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.

  • 2
    $\begingroup$ 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$\,\ldots\qquad$ $\endgroup$ Commented Jan 6, 2019 at 20:23
  • $\begingroup$ $\ldots\,$is that I probably wouldn't solve this problem by considering such integrals anyway. $\endgroup$ Commented Jan 6, 2019 at 20:23
  • 1
    $\begingroup$ What if X and Y are Bernoulli? Isn’t that a counterexample to P(X=Y)=0? $\endgroup$ Commented Jan 7, 2019 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
    Commented Jan 7, 2019 at 9:22

2 Answers 2


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
    Commented Jan 7, 2019 at 21:39
  • 1
    $\begingroup$ @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 $\Pr(Y\le x\ \&\ X\le y). \qquad$ $\endgroup$ Commented Jan 8, 2019 at 5:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.