Question

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

• A. \(\dfrac{1}{4}\)
• B. \(\dfrac{1}{2}\)
• C. \(\dfrac{3}{4}\)
• D. \(\dfrac{3}{8}\)

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.

Answer 1

The Correct Option is B

Step 1: Understand what the condition on \(F_{X,Y}\) means.
The condition
\[ F_{X,Y}(x,y)=F_X(x)F_Y(y) \quad \forall (x,y) \]
says the joint distribution function factors into the product of the marginals. This is exactly the definition of \(X\) and \(Y\) being independent random variables. So within each pair \((X_i,Y_i)\), the coordinate \(X_i\) is independent of the coordinate \(Y_i\).

Step 2: Note all four variables involved are mutually independent.
The pairs \((X_i,Y_i)\) and \((X_j,Y_j)\) are drawn independently of each other since they come from a random sample. Combined with Step 1, this means \(X_i\), \(Y_i\), \(X_j\), \(Y_j\) are four mutually independent random variables, each continuous.

Step 3: Study the sign of \(X_i-X_j\).
Since \(X_i\) and \(X_j\) are independent and identically distributed continuous random variables, the difference \(X_i-X_j\) is symmetric about \(0\), and \(P(X_i-X_j=0)=0\) because the distribution is continuous. So
\[ P(X_i-X_j>0)=P(X_i-X_j<0)=\frac{1}{2}. \]

Step 4: Study the sign of \(Y_i-Y_j\).
By exactly the same argument applied to \(Y_i\) and \(Y_j\),
\[ P(Y_i-Y_j>0)=P(Y_i-Y_j<0)=\frac{1}{2}. \]

Step 5: Use independence of the two signs.
Because \(X_i,X_j,Y_i,Y_j\) are mutually independent, the sign of \(X_i-X_j\) is independent of the sign of \(Y_i-Y_j\). The event \((X_i-X_j)(Y_i-Y_j)>0\) happens exactly when the two differences have the same sign, that is both positive or both negative. So
\[ P\left[(X_i-X_j)(Y_i-Y_j)>0\right]=P(X_i-X_j>0)P(Y_i-Y_j>0)+P(X_i-X_j<0)P(Y_i-Y_j<0) \]

Step 6: Compute the value.
\[ =\frac{1}{2}\times\frac{1}{2}+\frac{1}{2}\times\frac{1}{2}=\frac{1}{4}+\frac{1}{4}=\frac{1}{2}. \]

Final Answer:
When \(X\) and \(Y\) are independent, two independent pairs are just as likely to move together as to move apart, so the probability of concordance is exactly one half.\[ \boxed{\dfrac{1}{2}} \]