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From |
"JVerkuilen (Gmail)" <jvverkuilen@gmail.com> |

To |
statalist@hsphsun2.harvard.edu |

Subject |
Re: st: Poisson Two Level Random Intercept ICC |

Date |
Fri, 1 Mar 2013 18:55:32 -0500 |

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) * * For searches and help try: * http://www.stata.com/help.cgi?search * http://www.stata.com/support/faqs/resources/statalist-faq/ * http://www.ats.ucla.edu/stat/stata/

**Follow-Ups**:**Re: st: Poisson Two Level Random Intercept ICC***From:*Robert Sutter <rdsutterjr@gmail.com>

**References**:**st: Poisson Two Level Random Intercept ICC***From:*Robert Sutter <rdsutterjr@gmail.com>

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