Solve for probability with a transformed Beta distribution

Question

$X$ has a mean of $2$ and a variance of $4$. $X=a+Zb$ where $Z$ has a Beta distribution with $\alpha=\beta=2$.
Solve for $P(X>1)$ and $x$ such that $P(X>x)=0.25$.

I started out by calculating $a$ and $b$, which (I get) are $2-2\sqrt{5}$ and $4\sqrt{5}$. How shall I go from there?

Answer 1

$$\Pr[X > 1] = \Pr[a + b Z > 1] = \Pr\left[1 \ge Z > \frac{1 - a}{b} \right],$$ if $b > 0$. Then perform the integration.

For the quantile, we require $$0.25 = \Pr[X > x] = \Pr\left[1 \ge Z > \frac{x-a}{b}\right].$$ Unfortunately, this requires the solution to a cubic polynomial, which to express in radicals is cumbersome, but a numeric solution is $$\frac{x-a}{b} \approx 0.67364817766693034885.$$ But there are two solutions to $(a,b)$, both valid, so there will be two solutions to $x$ depending on which transformation is used.

Answer 2

You can rewrite $P(X > x)$ as $P(Z > (x-a)/b)$. The integral $P(Z>z) = \int_z^1 c t(1-t) \, dt$ is relatively easy to compute.

Answer 3

Starting from @heropup's answer, the cubic polynomial to be solved is $$8 x^3-12 x^2+3=0$$ Using the trigonometric method when the three roots are real, its solutions are given by $$x_k=\frac{1}{2}+\cos \left(\frac{2 \pi }{9}-\frac{2 k \pi }{3}\right)\qquad \text{with} \qquad k=0,1,2$$ and the solution looked for is $$x_1=\frac{1}{2}+\sin \left(\frac{\pi }{18}\right)$$ The problem is that the trigonometric functions of $\frac{\pi }{18}$ are given by infinite nested radicals (have a look here).