Re: st: Poisson Two Level Random Intercept ICC

["JVerkuilen (Gmail)" <jvverkuilen@gmail.com>]

On Fri, Mar 1, 2013 at 12:20 PM, Robert Sutter <rdsutterjr@gmail.com> wrote:
> How is the intraclass correlation coefficient calculated when using a
> two level random intercept model?
>
> Below is the output from a two level model for the count of excess
> deaths (oediff2) as the dependent variable and the clusters are
> attending physicians. sigma_u is the random intercept variance.
>
> In Rabe-Heskethʼs book the ICC for binary data is derived by the
> following formula:random intercept variance/( random intercept
> variance + π2/3).  π2/3 represents the variance of the logistic
> distribution.
>

Yes, this is one way to derive the ICC for binary data, working on the
scale of the linear predictor, which is the scale that the random
intercept variance exists on. There are others as the article I linked
to before by Harvey Goldstein and colleagues when this question was
posed.



> Can the same formula be used for count data by replacing π2/3  with
> the variance of oediff2?

I think the answer is no because the random effect variance isn't
directly comparable to the variance of the Poisson, because they are
on two different scales.



> xtpoisson oediff2, i(attending_phy_id) normal
>
> Random-effects Poisson regression               Number of obs      =        75
> Group variable: attending_ph~d                  Number of groups   =        75
>
> Random effects u_i ~ Gaussian                   Obs per group: min =         1
>                                                                avg =       1.0
>                                                                max =         1
>
>                                                 Wald chi2(0)       =         .
> Log likelihood  =  -213.3618                    Prob > chi2        =         .
>
> ------------------------------------------------------------------------------
>      oediff2 |      Coef.            Std. Err.      z    P>|z|
> [95% Conf. Interval]
> -------------+----------------------------------------------------------------
>        _cons |   2.617512   .0323429    80.93   0.000     2.554121    2.680903
> -------------+----------------------------------------------------------------
>     /lnsig2u |  -5.924562   4.653182    -1.27   0.203    -15.04463    3.195508
> -------------+----------------------------------------------------------------
>      sigma_u |   .0517009   .1202868                      .0005409     4.94192
> ------------------------------------------------------------------------------

And with these data it's irrelevant because the random effect variance
is essentially 0 anyway. But that's not surprising because your group
size is 1 so you have no real ability to estimate this anyway. I
suspect the model is identified only due to the assumption of the
distribution and that it works only because the Poisson doesn't have a
separate variance term.



-- 
JVVerkuilen, PhD
jvverkuilen@gmail.com

"It is like a finger pointing away to the moon. Do not concentrate on
the finger or you will miss all that heavenly glory." --Bruce Lee,
Enter the Dragon (1973)

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