# Calculating I-squared (in meta-analysis, given tau-squared)

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?

• 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? Commented Aug 23, 2013 at 23:54
• 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. Commented Aug 24, 2013 at 0:14
• 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? Commented Aug 24, 2013 at 14:54
• I don't have the values of Q. Commented Aug 26, 2013 at 18:08
• 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
– user55310
Commented Sep 5, 2014 at 22:46

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).
$$I^2=\frac{Q-(n-1)}{Q}$$
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.