Identically distributed vs P(X > Y) = P(Y > X)

Question

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. $p(x, y) = p(y, x)$

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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.

Answer 1

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.

Answer 2

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 $[X\in A], [Y\in B]$ are independent. In particular for any two numbers $x,y$ the events $[X\le x], [Y\le y]$ 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 $$ F_{X,Y}(x,y) = \Pr(X\le x\ \&\ Y\le y). $$

Therefore (the main point):

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