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

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!

• Yes, indeed. I thought about this as an extreme example when I first considered the problem. But does this answer the question immediately? What am I missing? Commented Sep 24, 2021 at 10:15
• An example where the probabilty is less than $\frac 1 2$ does not answer the question. Commented Sep 24, 2021 at 10:16
• @Snoop Sorry, edited the question for clarity. Commented Sep 24, 2021 at 10:27
• @Kavi Rama Murthy I have provided an answer that does not fit with the other (accepted) answer. I don't see where the "hiatus" is. May you have a look ? Commented Sep 25, 2021 at 6:31

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.

• really nice solution Commented Sep 24, 2021 at 12:22
• This is great, thanks! Commented Sep 24, 2021 at 12:37
• Thanks to you for this quite fun problem, I'll keep it in mind for courses on coupling techniques. Commented Sep 24, 2021 at 12:39
• very nice. Note that by your method, and by the fact that the uniform distribution has no atoms, you can use your technique to show that $\mathbb{P}(X+Y>1+\delta)=1-\epsilon$ provided $\epsilon\geq \delta.$ Commented Sep 25, 2021 at 2:40
• Which means, I suppose, $\mathbb{P}(X+Y>1+\delta)=1-\delta.$ Commented Sep 25, 2021 at 2:42

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 $$n$$.

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.

• This is really nice as well. Thanks for providing such a tangible discrete example to the original question. Commented Sep 25, 2021 at 7:54
• I wasn't able to introduce pictures directly in the comment, so I posted an extra answer above which I hope will answer your question. Commented Sep 25, 2021 at 15:13
• But the solution you offer does not exactly corresponds to the theoretical one I gave above. For instance, the whole point of the construction is to avoid the case $X = Y = 1/2$, i.e. the box at the center of the square should never be colored. Commented Sep 25, 2021 at 15:23
• @Gâteau-Gallois Thank you/Merci for your very interesting answer pointing to an issue I wasn't aware of. Commented Sep 25, 2021 at 17:54