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Pretty naive question here. If we have two paired large (>1000) samples and are interested in a paired Cohen's d then why not take the pairwise difference of the two samples and consider its mean divided by standard deviation? $$\frac{\overline{{X}_1 - {X}_2}}{SD_{X_1-X_2}}$$ To my meager knowledge this seems like a good paired version of Cohen's d.

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This is one way of computing a d-like measure for dependent samples. If the paired measurements are repeated measurements as in a pre-test post-test design, this is called the standardized mean change/gain and this uses change score standardization: $$d = \frac{\bar{D}}{SD_D} = \frac{\bar{X}_1 - \bar{X}_2}{SD_D}.$$ Alternatively, one can standardize the mean difference using one of the two SDs (typically the pre-test), so: $$d = \frac{\bar{D}}{SD_1} = \frac{\bar{X}_1 - \bar{X}_2}{SD_1}.$$ This is called the standardized mean change/gain using raw score standardization.

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  • $\begingroup$ This doesn't use pairedness. I added a formula to my question. $\endgroup$
    – Reza
    Apr 26 '17 at 16:16
  • $\begingroup$ $\bar{D}$ is the mean of the differences and $SD_D$ the SD of the differences, so this is the same as in your question. $\endgroup$
    – Wolfgang
    Apr 26 '17 at 16:42
  • $\begingroup$ OK great. Do you know of any reference on this subject (for citation)? Thank you. $\endgroup$
    – Reza
    Apr 26 '17 at 19:03
  • $\begingroup$ A place to start: Morris, S. B., & DeShon, R. P. (2002). Combining effect size estimates in meta-analysis with repeated measures and independent-groups designs. Psychological Methods, 7(1), 105-125. $\endgroup$
    – Wolfgang
    Apr 27 '17 at 19:07

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