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

probabilityrandom variablesuniform distribution

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!

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.