Uniform(0,1) Simplex Probability

Compute P(X₁ + X₂ + ··· + Xₙ ≤ 1)

Let X₁, …, Xₙ be i.i.d. Uniform(0,1). The event X₁ + X₂ + ··· + Xₙ ≤ 1 corresponds to an n-dimensional simplex inside the unit cube.

The Result

P(X₁ + X₂ + ··· + Xₙ ≤ 1) = 1 / n!
For n = 2: 1 / 2! = 1 / 2 = 0.5
n = 2
Factorial
2! = 2
Probability
1 / 2
Decimal
0.500000

Geometric Intuition

Since each Xᵢ is Uniform(0,1), the joint distribution is uniform on the unit cube [0,1]ⁿ. Therefore:

Probability = Volume of the region {x₁ + ··· + xₙ ≤ 1, xᵢ ≥ 0}

That region is the standard n-simplex, and its volume is:

Volume = 1 / n!

So the probability is exactly 1 / n!.

Visual for n = 2

When n = 2, the sample space is the unit square. The condition x₁ + x₂ ≤ 1 gives the triangle below.

x₁ x₂ 0 1 1 x₁ + x₂ = 1 valid region area = 1/2 sum > 1

The whole square has area 1, and the shaded triangle has area 1/2. Therefore, P(X₁ + X₂ ≤ 1) = 1/2.

Derivation

  1. The joint density of (X₁, …, Xₙ) is 1 on the unit cube [0,1]ⁿ.
  2. So
    P(X₁ + ··· + Xₙ ≤ 1) = ∫···∫x₁+···+xₙ≤1, xᵢ≥0 1 dx₁···dxₙ
  3. This is exactly the volume of the simplex
    Sₙ = {(x₁, …, xₙ) : xᵢ ≥ 0, x₁ + ··· + xₙ ≤ 1}
  4. The standard simplex has volume
    Vol(Sₙ) = 1 / n!
  5. Hence
    P(X₁ + X₂ + ··· + Xₙ ≤ 1) = 1 / n!

Recursive Proof of the Simplex Volume

Let Vₙ(t) be the volume of the region {x₁ + ··· + xₙ ≤ t, xᵢ ≥ 0}. Then:

Vₙ(t) = ∫₀ᵗ Vₙ₋₁(t - x) dx

If we assume Vₙ₋₁(t) = tⁿ⁻¹ / (n−1)!, then:

Vₙ(t) = ∫₀ᵗ (t - x)ⁿ⁻¹ / (n−1)! dx = tⁿ / n!

Setting t = 1 gives:

Vₙ(1) = 1 / n!