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!

## Best Answer

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.