In this answer to a question I asked (which derives the variance of Cohen's $d$), the approximation
$$\frac{\Gamma\left(\frac{n_T + n_C - 2}{2}\right)}{\sqrt{\frac{n_T+n_C-2}{2}}\Gamma\left(\frac{n_T+n_C-3}{2}\right)} \approx 1 - \frac{3}{4(n_T+n_C+2)-1}$$ is used. We can reasonably assume that $n_T, n_C> 0$ are integers.

How is this approximation derived? The answerer states:

I pulled it from the Hedges paper -- don't know its derivation at the moment but will think about it some more.

I wish I had more to contribute to this question than that, but the removal of the $\Gamma$ function I find completely baffling, and I wouldn't even know where to start.

Edit: Currently trying out Stirling's approximation, seeing if that leads me anywhere. And so far, I'm quite lost as to how to deal with the division by $2$ in the $\Gamma$ functions.

  • $\begingroup$ This hint might help. Consider $\Gamma(n/2)$ if $n$ is even then we have $\Gamma(2k/2) = \Gamma(k) = (k-1)!$. if $n$ is odd then $$\Gamma((2k+1)/2) = \Gamma(k+1/2)$$ Now you can use the duplication formula zaidalyafeai.files.wordpress.com/2015/09/… page 23 $\endgroup$ – Zaid Alyafeai Dec 7 '15 at 19:22
  • 2
    $\begingroup$ Unless I've mucked up the numerics this is an absolutely awful "approximation". The left-hand side is not close to the right-hand side at all. $\endgroup$ – Antonio Vargas Dec 7 '15 at 19:30
  • $\begingroup$ @AntonioVargas I will try this out myself via simulation, but I got my hands on the paper. "This approximation has the virtue that it can be computed algebraically when using packaged computer pgorams.... [it] has a maximum error of $0.007$ when $n_{T}+n_{C} = 2$ and is accurate to within $0.00033$ when $n_{T} + n_{C} \geq 10$. For $n_{T}+n_{C} > 50$, the error does not exceed $1.5 \times 10^{-5}$. I have double-checked the equation FYI $\endgroup$ – Clarinetist Dec 7 '15 at 19:42
  • $\begingroup$ @AntonioVargas AH, it looks like there's an error in the original answer. The numerator $\Gamma$ should have a division by $2$ in it. $\endgroup$ – Clarinetist Dec 7 '15 at 19:44
  • 1
    $\begingroup$ Right, so the new version can definitely be found using the first (and maybe second if you're feeling adventurous) higher order terms in Stirling's formula. $\endgroup$ – Antonio Vargas Dec 7 '15 at 19:47

Using Stirling's Approximation $$ \log(\Gamma(n))=n\log(n)-n-\frac12\log(n)+\frac12\log(2\pi)+\frac1{12n}+O\left(\frac1{n^3}\right) $$ we can compute $$ \log\left[\frac{\Gamma(n-1)}{\sqrt{n-1}\,\Gamma\left(n-\frac32\right)}\right]=-\frac3{8n}-\frac1{2n^2}+O\left(\frac1{n^3}\right) $$ and therefore, $$ \begin{align} \frac{\Gamma(n-1)}{\sqrt{n-1}\,\Gamma\left(n-\frac32\right)} &=1-\frac3{8n}-\frac{55}{128n^2}+O\left(\frac1{n^3}\right)\\ &=1-\frac3{8n-\frac{55}6}+O\left(\frac1{n^3}\right) \end{align} $$ Set $n=\frac{n_T+n_C}2$ and we get $$ \begin{align} \frac{\Gamma(n-1)}{\sqrt{n-1}\,\Gamma\left(n-\frac32\right)} &=1-\frac3{8n-\frac{55}6}+O\left(\frac1{n^3}\right)\\ &=1-\frac3{4\left(n_T+n_C-2\right)-\frac76}+O\left(\frac1{\left(n_T+n_C\right)^3}\right)\\ \end{align} $$ If I am correct, there seems to be a sign problem in $n_T+n_C\color{#FF0000}{-}2$, and $\frac76$ has been simplified to $1$.

  • $\begingroup$ Question: isn't that series up there for $\Gamma(n+1)$, rather than $\Gamma(n)$? See here for example. Also, $\log(2\pi n)$ instead of $\log(2\pi)$? $\endgroup$ – Clarinetist Dec 22 '15 at 18:25
  • $\begingroup$ @Clarinetist: nope on both counts. $\log(\Gamma(n+1))$ has $\color{#C00000}{+}\frac12\log(n)$. That, combined with the $\frac12\log(2\pi)$ would give $\frac12\log(2\pi n)$. $\endgroup$ – robjohn Dec 23 '15 at 3:04

Here is a simple, elementary derivation, using only the recursion relation $\Gamma(x+1)=x\ \Gamma(x)$ instead of Stirling's approximation. We define $$f(x) = \frac{\Gamma(x)}{\sqrt{x}\,\Gamma(x-\frac12)}\ ,$$ where $x = \frac{n_T+n_C-2}2$. Then what we want to prove is the following asymptotic behavior for large $x$: $$f(x)=1-\frac3{8x-\frac76}+O(x^{-3})\ .$$ The trick is to consider the product $f(x)f(x+\frac12)$. The above definition of $f(x)$ leads to $$f(x)f(x+{\small\frac12}) = \frac{\Gamma(x)}{\sqrt{x}\,\Gamma(x-\frac12)}\frac{\Gamma(x+\frac12)}{\sqrt{x+\frac12}\,\Gamma(x)} = \frac{x-\frac12}{\sqrt{x}\sqrt{x+\frac12}} = \frac{1-\frac1{2x}}{\sqrt{1+\frac1{2x}}} \ , $$ where in the 2nd equality we have used the recursion for $\Gamma$. Expanding the squareroot in the denominator yields $$f(x)f(x+{\small\frac12}) = 1-\frac3{4x}+\frac7{32x^2}+O(x^{-3})\ .$$ On the other hand, setting $$f(x)=1-\frac1{ax+b}+O(x^{-3})\ ,$$ we obtain another expansion $$f(x)f(x+{\small\frac12}) = \left( 1-\frac1{ax}\frac1{1+\frac b{ax}} \right) \left( 1-\frac1{ax}\frac1{1+\frac{\frac a2 +b}{ax}} \right)+O(x^{-3}) $$ $$ = 1-\frac2{ax}+\frac{1+\frac a2 +2b}{a^2x^2}+O(x^{-3})\ .$$ Comparison of the coefficients of the $x^{-1}$ and $x^{-2}$ terms in the two expansions finally gives $a=\frac83$ and $b=-\frac7{18}$, which concludes the proof.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.