Let \((X_1,Y_1),(X_2,Y_2),\ldots,(X_n,Y_n)\), \(n\geq2\), be a random sample from a continuous bivariate distribution with joint distribution function \(F_{X,Y}\). Further, \(F_X\) and \(F_Y\) are the marginal distribution functions of \(X\) and \(Y\), respectively. If \[ F_{X,Y}(x,y)=F_X(x)F_Y(y), \quad \forall (x,y), \]then, for any two independent pairs \((X_i,Y_i)\) and \((X_j,Y_j)\), \[ P\left[(X_i-X_j)(Y_i-Y_j)>0\right] \]equals
Show Hint
With \(X\) and \(Y\) independent, the four values \(X_i, Y_i, X_j, Y_j\) are mutually independent, so the signs of \(X_i-X_j\) and \(Y_i-Y_j\) are independent fair coin flips.
Step 1: Translate the quantity being asked into a familiar idea. The event $(X_i-X_j)(Y_i-Y_j)>0$ is what statisticians call a concordant pair: as $X$ moves from the smaller value to the larger, $Y$ moves the same way. This probability of concordance is tied to Kendall's measure of association, $\tau = 2P(\text{concordant})-1$, so finding this probability is the same as finding how strongly $X$ and $Y$ move together.
Step 2: Use the given factoring condition. We are told $F_{X,Y}(x,y)=F_X(x)F_Y(y)$ for every $(x,y)$. This is precisely the statement that $X$ and $Y$ are independent. So knowing $X$ tells us nothing about $Y$ inside a pair.
Step 3: Build four independent variables. The two pairs $(X_i,Y_i)$ and $(X_j,Y_j)$ come from independent draws of the sample. Along with the within pair independence from Step 2, all of $X_i, Y_i, X_j, Y_j$ act as four separate, mutually independent draws.
Step 4: Find the chance each difference is positive. $X_i$ and $X_j$ share the same continuous distribution, so by symmetry $X_i-X_j$ is just as likely to be positive as negative, and the chance of an exact tie is $0$. This gives $P(X_i-X_j>0)=\tfrac12$. The same reasoning for $Y_i-Y_j$ gives $P(Y_i-Y_j>0)=\tfrac12$.
Step 5: Combine using independence. Since the sign of $X_i-X_j$ does not depend on the sign of $Y_i-Y_j$, the product $(X_i-X_j)(Y_i-Y_j)$ is positive only when both differences share a sign. Adding the two matching cases, \[ P[(X_i-X_j)(Y_i-Y_j)>0] = \left(\frac12\right)\left(\frac12\right)+\left(\frac12\right)\left(\frac12\right) = \frac12. \]
Step 6: Read the answer through Kendall's tau. With independent $X$ and $Y$, Kendall's tau is $0$, and $\tau=2P(\text{concordant})-1=0$ gives $P(\text{concordant})=\tfrac12$, matching the direct computation. \[ \boxed{\dfrac{1}{2}} \]
