X,Y ~ Unif(0,1) not necessarily independent, can P(X+Y>1)>1/2?

Question

My set up is the following:

$X,Y \sim \text{Unif}(0,1)$ but their joint distribution is not constrained. My question is whether there exists a joint dependence between them (that preserves the marginals) such that $\operatorname{Prob}(X+Y>1)>1/2$.

I can show it is = 1/2 for independence (via integrating the joint PDF), but am wondering whether there is a simple argument or counterexample either way for the cases where the joint distribution is not constrained. I have tried conditioning on one of the variables but couldn't make progress. All the simiulation evidence I have suggests it is = 1/2 for a variety of dependencies. Thanks!

Answer 1

Here is an attempt, which suggests that for all $\epsilon > 0$ you can actually couple in such a way that $$ \mathbb{P}(X + Y > 1) = 1 - \epsilon. $$ Note first that $$ \mathbb{P}(X + Y > 1) = \mathbb{P}(X > 1 - Y)$$ and that $Z := 1-Y$ is also a uniform random variable other $[0,1]$. Hence we can reduce the problem to the coupling of $(X,Z)$ such that $$\mathbb{P}(X > Z) > \frac{1}{2}. $$ Stated this way, it is actually easy. Take $Z$ uniform on $[0,1]$ and set $$ X = Z + \epsilon \quad (\text{mod} 1). $$ Then, if $Z < 1 - \epsilon$, $X > Z$. This occurs with probability $1 - \epsilon$.

EDIT : Here is the solution.

Let $U$ be uniform over $[0,1]$, set $Y = 1-U$, $X = U + \epsilon \quad (\text{mod} 1)$. Note first that $X$ and $Y$ have the desired marginals. Then $X = U + \epsilon$ on the event $U \le 1 - \epsilon$. Therefore $$ \mathbb{P}(X + Y > 1) \le \mathbb{P}(X + Y > 1, U < 1 - \epsilon) =$$ $$= \mathbb{P}(U + \epsilon + (1 - U) > 1, U < 1 - \epsilon) = $$ $$ =\mathbb{P}(U < 1 - \epsilon) = 1 - \epsilon. $$ As mentioned above, this is much better than the $\frac{1}{2}$ bound mentioned in the original question, since actually any value in $(0,1)$ can be obtained.

Answer 2

I would like here to present a family of solutions that I consider (maybe hastily...) as a discrete version of the method used by @Gâteau-Gallois. This presentation will rely on graphical representation of the joint pdf $f_{(X,Y)}$, represented on this figure as a surface $z=f_{(X,Y)}(x,y)$ in the case of a subdivision of the square $[0,1] \times [0,1]$ into a $3 \times 3$ grid:

Fig. 1: Case of a $3 \times 3$ grid: The joint pdf with 5 cuboids following a staircase pattern + one isolated cuboid ; the idea being that the biggest part of the "mass" is grouped along the diagonal with equation $x+y=1$.

The green part (don't pay attention to the vertical faces) is a "plateau" situated at height $z=\frac32$ in order that the total volume under the surface is $\frac69 \times \frac32 = 1$. It is clear that the marginals are uniform.

A little calculation shows that:

$$\mathbb{P}(X+Y>1)=\frac{7}{12} \approx 0.583$$

which is larger than $\frac12$, giving a first explicit answer to your question. But there is more to say. See below.

Remark: The marginals $X$ and $Y$ are not independent. Here is a counterexample: $\mathbb{P}(X>2/3)=\mathbb{P}(Y>2/3)=\frac13 \ $, while $ \ \mathbb{P}(X>2/3 \ \text{and} \ (Y>2/3))=0 \ \ne \ \mathbb{P}(X>2/3).\mathbb{P}(Y>2/3)$.

Now, let us build other pdfs on the same model : instead of a $3 \times 3$ grid, one can take a $4 \times 4$ grid on which are placed 8 cuboids: 7 of them on a staircase pattern following the diagonal with equation $x+y=1$ and an 8th isolated cuboid near the origin. In this case, one obtains:

$$\mathbb{P}(X+Y>1)=\frac58 = 0.625$$

which is an improvement over the previous value.

More generally one can take an $n \times n$ subdivision having the same structure ($2n-1$ cuboids in the staircase pattern + $1$ isolated cuboid near the origin), giving the general value :

$$\text{general} \ n \times n \ \text{case}: \ \ \mathbb{P}(X+Y>1)=\dfrac{3n-2}{4n}$$

which tends to $\dfrac{3}{4}$ when $n \to \infty$. And not to $1$.

As a conclusion, though this general case can be considered as a discrete version of the method used by @Gâteau-Gallois (with $\varepsilon = \frac1n$), it is impossible to obtain in this way a value for $\mathbb{P}(X+Y>1)$ arbitrarily close to $1$... which gives me some doubt about the conclusion by Gâteau-Gallois: I would like to "see/understand" the "shapes" of the joint pdfs associated to these cases where $\mathbb{P}(X+Y>1)$ is arbitrarily close to $1$.

Edit: I have understood, thanks to the second version of the answer by Gâteau-Gallois, the source of the discrepancy between my results and his ones.

Answer 3

I think the idea of @Jean Marie of providing a visual solution is great. However, I think a closest ``discrete version'' of the answer can rather be seen on the 4x4 grind, which I think would clarify things.

In fact you only need to fill the first boxes over the diagonal and the bottom-left corner. Here is a view for a 4x4 grid :

And here it is for the 5x5 grid :

Clearly, both random variables are uniform (we have 1 filled square per line, and one per column). Moreover, the only square filled such that $X+Y \le 1$ represents $1/n$ of the colored area, when we consider a $n \times n$ grid. Hence we can find a decomposition arbitrarely close to $1$ by playing with $n$.