Let $X,Y$ be independent random variables the are distributed $\exp{(\alpha)}$ whereby $\alpha > 0$

Determine the Distribution Density of $\frac {X}{X+Y}$

My idea:

I believe it is too simple to suggest:

$$f_{ \frac {X}{X+Y}}(x,y)=\frac{\alpha\exp{(-\alpha x)}}{\alpha\exp{(-\alpha x)}+\alpha\exp{(-\alpha y)}}=\frac{\exp{(-\alpha x)}}{\exp{(-\alpha x)}+\exp{(-\alpha y)}}$$

But I fail to see another way of finding the joint PDF.

Any ideas?



Make the change of variables $W = X$ and $Z = \displaystyle\frac{X}{X+Y}$, where $W\geq 0$ and $0\leq Z\leq 1$. Since $X$ and $Y$ are independent, we obtain the following result \begin{align*} f_{W,Z}(w,z) = f_{X,Y}\left(w,\frac{w(1 - z)}{z}\right)|\det J(w,z)| = f_{X}(w)f_{Y}\left(\frac{w(1 - z)}{z}\right)|\det J(w,z)| \end{align*}

Once you have the expression of $f_{W,Z}(w,z)$ at hand, you determine its marginal distribution related to $Z$, which is the random variable whose distribution you are interested in. Can you proceed from here?


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