# Does $\Pr(\{X\leq x\})\geq\Pr(\{Y\leq x\})$ imply $\Pr(\{X\leq Y\})=1$?

Suppose that $(\Omega,\mathcal{F},P)$ is a probability space and $X,Y:\Omega\to\mathbb{R}$ are random variables satisfying $$P(\{X\leq x\})\geq P(\{Y\leq x\}),\quad\forall x\in\mathbb{R}.$$ (That is, $Y$ stochastically dominates $X$.) Does it follow that $$P(\{X\leq Y\})=1?$$ On a more general note, can you recommend a source that I can learn from to improve my ability to answer this sort of (simple-looking) probability theory questions?

Attempt: I tried playing around with the triangle inequality but got nothing.

$P(X\leq Y)=\int_{y}P(X\leq y)dF_Y(y)\geq \int_yP(Y\leq y)dF_Y(y)=\int_{-\infty}^\infty F_Y(y)dF(y)=\dfrac{1}{2}$. So the answer is NO, in general. You can only guarantee that $P(X\leq Y)\geq\dfrac{1}{2}$.

For example, consider $X\in N(0,1)$ and $Y\in N(1,1)$ then check that $P(X\leq x)\geq P(Y\leq x)$ for all $x\in\mathbb R$.

However, also check that $P(X\leq Y)=\int_{-\infty}^\infty\Phi(y)f_Y(y)dy\ne1$.

For reference, I can't suggest any book containing such arguments, because there isn't any. You won't find everything in books. With practice comes intuition and the ability to think and develop arguments. For practice, Hoel Port Stone or Feller are good resources.

• The computation showing that $P(X\leqslant Y)\geqslant\frac12$ assumes crucially that $X$ and $Y$ are independent. Hence one cannot "guarantee that $P(X\leqslant Y)\geqslant\frac12$". Actually, simple examples show that, for every positive $\varepsilon$, there exists random variables $X$ and $Y$ such that $Y$ dominates stochastically $X$ but $P(X\leqslant Y)\leqslant\varepsilon$. – Did Jul 17 '15 at 17:06
• Re references, textbooks explaining the coupling method are bound to deal with stochastic domination and, unsurprisingly Thorisson's and Lindvall's books do just that. – Did Jul 17 '15 at 17:08
• @Did Thank you for the references. If time permits, can you please elaborate on the example concerning $\varepsilon$? – Kim Jong Un Jul 17 '15 at 19:49
• @KimJongUn Hint: This works with $X$ uniform on $(0,1)$ and $Y$ uniform on $(\varepsilon,1+\varepsilon)$. – Did Jul 17 '15 at 20:55

Consider two normal distributions with the same variance and different means.

No, this is in general not correct. Just consider any two random variables $X$ and $Y$ such that $X \neq Y$, but $X = Y$ in distribution. Then $Y$ stochastically dominates $X$ (and vica versa), but we cannot expect $\mathbb{P}(X\leq Y)=1$.

(E.g. $X=1_{(0,1/2)}$ and $Y=1_{(1/2,1)}$ on $(0,1)$ endowed with the Lebesgue measure.)