Let $X, Y, Z$ - independent random events which have exponential distribution with parameter $1$.
Prove that:
a) $\frac{X}{Y}, X+Y$ are independent
b) $\frac{X}{X+Y}, \frac{X +Y}{X + Y +Z},X + Y +Z$ are independent
May I ask you for help how to deal with it? I have absolutely no idea how to approach this because calculations of this are very long and I think that is a better way...
One way to show this (perhaps not the most slick way) is to compute the joint distribution (CDF) and see that it factors as the product of the individual distributions. That is show that for $Z = X/Y$, $W = X+Y$, and for any $z,w \in \mathbb{R}$, $$\mathbb{P}(Z\le z, W\le w) = \mathbb{P}(Z\le z)\mathbb{P}(W\le w).$$ Let me just see if I can get you started to see who this might go. \begin{align} \mathbb{P}(Z\le z, W\le w) &= \mathbb{P}\left(\frac{X}{Y}\le z, X + Y\le w\right)\\ &= \int_0^\infty \mathbb{P}\left(\frac{X}{y}\le z, X + y\le w|Y=y\right)p_Y(y)dy\\ &= \int_0^\infty \mathbb{P}\left(X\le zy, X\le w-y\right)p_Y(y)dy\\ \end{align} The second equal sign is the Law of Total Probability, the second is because of the independence of $X$ and $Y$. The probability in the integral is just the integral of the pdf of x from 0 (since it's exponential) to $zy \wedge (w-y)$. Hopefully this gets us started in the right direction. The distribution of $W=X+Y$ will be the convolution of $p_Y$ and $p_X$. So, we will look for that as a factor.