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.

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    $\begingroup$ One interesting technical point is that if the probability of $X=c$ is $0$ for every value of $c,$ that's not enough to entail that probabilities are given by integrating a density function. The standard counterexample is the Cantor distribution. But more to the point$\,\ldots\qquad$ $\endgroup$ – Michael Hardy Jan 6 at 20:23
  • $\begingroup$ $\ldots\,$is that I probably wouldn't solve this problem by considering such integrals anyway. $\endgroup$ – Michael Hardy Jan 6 at 20:23
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    $\begingroup$ What if X and Y are Bernoulli? Isn’t that a counterexample to P(X=Y)=0? $\endgroup$ – The Laconic Jan 7 at 3:03
  • $\begingroup$ When using Fubini's theorem and a density $p(\cdot)$ against the Lebesgue measure, $\mathbb{P} (Y = X) = 0$, necessarily. $\endgroup$ – Xi'an Jan 7 at 9:22

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 $[X\in A], [Y\in B]$ 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 $(Y,X).$

  • $\begingroup$ I don't think "interchanging x and y in $F_{X,Y} (x,y)$ is the same as interchanging X and Y" because P(X ≤ x, Y ≤ y) ≠ P(X ≤ y, Y ≤ x) in general. Even if X and Y are independent, P(X ≤ x) P(Y ≤ y) ≠ P(X ≤ y) P(Y ≤ x) unless they are also identically distributed. $\endgroup$ – farmer Jan 7 at 21:39
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    $\begingroup$ @farmer : Start with $\Pr(X\le x\ \&\ Y\le y)$ and interchange $x$ and $y,$ and you get $\Pr(X\le y\ \&\ Y\le x).$ But if you start with the same thing and interchange $X$ and $Y,$ then you get $\Pr(Y\le x\ \&\ X\le y).$ The claim, then, is that $\Pr(X\le y\ \&\ Y\le x)$ is the same as $\Pr(Y\le x\ \&\ X\le y). \qquad$ $\endgroup$ – Michael Hardy Jan 8 at 5:28

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