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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.

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  • $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ 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
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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 $1$.

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  • $\begingroup$ 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
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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.

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