Question

Let {X_n}_n≥1 be a sequence of independent and identically distributed random variables having U(0, 1) distribution. Let Y_n = n min{X₁, X₂ , … , X_n}, n ≥ 1. If Y_n converges to Y in distribution, then the median of Y equals __________ (round off to 2 decimal places)

Answer 1

Correct Answer: 0.67

To solve this problem, we need to analyze the behavior of the sequence {Y_n} defined as Y_n = n min{X₁, X₂, …, X_n}, where {X_n} are i.i.d. random variables uniformly distributed over (0,1). We are interested in the distribution of Y_n as n approaches infinity.

1. Distribution of the Minimum: For i.i.d. random variables X₁, X₂, …, X_n, the probability that min{X₁, X₂, …, X_n} > x is (1-x)ⁿ for 0 ≤ x ≤ 1.

2. Distribution of n min: Then, P(n min{X₁, X₂, …, X_n} > y) = P(min{X₁, X₂, …, X_n} > y/n) = (1-y/n)ⁿ.

3. Limit Distribution: Consider the limit as n → ∞. Using the fact that (1-y/n)ⁿ approaches e^-y, the distribution of Y_n converges to an exponential distribution with λ = 1.

4. Median of Exponential Distribution: The median m of an exponential distribution with parameter λ is given by solving e^-m = 0.5, leading to m = ln(2)/λ.

5. Calculation: Since λ = 1, the median is m = ln(2) ≈ 0.6931, rounded to 0.69.

6. Validation: The computed median, 0.69