XTREG output interpretation

[Dino F Vitale]

I am asking help to get a better understand of
Code:
xtreg
output.
I am using this command to model a regression where the dependent variable is a measure of both eyes function in a group of subjects and, therefore, there is the need to take into account the correlation that exists between the 2 eyes on a single subject.
Absume the dependent variable name be 'meibo' to be regressed with 'age' and 'fabry' ( a 0/1 indicator of the presence of the Fabry disease) . After
Code:
xtset id
, where 'id' identify each subject, running
Code:
xtreg meibo i.fabry##c.age, re
gives the following output:
Code:

Random-effects GLS regression	Number of obs =	372
Group variable: id	Number of groups =	186
R-squared:	Obs per group:
Within = .	min =	2
Between = 0.8282	avg =	2.0
Overall = 0.7918	max =	2
	Wald chi2(3) =	877.31
corr(u_i, X) = 0 (assumed)	Prob > chi2 =	0.0000
meibo Coefficient Std. err.	z P>z [95% conf.	interval]
1.fabry .611988 .0507753	12.05 0.000 .5124701	.7115058
age .0052804 .0006747	7.83 0.000 .003958	.0066029
fabry#c.age
1 -.0035578 .0010202	-3.49 0.000 -.0055572	-.0015583
_cons .0562077 .0337378	1.67 0.096 -.0099171	.1223325
sigma_u .08600507
sigma_e .07248841
rho .58466635 (fraction	of variance due to u_i)

Questions:
1) Is it correct to say that the intraclass correlation coefficient (ICC) adjusted for the covariates relative to the 2 eyes measurement is 0.91, the square root of the 'R-squared_Between' (0.8282) ?
2) What relation exists (if any) between the named 'R-squared_Between' and 'rho', in the given example equal to .58466635?

Thank a lot in advance for the time spent to help me understand.
Best Regards.
DF Vitale

[Clyde Schechter]

1) No, the intraclass correlation is rho, in this case, .58 (to 2 decimal places).

2) There is no relationship between R²-between and the intraclass correlation.

[Carlo Lazzaro]

Dino:
as an aside to Clyde's helpful reply, as most of the patients have two eyes, you should cluster your standared errors on -id- if you collect right/left eye measures.
In addition, please use CODE delimiters (as per FAQ) to share what yoiu typed and what Stata gave you back. Thanks.

[Dino F Vitale]

Thank you both for your helpful reply. Clyde answer cleared a doubt I had.
As concerns the clustering on ‘id’ , I thought that ‘panelling’ on it and using xtreg would have correct for the eyes correlation in a single subject and ,hence, would have returned errors unbiased for the clustering.
However, since it is actually available a cluster option for xtreg, I am probably wrong and I am missing something.
Following I report the results obtained without an with cluster option.

. xtreg meibo i.fabry##c.age

Random-effects GLS regression                   Number of obs     =        372
Group variable: id                              Number of groups  =        186

R-squared:                                      Obs per group:
Within  =      .                                         min =          2
Between = 0.8282                                         avg =        2.0
Overall = 0.7918                                         max =          2

Wald chi2(3)      =     877.31
corr(u_i, X) = 0 (assumed)                      Prob > chi2       =     0.0000


meibo  Coefficient  Std. err.      z    P>z     [95% conf. interval]

1.fabry     .611988   .0507753    12.05   0.000     .5124701    .7115058
age    .0052804   .0006747     7.83   0.000      .003958    .0066029

fabry#c.age 
1    -.0035578   .0010202    -3.49   0.000    -.0055572   -.0015583

_cons    .0562077   .0337378     1.67   0.096    -.0099171    .1223325

sigma_u   .08600507
sigma_e   .07248841
rho   .58466635   (fraction of variance due to u_i) 


. xtreg meibo i.fabry##c.age ,re vce(cluster id)

Random-effects GLS regression                   Number of obs     =        372
Group variable: id                              Number of groups  =        186

R-squared:                                      Obs per group:
Within  =      .                                         min =          2
Between = 0.8282                                         avg =        2.0
Overall = 0.7918                                         max =          2

Wald chi2(3)      =    1014.58
corr(u_i, X) = 0 (assumed)                      Prob > chi2       =     0.0000

(Std. err. adjusted for 186 clusters in id)

Robust
meibo  Coefficient  std. err.      z    P>z     [95% conf. interval]

1.fabry     .611988   .0570159    10.73   0.000     .5002389     .723737
age    .0052804   .0006299     8.38   0.000     .0040459     .006515

fabry#c.age 
1    -.0035578   .0011651    -3.05   0.002    -.0058413   -.0012743

_cons    .0562077   .0264396     2.13   0.034      .004387    .1080284

sigma_u   .08600507
sigma_e   .07248841
rho   .58466635   (fraction of variance due to u_i)

There are indeed standard errors differences (even if small) so it would be safer to use clustering option.
Thank a lot again for the help