# [ADMB Users] Variance of ZINB - and partitioning variance

W Robert Long W.R.Long at leeds.ac.uk
Fri Jan 18 03:31:50 PST 2013

```I read in the Getting Started with glmmADMB guide that using
family="nbinom2" uses the
parameterization Var = mu(1+ (mu/k))

I assume mu is a mean parameter, but what is k ? And how do these relate
to the estimates from the model output  ?:

Negative binomial dispersion parameter: 3.9329 (std. err.: 0.69955)
Zero-inflation: 0.13191  (std. err.:  0.02779 )

I was asked to calculate a variance partition coefficient / intraclass
correlation coefficient. I have some experience of this with binary
outcomes and I know several ways of estimating the ICC. For example here
they discuss 4 different approaches.....
http://onlinelibrary.wiley.com/doi/10.1111/j.1467-985X.2004.00365.x/abstract

However I don't know anything about partitioning variance in a ZINB
model.  In my model output I have:

Group=Subject
Variance StdDev
(Intercept)    5.527  2.351

and
> var(fit1\$residuals)
[,1]
[1,] 68.95101

My naive attempt is:
> 5.527 / (5.527+68.95101)
 0.07420982

However I don't think this can be correct - it is a repeated measures
(x4) model with random intercepts for subjects:
fit <- glmmadmb(out ~ time*X1 + time*X2 +
( 1| Subject),
data=na.omit(dt.work),family="nbinom2",zeroInflation=TRUE)

I would expect much more variance to be between subjects, not within
subjects.....

Is there an adjustment to the ICC for ZINB outcomes ?

If I can digress a little, when I first came to help on this study,
lmer() from the lme4 package was being used:
fit.old <- lmer(mjm ~ time*X1 + time*X2 + (1 | Subject), data = dt.orig)

Of course the first thing I realised is that lme4 isn't suitable because
the outcome was very zero-inflated. It was also averaged over a week -
so I transformed the outcome into counts by multiplying it by 7 and
that's what I used in glmmADMB above. But the output from lmer() was

Random effects:
Groups   Name        Variance Std.Dev.
Subject  (Intercept) 2.8895   1.6999
Residual             1.6858   1.2984

which gives an ICC of around 0.6, markedly different from the 0.07
above, but within the expected ballpark.

Is my naive calculation of the ICC from the glmmADMB valid (I assume
not) or is there an alternative calculation that perhaps uses the
overdispersion parameter estimate ? Clearly the overdispersion is
increased by my scaling the original data by 7, so in my mind there
should be a way to adjust for this in the ICC calculation....

Thanks
Robert Long

Any advice would be most welcome.

Thanks !
Robert Long

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