0
$\begingroup$

Say we have a random variable $X$ with distribution $beta(\alpha,\beta)$. How can we derive the pdf of $U=X/(1-X)$?

Using $f_U(u) = f_X(x)|dx/du|$, we get $$f_U(u) = \frac{x^{\alpha-1}(1-x)^{\beta-1}}{B(\alpha,\beta)} \cdot \frac{1}{(u+1)^2}$$ $$= \frac{u^{\alpha-1}}{(u+1)^{\alpha+\beta}} \cdot \frac{1}{B(\alpha,\beta)}$$

Do you agree with this and if so, is there a way to simplify it further? Thanks.

$\endgroup$

1 Answer 1

Reset to default
0
$\begingroup$

That is it (for $u > 0$, of course; $f_U(u) = 0$ for $u \le 0$). No, there is no way to simplify it further.

$\endgroup$
2
  • $\begingroup$ Thanks! Also, how did you find the range of u values to be $u>0$? $\endgroup$
    – polly9900
    Oct 30, 2017 at 8:37
  • $\begingroup$ If $X$ has a beta distribution, $0 < X < 1$ with probability $1$. $\endgroup$
    – Robert Israel
    Oct 30, 2017 at 19:43

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .