# Distribution of $\frac{X}{X+Y}$ if $X,Y$ are independent Beta random variables

Let $$X \sim \text{Beta}(a,b)$$ and $$Y \sim \text{Beta}(b,a)$$ be independent random variables.

What is the distribution of $$\frac{X}{X+Y}$$? Could it be Beta itself?

• It is not a Beta distribution: experiment with $a_1=3,a_2=1$ and you will see something unlike any Beta distribution May 18, 2021 at 12:01
• It has a unique mode in $(0,1)$ and non-zero density at 0? May 18, 2021 at 12:23
• It can have as many as three modes. Try, e.g., $a=b=1/2.$ It can have zero density at $0$ and $1:$ try $a\ge 2$ and $b\ge 2.$ There's no "nice" closed formula for the density function.
– whuber
May 18, 2021 at 13:51
• I meant that experimenting showed a density shape which is clearly not a Beta distribution. @whuber 's example is even more obviously not a Beta distribution May 18, 2021 at 14:13
• Special case: math.stackexchange.com/questions/541322/… May 18, 2021 at 17:46

For the variable $$T = \frac{X_1}{X_1+X_2}$$ you get
$$f(t) = \begin{cases} t^{\alpha_1-1}(1-t)^{\alpha_1+1}\cdot B(\alpha_1+\alpha_2,\beta_2)\cdot {_2F_1}(\alpha_1+\alpha_2,1-\beta_1;\alpha_1+\alpha_2+\beta_2;\frac{t}{1-t})/A & \text{for 0 \leq t < 1/2} \\ t^{-(\alpha_2+1)}(1-t)^{\alpha_2-1}\cdot B(\alpha_1+\alpha_2,\beta_1)\cdot {_2F_1}(\alpha_1+\alpha_2,1-\beta_2;\alpha_1+\alpha_2+\beta_1;\frac{1-t}{t})/A & \text{for 1/2 \leq t \leq 1 } \end{cases}$$
with $$A = B(\alpha_1,\beta_1)\cdot B(\alpha_2,\beta_2)$$ and $${_2F_1}$$ a hypergeometric function.