PDF or CDF of $B^2-B$ for Beta-distributed $B$

Question

I'm looking for a way to compute the CDF (or PDF) of the following random variable:

Let $B\sim Beta(\alpha, \alpha)$ for some $\alpha \geq 1$. Then the random variable in question is $B^2-B$.

Trying to compute the CDF directly from the PDF of the beta distribution results in some nasty integration which I'm not sure how to complete. Ideas would be appreciated!

Answer 1

Let's set

$X\sim Beta(a;a)$ and $Y=X^2-X$

Y is defined on the support $y \in(-0.25;0]$

The transformation function is not monotonic but we can use the Fundamental transformation theorem getting Y density inverting the function as a piecewise one getting

$$f_Y(y)=\frac{\Gamma(2\alpha)}{\Gamma(\alpha)\Gamma(\alpha)}\Bigg[\Bigg(\frac{1}{2}+\sqrt{y+\frac{1}{4}}\Bigg)^{\alpha-1}\Bigg(\frac{1}{2}-\sqrt{y+\frac{1}{4}}\Bigg)^{\alpha-1}+\Bigg(\frac{1}{2}-\sqrt{y+\frac{1}{4}}\Bigg)^{\alpha-1}\Bigg(\frac{1}{2}+\sqrt{y+\frac{1}{4}}\Bigg)^{\alpha-1}\Bigg]\frac{1}{2\sqrt{y+\frac{1}{4}}}$$

$$ \bbox[5px,border:2px solid black] { f_Y(y)=\frac{\Gamma(2\alpha)}{\Gamma(\alpha)\Gamma(\alpha)}\frac{\Bigg(\frac{1}{2}+\sqrt{y+\frac{1}{4}}\Bigg)^{\alpha-1}\Bigg(\frac{1}{2}-\sqrt{y+\frac{1}{4}}\Bigg)^{\alpha-1}}{\sqrt{y+\frac{1}{4}}}\cdot\mathbb{1}_{(-0.25;0]}(y) \ } $$

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Examples:

• Fixed $\alpha=1$ you get

$$f_Y(y)=\frac{1}{\sqrt{y+\frac{1}{4}}}\cdot\mathbb{1}_{(-0.25;0]}(y)$$

• Fixed $\alpha=2$ you get

$$f_Y(y)=\frac{-6y}{\sqrt{y+\frac{1}{4}}}\cdot\mathbb{1}_{(-0.25;0]}(y)$$

You can easy check that the above expressions are nice densities