Identifying VARs with sign restrictions
-
fabio fornari
- Posts: 8
- Joined: Fri Mar 20, 2009 7:11 am
Re: Identifying VARs with sign restrictions
will see how it works with three shocks before moving on ... needs really lots of draws to get acceptances ...
will let you know
thanks f
Re: Identifying VARs with sign restrictions
Did anyone try to extend Uhilg_3 file for the paper published (2005) with penalty function to more than 6 variables? can anyone tell me how the penalty is going to change in this case?
Thanks for help
Re: Identifying VARs with sign restrictions
This uses an alternative and more flexible parameterization of the unit sphere than was used by Uhlig. (Note that it's really the parameterization that needs to change; the calculation of the penalty function is similar regardless). I didn't use it in the original uhlig3.prg because the %stereo function was added with version 6.35 of RATS. There are quite a few ways to parameterize the unit sphere. The stereoscopic projection used in the %stereo function takes the N-1 vector x and maps it to the N-vector formed bymmk wrote:Hi
Did anyone try to extend Uhilg_3 file for the paper published (2005) with penalty function to more than 6 variables? can anyone tell me how the penalty is going to change in this case?
Thanks for help
y(i)=2x(i)/(||x||^2+1)
y(N)=(||x||^2-1)/(||x||^2+1)
All such mappings have at least one minor flaw. In this case, the only way to make y(N)=0 is to make the x's very large. (Uhlig's mapping has the problem of non-uniqueness, as well as being rather difficult to generalize easily).
Code: Select all
*
* Replication File for Uhlig(2005), "What are the effects of monetary policy on
* output? Results from an agnostic identification procedure," Journal of Monetary
* Economics, 52, pp 381-419. Penalty function approach.
*
open data uhligdata.xls
calendar(m) 1965
data(format=xls,org=columns) 1965:1 2003:12 gdpc1 gdpdef cprindex totresns bognonbr fedfunds
*
set gdpc1 = log(gdpc1)*100.0
set gdpdef = log(gdpdef)*100.0
set cprindex = log(cprindex)*100.0
set totresns = log(totresns)*100.0
set bognonbr = log(bognonbr)*100.0
*
system(model=varmodel)
variables gdpc1 gdpdef cprindex fedfunds bognonbr totresns
lags 1 to 12
end(system)
estimate(noprint)
*
dec vect[strings] vl(6)
compute vl=||"real GDP","GDP price defl","Comm. Price Ind.",$
"Fed Funds Rate","Nonborr. Reserv.","Total Reserves"||
*
* Get the scale factors to be used for impulse responses
*
dec vect scales(6)
dec real func
compute fill=0
dofor i = gdpc1 gdpdef cprindex fedfunds bognonbr totresns
diff i / diff1
stats(noprint) diff1
compute fill=fill+1,scales(fill)=sqrt(%variance)
end dofor i
*
* ndraws is the desired number of draws from the posterior of the VAR
* nvar is the number of variables
* nstep is the number of IRF steps to compute
* KMAX is the "K" value for the number of steps constrained
*
*
compute ndraws=1000
compute nvar=6
compute nstep=60
compute KMAX=5
declare vect[rect] goodresp(ndraws)
declare vector ik a(nvar)
*
dec vect g(nvar-1)
compute g=%fill(nvar-1,1,1.0)
nonlin g
*
* This is the standard setup for MC integration of an OLS VAR
*
compute sxx =%decomp(%xx)
compute svt =%decomp(inv(%nobs*%sigma))
compute betaols=%modelgetcoeffs(varmodel)
compute ncoef =%rows(sxx)
compute wishdof=%nobs-ncoef
dec rect ranc(ncoef,nvar)
*
infobox(action=define,progress,lower=1,upper=ndraws) "Monte Carlo Integration"
*
* Plan for double the number of draws to allow for the ones which we
* reject.
*
compute accept=0
do draws=1,ndraws*2
*
* Make a draw from the posterior for the VAR and compute its impulse
* responses.
*
compute sigmad =%ranwisharti(svt,wishdof)
compute swish =%decomp(sigmad)
compute ranc =%ran(1.0)
compute betau =sxx*ranc*tr(swish)
compute betadraw=betaols+betau
compute %modelsetcoeffs(varmodel,betadraw)
impulse(noprint,model=varmodel,decomp=swish,results=impulses,steps=nstep)
*
* Minimize the penalty function, starting from the last set of minimizers
*
find(noprint) min func
compute a=%stereo(g)
compute func=0.0
do k=1,KMAX+1
compute ik=(%xt(impulses,k)*a)./scales
compute func=func+%if(ik(4)>0,-ik(4),-100*ik(4))+$
%if(ik(3)<0,ik(3),100*ik(3))+$
%if(ik(2)<0,ik(2),100*ik(2))+$
%if(ik(5)<0,ik(5),100*ik(5))
end do k
end find
compute testfunc=func,testbeta=%beta
*
* Try the minimization again, starting from a standard set of values
*
compute g=%fill(nvar-1,1,1.0)
find(noprint) min func
compute a=%stereo(g)
compute func=0.0
do k=1,KMAX+1
compute ik=(%xt(impulses,k)*a)./scales
compute func=func+%if(ik(4)>0,-ik(4),-100*ik(4))+$
%if(ik(3)<0,ik(3),100*ik(3))+$
%if(ik(2)<0,ik(2),100*ik(2))+$
%if(ik(5)<0,ik(5),100*ik(5))
end do k
end find
*
* If the two estimates don't match, reject the draw. If they do, copy out the
* impulse responses.
*
if abs(testfunc-func)<.01 {
compute accept=accept+1
dim goodresp(accept)(nstep,nvar)
ewise goodresp(accept)(i,j)=ik=%xt(impulses,i)*a,ik(j)
}
*
* If we've hit out desired number of accepted draws, break
*
if accept>=ndraws
break
infobox(current=accept)
end do draws
infobox(action=remove)
*
* Post-processing. Graph the mean of the responses along with the 16% and 84%-iles
*
clear upper lower resp
*
spgraph(vfields=3,hfields=2,hlabel="Figure 14. Impulse Responses with Penalty Function Approach")
do i=1,nvar
compute minlower=maxupper=0.0
smpl 1 accept
do k=1,nstep
set work = goodresp(t)(k,i)
compute frac=%fractiles(work,||.16,.84||)
compute lower(k)=frac(1)
compute upper(k)=frac(2)
compute resp(k)=%avg(work)
end do k
*
smpl 1 nstep
graph(ticks,number=0,picture="##.##",header="Impulse Responses for "+vl(i)) 3
# resp
# upper / 2
# lower / 2
end do i
*
spgraph(done)
Re: Identifying VARs with sign restrictions
Would please you kindly instruct me how to collect the value of the initial demand/monetary shock and how to show the variance of decomposition?
Re: Identifying VARs with sign restrictions
I tried to put the variance decomposition in the previous code with 2 shocks you posted few weeks ago.
and I got message as following,
## SX22. Expected Type REAL, Got MATRIX(REAL) Instead
>>>>%xt(impulses,t).^2<<<<
Would you kindly instruct me how to fix it?
Thank you very much!
Re: Identifying VARs with sign restrictions
Code: Select all
DEC SERIES[RECT] irfsquaredRe: Identifying VARs with sign restrictions
I tried to follow up your instruction in the latest RATS letter which demonstrateds the impact responses on variables 1 and 2 to be equal. But I failed. Perhaps I misunderstood your instruction.
If you could give me some guidances, I will be very appreciated.
Code: Select all
*
* Replication File for Uhlig (2005), "What are the effects of monetary policy on output?
* Results from an agnostic identification procedure." Journal of Monetary Economics, 52, pp
* 381-419. Pure sign restriction approach
*
open data levelsa1.xls
calendar 1970 1 4
data(format=xls,org=columns) 1970:01 2008:03 lrgspen1 lrntax lrgdp pgdpch rint lrpcon lrpinv
*
system(model=varmodel)
variables lrgspen1 lrntax lrgdp pgdpch rint lrpcon
lags 1 to 4
end(system)
estimate(noprint)
*
dec vect[strings] vl(6)
compute vl=||'Govt spending','net tax','real gdp','GDP deflator','real interest rate','Private onsumption'||
*
* n1 is the number of draws from the posterior of the VAR
* n2 is the number of draws from the unit sphere for each draw for the VAR
* nvar is the number of variables
* nstep is the number of IRF steps to compute
* KMAX is the "K" value for the number of steps constrained
*
compute n1=200
compute n2=200
compute nvar=6
compute nstep=60
compute KMAX=5
*
* This is the standard setup for MC integration of an OLS VAR
*
compute sxx =%decomp(%xx)
compute svt =%decomp(inv(%nobs*%sigma))
compute betaols=%modelgetcoeffs(varmodel)
compute ncoef =%rows(sxx)
compute wishdof=%nobs-ncoef
dec rect ranc(ncoef,nvar)
*
* Most draws are going to get rejected. We allow for up to 1000
* good ones. The variable accept will count the number of accepted
* draws. GOODRESP will be a RECT(nsteps,nvar) at each accepted
* draw.
*
declare vect[rect] goodresp(1000)
declare vector ik a(nvar)
*
compute accept=0
infobox(action=define,progress,lower=1,upper=n1) 'Monte Carlo Integration'
do draws=1,n1
*
* Make a draw from the posterior for the VAR and compute its impulse
* responses.
*
compute sigmad =%ranwisharti(svt,wishdof)
compute swish =%decomp(sigmad)
compute ranc =%ran(1.0)
compute betau =sxx*ranc*tr(swish)
compute betadraw=betaols+betau
compute %modelsetcoeffs(varmodel,betadraw)
impulse(noprint,model=varmodel,decomp=swish,results=impulses,steps=nstep)
*
* Do the subdraws over the unit sphere. These give the weights on the
* orthogonal components.
*
do subdraws=1,n2
compute rperp=$%perp(||1.0,-1.0,0.0,0.0,0.0,0.0||*swish)
compute a=rperp*%ranmat(5,1),a=a/sqrt(%normsqr(a))
*
* Check that the responses have the correct signs for steps 1 to KMAX+1
* (+1 because in the paper, the steps used 0-based subscripts, rather than
* the 1 based subscripts used by RATS).
*
do k=1,KMAX+1
compute ik=%xt(impulses,k)*a
if ik(1)<0
branch 105
end do k
*
* This is an accepted draw. Copy the information out. If we have our 1000
* good ones, drop out of the overall loop.
*
compute accept=accept+1
dim goodresp(accept)(nstep,nvar)
ewise goodresp(accept)(i,j)=ik=%xt(impulses,i)*a,ik(j)
if accept>=1000
break
:105
end do subdraws
if accept>=1000
break
infobox(current=draws)
end do draws
infobox(action=remove)
*
* Post-processing. Graph the mean of the responses along with the 16% and 84%-iles
*
clear upper lower resp
*
spgraph(vfields=3,hfields=2,hlabel='Figure 6. Impulse Responses with Pure-Sign Approach',subhea='monetary Shock')
do i=1,nvar
compute minlower=maxupper=0.0
smpl 1 accept
do k=1,nstep
set work = goodresp(t)(k,i)
compute frac=%fractiles(work,||.16,.84||)
compute lower(k)=frac(1)
compute upper(k)=frac(2)
compute resp(k)=%avg(work)
end do k
*
smpl 1 nstep
graph(ticks,number=0,picture='##.##',header='Impulse Responses for '+vl(i)) 3
# resp
# upper / 2
# lower / 2
end do i
*
spgraph(done)
Re: Identifying VARs with sign restrictions
The line I isolated has been garbled. (Probably didn't fit in one column in the newsletter). Get rid of the $ and it should be fine.lhlee0506 wrote:Hello,
I tried to follow up your instruction in the latest RATS letter which demonstrateds the impact responses on variables 1 and 2 to be equal. But I failed. Perhaps I misunderstood your instruction.
If you could give me some guidances, I will be very appreciated.
Code: Select all
compute rperp=$%perp(||1.0,-1.0,0.0,0.0,0.0,0.0||*swish)
Re: Identifying VARs with sign restrictions
Thank you for your instruction.
After removing $, the code seems to wotk.
But I am still confused about how to set up the sign restriction after putting variable1,2 to be equal.
This modification seems only make variable 1,2 to be equal for one period, how to make it last for four periods?
Hope to have your kind responses.
Thanks a lot!
Code: Select all
*
* Replication File for Uhlig (2005), "What are the effects of monetary policy on output?
* Results from an agnostic identification procedure." Journal of Monetary Economics, 52, pp
* 381-419. Pure sign restriction approach
*
open data levelsa1.xls
calendar 1970 1 4
data(format=xls,org=columns) 1970:01 2008:03 lrgspen1 lrntax lrgdp pgdpch rint lrpcon lrpinv
*
system(model=varmodel)
variables lrgspen1 lrntax lrgdp pgdpch rint lrpcon
lags 1 to 4
end(system)
estimate(noprint)
*
dec vect[strings] vl(6)
compute vl=||'Govt spending','net tax','real gdp','GDP deflator','real interest rate','Private onsumption'||
*
* n1 is the number of draws from the posterior of the VAR
* n2 is the number of draws from the unit sphere for each draw for the VAR
* nvar is the number of variables
* nstep is the number of IRF steps to compute
* KMAX is the "K" value for the number of steps constrained
*
compute n1=200
compute n2=200
compute nvar=6
compute nstep=60
compute KMAX=5
*
* This is the standard setup for MC integration of an OLS VAR
*
compute sxx =%decomp(%xx)
compute svt =%decomp(inv(%nobs*%sigma))
compute betaols=%modelgetcoeffs(varmodel)
compute ncoef =%rows(sxx)
compute wishdof=%nobs-ncoef
dec rect ranc(ncoef,nvar)
*
* Most draws are going to get rejected. We allow for up to 1000
* good ones. The variable accept will count the number of accepted
* draws. GOODRESP will be a RECT(nsteps,nvar) at each accepted
* draw.
*
declare vect[rect] goodresp(1000)
declare vector ik a(nvar)
*
compute accept=0
infobox(action=define,progress,lower=1,upper=n1) 'Monte Carlo Integration'
do draws=1,n1
*
* Make a draw from the posterior for the VAR and compute its impulse
* responses.
*
compute sigmad =%ranwisharti(svt,wishdof)
compute swish =%decomp(sigmad)
compute ranc =%ran(1.0)
compute betau =sxx*ranc*tr(swish)
compute betadraw=betaols+betau
compute %modelsetcoeffs(varmodel,betadraw)
impulse(noprint,model=varmodel,decomp=swish,results=impulses,steps=nstep)
*
* Do the subdraws over the unit sphere. These give the weights on the
* orthogonal components.
*
do subdraws=1,n2
compute rperp=%perp(||1.0,-1.0,0.0,0.0,0.0,0.0||*swish)
compute a=rperp*%ranmat(5,1),a=a/sqrt(%normsqr(a))
*
* Check that the responses have the correct signs for steps 1 to KMAX+1
* (+1 because in the paper, the steps used 0-based subscripts, rather than
* the 1 based subscripts used by RATS).
*
do k=1,KMAX+1
compute ik=%xt(impulses,k)*a
if ik(1)<0
branch 105
end do k
*
* This is an accepted draw. Copy the information out. If we have our 1000
* good ones, drop out of the overall loop.
*
compute accept=accept+1
dim goodresp(accept)(nstep,nvar)
ewise goodresp(accept)(i,j)=ik=%xt(impulses,i)*a,ik(j)
if accept>=1000
break
:105
end do subdraws
if accept>=1000
break
infobox(current=draws)
end do draws
infobox(action=remove)
*
* Post-processing. Graph the mean of the responses along with the 16% and 84%-iles
*
clear upper lower resp
*
spgraph(vfields=3,hfields=2,hlabel='Figure 6. Impulse Responses with Pure-Sign Approach',subhea='monetary Shock')
do i=1,nvar
compute minlower=maxupper=0.0
smpl 1 accept
do k=1,nstep
set work = goodresp(t)(k,i)
compute frac=%fractiles(work,||.16,.84||)
compute lower(k)=frac(1)
compute upper(k)=frac(2)
compute resp(k)=%avg(work)
end do k
*
smpl 1 nstep
graph(ticks,number=0,picture='##.##',header='Impulse Responses for '+vl(i)) 3
# resp
# upper / 2
# lower / 2
end do i
*
spgraph(done)
Re: Identifying VARs with sign restrictions
I use code for two shocks from the forum. In addition to the impulse responses, I would like to see how much of the variation is accounted for by the first shock and by the second shock (variance decomposition).
I am not quite sure whether fraction explained (variance decomposition) is correct. Could you please give me some suggestions and feedbacks whether my code is correct?
Thank you very much for your help and kindness.
Tim
Code: Select all
OPEN DATA uhligdata.xls
CALENDAR 1965 1 12
compute missc=1.0e+32
data(format=xlS,org=columns) 1965:1 2003:12 gdpc1 gdpdef cprindex totresns bognonbr fedfunds
set gdpc1 = log(gdpc1)*100.0
set gdpdef = log(gdpdef)*100.0
set cprindex = log(cprindex)*100.0
set totresns = log(totresns)*100.0
set bognonbr = log(bognonbr)*100.0
*
system(model=varmodel)
variables gdpc1 gdpdef cprindex fedfunds bognonbr totresns
lags 1 to 12
end(system)
estimate(noprint,resid=resids)
dec vect[strings] vl(6)
compute vl=||'real GDP','GDP price defl','Comm. Price Ind.',$
'Fed Funds Rate','Nonborr. Reserv.','Total Reserves'||
compute n1=4000
compute n2=4000
compute nvar=6
compute nstep=60
compute KMAX=5
* This is the standard setup for MC integration of an OLS VAR
*
dec symm s(6,6)
dec vect v1(6) ;* For the unit vector on the 1st draw
dec vect v2(5) ;* For the unit vector on the 2nd draw
dec vect v(6) ;* Working impulse vector
compute sxx =%decomp(%xx)
compute svt =%decomp(inv(%nobs*%sigma))
compute betaols=%modelgetcoeffs(varmodel)
compute ncoef =%rows(sxx)
compute wishdof=%nobs-ncoef
dec rect ranc(ncoef,nvar)
*
* Most draws are going to get rejected. We allow for up to 1000
* good ones. The variable accept will count the number of accepted
* draws. GOODRESP will be a RECT(nsteps,nvar) at each accepted
* draw.
*
declare vect[rect] goodresp(1000) goodfevd(1000)
declare vect[rect] goodrespa(1000) goodfevda(1000)
declare vector ik a(nvar) ones(nvar)
declare series[rect] irfsquared
compute ones=%const(1.0)
source forcedfactor.src
*
compute accept=0
infobox(action=define,progress,lower=1,upper=n1) 'Monte Carlo Integration'
do draws=1,n1
*
* Make a draw from the posterior for the VAR and compute its impulse
* responses.
*
compute sigmad =%ranwisharti(svt,wishdof)
compute p =%decomp(sigmad)
compute ranc =%ran(1.0)
compute betau =sxx*ranc*tr(p)
compute betadraw=betaols+betau
compute %modelsetcoeffs(varmodel,betadraw)
*
* This is changed to unit shocks rather than orthogonalized shocks.
*
impulse(noprint,model=varmodel,decomp=%identity(6),results=impulses,steps=nstep)
gset irfsquared 1 1 = %xt(impulses,t).^2
gset irfsquared 2 nstep = irfsquared{1}+%xt(impulses,t).^2
*
* Do the subdraws over the unit sphere. These give the weights on the
* orthogonal components.
*
do subdraws=1,n2
************************************************
* First Set of Restrictions - Unique Impulse Vector
************************************************
compute v1=%ran(1.0),v1=v1/sqrt(%normsqr(v1))
compute v=i1=p*v1
do k=1,KMAX+1
compute ik=%xt(impulses,k)*v
if ik(1)>0.or.ik(2)>0.or.ik(3)<0.or.ik(6)>0
branch 105
end do k
*
* Meets the first restriction
* Draw from the orthogonal complement of i1 (last five columns of the
* factor "f").
*
@forcedfactor(force=column) sigmad i1 f
compute v2=%ran(1.0),v2=v2/sqrt(%normsqr(v2))
compute v=i2=%xsubmat(f,1,6,2,6)*v2
*****************************************************
* Second Set of Restrictions - Demand Shock
*****************************************************
do k=1,KMAX+1
compute ik=%xt(impulses,k)*v
if ik(1)<0.or.ik(2)<0.or.ik(3)<0.or.ik(6)<0
branch 105
end do k
*
* Meets both restrictions
*
compute accept=accept+1
dim goodrespa(accept)(nstep,nvar) goodfevda(accept)(nstep,nvar)
dim goodresp(accept)(nstep,nvar) goodfevd(accept)(nstep,nvar)
ewise goodresp(accept)(i,j)=(ik=%xt(impulses,i)*i1),ik(j)
ewise goodfevd(accept)(i,j)=(ik=(irfsquared(i)*(i1.^2))./(irfsquared(i)*ones)),ik(j)
ewise goodrespa(accept)(i,j)=(ik=%xt(impulses,i)*i2),ik(j)
ewise goodfevda(accept)(i,j)=(ik=(irfsquared(i)*(i2.^2))./(irfsquared(i)*ones)),ik(j)
if accept>=1000
break
:105
end do subdraws
if accept>=1000
break
infobox(current=draws)
end do draws
infobox(action=remove)
*
* Post-processing. Graph the mean of the responses along with the 16% and 84%-iles
*
clear upper lower resp
*
spgraph(vfields=3,hfields=2,hlabel='Impulse Responses with Pure-Sign Approach',subhea='Subhea')
do i=1,nvar
compute minlower=maxupper=0.0
smpl 1 accept
do k=1,nstep
set work = goodresp(t)(k,i)
compute frac=%fractiles(work,||.16,.84||)
compute lower(k)=frac(1)
compute upper(k)=frac(2)
compute resp(k)=%avg(work)
display resp(k)
end do k
*
smpl 1 nstep
graph(pattern,ticks,number=0,picture='##.##',header='Impulse Responses for '+vl(i)) 3
# resp
# upper / 2
# lower / 2
end do i
*
spgraph(done)
smpl
*
clear upper lower resp
*
spgraph(vfields=3,hfields=2,hlabel='SECOND Impulse Responses with Pure-Sign Approach',subhea='Subhea')
do i=1,nvar
compute minlower=maxupper=0.0
smpl 1 accept
do k=1,nstep
set work = goodrespa(t)(k,i)
compute frac=%fractiles(work,||.16,.84||)
compute lower(k)=frac(1)
compute upper(k)=frac(2)
compute resp(k)=%avg(work)
display resp(k)
end do k
*
smpl 1 nstep
graph(pattern,ticks,number=0,picture='##.##',header='Impulse Responses for '+vl(i)) 3
# resp
# upper / 2
# lower / 2
end do i
*
spgraph(done)
smpl
*
clear upper lower resp
*
spgraph(vfields=3,hfields=2,hlabel='Fraction of Variance Explained with Pure-Sign Approach',subhea='Subhea')
do i=1,nvar
compute minlower=maxupper=0.0
smpl 1 accept
do k=1,nstep
set work = goodfevd(t)(k,i)
compute frac=%fractiles(work,||.16,.50,.84||)
compute lower(k)=frac(1)
compute upper(k)=frac(3)
compute resp(k)=frac(2)
display resp(k)
end do k
*
smpl 1 nstep
graph(pattern,ticks,number=0,min=0.0,picture="##.##",header="Fraction Explained for "+vl(i)) 3
# resp
# upper / 2
# lower / 2
end do i
*
spgraph(done)
smpl
*
clear upper lower resp
*
spgraph(vfields=3,hfields=2,hlabel='SECOND Fraction of Variance Explained with Pure-Sign Approach',subhea='Subhea')
do i=1,nvar
compute minlower=maxupper=0.0
smpl 1 accept
do k=1,nstep
set work = goodfevda(t)(k,i)
compute frac=%fractiles(work,||.16,.50,.84||)
compute lower(k)=frac(1)
compute upper(k)=frac(3)
compute resp(k)=frac(2)
display resp(k)
end do k
*
smpl 1 nstep
graph(pattern,ticks,number=0,min=0.0,picture="##.##",header="Fraction Explained for "+vl(i)) 3
# resp
# upper / 2
# lower / 2
end do i
*
spgraph(done)
smplRe: Identifying VARs with sign restrictions
This (the alpha weights on the "swish" matrix) will create responses for the first two variables which are equal to each other for the first four periods. Note that this severely restricts the possible shocks, so it may be difficult for the draws for shocks to satisfy any other constraints.lhlee0506 wrote:Hi~
Thank you for your instruction.
After removing $, the code seems to wotk.
But I am still confused about how to set up the sign restriction after putting variable1,2 to be equal.
This modification seems only make variable 1,2 to be equal for one period, how to make it last for four periods?
Hope to have your kind responses.
Thanks a lot!
Code: Select all
impulse(noprint,model=varmodel,decomp=swish,results=impulses,steps=nstep)
compute step0=%xt(impulses,1)
compute step1=%xt(impulses,2)
compute step2=%xt(impulses,3)
compute step3=%xt(impulses,4)
dec rect r(4,6)
compute %psubmat(r,1,1,||1.0,-1.0,0.0,0.0,0.0,0.0||*step0)
compute %psubmat(r,2,1,||1.0,-1.0,0.0,0.0,0.0,0.0||*step1)
compute %psubmat(r,3,1,||1.0,-1.0,0.0,0.0,0.0,0.0||*step2)
compute %psubmat(r,4,1,||1.0,-1.0,0.0,0.0,0.0,0.0||*step3)
compute rfperp=%perp(r)
compute alpha=rfperp*%ranmat(%cols(rfperp),1)
compute alpha=alpha/sqrt(%normsqr(alpha))Re: Identifying VARs with sign restrictions
That won't work without changing the way the impulse responses are computed. In the original Uhlig code, the impulse response and squares are computed by:TL wrote:Dear Tom,
I use code for two shocks from the forum. In addition to the impulse responses, I would like to see how much of the variation is accounted for by the first shock and by the second shock (variance decomposition).
I am not quite sure whether fraction explained (variance decomposition) is correct. Could you please give me some suggestions and feedbacks whether my code is correct?
Thank you very much for your help and kindness.
Tim
Code: Select all
impulse(noprint,model=varmodel,decomp=swish,results=impulses,steps=nstep)
gset irfsquared 1 1 = %xt(impulses,t).^2
gset irfsquared 2 nstep = irfsquared{1}+%xt(impulses,t).^2The shocks are computed using a factorization of the covariance matrix (decomp=swish). Because the shocks are orthogonal (across time by assumption, across variables by construction), the sums of squared impulse responses (in IRFSQUARED) provide all the information needed for computing the decomposition of variance. To get the variance produced by impulse vector swish * a (where ||a||=1) you just need the sum over j of irfsquared(i,j)*a(j)^2, which is what (irfsquared(i)*(a.^2)) is computing in:
Code: Select all
ewise goodfevd(accept)(i,j)=ik=(irfsquared(i)*(a.^2))./(irfsquared(i)*ones),ik(j)The code we wrote for doing multiple sign-restricted shocks uses:
Code: Select all
impulse(noprint,model=varmodel,decomp=%identity(nvar),results=impulses,steps=nstep)Code: Select all
gset irfsquared 1 1 = %xt(impulses,t).^2
gset irfsquared 2 nstep = irfsquared{1}+%xt(impulses,t).^2should have a big red X through it - the numbers produced are of almost no value; there are a whole bunch of non-zero covariances which this doesn't compute.
The solution to this is to return to the original base set of impulse responses, then back out the weights based upon those. That's really easy for the first one, since it just restores the original code:
Code: Select all
compute v1=%ransphere(6)
compute i1=p*v1
do k=1,KMAX+1
compute ik=%xt(impulses,k)*v1
if ik(1)>0.or.ik(2)>0.or.ik(3)<0.or.ik(6)>0
branch 105
end do kCode: Select all
@forcedfactor(force=column) sigmad i1 f
compute v2=%ransphere(5)
compute i2=%xsubmat(f,1,6,2,6)*v2
compute v2=inv(p)*i2
*****************************************************
* Second Set of Restrictions - Demand Shock
*****************************************************
do k=1,KMAX+1
compute ik=%xt(impulses,k)*v2
if ik(1)<0.or.ik(2)<0.or.ik(3)<0.or.ik(6)<0
branch 105
end do kv1 and v2 are the two weight vectors on the "SWISH" factored draws. The FEVD's are computed with:
Code: Select all
ewise goodfevd(accept)(i,j)=(ik=(irfsquared(i)*(v1.^2))./(irfsquared(i)*ones)),ik(j)
ewise goodfevda(accept)(i,j)=(ik=(irfsquared(i)*(v2.^2))./(irfsquared(i)*ones)),ik(j)Code: Select all
OPEN DATA uhligdata.xls
CALENDAR 1965 1 12
compute missc=1.0e+32
data(format=xlS,org=columns) 1965:1 2003:12 gdpc1 gdpdef cprindex totresns bognonbr fedfunds
set gdpc1 = log(gdpc1)*100.0
set gdpdef = log(gdpdef)*100.0
set cprindex = log(cprindex)*100.0
set totresns = log(totresns)*100.0
set bognonbr = log(bognonbr)*100.0
*
system(model=varmodel)
variables gdpc1 gdpdef cprindex fedfunds bognonbr totresns
lags 1 to 12
end(system)
estimate(noprint,resid=resids)
dec vect[strings] vl(6)
compute vl=||'real GDP','GDP price defl','Comm. Price Ind.',$
'Fed Funds Rate','Nonborr. Reserv.','Total Reserves'||
compute n1=1000
compute n2=1000
compute nvar=6
compute nstep=60
compute KMAX=5
* This is the standard setup for MC integration of an OLS VAR
*
dec symm s(6,6)
dec vect v1(6) ;* For the unit vector on the 1st draw
dec vect v2(5) ;* For the unit vector on the 2nd draw
dec vect v(6) ;* Working impulse vector
compute sxx =%decomp(%xx)
compute svt =%decomp(inv(%nobs*%sigma))
compute betaols=%modelgetcoeffs(varmodel)
compute ncoef =%rows(sxx)
compute wishdof=%nobs-ncoef
dec rect ranc(ncoef,nvar)
*
* Most draws are going to get rejected. We allow for up to 1000
* good ones. The variable accept will count the number of accepted
* draws. GOODRESP will be a RECT(nsteps,nvar) at each accepted
* draw.
*
declare vect[rect] goodresp(1000) goodfevd(1000)
declare vect[rect] goodrespa(1000) goodfevda(1000)
declare vector ik a(nvar) ones(nvar)
declare series[rect] irfsquared
compute ones=%const(1.0)
source forcedfactor.src
*
compute accept=0
infobox(action=define,progress,lower=1,upper=n1) 'Monte Carlo Integration'
do draws=1,n1
*
* Make a draw from the posterior for the VAR and compute its impulse
* responses.
*
compute sigmad =%ranwisharti(svt,wishdof)
compute p =%decomp(sigmad)
compute ranc =%ran(1.0)
compute betau =sxx*ranc*tr(p)
compute betadraw=betaols+betau
compute %modelsetcoeffs(varmodel,betadraw)
*
* This is changed to unit shocks rather than orthogonalized shocks.
*
impulse(noprint,model=varmodel,decomp=p,results=impulses,steps=nstep)
gset irfsquared 1 1 = %xt(impulses,t).^2
gset irfsquared 2 nstep = irfsquared{1}+%xt(impulses,t).^2
*
* Do the subdraws over the unit sphere. These give the weights on the
* orthogonal components.
*
do subdraws=1,n2
************************************************
* First Set of Restrictions - Unique Impulse Vector
************************************************
compute v1=%ransphere(6)
compute i1=p*v1
do k=1,KMAX+1
compute ik=%xt(impulses,k)*v1
if ik(1)>0.or.ik(2)>0.or.ik(3)<0.or.ik(6)>0
branch 105
end do k
*
* Meets the first restriction
* Draw from the orthogonal complement of i1 (last five columns of the
* factor "f").
*
@forcedfactor(force=column) sigmad i1 f
compute v2=%ransphere(5)
compute i2=%xsubmat(f,1,6,2,6)*v2
compute v2=inv(p)*i2
*****************************************************
* Second Set of Restrictions - Demand Shock
*****************************************************
do k=1,KMAX+1
compute ik=%xt(impulses,k)*v2
if ik(1)<0.or.ik(2)<0.or.ik(3)<0.or.ik(6)<0
branch 105
end do k
*
* Meets both restrictions
*
compute accept=accept+1
dim goodrespa(accept)(nstep,nvar) goodfevda(accept)(nstep,nvar)
dim goodresp(accept)(nstep,nvar) goodfevd(accept)(nstep,nvar)
ewise goodresp(accept)(i,j)=(ik=%xt(impulses,i)*v1),ik(j)
ewise goodfevd(accept)(i,j)=(ik=(irfsquared(i)*(v1.^2))./(irfsquared(i)*ones)),ik(j)
ewise goodrespa(accept)(i,j)=(ik=%xt(impulses,i)*v2),ik(j)
ewise goodfevda(accept)(i,j)=(ik=(irfsquared(i)*(v2.^2))./(irfsquared(i)*ones)),ik(j)
if accept>=1000
break
:105
end do subdraws
if accept>=1000
break
infobox(current=draws)
end do draws
infobox(action=remove)
*
* Post-processing. Graph the mean of the responses along with the 16% and 84%-iles
*
clear upper lower resp
*
spgraph(vfields=3,hfields=2,hlabel='Impulse Responses with Pure-Sign Approach',subhea='Subhea')
do i=1,nvar
compute minlower=maxupper=0.0
smpl 1 accept
do k=1,nstep
set work = goodresp(t)(k,i)
compute frac=%fractiles(work,||.16,.84||)
compute lower(k)=frac(1)
compute upper(k)=frac(2)
compute resp(k)=%avg(work)
end do k
*
smpl 1 nstep
graph(pattern,ticks,number=0,picture='##.##',header='Impulse Responses for '+vl(i)) 3
# resp
# upper / 2
# lower / 2
end do i
*
spgraph(done)
smpl
*
clear upper lower resp
*
spgraph(vfields=3,hfields=2,hlabel='SECOND Impulse Responses with Pure-Sign Approach',subhea='Subhea')
do i=1,nvar
compute minlower=maxupper=0.0
smpl 1 accept
do k=1,nstep
set work = goodrespa(t)(k,i)
compute frac=%fractiles(work,||.16,.84||)
compute lower(k)=frac(1)
compute upper(k)=frac(2)
compute resp(k)=%avg(work)
end do k
*
smpl 1 nstep
graph(pattern,ticks,number=0,picture='##.##',header='Impulse Responses for '+vl(i)) 3
# resp
# upper / 2
# lower / 2
end do i
*
spgraph(done)
smpl
*
clear upper lower resp
*
spgraph(vfields=3,hfields=2,hlabel='Fraction of Variance Explained with Pure-Sign Approach',subhea='Subhea')
do i=1,nvar
compute minlower=maxupper=0.0
smpl 1 accept
do k=1,nstep
set work = goodfevd(t)(k,i)
compute frac=%fractiles(work,||.16,.50,.84||)
compute lower(k)=frac(1)
compute upper(k)=frac(3)
compute resp(k)=frac(2)
end do k
*
smpl 1 nstep
graph(pattern,ticks,number=0,min=0.0,picture="##.##",header="Fraction Explained for "+vl(i)) 3
# resp
# upper / 2
# lower / 2
end do i
*
spgraph(done)
smpl
*
clear upper lower resp
*
spgraph(vfields=3,hfields=2,hlabel='SECOND Fraction of Variance Explained with Pure-Sign Approach',subhea='Subhea')
do i=1,nvar
compute minlower=maxupper=0.0
smpl 1 accept
do k=1,nstep
set work = goodfevda(t)(k,i)
compute frac=%fractiles(work,||.16,.50,.84||)
compute lower(k)=frac(1)
compute upper(k)=frac(3)
compute resp(k)=frac(2)
end do k
*
smpl 1 nstep
graph(pattern,ticks,number=0,min=0.0,picture="##.##",header="Fraction Explained for "+vl(i)) 3
# resp
# upper / 2
# lower / 2
end do i
*
spgraph(done)
smplRe: Identifying VARs with sign restrictions
One problem which I am always confused about is what is the rule to set up sign restriction.
say, if we would like to disscuss positive government spending shock and it is put as the variable 1,
then we restric ik(1)<0 in the loop.
Is it correct?
Re: Identifying VARs with sign restrictions
Re: Identifying VARs with sign restrictions
Thank you for your kind instruction.
I tried to apply your code into year delayed postivative shock for 4 periods, and it works.
I also tried to extend it to 3shocks case, but I think I could get wrong with something.
If I could have your guidance, it will ber very helpful.
@forcedfactor(force=column) sigmad i1~i2 f
compute step0=%xt(impulses,1)
compute step1=%xt(impulses,2)
compute step2=%xt(impulses,3)
compute step3=%xt(impulses,4)
dec rect v3(4,6)
compute %psubmat(v3,1,1,||0.0,1.0,0.0,0.0,0.0,0.0||*step0)
compute %psubmat(v3,2,1,||0.0,1.0,0.0,0.0,0.0,0.0||*step1)
compute %psubmat(v3,3,1,||0.0,1.0,0.0,0.0,0.0,0.0||*step2)
compute %psubmat(v3,4,1,||0.0,1.0,0.0,0.0,0.0,0.0||*step3)
compute rfperp=%perp(V3)
compute v3=rfperp*%ranmat(%cols(rfperp),1)
compute v3=v3/sqrt(%normsqr(v3))
compute v=i3=f*v3