# 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. Then,

$$\begin{eqnarray*} F_{X}(z)=P(X\le z) & = & P(X\le z,X

Can anyone help from here?

• This isn't true. Does the question ask you to assume $X$ and $Y$ are independent? – whuber Apr 15 '19 at 20:12
• @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. – StatCurious Apr 15 '19 at 21:08
• 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. – whuber Apr 15 '19 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 $$\mathbb{P}(X 0$$.

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

• 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! – StatCurious Apr 16 '19 at 0:55

Under the assumption that $$X$$ and $$Y$$ are independent and continuous, \begin{align*}\Bbb P(X\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, $$\int_{\Bbb R} F_Y(y) \,\text{d}F_Y(y)=\int_{\Bbb R} \Bbb P(Y' when $$Y'\sim F_Y(\cdot)$$, or $$\int_{\Bbb R} F_Y(y) \,\text{d}F_Y(y)=\Bbb P(Y' when $$Y,Y'\stackrel{\text{iid}}{\sim} F_Y(\cdot)$$, implying $$\int_{\Bbb R} F_Y(y) \,\text{d}F_Y(y)=1/2$$

• 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. – whuber Apr 15 '19 at 20:42
• Sorry for any confusion. I cannot assume independence. I've updated the question to make it clear. – StatCurious Apr 15 '19 at 21:12
• @whuber: right and right. I assumed continuity in addition to independence. – Xi'an Apr 16 '19 at 4:28