# Variance of Cohen's $d$ statistic

Cohen’s $d$ is one of the most common ways we measure the size of an effect (see Wikipedia). It simply measures the distance between two means in terms of the pooled standard deviation. How can we derive the mathematical formula of variance estimate of Cohen's $d$?

December 2015 edit: Related to this question is the idea of calculating confidence intervals around $d$. This article states that

$$\sigma_{d}^2 = \dfrac{n_{+}}{n_{\times}} + \dfrac{d^2}{2n_{+}}$$

where $n_{+}$ is the sum of the two sample sizes and $n_{\times}$ is the product of the two sample sizes.

How is this formula derived?

• @Clarinetist: It is somewhat controversial to edit another person's question to add more substance and more questions to it (as opposed to improving the wording). I took the liberty to approve your edit (given that you placed a generous bounty and that I think your edit does improve the question), but others might decide to roll back. Dec 3, 2015 at 17:31
• @amoeba No problem. As long as the formula is there for $\sigma^2_d$ (which wasn't there before) and it's clear that we're looking for a mathematical derivation of the formula, that's fine. Dec 3, 2015 at 17:32
• I think the denominator of the second fraction should be $2(n_{+}-2)$. See my answer below.
– user5594
Dec 7, 2015 at 0:54

Note that the variance expression in the question is an approximation. Hedges (1981) derived the large sample variance of $d$ and approximation in a general setting (i.e. multiple experiments/studies), and my answer pretty much walks through the derivations in the paper.

First, the assumptions we'll utilize are the following:

Let's assume we have two independent treatment groups, $T$ (treatment) and $C$ (control). Let $Y_{Ti}$ and $Y_{Cj}$ be the scores/responses/whatever from subject $i$ in group $T$ and subject $j$ in group $C$, respectively.

We assume the responses are normally distributed and the treatment and control groups share a common variance, i.e.

\begin{align*} Y_{Ti} &\sim N(\mu_T, \sigma^2), \quad i = 1, \dots n_T \\ Y_{Cj} &\sim N(\mu_C, \sigma^2), \quad j = 1, \dots n_C \end{align*}

The effect size we're interested in estimating in each study is $\delta = \frac{\mu_T - \mu_C}{\sigma}$. The estimator of the effect size we'll use is \begin{equation*} d = \frac{\bar{Y}_T - \bar{Y}_C}{\sqrt{\frac{(n_T - 1)S_T^2 + (n_C - 1)S_C^2}{n_T + n_C - 2}}} \end{equation*} where $S_k^2$ is the unbiased sample variance for group $k$.

Let's consider the large-sample properties of $d$.

First, note that: \begin{equation*} \bar{Y}_T - \bar{Y}_C \sim N \Bigg( \mu_T - \mu_C, \,\sigma^2\frac{n_T + n_C}{n_T n_C} \Bigg) \end{equation*} and (being loose with my notation): \begin{equation} \frac{(n_T - 1)S_T^{2}}{\sigma^2(n_T + n_C - 2)} = \frac{1}{n_T + n_C - 2}\frac{(n_T - 1)S_T^{2}}{\sigma^2} \sim \frac{1}{n_T + n_C- 2}\chi_{n_T - 1}^2 \tag{1} \end{equation} and \begin{equation} \frac{(n_C - 1)S_C^{2}}{\sigma^2(n_T + n_C - 2)} = \frac{1}{n_T + n_C - 2}\frac{(n_C - 1)S_C^{2}}{\sigma^2} \sim \frac{1}{n_T + n_C- 2}\chi_{n_C - 1}^2 \tag{2} \end{equation}

Equations (1) and (2) lead to the fact that (again, being loose with my notation): \begin{equation*} \frac{1}{\sigma^2}\frac{(n_T - 1)S_T^{2} + (n_C - 1)S_C^{2}}{n_T + n_C - 2} \sim \frac{1}{n_T + n_C - 2}\chi_{n_T + n_C - 2}^2 \end{equation*}

Now, some clever algebra: \begin{align*} d &= \frac{\bar{Y}_T - \bar{Y}_C}{\sqrt{\frac{(n_T - 1)S_T^2 + (n_C - 1)S_C^2}{n_T + n_C - 2}}} \\\\ &= \frac{\left(\sigma\sqrt{\frac{n_T + n_C}{n_T n_C}}\right)^{-1}(\bar{Y}_T - \bar{Y}_C)}{\left(\sigma\sqrt{\frac{n_T + n_C}{n_T n_C}}\right)^{-1}\sqrt{\frac{(n_T - 1)S_T^2 + (n_C - 1)S_C^2}{n_T + n_C - 2}}} \\\\ &= \frac{\frac{(\bar{Y}_T - \bar{Y}_C) - (\mu_T - \mu_C)}{\sigma\sqrt{\frac{n_T + n_C}{n_T n_C}}} + \frac{\mu_T - \mu_C}{\sigma\sqrt{\frac{n_T + n_C}{n_T n_C}}}}{\left(\sqrt{\frac{n_T + n_C}{n_T n_C}}\right)^{-1}\sqrt{\frac{(n_T - 1)S_T^2 + (n_C - 1)S_C^2}{\sigma^2(n_T + n_C - 2)}}} \\\\ &= \sqrt{\frac{n_T + n_C}{n_T n_C}}\left(\frac{\theta + \delta\sqrt{\frac{n_T n_C}{n_T + n_C}}}{\sqrt{\frac{V}{\nu}}}\right) \end{align*} where $\theta \sim N(0,1)$, $V \sim \chi^2_{\nu}$, and $\nu = n_T+n_C-2$. Thus, $d$ is $\sqrt{\frac{n_T + n_C}{n_T n_C}}$ times a variable which follows a non-central t-distribution with $n_T + n_C - 2$ degrees of freedom and non-centrality parameter of $\delta\sqrt{\frac{n_T n_C}{n_T + n_C}}$.

Using the moment properties of the non-central $t$ distribution, it follows that: \begin{equation*} \mathrm{Var}(d) = \frac{(n_T + n_C - 2)}{(n_T + n_C - 4)}\frac{(n_T + n_C)}{n_T n_C}(1+ \delta^2\frac{n_T n_C}{n_T + n_C}) - \frac{\delta^2}{b^2} \tag{3} \end{equation*} where \begin{equation*} b = \frac{\Gamma\left(\frac{n_T + n_C - 2}{2}\right)}{\sqrt{\frac{n_T+n_C-2}{2}}\Gamma\left(\frac{n_T+n_C-3}{2}\right)} \approx 1 - \frac{3}{4(n_T+n_C-2)-1} \end{equation*}

So Equation (3) provides the exact large sample variance. Note that an unbiased estimator for $\delta$ is $b d$, with variance:

\begin{equation*} \mathrm{Var}(bd) = b^2\frac{(n_T + n_C - 2)}{(n_T + n_C - 4)}\frac{(n_T + n_C)}{n_T n_C}(1+ \delta^2\frac{n_T n_C}{n_T + n_C}) - \delta^2 \end{equation*}

For large degrees of freedom (i.e. large $n_T+n_C-2$), the variance of a non-central $t$ variate with $\nu$ degrees of freedom and non-centrality parameter $p$ can be approximated by $1 + \frac{p^2}{2\nu}$ (Johnson, Kotz, Balakrishnan, 1995). Thus, we have: \begin{align*} \mathrm{Var}(d) &\approx \frac{n_T + n_C}{n_T n_C}\left(1 + \frac{\delta^2\left(\frac{n_T n_C}{n_T + n_C}\right)}{2(n_T+n_C-2)}\right) \\\\ &= \frac{n_T + n_C}{n_T n_C} + \frac{\delta^2}{2(n_T+n_C-2)} \end{align*}

Plug in our estimator for $\delta$ and we're done.

• Very, very nice derivation. Just a few questions: 1) could you clarify what the notation $\bar{Y}^{T}_{i} - \bar{Y}^{C}_{i}$ means (I know it's something to do with difference of sample means, but how can they both have the same index?)? 2) could you clarify how the approximation for $b$ is done (I don't need all of the details, a source is fine and maybe a brief explanation)? Otherwise, I'm quite pleased with this. (+1) This also agrees with the observation that I've made that $d$ doesn't follow a normal distribution, contrary to the explanation in the linked article in the OP. Dec 7, 2015 at 1:13
• @Clarinetist Thanks! 1) How can they have the same index? Typo, that's how! :P They're an artifact of my first draft of the answer. I'll fix that. 2) I pulled it from the Hedges paper -- don't know its derivation at the moment but will think about it some more.
– user5594
Dec 7, 2015 at 1:26
• I'm looking into the derivation now, but FYI, the numerator of $b$ should be $\Gamma\left(\dfrac{n_T+n_C-2}{2}\right)$. Dec 7, 2015 at 19:46
• Derivation provided for reference: math.stackexchange.com/questions/1564587/… . Turns out there's likely a sign error. Dec 31, 2016 at 5:53
• @mike : very impressing answer. Thanks for taking the time to share it with us. Jan 2, 2017 at 21:57