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I'm working on following problem:

$X,Y \sim Exp(\lambda)$ and are independent. Define $U=\min(X,Y)$ and $V=|X-Y|.$ Prove that $U,V$ are independent as well.

I managed to find that $U\sim Exp(2\lambda)$ and $V~\sim Exp(\lambda)$. Now I want to use random vector $Z=(U,V)$ and prove that $F_{Z}(s,t)=F_{U}(s)F_{V}(t)$ but I struggle to define integral since I do not have explicit joint distribution function.

I'm not looking for a solution but a hint to move from where I am now.

borg
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1 Answers1

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HINT: Your strategy to use distribution functions is good. You actually do have enough to set up integrals there -- you just need to relate those integrals back to the original variables $X, Y$. For example,

$$F_V(t) = \mathbb P(|X-Y| \leq t) = \iint_D f_X(x) f_Y(y) \, \textrm d y \, \textrm d x$$

where $f_X$, $f_Y$ are the densities of $X, Y$ respectively, and $D$ is the region on the plane that satisfies $|x-y|<t$ (and $x > 0, y > 0$).

Aaron Montgomery
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    Thanks for the hint! – borg Jun 24 '22 at 19:33
  • So what I need to do is to evaluate $$\iint_{D_{1} \cup D_{2}} f_X(x) f_Y(y) , \textrm d y , \textrm d x$$ where $D_{1} = { (x,y) \in \mathbb{R}^2: 0 \leq x-y \leq t\ \wedge 0 \leq y \leq s}$ and $D_{2} = { (x,y) \in \mathbb{R}^2: 0 \leq y-x \leq t\ \wedge 0 \leq x \leq s}$ - right? – borg Jun 24 '22 at 19:47
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    That term is for $F_Z(s, t)$, right? I think you've got it correct. – Aaron Montgomery Jun 24 '22 at 19:59