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If the gamma function is given by

$$\Gamma(\alpha) = \int_0^{+\infty}t^{\alpha-1}e^{-t}\text dt$$

and the lower incomplete gamma function by

$$\gamma(\alpha,x) = \int_{0}^{x}t^{\alpha-1}e^{-t}\text dt$$

Is it possible to derive $\gamma(2\alpha,x)$ from $\gamma(\alpha,x)$, $\Gamma(\alpha)$, and $\Gamma(2\alpha)$? The reason why I believe a relationship like that must hold is that, when adding two random variables distributed according to a truncated $\Gamma(\alpha,\beta)$ distribution with support on $[0,w)$

$$f(x;\alpha,\beta)=\mathbb{1}_{0\leq x < w}\frac{\beta^\alpha x^{\alpha-1} e^{-x\beta}}{\gamma(\alpha,w\beta)}$$

I get a term that goes like

$$\mathbb{1}_{0\leq x < w}\frac{\beta^{2\alpha} x^{2\alpha-1} e^{-x\beta}}{\gamma(\alpha,w\beta)^2} \frac{\Gamma(\alpha)^2}{\Gamma(2\alpha)}$$

which looks a lot like the truncated gamma function with parametre $2\alpha$

$$\mathbb{1}_{0\leq x < w}\frac{\beta^{2\alpha} x^{2\alpha-1} e^{-x\beta}}{\gamma(2\alpha,w\beta)}$$

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  • $\begingroup$ If you meant $x$ is fixed then no, because $f_{X+Y}(x_0)$ ($X,Y$ independent) doesn't depend only on $f_X(x_0)$ and $f_Y(x_0)$ but on $f_X(x)$ and $f_Y(x_0-x)$ for every $x$ $\endgroup$
    – reuns
    Sep 26, 2016 at 17:52
  • $\begingroup$ I don't understand how this relates to my question. $\endgroup$
    – Red
    Sep 26, 2016 at 18:02
  • $\begingroup$ I explained why your argument "adding two random variables distributed according to a truncated Γ(α,β) distribution" is flawed : adding two (independent) r.v. means convoluting their pdf, and for this you need to know their whole pdf, not only their value at $x_0$ $\endgroup$
    – reuns
    Sep 26, 2016 at 18:14
  • $\begingroup$ I didn't understand your explanation, then; could you elaborate on it? $\endgroup$
    – Red
    Sep 26, 2016 at 18:15
  • $\begingroup$ Can you write the pdf of $X+Y$ in term of the pdf of $X$ and $Y$ (when $X,Y$ are independent) ? $\endgroup$
    – reuns
    Sep 26, 2016 at 18:16

1 Answer 1

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Notice that

$$\gamma(s, x) + \Gamma(s, x) = \Gamma(s)$$

Now if $s = 2\alpha$...

Notice also that

$$\Gamma(2\alpha) = \frac{2^{2\alpha - 1}}{\sqrt{\pi}}\Gamma\left(\alpha + \frac{1}{2}\right)\Gamma(\alpha)$$

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  • $\begingroup$ Finding $\gamma(2\alpha,x)$ from $\Gamma(2\alpha)$ and $\Gamma(2\alpha, x)$ is indeed straightforward but I don't have $\Gamma(2\alpha, x)$, I have instead $\Gamma(\alpha)$ and $\gamma(\alpha, x)$. $\endgroup$
    – Red
    Sep 26, 2016 at 18:10
  • $\begingroup$ @PedroCarvalho Use the duplication formula for $\Gamma(s)$ (I'm going to edit the answer). $\endgroup$
    – Enrico M.
    Sep 26, 2016 at 18:15
  • $\begingroup$ You might find this article useful, maybe. arxiv.org/pdf/0712.0292.pdf $\endgroup$
    – Enrico M.
    Sep 26, 2016 at 18:22
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    $\begingroup$ The problem there is that I have to subtract that value from somewhere not multiply. In the ideal world I'd get a duplication formula for $\gamma(\alpha,x)$ but it would seem one doesn't exist. $\endgroup$
    – Red
    Sep 26, 2016 at 19:00
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    $\begingroup$ @PedroCarvalho Interesting! I will think about!! $\endgroup$
    – Enrico M.
    Sep 26, 2016 at 19:01

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