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RE: st: RE: Hannan-Quinn Criterion for nonstationary series


From   "Sean MC allister" <[email protected]>
To   [email protected]
Subject   RE: st: RE: Hannan-Quinn Criterion for nonstationary series
Date   Thu, 10 Feb 2005 14:12:08 +0000

Scott,
It works well.
Thank you for your kind help.
Sean


From: "Scott Merryman" <[email protected]>
Reply-To: [email protected]
To: <[email protected]>
Subject: st: RE: Hannan-Quinn Criterion for nonstationary series
Date: Tue, 8 Feb 2005 18:03:39 -0600

Sean,

I would think you would be able to use -varsoc- for each cross-section to
find the optimal lag length.

Below, is an example where -varsoc- is used to determine the optimal lag
length for each county and then these values are fed into -levinlin-.

Hope this helps,
Scott

use http://fmwww.bc.edu/ec-p/data/hayashi/sheston91.dta,clear
qui {
keep if country < 13
preserve
levels country, local(levels)
foreach l of local levels {
varsoc rgdp if country == `l'
matrix A = r(stats)
if `l' > 1 {
matrix C = C \ A
}
else {
matrix C = A
}
}
svmat C, name(col)
keep if HQ < .

egen id = fill(1 1 1 1 1 2 2 2 2 2)

egen minh = min(H), by(id)
gen optimal_lag = lag if minh == HQ

sort id opt
forv i = 1(5)60 {
local lag = opt[`i']
local opt_lag "`opt_lag' `" "' `lag'"
}

restore
}
levinlin rgdp, lag(`opt_lag')


> -----Original Message-----
> From: [email protected] [mailto:owner-
> [email protected]] On Behalf Of Sean MC allister
> Sent: Tuesday, February 08, 2005 4:35 AM
> To: [email protected]
> Subject: st: Hannan-Quinn Criterion for nonstationary series
>
> Hi
> I'm performing a convergence test on panel data based on the unit root
> test
> of Levin and Lin. I face the problem of selecting, for each country, the
> lag's length which eliminates residual autocorrelation. I wanted to use
> the
> Hannan-Quinn's criterion but the only command I found to get it is
> "varsoc"
> which is used for VAR. I worry because my sample is very small (T=11 and
> N=15). Is there a command that give this criterion for short nonstationary
> time series?
> I hope you could help me and I thank you in advance for your answer.
> Yours sincerely,
> Sean



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