Probability that $\displaystyle \vert x\vert +\vert y\vert +\vert z\vert +\vert x+y+z\vert=\vert x+y\vert +\vert x+z\vert +\vert y+z\vert$

Real numbers $x, y$, and $z$ are chosen from the interval $[−1, 1]$ independently and uniformly at random. What is the probability that $$\vert x\vert +\vert y\vert +\vert z\vert +\vert x+y+z\vert=\vert x+y\vert +\vert x+z\vert +\vert y+z\vert$$

Now if all of $x, y, z$ are positive or all negative then the equation is of course satisfied. Hence if we consider a 3D space , it denotes two unit cubes, one in the first octant centred at $\left( \frac 12,\frac 12,\frac 12\right)$ and the other in seventh octant centred at $\left( -\frac 12,-\frac 12,-\frac 12\right)$.

The total measure of universal set is the cube with edge length $2$ centred at origin.

But now I have a problem about what if any two of $x, y, z$ are positive while the other remaining be negative or the other way around. Even if I try to make cases it seems to be quite a cumbersome task to approach since we will also need to check signs of $\vert x+y\vert$ and similarly others as well as that of $\vert x+y+z\vert $

I also thought to give a shot using vectors but didn't reach any specific result.

Any help would be quite beneficial.

Edit:

I would also be happy to see a geometrical intuitive way to attack the problem.

Solution 1:

Suppose $x$ and $y$ are positive and consider the possible values of $z$. Because $|x|+|y|=|x+y|$ here, we want the equation $$ |z| + |x+y+z| = |x+z| + |y+z| $$ to hold.

Assume $x \le y$; in this case, we have $z \le x+z \le y+z \le x+y+z$, and so we can consider five possibilities based on which of these are positive.

1. $0 \le z \le x+z \le y+z \le x+y+z$. Then the equation holds. You already know this case.
2. $z \le 0 \le x+z \le y+z \le x+y+z$. Then the equation simplifies to $x+y=x+y+2z$, which has probability $0$.
3. $z \le x+z \le 0 \le y+z \le x+y+z$. Then the equation simplifies to $x+y=y-x$, which has probability $0$.
4. $z \le x+z \le y+z \le 0 \le x+y+z$. Then the equation simplifies to $x+y = -x-y-2z$, which has probability $0$.
5. $z \le x+z \le y+z \le x+y+z \le 0$. Then the equation holds. This case is new.

The same thing happens when $x \ge y$, so that doesn't need to be considered separately.

So we see that when $x$ and $y$ are positive, we want either $z$ to be positive as well, or we want $x+y+z$ to be negative.

By symmetry, this covers all the possibilities. The equation holds when:

• All three of $x,y,z$ are positive;
• Two of $x,y,z$ are positive, but $x+y+z$ is negative;
• Two of $x,y,z$ are negative, but $x+y+z$ is positive;
• All three of $x,y,z$ are negative.

The regions inside $[-1,1]^3$ where these hold have volume respectively:

• $1$ (it's a cube of side length $1$);
• $\frac12$ (it's three pyramids that form a corner of a cube with $\frac16$ the volume);
• $\frac12$;
• $1$.

So total volume $3$ (out of $8$), so the equation holds with probability $\frac38$.

Solution 2:

A simulation using R statistical software, for those interested, agrees with the theoretical answer.

> x<-runif(10^7,-1,1)
> y<-runif(10^7,-1,1)
> z<-runif(10^7,-1,1)
> mean(abs(x)+abs(y)+abs(z)+abs(x+y+z)==abs(x+y)+abs(x+z)+abs(y+z))
[1] 0.3749906