# Is a biased or unbiased estimator used for pooled SD in calculating Cohen's d?

When calculating Cohen's $d$ for independent samples, you must use a pooled $SD$. However, I have seen both of these:

$$SD_{\text{pooled1}} = \sqrt{\frac{ (n_1 - 1)s_1^2 + (n_2 - 1)s_2^2}{n_1 + n_2}}$$

vs.

$$SD_{\text{pooled2}} = \sqrt{\frac{ (n_1 - 1)s_1^2 + (n_2 - 1)s_2^2}{n_1 + n_2 -2}}$$

Supporting the use of $SD_{\text{pooled1}}$ , some website have $SD_{\text{pooled}}$ listed as $\sqrt{\frac{s_1^2 + s_2^2}{2}}$, which is the same as $SD_{\text{pooled1}}$ when sample sizes are equal.

From online discussions, it sounds like $SD_{\text{pooled1}}$ is a "biased" estimator of $SD$, and $SD_{\text{pooled2}}$ is less biased, that is, I think it means $SD_{\text{pooled1}}$ underestimates $SD$. In addition, some sites suggest that some effect size metrics use $SD_{\text{pooled1}}$ (Cohen's $d$), and others $SD_{\text{pooled2}}$ (Hedges' $g$).

Is this true? And if so, why would one effect size metric (Hedges' $g$) use the unbaised estimator, $SD_{\text{pooled2}}$ , while the other (Cohen's $d$) not?

• Closely related: stats.stackexchange.com/questions/1850/… That said, I believe the initial definition for Cohen's d didn't specify how to compute the SD, so the de facto formula (with the biased version) might just be a historical accident. – Matt Krause Jun 14 '12 at 8:46
• Thanks for editing my question so the formatting is much more clear, Jeromy. – Alon Jun 21 '12 at 20:48