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st: RE: RE: Optimal AR-structure based on AIC - "initial values not feasible" problem


From   "Hannes Stieperaere" <hannes.stieperaere@ugent.be>
To   <statalist@hsphsun2.harvard.edu>
Subject   st: RE: RE: Optimal AR-structure based on AIC - "initial values not feasible" problem
Date   Tue, 1 Oct 2013 11:14:12 +0200

Thanks for your response.

Your code works, but there is 1 problem.
In your loop, the only models that are estimated, are the ones that
incorporate all the previous lags.
If an AR(5) is estimated, the model takes automatically lag 1 to 4 into
account as well.
My purpose is to find a way to estimate all models, even the ones that not
contain smaller lags then the highest in the model.

For example, a model with only lag 4 must be possible (without incorporating
lag 1, 2 and 3), or a model with lag 5 en 2, without lag 1, 3 and 4.
So if I allow 5 lags, 32 (2^5) models should be estimated to account for all
possible combinations. 

Thanks in advance for any help.

Kind regards,
Hannes Stieperaere
PhD Student
Hannes.Stieperaere@ugent.be




-----Oorspronkelijk bericht-----
Van: owner-statalist@hsphsun2.harvard.edu
[mailto:owner-statalist@hsphsun2.harvard.edu] Namens hasnain.yousaf
Verzonden: dinsdag 1 oktober 2013 7:25
Aan: statalist@hsphsun2.harvard.edu
Onderwerp: st: RE: Optimal AR-structure based on AIC - "initial values not
feasible" problem

I am not aware how IC results are stored in matrix (trying scalar didn't
work for me). You can write a smaller loop for getting the results table for
AR structures here is how i did it
1- for strictly AR models upto ARIMA(5,1,0) for one variable forv i=1(1)5{
quietly arima variable,arima(`i',1,0) estat ic }
2- for strictly AR models upto ARIMA(5,1,0) for multiple  variables foreach
v of varlist variable1 variable 2{ forv i=1(1)5 { quietly arima
`v',arima(`i',1,0) estat ic } }
3- for  ARMA (Including AR models in case 1) models upto ARIMA(5,1,5) for
multiple  variables foreach v of varlist variable1 variable 2{ forv i=1(1)5
{ for k=0(1)5{ quietly arima `v',arima(`i',1,`k') estat ic } } } See if it
works

________________________________________
From: owner-statalist@hsphsun2.harvard.edu
<owner-statalist@hsphsun2.harvard.edu> on behalf of Hannes Stieperaere
<hannes.stieperaere@ugent.be>
Sent: Monday, September 30, 2013 7:59 PM
To: statalist@hsphsun2.harvard.edu
Subject: st: Optimal AR-structure based on AIC - "initial values not
feasible" problem

Hello,

I'm having a problem using loops to get the best AR-structure based on AIC
(Akaike Information Criterion).
I use Stata 12.1.

I want to use Stata to compare the AIC values of AR models. It is not
necessary that all previous lags are included. C.f. an AR(1 3 4) without the
second lag must be possible.
The code below works fine until I adjust the code to include 5 lags or more.
In the end I want to include 12 possible lags.
When there are 5 possible lags or more, Stata returns the error: "initial
values not feasible r(1400)" after the phase I estimate the models by using
loops.
I know that the possible models Stata has to estimate rise exponentially,
but estimating 32 models doesn't seem problematic to me.

If we look at the code, the problem is situated in the first part between
the ---- brackets. Here, all the possible models are estimated and the AIC
are stored.
I use a loop that makes it possible per lag to be "number of lag" or "". The
latter is done to have the possibility that stata does not add a lag for a
specific lag (like in the AR(1 3 4) model, the second value would be "" or
no lag).

I am relatively new to Stata so maybe this is not the correct and fasted way
to program it.
Thanks in advance for any help on this code, or on sharing a faster way to
select an AR model based on AIC.

Stata Code:

The code that generates the error:
----------
local ok_1_1 = ""
local ok_1_2 = 1
local ok_2_1 = ""
local ok_2_2 = 2
local ok_3_1 = ""
local ok_3_2 = 3
local ok_4_1 = ""
local ok_4_2 = 4
local ok_5_1 = ""
local ok_5_2 = 5

forvalues lag5 = 1/2 {
forvalues lag4 = 1/2 {
forvalues lag3 = 1/2 {
forvalues lag2 = 1/2 {
forvalues lag1 = 1/2 {
qui arima l_regr in 17/92, ar(`ok_1_`lag1'' `ok_2_`lag2'' `ok_3_`lag3''
`ok_4_`lag4'' `ok_5_`lag5'')
qui estat ic
matrix xs`lag1'`lag2'`lag3'`lag4'`lag5' = r(S) svmat
xs`lag1'`lag2'`lag3'`lag4'`lag5',
name(xs`lag1'`lag2'`lag3'`lag4'`lag5')
}
}
}
}
}
--------


Rest of the code that must return the model with the lowest AIC-value:
----------
qui mkmat xs*5 in 1, matrix(AIC) obs nchar(1) qui matrix list AIC qui egen
min_aic = rowmin(xs*5) qui local names : colnames AIC qui generate model =
"`names'"
qui split(model)
qui forvalues i = 1/32 {
qui split model`i', parse(s) generate(modelaic`i') } qui destring modelaic*2
, replace float qui mkmat modelaic*2 in 1, matrix(models) obs nchar(1) qui
svmat AIC, name(AIC) qui matrix lag_aic = [models \ AIC] qui forvalues i =
1/32 { qui gen opt_lag`i' = lag_aic[1 , (`i')]  if lag_aic[2 , (`i')] ==
min_aic } qui egen opt_lag_aic = rowmax(opt_lag*) qui display opt_lag_aic
qui tostring opt_lag_aic, replace qui forvalues j = 1/5 { qui gen ar`j' =
substr(opt_lag_aic,`j',1) } qui destring ar* , replace float qui forvalues j
= 1/5 { qui if ar`j' == 1 gen ar_`j' = ""
qui if ar`j' == 2 gen ar_`j' = `j'
}
qui forvalues j = 1/5 {
qui local ar_`j' = ar_`j'
}
arima l_regr in 17/92, ar(`ar_1' `ar_2' `ar_3' `ar_4' `ar_5')
------


Kind regards,
Hannes Stieperaere
PhD Student Ghent University, Belgium


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