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I am getting the following, using the property $\Gamma(z) = z\Gamma(z-1)$

$\frac{1}{2}\frac{\Gamma(\frac{v}{2}-1)}{\Gamma(\frac{v}{2})} = \frac{1}{2}\frac{\Gamma(\frac{v}{2}-1)}{\frac{v}{2}\times\Gamma(\frac{v}{2}-1)} = \frac{1}{v}$

But the answer is supposed to be $\frac{1}{v-2}$

Arnold Doveman
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1 Answers1

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The correct recurrence for the gamma function is $\Gamma(x+1)=x\Gamma(x)$ or $\Gamma(x)=(x-1)\Gamma(x-1)$ so the $\frac v2$ in the denominator should be $\frac v2-1$

Ross Millikan
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