# 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 $E(Y_2 - Y_1 \mid X_2 > X_1) > 0$?

• OK, it was easy enough to come up with a counterexample for the probability case: if $A$ is a discrete equidistributed space of size 4 and $(X, Y)$ takes values $(-3, -3)$, $(1, 11)$, $(1, -4)$, $(1, -4)$, then $\bar X = \bar Y = 0$, so the covariance is $\frac{1}{4} \sum_{i=1}^4 X_i Y_i = 3$ which implies the correlation is positive, but $p(Y_2 > Y_1 | X_2 > X_1) = \frac{1}{3}$. On the other hand, $E(Y_2 - Y_1 | X_2 > X_1) = 4 > 0$ so that part of the question remains open. Commented Jul 27, 2017 at 17:44
• The notation $p(Y_2 > Y_1 \mid X_2 > X_1)$ suggests that four different random variables are involved, whereas correlation is between only two random variables. This leaves the precise meaning of the question unclear. Commented Jul 27, 2017 at 17:52
• @MichaelHardy That was what I was trying to specify with the definitions of $X_i = X \circ \pi_i$, $Y_i = Y \circ \pi_i$ on $A \otimes A$. I essentially want $(X_1, Y_1)$ and $(X_2, Y_2)$ to represent two independent samples of the joint distribution of $(X, Y)$. Commented Jul 27, 2017 at 17:55
• Oh, and if I perturb the previous example a bit to $(-3 - 2 \epsilon, -3)$, $(1, 11)$, $(1 + \epsilon, -4)$, $(1 + \epsilon, -4)$ with $\epsilon > 0$ small enough that the covariance remains positive, then the values of $Y_2 - Y_1$ where $X_2 > X_1$ are $14, -1, -1, -15, -15$ which gives $E(Y_2 - Y_1 \mid X_2 > X_1) = -\frac{18}{5} < 0$... Commented Jul 27, 2017 at 18:03
• I wonder if the moral of the example should be that the covariance is related to a weighted average of $Y_2 - Y_1$ weighted according to the difference $X_2 - X_1$ given $X_2 > X_1$. Commented Jul 27, 2017 at 18:08

Summary: Here is a case in which $$\operatorname{corr}(X,Y) \approx 0.99999972 \text{ and } \Pr(Y_1<Y_2 \mid X_1<X_2) = \dfrac{57}{253} < \dfrac 1 2.$$

Suppose $(X_1,Y_1), (X_2,Y_2),(X_3,Y_3),\ldots$ are independent and all belong to the same bivariate distribution and $\operatorname{corr}(X_1,Y_1)>0.$

Can we conclude that $\Pr(Y_2 > Y_1 \mid X_2 > X_1) > \dfrac 1 2 \text{?}$

Suppose $$(X_1,Y_1) = \begin{cases} (10000,10000) & \text{with probability } 1/30, \\ (-10000,-10000) & \text{with probability } 1/30, \\ (1,-1) & \text{with probability } 14/30, \\ (-1,1) & \text{with probability } 14/30. \end{cases}$$ Let us find $\Pr(Y_2>Y_1\mid X_2>X_1).$ I'm getting $\operatorname{corr}(X_1,Y_1) \approx 0.99999972.$

First look at the space on which we are conditioning: $X_2>X_1.$ $$(X_1,X_2) = \begin{cases} (X_1,X_1) & \text{probability} \\[12pt] (-10000,-1) & 14/30^3, \\ (-10000,1) & 14/30^2, \\ (-10000,10000) & 1/30^2, \\ (-1,1) & 14^2/30^2, \\ (-1,10000) & 14/30^2, \\ (1,10000) & 14/30^2. \end{cases}$$ We have $14 + 14 + 1 + 14^2 + 14 + 14 = 253.$ So conditional probabilities given this event are $$(X_1,X_2) = \begin{cases} (X_1,X_1) & \text{probability} \\[12pt] (-10000,-1) & 14/253, \\ (-10000,1) & 14/253, \\ (-10000,10000) & 1/253, \\ (-1,1) & 14^2/253, \\ (-1,10000) & 14/253, \\ (1,10000) & 14/253. \end{cases}$$ In which of these cases where $X_1<X_2$ do we have $Y_1<Y_2\text{?}$ $$(X_1,X_2) = \begin{cases} (X_1,X_1) & Y_1<Y_2\text{ ?} \\[12pt] (-10000,-1) & \text{true} \\ (-10000,1) & \text{true} \\ (-10000,10000) & \text{true} \\ (-1,1) & \text{false} \\ (-1,10000) & \text{true} \\ (1,10000) & \text{true} \end{cases}$$ Thus $\Pr(Y_1<Y_2 \mid X_1<X_2) = \dfrac{57}{253} < \dfrac 1 2.$

So the order in which $X,Y$ appear, i.e. $X<Y$ or $X>Y,$ is not the only thing that matters: the absolute size of the numbers also matters.

• I like this example better than mine for the probability statement. Do you have a similar example for the conditional expectation statement? I think in this example, $Y_2 - Y_1 \ge 9999$ with conditional probability $\frac{57}{253}$ and in the other case $Y_2 - Y_1 = -2$, so $E(Y_2 - Y_1 \mid X_2 > X_1) > 0$. Commented Jul 27, 2017 at 18:58
• I think if I modify the points to $(-10000, -10000), (-1, 9999), (1, -9999), (10000, 10000)$ with the same probabilities then still the correlation is positive, but $E(Y_2 - Y_1 \mid X_2 > X_1) < 0$. Commented Jul 28, 2017 at 18:18

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.