X and Y are independent random variables. If P(X>u)>=P(Y>U) for all u, then P(X>Y)>=1/2
Thanks a lot
Assume X and Y are continuous random variables with pdf f(x) and g(y) (the argument is easily modified for the discrete case). Then the joint pdf is f(x) g(y) due to independence of X and Y.
What my argument boils down to is that in the XY-plane, the 'weighted area' below the line Y = X is greater that the 'weighted area' above the line Y = X since $\displaystyle \Pr(X > u) \geq \Pr(Y > u)$:
1. $\displaystyle \Pr(X > Y) = \int_{y=-\infty}^{\infty} g(y) \int_{x = y}^{\infty} f(x) \, dx \, dy = \int_{u=-\infty}^{\infty} g(u) \Pr(X > u) \, du$.
2. $\displaystyle \Pr(X < Y) = \Pr(Y > X) = \int_{x=-\infty}^{\infty} f(x) \int_{y = x}^{\infty} g(y) \, dy \, dx = \int_{u=-\infty}^{\infty} f(u) \Pr(Y > u) \, du$.
3. $\displaystyle \Pr(X > Y) + \Pr(X < Y) = 1$.
Now use 1., 2., 3. and $\displaystyle \Pr(X > u) \geq \Pr(Y > u)$ to show that $\displaystyle \Pr(X > Y) + \Pr(X > Y) \geq 1 \Rightarrow \Pr(X > Y) \geq \frac{1}{2}$.