Given two identically distributed random variables, must the probability that the first is larger be 1/2? Let $(X_1 , X_2)$ be two real valued random variables with the same distribution, defined on the same probability space but not necessarily independent. Suppose that $X_1 \not=X_2$ almost surely. Must it hold that $\mathbb P(X_1 > X_2) = 1/2$? I know this is equivalent to the statement that the median of $X_2 - X_1$ is 0.  However, I can't quite seem to prove it, or find a counterexample. Does anyone have any suggestions?
To prove the statement, I tried assuming otherwise for contradiction and showing that $\mathbb P(X_2 > a) > \mathbb P(X_1 > a)$ for some $a$. I also tried assuming the variables had finite mean and calculating various expectations. However, I was not successful. 
 A: Let $X_1$ be uniform on $[0,1]$ and set $X_2=X_1+0.1 \pmod{1}$.  $X_2$ is also uniform on $[0,1]$ but $P(X_2>X_1)=0.9$.
A: In general, this is not true. It is true if the variables are independent. Yes, if the variables are independent. Let $Z = Y - X$, then:
$$\mathbb{P}(Y>X) = \mathbb{P}(Z = Y-X >0)$$
To demonstrate $\mathbb{P}(Y>X) = 1/2$ in the continuous case, we compute de Probability Density Function of $Z$, which is given by:
$$f_Z(z) = \int_{-\infty}^{\infty} f_{X}(x)f_{Y}(z+x)\ \text{d}x = \int_{-\infty}^{\infty} f_{X}\left(x-\frac{z}{2}\right)f_{X}\left(x+\frac{z}{2}\right)\ \text{d}x $$
Thus, we have that $f_Z$ is an even function [$f_Z(z) = f_Z(-z)$]. Now we have:
$$ 1 = \int_{-\infty}^{\infty} f_Z(z)\text{d}z = \int_{-\infty}^0 f_Z(z)\text{d}z + \int_0^{\infty} f_Z(z)\text{d}z = -\int_{\infty}^0 f_Z(-z)\text{d}z + \int_0^{\infty} f_Z(z)\text{d}z$$
because $f_Z(z) = f_Z(-z)$, we conclude that:
$$ \frac{1}{2} = \int_0^{\infty} f_Z(z)\text{d}z = \mathbb{P}(Z>0)=\mathbb{P}(Y>X)$$
