Conditional density of $(X,Y)$ given that $X>Y$, where $X,Y$ are i.i.d $U(0,1)$

Question

I was told that my answer to this question was wrong, and I can't figure out why.

Suppose you have uniformly distributed i.i.d variables $X$ and $Y$. Find the conditional pdf of the joint distribution, given that $X>Y$, where $X,Y\sim U(0,1)$.

By definition, $$f(x,y\mid x>y) = \frac{f(x,y)}{P(X>Y)}$$

The numerator is just $f(x,y) = f(x)f(y) = 1$, and the denominator is

$$P(X>Y) = \int_{y}^{1}\left(\int_{0}^{1}f(y)\,dy\right)dx = \int_{y}^{1}1\,dx = 1 - y$$

Therefore, $$f(x,y\mid x>y) = \frac1{1-y}$$

To verify that this is a pdf, I did:

$$\int_{y}^{1}\frac1{1-y}\,dx = \frac1{1-y}[x]^{1}_{y} = \frac{1-y}{1-y}= 1\,,$$

since $0<y<x<1$.

Answer 1

As @pre-kidney pointed out, the correct denominator is:

$\mathbb P(X>Y)= \int_{0}^{1}\left(\int_{y}^{1}f(x)\,dx\right)dy = \int_{0}^{1}(1-y)\,dy = [y - \frac{y^{2}}{2}]_0^1=1 - \tfrac12 = \tfrac12$

The numerator is just $1$.

Therefore,

$f(x,y\mid x>y) = \frac{f(x,y)}{P(X>Y)} = 2$, so that $\int_{0}^{1}\left(\int_{y}^{1}2\,dx\right)dy = 1$

Answer 2

Thinking geometrically, observe that before the conditioning, $(X,Y)$ is uniformly distributed on the unit square $[0,1]^2$. After conditioning, it is uniformly distributed on the triangle $\{(x,y)\in [0,1]^2\colon x>y\}$. Since the area of the triangle is $\tfrac12$, this means that the density must be constant and equal to $2$ in order to make the probabilities add up to $1$.

Your mistake was in your computation of $\mathbb P(X>Y)$, you reversed the limits in the integral and obtained a non-sensical answer, since the the probability is a number that can't depend on any free variables (like $x$ or $y$). It means that the outermost integral's bounds cannot involve any variables, those must remain on the inner integral. If you correct your integral calculation, you will see that $\mathbb P(X>Y)=\tfrac12$, which is clear intuitively since there is an equal chance which of $X$ or $Y$ is the larger of the two.