How to calculate the probability density for $W=\frac{Y}{\max(X,Y)}$?

Question

Assume that $X,Y$ are two independent exponential random variables with mean $\lambda$. How can I calculate the probability density function of $W=\frac{Y}{\max(X,Y)}$?

I know how to calculate the probability density function or $\max(X,Y)$ of $\min(X,Y)$, which has been asked before here many times. But when I wanted to use the same way for this, I got confuses at the beginning.

Is it a good way to condition on $X\geq Y$?

Another question is how to compute the probability density function of $Z=\frac{X}{\min(X,2Y)}$?

Answer 1

Hints: $W=1$ if $X \leq Y$ and $W=\frac Y X$ if $X>Y$. To find $P(W\leq w)$ split this into $P(W\leq w, X \leq Y)$ and $P(W\leq w, X>Y)$.

Similarly, $Z=1$ if $X \leq 2Y$ and $Z=\frac X{2Y}$ if $X>2Y$.

Answer 2

@b.j: following the good hints by @Kavi Rama Murthy you will easily realize that the density you are searching for is not absolutely continuous...(just FYK here are my results)

$f_W(w)=\frac{1}{(w+1)^2}\mathbb{1}_{[0;1)}(w)+\frac12\mathbb{1}_{[1]}(w)$

$F_W(w)=[1-\frac{1}{w+1}]\mathbb{1}_{[0;1)}(w)+\mathbb{1}_{[1;\infty)}(w)$

as you can see, the pdf has a positive probability mass in $w=1$ as the corresponding CDF has a "jump" in $w=1$

If the procedure is not clear, please post you efforts and we will take you hand in hand to the solution

bye bye