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Is there any property due to which we can write $$\frac{\Gamma(\frac{n}{2},\frac{x}{2})}{\Gamma(\frac{n}{2})}=\frac{\Gamma(n,x)}{\Gamma(n)}$$ where $n$ is an integer. Thanks in advance.

Frank Moses
  • 2,718
  • If you set, for example, $n = x = 4$ you can immediately see that your identity doesn't hold. – Enrico M. Aug 29 '16 at 09:35
  • @Beta thanks for your comment. In one of the paper it is written that the CDF of chi square random variable with degree of freedom being $M$ is $F(x)=1-\frac{\Gamma(M,x)}{\Gamma(M)}$. But in many other references I found the CDF to be $\frac{\gamma(\frac{M}{2},\frac{x}{2})}{\Gamma{\frac{M}{2}}}$. Now these quantities become equal only if the above identity is true. (the paper that I mention is "Secure D2D Communication in large scale cognitive cellular networks: A wireless power transfer model", Equation no (19) you can check by yourself.) – Frank Moses Aug 29 '16 at 09:44
  • Beware of the SYMBOLS!! $$\Gamma(n, x) \neq \gamma(n, x)$$

    They are two different functions!

    https://en.wikipedia.org/wiki/Incomplete_gamma_function

    In that case one shall examine the identity!

     – Enrico M. Aug 29 '16 at 09:46
  • @Beta I know that but I used the property $\Gamma(M)=\gamma(M,x)+\Gamma(M,x)$ – Frank Moses Aug 29 '16 at 09:48
  • I think that the only true identity, in this case, may be:

    $$1 - \frac{\Gamma(M, x)}{\Gamma(M)} = 1 - \frac{\Gamma(M) - \gamma(M, x)}{\Gamma(M)} = \frac{\gamma(M, x)}{\Gamma(M)}$$

     – Enrico M. Aug 29 '16 at 09:51

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