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Mar 2, 2017 at 18:26 comment added soakley Yes, I have added the development and changed the beta cdf to be designated as $H$ in order to avoid confusion with an $F$ distribution.
Mar 2, 2017 at 18:24 history edited soakley CC BY-SA 3.0
Added development of the result presented in the initial answer
Mar 2, 2017 at 18:20 comment added A. Webb @whuber soakley I believe this goes something like: $P[X>Y] = P[Y^* < \frac{\beta_2}{\beta_1} X^*]$ where $X^* \sim \rm{Gamma}(\alpha_1,1)$ and $Y^* \sim \rm{Gamma}(\alpha_2,1)$ are unit scaled versions of $X$ and $Y$. Using $\frac{Y^*}{Y^*+X^*} \sim \rm{Beta}(\alpha_2,\alpha_1)$ (Wikipedia), and noting that $P[\frac{Y^*}{Y^*+X^*} < c] = P[Y^* < \frac{c}{1-c}X^*]$, we set $\frac{c}{1-c} = \frac{\beta_2}{\beta_1}$ giving $c = \frac{\beta_1}{\beta_1 + \beta_2}$.
Mar 2, 2017 at 15:02 comment added whuber Sorry, I overlooked the "beta" part and assumed you were referring to the (closely related) F ratio distribution! Thank you for explaining.
Mar 2, 2017 at 14:38 comment added soakley Where do you get $F_{1,1}(1)=0.391826$? As stated, $F$ is the cdf of a beta random variable. So $F_{1,1}(1)=1.$ I see how the notation could be misleading, though. I will add some development to show the result if I can make time.
Mar 2, 2017 at 0:34 comment added whuber Could you demonstrate this result or cite an accessible reference? I'm having a hard time seeing why it should be true. For instance, let $\alpha_1=\alpha_2=\beta_2=1$ and consider what happens as $\beta_1$ grows large. The right hand side approaches $F_{1,1}(1) = 0.391826\ldots$ from below, whereas the left hand side ought to approach $1$.
Mar 1, 2017 at 23:44 history edited soakley CC BY-SA 3.0
added 12 characters in body
Mar 1, 2017 at 23:37 history answered soakley CC BY-SA 3.0