Let $x \sim U[0, 1]$, $y \sim U[0, 1]$. Let $M_1$ = $\max(1-2x,y)$, $M_2 = \max(1-2y,x)$. Find values of $a$ so that events $M_1 \leq a$ and $M_2 \leq a$ are independent.
This question has already been asked, and no answer was provided. I decided to tackle it myself and bring the attention.
My solution
Recall that $M_1 \leq a$ and $M_2 \leq a$ are independent iff $$P(M_1\leq a)P(M_2 \leq a) = P(M_1 \leq a, M_2 \leq a)$$ Since $x$ and $y$ are nonnegative, $M_1$ and $M_2$ are also nonnegative, therefore $$P(M_1 \leq a)=P(M_2\leq a) = P(M_1 \leq a, M_2 \leq a) = 0 \space \text{if a} \leq0$$ On the other hand, $M_1$ and $M_2$ do not exceed $1$, so $$P(M_1 \leq a) = P(M_2 \leq a) = P(M_1\leq a, M_2 \leq a) = 1 \space \text{if a} \geq 1$$ What if $a \in [0, 1]$?
Consider $P(M_1 < a, M_2 < a)$, which equals to: $$P(1-2x<a, y<a, 1-2y<a, x<a) = P\bigg(\frac{1-a}{2} < x < a\bigg) \cdot P\bigg(\frac{1-a}{2} < y < a\bigg) = P^{2}\bigg(\frac{1-a}{2} < x < a\bigg) = \bigg(a - \frac{1-a}{2}\bigg)^2 = \bigg(\frac{3a-1}{2}\bigg)^2$$ $\tag{1}$
On the other hand, $$P(M_1 < a) \cdot P(M_2<a) = P(x>\frac{1-a}{2}, \space y < a) \cdot P(y > \frac{1-a}{2}, \space x < a) = P^{2}\bigg(x > \frac{1-a}{2}, \space y < a \bigg) = \bigg(\frac{a(a+1)}{2}\bigg)^2$$ $\tag{2}$
Putting $(1)$ and $(2)$ together, we obtain an equation $$\bigg(\frac{3a-1}{2}\bigg)^2 = \bigg(\frac{a(a+1)}{2}\bigg)^2$$ Note that those are the squares of probabilities, so we can drop the squares and only look for roots of $$\frac{3a-1}{2} = \frac{a(a+1)}{2}$$ such that the both sides are in $[0, 1]$ interval. Only $a = 1$ satisfies. Thus, the final answer is $$a \leq 0, a \geq 1$$