Inequality of independent random variables.

Question

In the following $P$, $E$ and $F$ respectively denote the probability, expectation operator and CDF.

Let $X$, $Y$ and $Z$ be three independent random variables. Let $x$, $y$ and $z$ be the instances of these random variables. Then is the following statement true?

$$ P(x>y,x>z) = P(x>y)P(x>z) \qquad \qquad\qquad\qquad (1) $$

I understand that we can resolve it into the following form:

$$ P(x>y,x>z) = P(x>y|y>z)P(y>z) + P(x>z|z>y)P(z>y) \\ =P(x>y,y>z) + P(x>z,z>y) \\ = P(x>y>z) + P(x>z>y)\\ = E_{x,z}[F_y(x) - F_y(z)] + E_{x,y}[F_z(x) - F_z(y)] $$

And maybe the last term is solvable. However, is there an easier way to solve the LHS of (1)? Particularly, is (1) true?

Answer 1

No. The events $x>y$ and $x>z$ are not, in general, independent.

An explicit example: Suppose $X=\{0,10\}$, $Y=\{2,3\}$, $Z=\{4,5\}$. Say that in each case the probability of each outcome is $\frac 12$.

Then $P(x>y)=\frac 12=P(x>z)$ but $P(x>y,x>z)$ is also $\frac 12$.