# Beta distribution: find the parameter $\alpha$ of $\mathcal{B}e(\alpha,\frac{1}{3})$

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!!

Maple gives the CDF for $$\mathcal{B}e(\alpha,\beta)$$ as $${\frac {\Gamma \left( \alpha+\beta \right) {y}^{\alpha} {\mbox{_2F_1}(\alpha,1-\beta;\,1+\alpha;\,y)}}{\Gamma \left( \alpha \right) \Gamma \left( \beta \right) \alpha}}$$ In any case, you need to use numerical methods to solve $$F(0.416) = 0.2$$. Maple says $$\alpha = 0.8563203833$$.
• 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 – andrew Dec 13 '18 at 15:25
• ${}_2F_1(a,b; c; z)$ is a Gaussian hypergeometric function. The value of $\alpha$ was obtained using a numerical solver (fsolve in Maple). – Robert Israel Dec 13 '18 at 18:55