Independence between events involving three random variables

Question

Let $X,Y,Z$ be three independent random variables, we want to find out if the following holds: $$P(X\geq Y,X\geq Z) = P(X\geq Y)P(X\geq Z)$$ that is, if $X\geq Y$ and $X\geq Z$ are independent events. I first tried to using the law of total probability to split up the RHS but that got complicated really quickly and I don't think that's the best route. I tried the following as well: $$P(X\geq Y, X\geq Z) = P(Y\leq X, Z\leq X) = P(Y-X\leq0,Z-X\leq0)$$ Now, we know that $Y,Z$ are independent and thus for $k\in\mathbb{R}$, we have $$P(Y\leq k,Z\leq k) = P(Y\leq k)P(Z\leq k)$$ But here, $X$ is not a real number but a random variable, so it boils down to the fact of whether the random variables, $Y-X$ and $Z-X$ are independent. I started to solve this by setting $\xi = -X$ such that $f_{\xi}(c) = f_X(-c)$, by my computations. So now we just want to find out whether $$P(Y+\xi\leq k,Z+\xi\leq k) = P(Y+\xi\leq k)P(Z+\xi\leq k)$$ by definition of independence. I'm not quite sure if this is the best way to move forward, especially that it will require convolutions on the random variable's density functions. If anybody has seen something similar to this, please let me know, thanks!

Answer 1

Let $X$, $Y$ and $Z$ be identically distributed, taking the values $0$ and $1$ with probability $1/2$. The event $\{X\geqslant Y\}\cap \{X\geqslant Z\}$ is $\{X=1\}\cup\left(\{X=0\}\cap\{Y=0\}\cap\{Z=0\}\right)$ and since the union is disjoint, $$ \mathbb P\left(\{X\geqslant Y\}\cap \{X\geqslant Z\}\right)=\mathbb P\{X=0\}+ \mathbb P\left(\{X=0\}\cap\{Y=0\}\cap\{Z=0\}\right) $$ and by independent of $(X,Y,Z)$, it follows that $$ \mathbb P\left(\{X\geqslant Y\}\cap \{X\geqslant Z\}\right)=1/2+1/8=\frac 58. $$ The event $\{X\geqslant Y\}$ can be rewritten as $\{X=1\}\cup \left(\{X=0\}\cap\{Y=0\}\right)$ hence by the same reasoning as before, its probability is $1/2+1/4=3/4$. Finally, $$ \mathbb P\left(\{X\geqslant Y\}\cap \{X\geqslant Z\}\right)=\frac{10}{16}\neq\frac 9{16}= \mathbb P\left(\{X\geqslant Y\} \right)\mathbb P\left( \{X\geqslant Z\}\right). $$