If $F_X(z) > F_Y (z)$ for all $z\in \mathbb{R}$ then $P(X < Y ) > 0$? I came across this question in a review of an old exam I took.  I didn't get the answer correctly then, and I'm struggling to figure the answer out now.  Can anyone help me reason through this?

Prove or Disprove that if $F_X(z) > F_Y (z)$ for all $z\in \mathbb{R}$ then $P(X < Y ) > 0$.  We may not assume independence.

Here is what I attempted:
I figured I might be able to approach this by proving this through contradiction.  I started by assuming $P(X<Y)=0$.  Then, 
\begin{eqnarray*}
F_{X}(z)=P(X\le z) & = & P(X\le z,X<Y)+P(X\le z,X\ge Y)\\
 & = & 0+P(X\le z,X\ge Y)
\end{eqnarray*}
Can anyone help from here?
 A: Firstly, it is worth noting that the antecedent condition in your conjecture is a slightly stronger version of the condition for strict first-order stochastic dominance (FSD) $X \ll Y$, so it implies this stochastic dominance relationship.  This condition is much stronger than what you actually need to get the result in the conjecture, so I will give you a proof for a stronger result (same implication but with a weaker antecedent condition).  Your chosen method of proof is a good one, and you are almost there - just one more step to go!


Theorem: If $F_X(z) > F_Y(z)$ for some $z \in \mathbb{R}$ then $\mathbb{P}(X<Y) > 0$.



Proof: We will proceed using a proof-by-contradiction.  Contrary to the result in the theorem, suppose that $\mathbb{P}(X<Y)=0$.  Then for all $z \in \mathbb{R}$ you have:
  $$\begin{equation} \begin{aligned}
F_X(z) = \mathbb{P}(X \leqslant z) 
&= \mathbb{P}(X \leqslant z, X < Y) + \mathbb{P}(Y \leqslant X \leqslant z) \\[6pt]
&= 0 + \mathbb{P}(Y \leqslant X \leqslant z) \\[6pt]
&= \mathbb{P}(Y \leqslant X \leqslant z) \\[6pt]
&\leqslant \mathbb{P}(Y \leqslant z) = F_Y(z), \\[6pt]
\end{aligned} \end{equation}$$
  which contradicts the antecedent condition for the theorem.  This establishes the theorem by contradiction.  $\blacksquare$

A: Under the assumption that $X$ and $Y$ are independent and continuous,
\begin{align*}\Bbb P(X<Y)&=\Bbb E^Y[\Bbb I_{X<Y}\mid Y]\\ &=\Bbb E^Y[F_X(Y)]\\&>\Bbb E^Y[F_Y(Y)]\\ &=\int_{\Bbb R} F_Y(y) \, \text{d}F_Y(y) \\&= \frac{1}{2} \int_{\Bbb R} \, \text{d}F_Y^2(y)\\&=\frac{1}{2}F_Y^2(\infty)-\frac{1}{2}F_Y^2(-\infty)\\&=1/2\end{align*}
Further,
$$\int_{\Bbb R} F_Y(y) \,\text{d}F_Y(y)=\int_{\Bbb R} \Bbb P(Y'<y) \,\text{d}F_Y(y)$$
when $Y'\sim F_Y(\cdot)$, or
$$\int_{\Bbb R} F_Y(y) \,\text{d}F_Y(y)=\Bbb P(Y'<Y)$$
when $Y,Y'\stackrel{\text{iid}}{\sim} F_Y(\cdot)$, implying
$$\int_{\Bbb R} F_Y(y) \,\text{d}F_Y(y)=1/2$$
