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

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.

• 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$\,\ldots\qquad$ – Michael Hardy Jan 6 at 20:23
• $\ldots\,$is that I probably wouldn't solve this problem by considering such integrals anyway. – Michael Hardy Jan 6 at 20:23
• What if X and Y are Bernoulli? Isn’t that a counterexample to P(X=Y)=0? – The Laconic Jan 7 at 3:03
• When using Fubini's theorem and a density $p(\cdot)$ against the Lebesgue measure, $\mathbb{P} (Y = X) = 0$, necessarily. – Xi'an Jan 7 at 9:22

## 2 Answers

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$$\mathbb{P}(Xis that $$X$$ and $$Y$$ are exchangeable, that is, that $$(X,Y)$$ and $$(Y,X)$$ have the same joint distribution. And obviously $$\mathbb{P}(Xsince 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 $$(Y,X).$$

• 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. – farmer Jan 7 at 21:39
• @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 $\Pr(Y\le x\ \&\ X\le y). \qquad$ – Michael Hardy Jan 8 at 5:28