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Re: st: Poisson Two Level Random Intercept ICC


From   Robert Sutter <rdsutterjr@gmail.com>
To   "statalist@hsphsun2.harvard.edu" <statalist@hsphsun2.harvard.edu>
Subject   Re: st: Poisson Two Level Random Intercept ICC
Date   Sun, 3 Mar 2013 15:41:01 -0600

Thanks!

Sent from my iPad

On Mar 1, 2013, at 5:55 PM, "JVerkuilen (Gmail)" <jvverkuilen@gmail.com> wrote:

> 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-Hesketh’s 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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