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I have this variable with beta distribution : $Y \sim \mathcal{B}e(\alpha,\frac{1}{3})$.

I have to find the value of $\alpha$ such as : $P(Y \leq 0.416) =0.2 $

Formally for $\alpha \geq 0$ , $\beta \geq 0$ and $0 \leq y \leq 1$ the CDF function of $Y$ at 0.416 is:

$P(Y \leq 0.416) = \frac{\Gamma(\alpha + \beta)}{\Gamma(\alpha)\Gamma(\beta)} \int_0^{0.416} t^{\alpha-1} (1-t)^{\beta-1} dt=0.2$

I am not sure how to proceed. Thanks for the help in advance!!

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Maple gives the CDF for $\mathcal{B}e(\alpha,\beta)$ as $$ {\frac {\Gamma \left( \alpha+\beta \right) {y}^{\alpha} {\mbox{$_2$F$_1$}(\alpha,1-\beta;\,1+\alpha;\,y)}}{\Gamma \left( \alpha \right) \Gamma \left( \beta \right) \alpha}} $$ In any case, you need to use numerical methods to solve $F(0.416) = 0.2$. Maple says $\alpha = 0.8563203833$.

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  • $\begingroup$ The question is stated as though an analytical solution is expected...but I agree that numerical methods are the way to proceed here. $\endgroup$ – Math1000 Dec 13 '18 at 15:13
  • $\begingroup$ thank you very much for the kind answer. But i don't understand the meaning of this quantity: ${\mbox{$_2$F$_1$}(\alpha,1-\beta;\,1+\alpha;\,y)}$. And how did you obtain the value of $\alpha$? Thank you in advance @Robert Israel $\endgroup$ – andrew Dec 13 '18 at 15:25
  • $\begingroup$ Can i get a solution with the software R? $\endgroup$ – andrew Dec 13 '18 at 16:40
  • $\begingroup$ ${}_2F_1(a,b; c; z)$ is a Gaussian hypergeometric function. The value of $\alpha$ was obtained using a numerical solver (fsolve in Maple). $\endgroup$ – Robert Israel Dec 13 '18 at 18:55

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