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I was looking at a proof of the following fact

Let $X \sim \mbox{Gamma}(\alpha, 1)$ and $Y \sim \mbox{Gamma}(\beta, 1)$ where the paramaterization is such that $\alpha$ is the shape parameter. Then
$$ \frac{X}{X + Y} \sim \mbox{Beta}(\alpha, \beta). $$

found here in the second answer.

In the link I gave, when the author of the answer writes "To prove this, write the joint pdf $f_{X, Y} (x, y) = \frac{1}{\Gamma(\alpha) \Gamma(\beta)} x^{\alpha - 1} y^{\beta - 1} e^{-(x + y)}$ ", does this mean $X$ and $Y$ are to be considered independent?

Also the kind reader that may wish to give their proof of the above fact is very welcome to do so.

EDIT: when I copied pasted the linked answer I did not notice that the Lemma I would like to prove is stated a bit differently:

If $Z \sim Gamma(\alpha, \theta)$ and $W \sim(\theta, \beta)$, then $A = Z / (W+Z) \sim Beta(\alpha, \beta)$.

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    $\begingroup$ I am sure you can find a proof on the internet, this is pretty standard. $\endgroup$ – JohnK Jun 8 '15 at 22:44
  • $\begingroup$ @JohnK ok I will look for one better and link it if I find it. $\endgroup$ – Monolite Jun 8 '15 at 22:45
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    $\begingroup$ Here is one link. math.uah.edu/stat/special/Beta.html . Scroll down and open up the proof of 25. $\endgroup$ – JohnK Jun 8 '15 at 22:51
  • $\begingroup$ Yes, you will need independence $\endgroup$ – kjetil b halvorsen Dec 18 '18 at 21:26

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