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?

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    $\begingroup$ This isn't true. Does the question ask you to assume $X$ and $Y$ are independent? $\endgroup$ – whuber Apr 15 at 20:12
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    $\begingroup$ @whuber, sorry, I wasn't clear about this earlier. The question asks to prove OR disprove that this is true. We may NOT assume independence. $\endgroup$ – StatCurious Apr 15 at 21:08
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    $\begingroup$ I found counterexamples by considering discrete variables supported on $\{0,1,2\}$ and constructing a table with given marginals (nearly satisfying your requirements, for of course the two distribution functions must agree on $(-\infty,0)$ and $(,\infty)$) and distributing the probability within the tables to make $\Pr(X\lt Y)$ very small. This provides the insight; such near-counterexamples are readily modified into genuine counterexamples. $\endgroup$ – whuber Apr 15 at 21:53

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. $\blacksquare$

  • $\begingroup$ Ah, yes of course! I don't know how I didn't see those last few steps earlier! Thank you, @Ben. This is very helpful. I appreciate the FSD info too! $\endgroup$ – StatCurious Apr 16 at 0:55

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

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    $\begingroup$ You could have stopped at the first line by noting that $F_Y(y) \ge 0$ for all $y\in \mathbb R$ implies $\mathbb{E}^Y[F_Y(Y)] \gt 0.$ I don't fully believe the rest of the calculation because I can find discrete variables $Y$ where this expectation does not equal $1/2.$ A uniform distribution on $\{0,1,2\}$ will work as a counterexample. $\endgroup$ – whuber Apr 15 at 20:42
  • $\begingroup$ Sorry for any confusion. I cannot assume independence. I've updated the question to make it clear. $\endgroup$ – StatCurious Apr 15 at 21:12
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    $\begingroup$ @whuber: right and right. I assumed continuity in addition to independence. $\endgroup$ – Xi'an Apr 16 at 4:28

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