Squares of Random Numbers - 3D

Choosing three real numbers randomly uniformly and independently from $[0, 1]$ is equivalent to choosing a point $(x, y, z)$ randomly and uniformly in 3D space where $0 \leq x, y, z \leq 1$ . In other words, we have to choose a random point in a cube of side length $1$ with one vertex at the origin, and three edges along the coordinate axes. We will use the principles of geometric probability, $P(X) = \frac{\text{desired volume}}{\text{total volume}}$ . Since the total volume in this case is $1$ , the probability of any event is essentially equal to its volume.

We are to find the probability that the square of one of the numbers will be greater than the sum of the squares of the other two numbers. We will call this event $A$ . This can happen in three ways $x^2 > y^2 + z^2$ ; $y^2 > z^2 + x^2;$ and $z^2 > x^2 + y^2$ . Since the probability distribution is symmetric in $x, y$ and $z$ , each region will have the same volume, say $V$ . These are mutually exclusive events, so there is no point in common between any two regions and the total volume of the union of these regions is $3V$ . Therefore $P(A)$ can be written as sum of three probabilities: $P(x^2 > y^2 + z^2 )$ $+$ $P(y^2 > z^2 + x^2 )$ $+$ $P(z^2 > x^2 + y^2)$ . Since it is symmetric in $x, y,$ and $z$ , we can rewrite it as $P(A)$ $=$ $3 P(z^2 > x^2 + y^2)$ $=$ $3V$

We will first find out $V$ , i.e. the volume of the space $z^2 > x^2 + y^2$ that lies inside the cube.

Here are two different ways to calculate $V$ .

Method 1: Note that the volume $V$ is in fact a quarter cone. Its height is $1$ , and radius is also $1$ .

$V = \frac{1}{4} \times \frac{1}{3} \pi r^{2} h = \frac{\pi}{12}$

Method 2: It turns out that it is easier to find the complement volume $1 - V$ using double integrals.

$1 - V = \iint_{D} \min \left(\sqrt{ x^{ 2 } + y^{ 2 } }, 1 \right) \, dA$

where $D = \{(x, y) \in \mathbb{R^{2}} \ \vert \ 0 \leq x, y \leq 1 \}$ .

We can split the $\min$ function:

$1 - V = \iint_{D_{1}} \sqrt{ x^{ 2 } + y^{ 2 } } \, dA + \iint_{D_{2}} \, dA$

where $D_{1} = \{(x, y) \in \mathbb{R^{2}} \ \vert \ 0 \leq x \leq 1, 0 \leq y \leq \sqrt{1 - x^2} \}$ and $D_{2} = \{(x, y) \in \mathbb{R^{2}} \ \vert \ 0 \leq x \leq 1, \sqrt{1 - x^2} \leq y \leq 1 \}$

$V$ evaluates to $\frac{\pi}{12} = 0.261799...$
Thus we get $P(A) = 3V = \frac{\pi}{4}\approx \boxed{0.785397}$ $\square$

Since there are 3 choices to make the probabilities can be represented on a 3D sample space. It is easy to figure out intuitively that the sample space is a unit cube since each of our 3 choices : x, y and z can vary between the interval 0 and 1. Now lets consider each of the cases step by step. 1) $x^{2}$ + $y^{2}$ = $z^{2}$ Either intuitively or through some Graphing calculators it is easy to observe that x^2 + y^2 = z^2 graphs out two inverted cones with common vertex at (0,0,0). All the points with $x^{2}$ + $y^{2}$ < $z^{2}$ lie within the cones. This figure and our unit cube have a common volume that is equal to a quarter cone with radius = 1 and height = 1. This Common Volume has all the points that have x, y, z in the interval [0,1] and obey the condition $x^{2}$ + $y^{2}$ < $z^{2}$ .

Common Volume = Volume of Quarter Cone = 1/4 × 1/3 × $πr^{2}$ h = 0.25 × 1/3 × $π × 1^{2}$ × 1 = π/12

Since including this there are a total of three cases that have same probability of occurring, the

Total Probability = $\frac{3×Volume of Quarter Cone}{Volume of Unit Cube}$ = $\frac{3×π/12}{1}$ ≈ 0.7854