7
$\begingroup$

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?

$\endgroup$
6
  • 3
    $\begingroup$ It is not a Beta distribution: experiment with $a_1=3,a_2=1$ and you will see something unlike any Beta distribution $\endgroup$
    – Henry
    May 18, 2021 at 12:01
  • $\begingroup$ It has a unique mode in $(0,1)$ and non-zero density at 0? $\endgroup$
    – econ86
    May 18, 2021 at 12:23
  • 2
    $\begingroup$ 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. $\endgroup$
    – whuber
    May 18, 2021 at 13:51
  • 3
    $\begingroup$ 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 $\endgroup$
    – Henry
    May 18, 2021 at 14:13
  • 2
    $\begingroup$ Special case: math.stackexchange.com/questions/541322/… $\endgroup$
    – Peter O.
    May 18, 2021 at 17:46

1 Answer 1

6
$\begingroup$

This situation is described by T. Pham-Gia in 'Distributions of the ratios of independent beta variables and applications' in Communications in Statistics - Theory and Methods Volume 29, 2000 - Issue 12 https://doi.org/10.1080/03610920008832632

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.