I've two related propositions which seem correct intuitively, but I struggle to prove them properly.
Question 1
Prove or disprove: If $X$ and $Y$ are independent and have identical marginal distributions, then $\mathbb{P} (Y > X) = \mathbb{P} (X > Y) = 1/2$
Due to independence, the joint PDF of $X$ and $Y$ is the product of their marginal PDF:
$$ \begin{align} \mathbb{P} (Y > X) &= \int_{-\infty}^\infty \int_x^\infty p(x) \, p(y) \, dy \, dx \\ \mathbb{P} (X > Y) &= \int_{-\infty}^\infty \int_y^\infty p(x) \, p(y) \, dx \, dy = \int_{-\infty}^\infty \int_x^\infty p(y) \, p(x) \, dy \, dx \end{align} $$
The last step is based on the fact that the integral won't change if we simply rename the integration parameters $x$ and $y$ consistently. So we have shown that $\mathbb{P} (Y > X) = \mathbb{P} (X > Y)$
Side note: Even if $X$ and $Y$ are dependent, this result still holds so long as their joint PDF is exchangeable i.e. $p(x, y) = p(y, x)$
Let $u = y - x$ so that
$$ \mathbb{P} (Y > X) = \int_{-\infty}^\infty \int_0^\infty p(x) \, p(u + x) \, du \, dx $$
I thought of applying Fubini's theorem but it doesn't help to show that it's equal to 1/2, so maybe it's not 1/2?
Alternatively, consider that
$$ \mathbb{P} (Y > X) + \mathbb{P} (X > Y) + \mathbb{P} (Y = X) = 1 $$
If we assume that $\mathbb{P} (Y = X) = 0$ then we can conclude that $\mathbb{P} (Y > X) = 1/2$. But is this assumption justified?
Question 2
Prove or disprove: If $X$ and $Y$ are independent and $\mathbb{P} (Y > X) = \mathbb{P} (X > Y)$, then they have identical marginal distributions. If this statement is true, then is it still true if $X$ and $Y$ are dependent?
@Xi'an provided a counter-example. Suppose that
$$ \begin{bmatrix} X \\ Y \end{bmatrix} \sim \mathcal{N} \left( \begin{bmatrix} \mu \\ \mu \end{bmatrix}, \begin{bmatrix} \sigma_1^2 & c \\ c & \sigma_2^2 \end{bmatrix} \right)$$
Then $X-Y$ and $Y-X$ have the same distribution: $\mathcal{N} \left(0, \sigma_1^2 + \sigma_2^2 - 2c \right)$ and hence $\mathbb{P} (Y - X > 0) = \mathbb{P} (X - Y > 0)$
However the marginal distributions of $X \sim \mathcal{N} \left(\mu, \sigma_1^2\right)$ and $Y \sim \mathcal{N} \left(\mu, \sigma_2^2\right)$ may be different. This result holds regardless of whether $X$ and $Y$ are independent.