Combined distribution of beta and uniform variables

Question

Given

$$X \sim \text{Beta}(\alpha,\beta)$$ (where $\alpha=\beta$, if that helps) and

$$\theta \sim \text{Uniform}(0, \pi/2).$$

I'm trying to find a formula for $P(Y)$ (or even the cdf) of

$$Y = X + (1-2X)\cos(\theta)$$

on the domain $(0,1)$.

I know from here that given $C = \cos(\theta),$ its PDF is

$$P(C) = \frac{2}{\pi\sqrt{1-C^2}}$$

and of course the pdf of a beta distribution is

$$P(X) = \frac{X^{\alpha-1}(1-X)^{\beta-1}}{\mathrm{Beta}(\alpha, \beta)}.$$

But combining them is getting beyond my skills as an engineer.

EDIT: As it seems a closed form is not possible for this, could someone please show me how to formulate the integral to calculate this sort of PDF? There may be some way I can reformulate the problem to be more solution-friendly if I could wrap my head around how compound distributions of this type are built mathematically.

Answer 1

There is no closed form for the density. Its integral form can be obtained as follows. If we condition on $X=x$ we have $Y = x + (1-2x) C$ where $C$ ranges over the unit interval. The support under this condition is:

$$\text{supp}(Y|X=x) = \begin{cases} [x,1-x] & & & \text{for } 0 \leqslant x < \tfrac{1}{2}, \\[6pt] [1-x,x] & & & \text{for } \tfrac{1}{2} < x \leqslant 1. \\[6pt] \end{cases}$$

(We can ignore the case where $x=\tfrac{1}{2}$ since this occurs with probability zero.) Over this support we have the conditional density:

$$\begin{aligned} p_{Y|X}(y|x) &= \frac{1}{|1-2x|} \cdot p_C \bigg( \frac{y-x}{1-2x} \bigg) \\[6pt] &= \frac{1}{|1-2x|} \cdot \frac{2}{\pi} \bigg( 1 - \bigg( \frac{y-x}{1-2x} \bigg)^2 \bigg)^{-1/2} \\[6pt] &= \frac{1}{|1-2x|} \cdot \frac{2}{\pi} \bigg( \frac{(1-2x)^2 - (y-x)^2}{(1-2x)^2} \bigg)^{-1/2} \\[6pt] &= \frac{1}{|1-2x|} \cdot \frac{2}{\pi} \bigg( \frac{(1-4x+4x^2) - (y^2-2xy+x^2)}{(1-2x)^2} \bigg)^{-1/2} \\[6pt] &= \frac{1}{|1-2x|} \cdot \frac{2}{\pi} \bigg( \frac{1 - 4x + 3x^2 + 2xy - y^2}{(1-2x)^2} \bigg)^{-1/2} \\[6pt] &= \frac{1}{|1-2x|} \cdot \frac{2}{\pi} \cdot \frac{|1-2x|}{\sqrt{1 - 4x + 3x^2 + 2xy - y^2}} \\[6pt] &= \frac{2}{\pi} \cdot \frac{1}{\sqrt{1 - 4x + 3x^2 + 2xy - y^2}}. \\[6pt] \end{aligned}$$

Inverting the support we have $\text{supp}(X|Y=y) = [0,\min(y,1-y)] \cap [\max(y,1-y),1]$. Thus, applying the law of total probability then gives you:

$$\begin{aligned} p_Y(y) &= \int \limits_0^1 p_{Y|X}(y|x) p_X(x) \ dx \\[6pt] &= \int \limits_0^{\min(y,1-y)} p_{Y|X}(y|x) p_X(x) \ dx + \int \limits_{\max(y,1-y)}^1 p_{Y|X}(y|x) p_X(x) \ dx \\[6pt] &= \frac{2}{\pi} \frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha) \Gamma(\beta)} \Bigg[ \quad \int \limits_0^{\min(y,1-y)} \frac{x^{\alpha-1} (1-x)^{\beta-1}}{\sqrt{1 - 4x + 3x^2 + 2xy - y^2}} \ dx \\ &\quad \quad \quad \quad \quad \quad \quad \quad + \int \limits_{\max(y,1-y)}^1 \frac{x^{\alpha-1} (1-x)^{\beta-1}}{\sqrt{1 - 4x + 3x^2 + 2xy - y^2}} \ dx \Bigg]. \\[6pt] \end{aligned}$$

There is no closed form for this integral so it must be evaluated using numerical methods.

Answer 2

First of all note that given the support of $\theta$, the function $\theta\to\cos(\theta)=:C$ is a bijection. So once you fix $Y=y,X=x$, your variable $C$ has value $\frac{y-x}{1-2x}$ and there is a unique $\theta$ giving such value of $C$.

Given $Y:=X+(1-2X)C$ you can see that the support of $Y$ given $X=x$ is $[1-x,1+x]$ (since $X>0$ and $C\in[0,1]$). Conversely, fixing $Y=y$,

$X\in\begin{cases}[1-y,1+y]&\text{if $y>0$}\\ [1+y,1-y]&\text{if $y<0$}\end{cases}$

So

$p_Y(y)=\begin{cases}\int_{1-y}^{1+y}p_X(x)p_C\left(\frac{y-x}{1-2x}\right)dx&\text{if $y>0$}\\ \int_{1+y}^{1-y}p_X(x)p_C\left(\frac{y-x}{1-2x}\right)dx&\text{if $y<0$}\end{cases}$