To prove the independency of two random variables

Question

Suppose two random variables $X_1$ and $X_2$ are of identical independent distribution, with the same PDF $f(x) = e^{-x}, \space x>0$. Now, we have $$Y_1=\min(X_1, X_2)$$ $$Y_2=\max(X_1, X_2)$$ $$Y_3=Y_2 - Y_1$$ The problem is to determine if $Y_1$ and $Y_3$ are independent, and prove why. Unfortunately, I have no idea how to prove it. I only have the notion of $f(X_1X_2)=f(X_1)f(X_2)$ for $X_1$ and $X_2$ independent. Please help me.

Answer 1

The simplest way to show independence is to compute the joint complementary cumulative distribution function and that it factors: $$\begin{eqnarray} \Pr(Y_1 > y_1, Y_3 > y_3) &=& \Pr(\min(X_1,X_2) > y_1, \max(X_1,X_2) > y_3 + \min(X_1,X_2)) \\ &=& \Pr(X_1>y_1, X_2 > y_3+X_1 , X_2>X_1) + \\ && \Pr(X_2>y_1, X_1 > y_3+X_2, X_1\geqslant X_2) \end{eqnarray} $$ where disjointness of events $\{X_2 > X_1\}$ and $\{X_1 \geqslant X_2\}$ was used in the last step. Assuming $y_3 > 0$ and $y_1>0$, $$\begin{eqnarray} \Pr(X_1>y_1, X_2 > y_3+X_1, X_2>X_1) &=& \Pr(X_1>y_1, X_2 > y_3+X_1) \\ &=& \int_{y_1}^\infty \mathrm{d}x_1 \exp(-x_1) \int_{y_3+x_1}^\infty \exp(-x_2) \mathrm{d}x_2 \\ &=& \int_{y_1}^\infty \mathrm{d}x_1 \exp(-2 x_1 - y_3) \\ &=& \frac{1}{2} \exp(-y_3-2y_1) \end{eqnarray} $$ By symmetry $\Pr(X_2>y_1, X_1 > y_3+X_2, X_1\geqslant X_2)$ has the same value. Hence $$ \Pr(Y_1 > y_1, Y_3 > y_3) = \exp(-y_3) \cdot \exp(-2y_1) $$ and the independence follows.