Maximum & Minimum of Uniform(0,1) Samples

Let X1,…,Xn be i.i.d. Uniform(0,1). Define Zn = max(X1,…,Xn) and Yn = min(X1,…,Xn).
1 5 20
One random sample
Each dot is one Xi. The extremes are Yn and Zn.
Yn = min: —  ·  Zn = max: —
Part 1: Maximum Zn
For the maximum to be at most z, every sample must be at most z.
FZₙ(z) = P(Zₙ ≤ z) = P(X₁ ≤ z, …, Xₙ ≤ z) = zⁿ
fZₙ(z) = d/dz FZₙ(z) = nzⁿ⁻¹
E[Zₙ] = ∫₀¹ z · nzⁿ⁻¹ dz = n/(n+1)
Part 2: Minimum Yn
For the minimum, use the complement: Yn > y only if every sample is greater than y.
FYₙ(y) = 1 − P(Yₙ > y) = 1 − (1−y)ⁿ
fYₙ(y) = n(1−y)ⁿ⁻¹
E[Yₙ] = ∫₀¹ y · n(1−y)ⁿ⁻¹ dy = 1/(n+1)
Interview intuition
Maximum → AND condition
Zn ≤ z means X₁ ≤ z AND X₂ ≤ z AND … AND Xₙ ≤ z. Since the Xᵢ are independent, multiply probabilities: z × z × … × z = zⁿ.
Minimum → complement
Yn > y means X₁ > y AND … AND Xₙ > y. That has probability (1−y)ⁿ, so FYₙ(y)=1−(1−y)ⁿ.
As n grows, the maximum moves toward 1 and the minimum moves toward 0. Their expected values are symmetric: E[Zn] = n/(n+1), E[Yn] = 1/(n+1), so E[Yn] + E[Zn] = 1.