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$$\frac{\left(\beta_{2}+\beta_{1}\right)^{\alpha_{1}+\alpha_{2}-1}B\left(\alpha_{1},\alpha_{2}\right)}{\beta_{1}^{\alpha_{1}}\beta_{2}^{\alpha_{2}}}=\frac{1}{\alpha_{1}\beta_{2}}{_{2}F_{1}}\left({1-\alpha_{2},1\atop 1+\alpha_{1}};-\tfrac{\beta_{1}}{\beta_{2}}\right)+\frac{1}{\alpha_{2}\beta_{1}}{_{2}F_{1}}\left({1-\alpha_{1},1\atop 1+\alpha_{2}};-\tfrac{\beta_{2}}{\beta_{1}}\right)$$
Where $B(\alpha_{1},\alpha_{2})=\frac{\Gamma\left(\alpha_{1}\right)\cdot\Gamma\left(\alpha_{2}\right)}{\Gamma\left(\alpha_{1}+\alpha_{2}\right)}$ is the Beta function
To me it's interesting because:

  1. It has 4 independent parameters/variables.

  2. It relates $\frac{\beta_{1}}{\beta_{2}} and \frac{\beta_{2}}{\beta_{1}}$.

  3. The 1 in the upper term can be manipulated by Hypergeometric relations or Generalized Hypergeometric relations:

• DLMF 15.5: https://dlmf.nist.gov/15.5

• DLMF 16.3(i): https://dlmf.nist.gov/16.3.i
If this type of identity is known I would be interested in any information. Is it possible that the original probability problem (see below) could be a source of new Hypergeometric identities?

Background:
While kibitzing on a paper by Aaron Hendrickson I noticed an equivalent form arising from the calculation of conditional probabilities for the gamma-difference distribution. Specifically, for $Y\sim\mathcal{GD}(\alpha_{1},\alpha_{2},\beta_{1},\beta_{2})$ which is the distribution of the difference of independent gamma variables we have the following results. $$\operatorname{pr}(Y\leq0)=C_{Y}\frac{\Gamma(\alpha_{1}+\alpha_{2})}{\beta_{2}\alpha_{1}}{_{2}F_{1}}\left({1-\alpha_{2},1\atop 1+\alpha_{1}};-\tfrac{\beta_{1}}{\beta_{2}}\right),$$
$$\operatorname{pr}(Y\geq0)=C_{Y}\frac{\Gamma(\alpha_{1}+\alpha_{2})}{\beta_{1}\alpha_{2}}{_{2}F_{1}}\left({1-\alpha_{1},1\atop 1+\alpha_{2}};-\tfrac{\beta_{2}}{\beta_{1}}\right),$$
where
$$C_{Y}=\beta_{1}^{\alpha_{1}}\beta_{2}^{\alpha_{2}}(\beta_{1}+\beta_{2})^{1-\alpha_{1}-\alpha_{2}},$$
for $\alpha_{1},\alpha_{2},\beta_{1},\beta_{2}>0$. If true then it follows that $\operatorname{pr}(Y\leq0)+\operatorname{pr}(Y\geq0)=1$ (with an overlap of measure zero) and the equation is true.
As far as I can tell the statement is true numerically but I have no idea how to prove it in terms of Hypergeometric functions.

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1 Answer 1

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This expression can be obtained from the connection formula \begin{equation} w_{1}(z)=\frac{\Gamma\left(c\right)\Gamma\left(b-a\right)}{\Gamma\left(b\right)\Gamma\left(c-a\right)}w_{5}(z)+\frac{\Gamma\left(c\right)\Gamma\left(a-b% \right)}{\Gamma\left(a\right)\Gamma\left(c-b\right)}w_{6}(z) \end{equation} where \begin{align} w_{1}(z)&={_{2}F_{1}}\left({a,b\atop c};z\right)\\ w_{5}(z)&=e^{a\pi\mathrm{i}}z^{-a}{_{2}F_{1}}\left({a,a-c+1\atop a-b+1};\frac{1}{z}\right)\\ w_{6}(z)&=e^{b\pi\mathrm{i}}z^{-b}{_{2}F_{1}}\left({b,b-c+1\atop b-a+1};\frac{1}{z}\right) \end{align} Choosing $a=1,b=1-\alpha_2,c=1+\alpha_1$ and $z=-\beta_1/\beta_2$, we have \begin{align} w_{1}(z)&={_{2}F_{1}}\left({1,1-\alpha_2\atop 1+\alpha_1};-\frac{\beta_1}{\beta_2}\right)\\ w_{5}(z)&=\frac{\beta_2}{\beta_1}{_{2}F_{1}}\left({1,1-\alpha_1\atop 1+\alpha_2};-\frac{\beta_2}{\beta_1}\right)\\ w_{6}(z)&=e^{(1-\alpha_2)\pi\mathrm{i}}\left( -\frac{\beta_1}{\beta_2} \right)^{-(1-\alpha_2)}{_{2}F_{1}}\left({1-\alpha_2,1-\alpha_1-\alpha_2\atop 1-\alpha_2};-\frac{\beta_2}{\beta_1}\right) \end{align} The last expression can be simplified as (see here) \begin{equation} w_{6}(z)=\left( \frac{\beta_1}{\beta_2} \right)^{-(1-\alpha_2)}(1+\frac{\beta_2}{\beta_1})^{\alpha_1+\alpha_2-1} \end{equation} The connection formula reads thus \begin{align} {_{2}F_{1}}\left({1,1-\alpha_2\atop 1+\alpha_1};-\frac{\beta_1}{\beta_2}\right)=& \frac{\Gamma(1+\alpha_1)\Gamma(-\alpha_2)}{\Gamma(1-\alpha_2)\Gamma(\alpha_1)}\frac{\beta_2}{\beta_1}{_{2}F_{1}}\left({1,1-\alpha_1\atop 1+\alpha_2};-\frac{\beta_2}{\beta_1}\right)\\ &+\frac{\Gamma(1+\alpha_1)\Gamma(\alpha_2)}{\Gamma(1)\Gamma(\alpha_1+\alpha_2)}\left( \frac{\beta_1}{\beta_2} \right)^{-(1-\alpha_2)}(1+\frac{\beta_2}{\beta_1})^{\alpha_1+\alpha_2-1} \end{align} which after some simplifications leads to \begin{align} {_{2}F_{1}}\left({1,1-\alpha_2\atop 1+\alpha_1};-\frac{\beta_1}{\beta_2}\right)=& -\frac{\alpha_1}{\alpha_2}\frac{\beta_2}{\beta_1}{_{2}F_{1}}\left({1,1-\alpha_1\atop 1+\alpha_2};-\frac{\beta_2}{\beta_1}\right)\\ &+\alpha_1 B(\alpha_1,\alpha_2)\left( \beta_1+\beta_2 \right)^{\alpha_1+\alpha_2-1}\beta_1^{-\alpha_1}\beta_2^{1-\alpha_2} \end{align} or to \begin{equation} \frac{1}{\alpha_1\beta_2}{_{2}F_{1}}\left({1,1-\alpha_2\atop 1+\alpha_1};-\frac{\beta_1}{\beta_2}\right)+ \frac{1}{\alpha_2\beta_1}{_{2}F_{1}}\left({1,1-\alpha_1\atop 1+\alpha_2};-\frac{\beta_2}{\beta_1}\right) = B(\alpha_1,\alpha_2)\left( \beta_1+\beta_2 \right)^{\alpha_1+\alpha_2-1}\beta_1^{-\alpha_1}\beta_2^{-\alpha_2} \end{equation} as expected.

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    $\begingroup$ Thanks! Neat, clean, and instructive :) Is there some way to give you extra credit? I've basically ignored Kummer's solution set (20); which I now see was obviously foolish. $\endgroup$
    – rrogers
    Jun 4, 2018 at 18:42
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    $\begingroup$ You're welcome! That was a nice exercise for me... $\endgroup$
    – Paul Enta
    Jun 4, 2018 at 18:48

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