estimating I(1) factors, where g1_standnonstat1 is our group of nonstationary standardized variables and the subscript denotes the I(1) property.

Code: Select all

'Bai (2004) Code

stom(g1_standnonstat1,i_X)
scalar i_T = @rows(i_X)
scalar i_N = @columns(i_X)
scalar i_k_max = 4
  sym i_XX = i_X*@transpose(i_X)
'Note the difference with the stationary routine here:
           scalar i_il = T-k_max+1
           scalar i_iu = T
matrix i_va = @eigenvectors(i_XX)
vector i_ve = @eigenvalues(i_XX)
for !i = i_il to i_iu
'Again, note the difference with the stationary routine here:
vector i_factor!i = (T)*@columnextract(i_va,!i)
matrix(i_T,i_k_max) i_factors
colplace(i_factors,i_factor!i,!i-i_il+1)
next
matrix i_loadings = @transpose(i_factors)*X/(i_T^2)
'normalize
scalar i_norm = i_loadings(i_k_max,1)
i_factors = i_factors*i_norm
i_loadings = i_loadings/i_norm
'Re-extract normalized factors from this matrix, in reverse order: i.e. the factor associated with the largest eigenvalue is if1, second is i_F_2,...,i_F_km_max
mtos(i_factors,i_factorsgroup1)
for !i = 1 to i_k_max
scalar temp = i_k_max-!i+1
rename ser0{temp} if1_!i_!j
next
delete temp
delete i_factorsgroup1



Best wishes
Last edited by CharlieEVIEWS on Sun Feb 22, 2015 8:01 am, edited 1 time in total.

CharlieEVIEWS
Posts: 202
Joined: Tue Jul 17, 2012 9:47 am

Re: I(0) and I(1) Factors

Postby CharlieEVIEWS » Fri Sep 20, 2013 5:16 am

Where T>N:

Code: Select all

%start = "1960q1"
   %endminusone = "2002q4-1"
   !noiterations = 20   
for !j = 1 to !noiterations
smpl {%start}+1 {%endminusone}+!j

'Stock and Watson (2002) - I(0) factors where T>N

stom(g1_standstat1,X)
scalar T = @rows(X)
scalar N = @columns(X)
'maximum number of factors:
scalar k_max = 4
  sym XX = @transpose(X)*X
           scalar il = N-k_max+1
           scalar iu = N
matrix va = @eigenvectors(XX)
vector ve = @eigenvalues(XX)
for !i = il to iu
vector loadings!i = (N)^(0.5)*@rowextract(@transpose(va),!i)
matrix(k_max,N) loadings
rowplace(loadings,loadings!i,!i-il+1)
next
matrix factors = ((X*@transpose(loadings))/N)
'For a normalization, multiply/divide by the element [kmax,1] of the loadings matrix.
scalar norm = loadings(k_max,1)
factors = factors*norm
loadings = loadings/norm
'Re-extract normalized factors from this matrix as series for future subroutines, in reverse order:
'i.e. the factor associated with the largest eigenvalue is F_1, second is F_2,...,F_km_max
mtos(factors,factorgroup1)
for !i = 1 to k_max
scalar temp = k_max-!i+1
rename ser0{temp} f_1_!i_!j
next
delete temp
delete factorgroup1
for !i = il to iu
delete loadings!i
next


And again, for the case where we want to extract I(1) factors:

Code: Select all



'Factors extracted as per Bai (2004): with a subscript i_ denoting integrated variables.

stom(g1_standnonstat1,i_X)
scalar i_T = @rows(i_X)
scalar i_N = @columns(i_X)
scalar i_k_max = 4
sym i_XX = @transpose(i_X)*i_X
           scalar i_il = N-i_k_max+1
           scalar i_iu = N
matrix i_va = @eigenvectors(i_XX)
vector i_ve = @eigenvalues(i_XX)
for !i = i_il to i_iu
vector i_loadings!i = (i_N)^(0.5)*@rowextract(@transpose(i_va),!i)
matrix(i_k_max,i_N) i_loadings
rowplace(i_loadings,i_loadings!i,!i-il+1)
next
matrix i_factors = ((i_X*@transpose(i_loadings))/N)
'Normalize
scalar i_norm = i_loadings(i_k_max,1)
i_factors = i_factors*i_norm
i_loadings = i_loadings/i_norm
'Re-extract normalized factors from this matrix, in reverse order: i.e. the factor associated with the largest eigenvalue is if1, second is i_F_2,...,i_F_km_max
mtos(i_factors,i_factorsgroup1)
for !i = 1 to i_k_max
scalar temp = i_k_max-!i+1
rename ser0{temp} if1_!i_!j
next
delete temp
delete i_factorsgroup1
for !i = i_il to i_iu
delete i_loadings!i
next



Warmest regards,

Charlie

hamidlalkhezri
Posts: 7
Joined: Sat Mar 25, 2017 11:54 pm

Re: I(0) and I(1) Factors

Postby hamidlalkhezri » Sun Mar 26, 2017 8:59 pm

hi
im working favar model with Eviews and I thank you for your guidance.
I have a few questions
1- what do you mean from I(0),I(1) factors?You're talking about stability?
2-when I run program[ 'Stock and Watson (2002) - I(0) factors where T>N] "for statement unterminated in "for !J=1 to 20"
please guide me

hamidlalkhezri
Posts: 7
Joined: Sat Mar 25, 2017 11:54 pm

Re: I(0) and I(1) Factors

Postby hamidlalkhezri » Sun Mar 26, 2017 8:59 pm

hi
im working favar model with Eviews and I thank you for your guidance.
I have a few questions
1- what do you mean from I(0),I(1) factors?You're talking about stability?
2-when I run program[ 'Stock and Watson (2002) - I(0) factors where T>N] "for statement unterminated in "for !J=1 to 20"
please guide me

hamidlalkhezri
Posts: 7
Joined: Sat Mar 25, 2017 11:54 pm

Re: I(0) and I(1) Factors

Postby hamidlalkhezri » Sun Apr 09, 2017 9:32 pm

CharlieEVIEWS wrote:Where T>N:

Code: Select all

%start = "1960q1"
   %endminusone = "2002q4-1"
   !noiterations = 20   
for !j = 1 to !noiterations
smpl {%start}+1 {%endminusone}+!j

'Stock and Watson (2002) - I(0) factors where T>N

stom(g1_standstat1,X)
scalar T = @rows(X)
scalar N = @columns(X)
'maximum number of factors:
scalar k_max = 4
  sym XX = @transpose(X)*X
           scalar il = N-k_max+1
           scalar iu = N
matrix va = @eigenvectors(XX)
vector ve = @eigenvalues(XX)
for !i = il to iu
vector loadings!i = (N)^(0.5)*@rowextract(@transpose(va),!i)
matrix(k_max,N) loadings
rowplace(loadings,loadings!i,!i-il+1)
next
matrix factors = ((X*@transpose(loadings))/N)
'For a normalization, multiply/divide by the element [kmax,1] of the loadings matrix.
scalar norm = loadings(k_max,1)
factors = factors*norm
loadings = loadings/norm
'Re-extract normalized factors from this matrix as series for future subroutines, in reverse order:
'i.e. the factor associated with the largest eigenvalue is F_1, second is F_2,...,F_km_max
mtos(factors,factorgroup1)
for !i = 1 to k_max
scalar temp = k_max-!i+1
rename ser0{temp} f_1_!i_!j
next
delete temp
delete factorgroup1
for !i = il to iu
delete loadings!i
next


And again, for the case where we want to extract I(1) factors:

Code: Select all



'Factors extracted as per Bai (2004): with a subscript i_ denoting integrated variables.

stom(g1_standnonstat1,i_X)
scalar i_T = @rows(i_X)
scalar i_N = @columns(i_X)
scalar i_k_max = 4
sym i_XX = @transpose(i_X)*i_X
           scalar i_il = N-i_k_max+1
           scalar i_iu = N
matrix i_va = @eigenvectors(i_XX)
vector i_ve = @eigenvalues(i_XX)
for !i = i_il to i_iu
vector i_loadings!i = (i_N)^(0.5)*@rowextract(@transpose(i_va),!i)
matrix(i_k_max,i_N) i_loadings
rowplace(i_loadings,i_loadings!i,!i-il+1)
next
matrix i_factors = ((i_X*@transpose(i_loadings))/N)
'Normalize
scalar i_norm = i_loadings(i_k_max,1)
i_factors = i_factors*i_norm
i_loadings = i_loadings/i_norm
'Re-extract normalized factors from this matrix, in reverse order: i.e. the factor associated with the largest eigenvalue is if1, second is i_F_2,...,i_F_km_max
mtos(i_factors,i_factorsgroup1)
for !i = 1 to i_k_max
scalar temp = i_k_max-!i+1
rename ser0{temp} if1_!i_!j
next
delete temp
delete i_factorsgroup1
for !i = i_il to i_iu
delete i_loadings!i
next



Warmest regards,

Charlie

hi
im working favar model with Eviews and I thank you for your guidance.
I have a few questions
1- what do you mean from I(0),I(1) factors?You're talking about stability?
2-when I run program[ 'Stock and Watson (2002) - I(0) factors where T>N] "for statement unterminated in "for !J=1 to 20"
please guide me


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