-2
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

HINT

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?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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