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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.

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  • $\begingroup$ I have to admit nothing comes quickly to mind. I did a quick simulation, and the probability (assuming I haven't made any mistakes) comes out to something close to $0.3489$. Doesn't ring any bells yet. I'll keep thinking about it. Interesting problem. $\endgroup$ – Brian Tung Feb 17 at 5:30
  • $\begingroup$ @BrianTung Thanks and by the way the answer is $3/8$ as stated by my professor. $\endgroup$ – Rohan Shinde Feb 17 at 5:33
  • $\begingroup$ Weird. I must have done something wrong. Let me make sure I've picked the right PRNG. $\endgroup$ – Brian Tung Feb 17 at 5:36
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    $\begingroup$ @Digamma: Yeah, it was all roundoff error. I get $3/8$ now, too. $\endgroup$ – Brian Tung Feb 17 at 5:39
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    $\begingroup$ There is a natural $n$-d generalization (even length sums = odd length sums), and the 4-d probability is 13/48. As in the 3-d case, the measure in two orthants is 1, and the measure in all others is 1/6. My simple symbolic integration routine crashed Mathematica for the 5-d case, however. $\endgroup$ – Jim Ferry Feb 20 at 5:21
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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 $\frac38$.

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  • $\begingroup$ Can you try to generalize it for n-dimensional space please? $\endgroup$ – Rohan Shinde Feb 20 at 9:30
  • $\begingroup$ Not with the same approach; the amount of casework grows exponentially. $\endgroup$ – Misha Lavrov Feb 20 at 14:32
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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
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The following simulation in Matlab gives simulated probabilities, $10^7$ results for n from 2 to 10. Things written after percent signs are comments.

for n=2:10 % dimension n
b=cell(1,n);
[b{:}]=ndgrid(0:1); % n different 2x2x2x...x2 arrays
A=cat(20,b{:});
B=reshape(A,2^n,n); % array contains 0s and 1s to define all 2^n sums
C=(-1).^sum(B,2); % array contains +1 or -1 for each sum
for m=1:10000 % 10000 times 1000 trials is 10^7
D=(2*rand(1000,n)-1)*B'; % 1000 trials, all 2^n sums
E=abs(D)*C; % all 1000 combined sums
k(n)=k(n)+sum(abs(E)<1e-8); % count combined sums =0, give or take underflow
end;
disp(num2str([n k(n)/10000000]));
end;

2.0000    0.4998
3.0000    0.3748
4.0000    0.2710
5.0000    0.1900
6.0000    0.1240
7.0000    0.0815
8.0000    0.0494
9.0000    0.0297
10.0000    0.0168
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