I have results of meta-analysis, which have been carried out and reported. These analyses report tau-squared as a measure of heterogeneity (and the other details from the analysis). I've been asked to report the I-squared.

Is it possible to calculate I-squared from the reported results, or would I need to track down the original files to rerun the meta-analysis?

  • 1
    $\begingroup$ It would be helpful to tell us what tau-squared & I-squared are, apart from Greek & Latin letters. And don't you mean "Calculating I-squared [...] given tau-squared" in the title? $\endgroup$ Commented Aug 23, 2013 at 23:54
  • 4
    $\begingroup$ They are measures of heterogeneity that are used in meta-analysis. I-squared is the percentage of total variation across studies that is due to heterogeneity rather than chance. It's calculated using 100%×(Q - df)/Q where Q is the Cochran's heterogeneity statistic, which is chi-square distributed. Tau-squared as an absolute measure of heterogeneity, it's square root is a measure of the standard deviation of effect sizes across studies. My assumption was that someone unfamiliar with the terms as used in meta-analysis would not know answer the question, but that might have been incorrect. $\endgroup$ Commented Aug 24, 2013 at 0:14
  • 2
    $\begingroup$ Most reports of meta-analyses would include the value of $Q$ and its $df$. So why don't you compute $I^2$ based on that? $\endgroup$
    – Wolfgang
    Commented Aug 24, 2013 at 14:54
  • $\begingroup$ I don't have the values of Q. $\endgroup$ Commented Aug 26, 2013 at 18:08
  • 1
    $\begingroup$ Also you might want to take a look at: - Multivariate random-effects meta-regression: Updates to mvmeta (2011) STATA Journal by Ian R. White, MRC Biostatistics Unit, Cambridge, UK $\endgroup$
    – user55310
    Commented Sep 5, 2014 at 22:46

1 Answer 1


In the random-effects model the method-of-moments (DerSimonian–Laird) estimator of the between-study variance for $n$ studies is given by

$$ \hat{\tau}^2=\frac{Q-(n-1)}{\sum{w_i}-\frac{\sum{w_i^2}}{\sum{w_i}}}$$

where Cochran's $Q=\sum{w_i}\left(y_i-\frac{\sum{w_i y_i}}{\sum{w_i}}\right)^2$ (with $y_i$ & $w_i$ as the effect size & the reciprocal of the within-study variance respectively, estimated from the $i$th study).

The total proportion of variance owing to heterogeneity is


So from $\hat{\tau}^2$ alone you can't calculate $I^2$; you also need the number of studies, the sum of the weights, & the sum of the squared weights.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.