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$X,Y\sim\mathscr{E}(1)$ (exp. with parameter $1$) and independent. I'd like to show that $\min\{X,Y\}$ and $|X-Y|$ are independent.

Let $Z=\min\{X,Y\}$ and $W=|X-Y|$. The transformation gives a Jacobian determinant of $1$.

$$\begin{aligned}f_{Z,W}(z,w)=f_{X,Y}(x,y) &=f_{X,Y}(z,w+z)+f_{X,Y}(w+z,z)\\&=f_X(z)f_Y(w+z)+f_Y(z)f_X(w+z)\\&=2e^{-w-2z}\\&=\underbrace{2e^{-w}}_{f_W?}\,f_Z(z)\end{aligned}$$

  • are these manipulations legitimate? I am worried about the end of the first line: I split it thinking: if $X=Z$ then $W=Y-X$ and vice versa, but I am not sure whether this is legal.
  • I've shown that $f_Z(z)=e^{-2z}$, so hopefully $f_W(w)=2e^{-w}$. Would I be correct? If so, could someone offer a hint as to how I could prove this?
anaconda
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1 Answers1

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Yes these manipulations are legitimate, and they show that $$ f_{Z,W}(z,w)=2\,\mathrm e^{-w-2z}\,\mathbf 1_{z\gt0}\,\mathbf 1_{w\gt0}. $$ But, oddly enough, starting from this correct point, you think that $f_Z(z)=\mathrm e^{-2z}\,\mathbf 1_{z\gt0}$ although the integral of this function is not even $1$. To be sure of what $f_Z$ is, note that $$ f_Z(z)=\int_\mathbb R f_{Z,W}(z,w)\,\mathrm dw, $$ which should lead you pretty quickly to $$ f_Z(z)=2\,\mathrm e^{-2z}\,\mathbf 1_{z\gt0}. $$ Likewise, the identity $$ f_W(w)=\int_\mathbb R f_{Z,W}(z,w)\,\mathrm dz $$ leads to $$ f_W(w)=\mathrm e^{-w}\,\mathbf 1_{w\gt0}. $$

Did
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