Let's $\xi, \eta$ - independent random values. $\xi \thicksim F(x) = 1 - e^{-\alpha x}$ and $\eta \thicksim F(x) = 1 - e^{-\alpha x}$. How to find distribution function and distribution density of $|\xi - \eta|$?
As I know I should find integral $\int\limits_{0}^{+\infty} P(|\xi - \eta| \leq x)|_{\eta = y} f(y) dy$, but how to expand the modulo correctly?
Due to symmetry in the second equation we have $$\Bbb P (\vert \xi - \eta \vert \leq x) = \Bbb P (\eta - \xi \leq x , \xi < \eta ) +\Bbb P (\xi - \eta \leq x , \eta < \xi)\\ = 2\Bbb P (\eta - \xi \leq x , \xi < \eta ) = 2 \Bbb P (\eta \leq x + \xi, \xi < \eta )\\ = 2\int_0^\infty \Bbb P (t \leq x +\xi , \xi < t) \alpha e^{-\alpha t} \text d t= 2\int_0^\infty \Bbb P (t - x \leq \xi < t) \alpha e^{-\alpha t} \text d t\\ = 2\int_0^\infty \Bbb P (\max (t - x,0) \leq \xi < t) \alpha e^{-\alpha t} \text d t = 2\int_0^\infty( F(t) - F(\max (t-x,0))) \alpha e^{-\alpha t} \text d t\\ = 2\int_0^\infty( e^{-\alpha \max(t-x,0)} - e^{-\alpha t}) \alpha e^{-\alpha t} \text d t = 2 \int_0^x \alpha e^{-\alpha t}\text d t + 2 \int_x^\infty e^{-\alpha (t-x)} \alpha e^{-\alpha t}\text d t - \int_0^\infty 2 \alpha e^{-2\alpha t} \text d t\\ =2F(x) + e^{\alpha x} \int_x^\infty 2\alpha e^{-2\alpha t} \text d t - \int_0^\infty 2 \alpha e^{-2\alpha t} \text d t\\ = 2F(x) + e^{\alpha x } e^{-2\alpha x} -1 = 2F(x) - F(x) = F(x).$$
