Does uniform distribution on every square $[0,a]^2$ along diagonal imply uniform CDF on the entire $ [0,1]^2$

Question

Let $F: [0,1]^2\to R $ be a continuous cdf with uniform marginals, i.e., $F(x,1)=x$ and $F(1,y)=y$. Suppose $F$ is symmetric, i.e., $F(x,y)=F(y,x)$.

Suppose we also know that $F(a,a)=a^2$ for all $a\in [0,1]$.

Can we conclude that $F$ is uniform distribution on $[0,1]^2$?

Thanks.

Answer 1

Your conditions on $F$ are really loose.

Here’s a counterexample: $F(x,y)=xy+.01xy(1-x)(1-y)(x-y)^2$.

Answer 2

Here is an example (which @Eric points out is not symmetric, but I deal with that later), where the joint density is $2$ in the red shaded areas and $0$ elsewhere

So for example $$\mathbb P(X \le 0.2, Y \le 0.1) = 0.03 \not= 0.02 = \mathbb P(X \le 0.2)\mathbb P(Y \le 0.1)$$ giving the desired counter example.

You can check the marginal density visually: any vertical or horizontal line has red-shaded length $\frac12$ so the marginal density is $1$, and the red-shaded area of a square with opposite vertices at $(0,0)$ and $(a,a)$ is $\frac12a^2$ so, remembering the red joint density is $2$, this meets the desired conditions.

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Added

To handle the lack of symmetry, take two of these away from the diagonal following @Trailblazer's idea, so consider the next symmetric picture where the joint density is $2$ in the red shaded areas, $1$ in the blue shaded areas and $0$ elsewhere.

and now a possible calculation is $$\mathbb P(X \le 0.7, Y \le 0.1) = 1 \times 0.5 \times 0.1+2\times \tfrac12 \times 0.1^2=0.06 \\ \not= 0.07 = \mathbb P(X \le 0.7)\,\mathbb P(Y \le 0.1)$$