If $F_X(z) > F_Y (z)$ for all $z\in \mathbb{R}$ then $P(X < Y ) > 0$?

Question

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?

Answer 1

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!

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

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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. $\blacksquare$

Answer 2

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 $$\int_{\Bbb R} F_Y(y) \,\text{d}F_Y(y)=1/2$$