# [R-meta] Metafor Fixed Effects giving I2- why?

Gerta Ruecker ruecker @end|ng |rom |mb|@un|-|re|burg@de
Tue Jul 30 16:15:42 CEST 2019

Thank you, Wolfgang!

The difference is between the *parameter* tau^2, which is zero if we
believe the fixed effect model, and the estimate \hat tau^2, which in
your example is likewise positive (all methods), though (of course) small.

Best,

Gerta

Am 30.07.2019 um 15:28 schrieb Viechtbauer, Wolfgang (SP):
> My understanding of Arthur's question was that he is wondering why the following code yields a value of I^2 = 40% given that this fits a fixed-effects model where tau^2 = 0 by definition.
>
> library(metafor)
>
> yi <- c(.2, .3, .4, .5)
> vi <- rep(.01, 4)
>
> res <- rma(yi, vi, method="FE")
> res\$I2
>
> (and in the 'devel' version, the value of I^2 is shown explicitly in the output also for a FE model).
>
> Best,
> Wolfgang
>
> -----Original Message-----
> From: Gerta Ruecker [mailto:ruecker using imbi.uni-freiburg.de]
> Sent: Tuesday, 30 July, 2019 10:37
> To: Viechtbauer, Wolfgang (SP); Aman Dheri; r-sig-meta-analysis using r-project.org
> Subject: Re: [R-meta] Metafor Fixed Effects giving I2- why?
>
> Hi all,
>
> The question whether it is possible in the fixed effect model that tau^2
> = 0 though I^2 > 0 was not answered in Wolfgang's reply. Following his
> response, in this case metafor calculates
>
> I^2 = (Q - (df)) / Q,
>
> which probably means
>
> I^2 = max(0, (Q - (df)) / Q).
>
> Now it depends on the way tau^2 is calculated. For the DerSimonian-Laird
> estimate, \hat tau^2 has the same numerator as I^2, that is, if \hat
> tau^2 = 0, then I^2 = 0.
>
> Thus the case Arthur describes can only, if at all, occur for other
> tau^2 estimates. Thus, I repeat his question: Is it possible and under
> which circumstances?
>
> Best,
>
> Gerta
>
> Am 29.07.2019 um 19:56 schrieb Viechtbauer, Wolfgang (SP):
>> Hi Arthur,
>>
>> Indeed, for random/mixed-effects models, I^2 is computed with tau^2 / (tau^2 + v), where tau^2 is the estimate of (residual) heterogeneity and v is the 'typical' sampling variance. So, if tau^2 = 0, then I^2 = 0.
>>
>> Since I^2 is a useful statistic also for fixed-effects models, I^2 is also computed then, but using (Q - (df)) / Q, where Q is the test statistic for (residual) heterogeneity and df = degrees of freedom of the test (k-1 for a model without moderators). But this has no effect on the rest of the computations (such as the weights).
>>
>> Best,
>> Wolfgang
>>
>> -----Original Message-----
>> From: R-sig-meta-analysis [mailto:r-sig-meta-analysis-bounces using r-project.org] On Behalf Of Aman Dheri
>> Sent: Monday, 29 July, 2019 19:11
>> To: r-sig-meta-analysis using r-project.org
>> Subject: [R-meta] Metafor Fixed Effects giving I2- why?
>>
>> Hello,
>>
>> I am doing a fixed effects meta-analysis in Metafor and I see that it returns a non-zero I^2 value when tau^2 is always zero. According to Metafor documentation the fixed effects model in rma.uni should be 0 because of its calculation from the tau^2. Does any one know why this is happening, and if it is changing the weighting during the meta-analysis vs if the I^2 was 0?
>>
>> I assumed my code wasn’t needed for the answer to this but please let me know if I’m wrong!
>>
>> Thank you!
>> Arthur

--

Dr. rer. nat. Gerta Rücker, Dipl.-Math.

Institute of Medical Biometry and Statistics,
Faculty of Medicine and Medical Center - University of Freiburg

Stefan-Meier-Str. 26, D-79104 Freiburg, Germany

Phone:    +49/761/203-6673
Fax:      +49/761/203-6680
Mail:     ruecker using imbi.uni-freiburg.de
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