interpretation of sigma_e in xtregar, re

[Pierre Paul]

Hello everyone,

I have a question about the interpretation of the sigma_e estimate from xtregar, re.
Here is an example of such an outpput, taking from the manual :

----------------------------------------------------------------------------------------------------------------------------------
. xtregar invest mvalue kstock if year !=1934 & year !=1944, re lbi
RE GLS regression with AR(1) disturbances Number of obs = 190
Group variable: company Number of groups = 10
R-sq: within = 0.7707 Obs per group: min = 19
between = 0.8039 avg = 19.0
overall = 0.7958 max = 19
Wald chi2(3) = 351.37
corr(u_i, Xb) = 0 (assumed) Prob > chi2 = 0.0000
invest
Coef. Std. Err. z P>|z| [95% Conf. Interval]
mvalue
.0947714 .0083691 11.32 0.000 .0783683 .1111746
kstock
.3223932 .0263226 12.25 0.000 .2708019 .3739845
_cons
-45.21427 27.12492 -1.67 0.096 -98.37814 7.949603
rho_ar
.6697198 (estimated autocorrelation coefficient)
sigma_u
74.662876
sigma_e
42.253042
rho_fov
.75742494 (fraction of variance due to u_i)
theta
.66973313
modified Bhargava et al. Durbin-Watson = .71380994
Baltagi-Wu LBI = 1.0134522
----------------------------------------------------------------------------------------------------------------------------------

The model can be written as follows :
y_it = x_it * beta + u_i + eta_it
where eta_it follows an AR(1) process : eta_it = rho * eta_i,t-1 + nu_it

Is sigma_e an estimate of the standard deviation of eta_it or of nu_it ?
Since rho_fov = sigma_u ^2/(sigma_u ^2 + sigma_e^2), I would say eta_it but the manual (page 9) says it's the standard deviation of nu_it. I am a little bit confused.

I would appreciate your help.
Thanks.

[Hugo Terrisse]

Hello,

I was confused by this as well, so I tried to simulate some data to check what sigma_e was. I simulated a dataset close to this example, with 10 groups and 19 observations per group. I simulated only the random part, so y = u_i + eta_it. To have eta_it follow an AR(1) process, I used the R function arima.sim. I used 75 for the standard deviation of u_i, 42 for the standard deviation of nu_it and 0.9 as the autocorrelation coefficient. So if sigma_e is the standard deviation of nu_it, it should be close to 42. Else if it is the standard deviation of eta_it, it should be close to sqrt(42 * 42 / (1 - 0.9 * 0.9)) = 96.4

Here are the results :

. xtregar y 

RE GLS regression with AR(1) disturbances       Number of obs     =    190
Group variable: id                              Number of groups  =    10

R-sq:                                           Obs per group:
within  =      .                                         min =    19
between =      .                                         avg =    19.0
overall =      .                                         max =    19

chi2(1)          =    .
corr(u_i, Xb)      = 0 (assumed)                Prob > chi2       =    .

    
y       Coef.   Std. Err.      z    P>z     [95% Conf.    Interval]
    
_cons    12.94061   32.21662     0.40   0.688     -50.2028    76.08401
    
rho_ar   .76098172   (estimated autocorrelation coefficient)
sigma_u   96.967043
sigma_e    36.55149
rho_fov   .87558815   (fraction of variance due to u_i)
theta   .70118624

sigma_e is closer to 42 than to 96.4, so it seems like sigma_e is the standard deviation of nu_it.

On a related matter, we can fit the same model but with a different method using the command mixed.

. mixed y ||  id : , reml residuals(ar , t(time)) stddeviations

Obtaining starting values by EM: 

Performing gradient-based optimization: 

Iteration 0:   log restricted-likelihood = -1046.1921  (not concave)
Iteration 1:   log restricted-likelihood = -964.56647  
Iteration 2:   log restricted-likelihood = -963.90686  
Iteration 3:   log restricted-likelihood = -963.70412  
Iteration 4:   log restricted-likelihood = -963.54944  
Iteration 5:   log restricted-likelihood = -963.54889  
Iteration 6:   log restricted-likelihood = -963.54889  

Computing standard errors:

Mixed-effects REML regression                   Number of obs     =        190
Group variable: id                              Number of groups  =         10

Obs per group:
min =         19
avg =       19.0
max =         19

Wald chi2(0)      =          .
Log restricted-likelihood = -963.54889          Prob > chi2       =          .


y       Coef.   Std. Err.      z    P>z     [95% Conf. Interval]

_cons    12.42014   34.35683     0.36   0.718    -54.91801     79.7583



Random-effects Parameters     Estimate   Std. Err.     [95% Conf. Interval]

id: Identity                 
sd(_cons)    96.77283   29.68368      53.04684    176.5417

Residual: AR(1)              
rho    .8697427   .0581564      .6983155      .94681
sd(e)    74.14402   16.37372      48.09495    114.3017

LR test vs. linear model: chi2(2) = 403.52                Prob > chi2 = 0.0000

Note: LR test is conservative and provided only for reference.

Here, sd(e) seems to be the standard deviation of eta_it.

I hope this helps. I am not 100 % confident in that little "proof", so if anyone has a better answer or finds a mistake, you're welcome !