# Exchangeable random variables $X$, $Y$, then $P(X<Y) = P(X>Y) = 1/2$ and $P(X=Y)=0$.

Two random variables $X$ and $Y$ are exchangeable if $(X,Y)\sim(Y,X)$, i.e. if the random vectors $(X,Y)$ and $(Y,X)$ have same joint distribution. Let $X$ and $Y$ have a joint probability density function $f(x,y)$. Prove that $P(X<Y) = P(X>Y) = 1/2$ and $P(X=Y)=0$.

What I tried

So if I'm right, exchangeable $X$ and $Y$ means $F_{X,Y}(a,b) = P(X\leq a, Y\leq b) = P(Y\leq a, X\leq b) = F_{Y,X}(a,b)$, or

$$\int_{-\infty}^{a} \int_{-\infty}^{b} f(x,y) dydx = \int_{-\infty}^{b} \int_{-\infty}^{a} f(x,y) dydx$$ where $f(x,y)$ is the joint density of $X$ and $Y$. Is there any property of multivariate integrals that I need to solve the problem?

• Just write $P(X=Y)$ as a double integral, and notice that for any $x$ the integrand is $0$ when $y \ne x$. – Robert Israel Sep 19 '16 at 21:18
• If $X=7$ with probability $1$ and $Y=7$ with probability $1$, are they not exchangeable? – Greg Martin Sep 19 '16 at 21:56
• But they would be discrete random variables instead of continuous. – Michael Sep 19 '16 at 22:01
• Example: Here's an exchangeable pair $(X,Y)$ with unit-rate exponential marginals and $\Bbb P[X=Y]>0$. Let $S,T$ be iid unit-rate exponential random variables, and let $B$ be a Bernoulli random variable ($\Bbb P[B=0]=1/2$) independent of $(S,T)$. Define $X:=S$ and $Y:=T$ on the event $\{B=0\}$; and $X:=S, Y:=S$ on the event $\{B=1\}$. This shows that the desired conclusion ($\Bbb P[X=Y]=0$) requires more than "continuous marginals" of the exchangeable pair $(X,Y)$. – John Dawkins Sep 19 '16 at 22:15