Solved – Is a biased or unbiased estimator used for pooled SD in calculating Cohen’s d

cohens-deffect-sizestandard deviationunbiased-estimator

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?

Best Answer

Both estimates are biased. The square of the second one is an unbiased estimator of the common variance. It is not clear that taking the square root of an unbiased estimator makes the estimate of standard deviation better than taking the square root of a biased estimator of variance.