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Re: st: Poisson Two Level Random Intercept ICC
From
Robert Sutter <[email protected]>
To
"[email protected]" <[email protected]>
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)" <[email protected]> wrote:
> On Fri, Mar 1, 2013 at 12:20 PM, Robert Sutter <[email protected]> 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
> [email protected]
>
> "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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