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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 $X sim mathcal{N} left(mu, sigma_1^2right)$ and $Y sim mathcal{N} left(mu, sigma_2^2right)$ may be different. This result holds regardless of whether $X$ and $Y$ are independent.

This answer is written under the assumption that $mathbb{P}(Y=X)=0$
which was part of the original wording of the question.

Question 1: A sufficient condition for$$mathbb{P}(X<Y)=mathbb{P}(Y<X)tag{1}$$is that $X$ and $Y$ are exchangeable, that is, that $(X,Y)$ and $(Y,X)$ have the same joint distribution. And obviously
$$mathbb{P}(X<Y)=mathbb{P}(Y<X)=1/2$$since they sum up to one. (In the alternative case that $mathbb{P}(Y=X)>0$ this is obviously no longer true.)

Question 2: Take a bivariate normal vector $(X,Y)$ with mean $(mu,mu)$. Then $X-Y$ and $Y-X$ are identically distributed, no matter what the correlation between $X$ and $Y$, and no matter what the variances of $X$ and $Y$ are, and therefore (1) holds. The conjecture is thus false.

I will show that the distribution of the pair $(X,Y)$ is the same as the distribution of the pair $(Y,X).$

That two random variables $X,Y$ are independent means that for every pair of measurable sets $A,B$ the events $[Xin A], [Yin B]$ are independent. In particular for any two numbers $x,y$ the events $[Xle x], [Yle y]$ are independent, so $F_{X,Y}(x,y) = F_X(x)cdot F_Y(y).$

And the distribution of the pair $(X,Y)$ is completely determined by the joint c.d.f.

Since it is given that $F_X=F_Y,$ we can write $F_{X,Y}(x,y) = F_X(x)cdot F_X(y).$

This is symmetric as a function of $x$ and $y,$ i.e. it remains the same if $x$ and $y$ are interchanged.

But interchanging $x$ and $y$ in $F_{X,Y}(x,y)$ is the same as interchanging $X$ and $Y,$ since
$$
F_{X,Y}(x,y) = Pr(Xle x & Yle y).
$$

Therefore (the main point):

The distribution of the pair $(X,Y)$ is the same as the distribution of the pair $(Y,X).$