Relation between positive correlation and $p(Y_2 > Y_1 \mid X_2 > X_1) > \frac{1}{2}$? Looking at another question regarding "intuition" on the sign of the correlation, I was thinking to say positive correlation $\rho(X, Y) > 0$ roughly means if $X$ increases, then $Y$ is more likely than not to increase also.  But then I realized the latter could be made precise using a conditional probability: suppose $X$ and $Y$ are random variables on a probability space $A$, and for $i \in \{ 1, 2 \}$ we let $X_i = X \circ \pi_i$, $Y_i = Y \circ \pi_i$ on the product probability space $A \times A$.  Then we want to know whether $p(Y_2 > Y_1 \mid X_2 > X_1) > \frac{1}{2}$.  And I'm not sure if there might be situations where the correlation is positive, but the conditional probability is strictly less than $\frac{1}{2}$.
So, the question is: is there any implication one way or the other between these two statements?  Or, if not, what about the similar idea $E(Y_2 - Y_1 \mid X_2 > X_1) > 0$?

Consider the case where $A = \{ 1, 2, 3, 4 \}$ with equidistributed probability, and the values of $(X, Y)$ are $(-3.2, -3)$, $(1, 11)$, $(1.1, -4)$, $(1.1, -4)$.  Then $\bar X = \bar Y = 0$ so the covariance is equal to $\frac{1}{4} \sum_{i=1}^4 X(i) Y(i) = 2.95 > 0$, which implies the correlation is positive.  However, the combinations of $(i, j)$ such that $X(i) < X(j)$ are $(1, 2)$, $(1, 3)$, $(1, 4)$, $(2, 3)$, $(2, 4)$.  Out of these, the only combination where $Y(i) < Y(j)$ is $(1, 2)$, so $p(Y_2 > Y_1 \mid X_2 > X_1) = \frac{1}{5}$.
Similarly, $E(Y_2 - Y_1 \mid X_2 > X_1) = \frac{1}{5} (14 + (-1) + (-1) + (-15) + (-15)) = -3.6 < 0$.
Therefore, there is no implication in general between positive correlation and either the conditional probability statement or the conditional expectation statement.
