Demonstrate the conditional independence of order statistics given a specific counting condition | Step-by-Step Solution

What We're Solving:

We need to prove that when we have i.i.d. random variables X₁,...,Xₙ and we condition on exactly d of them being ≤ T, the smallest d order statistics and the largest (n-d) order statistics become independent. Think of it like this: once we know how many values fall below our threshold, the "small" and "large" groups don't influence each other anymore!

The Approach:

The key insight is that conditioning on D=d gives us crucial information that "separates" our order statistics into two independent groups. We'll use the definition of conditional independence and properties of order statistics to show this mathematically. The continuous CDF assumption will be essential for avoiding ties.

Step-by-Step Solution:

Step 1: Set up the notation clearly

• X₍₁₎ₙ ≤ X₍₂₎ₙ ≤ ... ≤ X₍ₙ₎ₙ are our order statistics
• D = Σᵢ₌₁ⁿ I(Xᵢ ≤ T) counts how many original variables are ≤ T
• We want to show: (X₍₁₎ₙ,...,X₍ᵈ₎ₙ) ⊥ (X₍ᵈ₊₁₎ₙ,...,X₍ₙ₎ₙ) | D = d

Step 2: Use the key insight about what D=d tells us When D=d, we know that:

• Exactly d of our original variables are ≤ T
• The remaining (n-d) variables are > T
• Since F is continuous, P(X = T) = 0, so we have strict inequality

Step 3: Connect this to order statistics Given D=d, we can deduce:

• X₍₁₎ₙ,...,X₍ᵈ₎ₙ ≤ T (the d smallest values)
• X₍ᵈ₊₁₎ₙ,...,X₍ₙ₎ₙ > T (the (n-d) largest values)

Step 4: Apply the conditional distribution argument The crucial step: Given D=d, we can think of our problem as having two separate groups:

• The d values ≤ T follow the conditional distribution F(x|X ≤ T) = F(x)/F(T) for x ≤ T
• The (n-d) values > T follow the conditional distribution F(x|X > T) = (F(x)-F(T))/(1-F(T)) for x > T

Step 5: Use the independence of the original variables Since X₁,...,Xₙ are i.i.d., and we're conditioning on an event that only specifies which "side" of T each falls on (not their exact values), the order statistics within each group maintain their independence from the other group.

Step 6: Formalize with joint distributions For any measurable sets A and B: P((X₍₁₎ₙ,...,X₍ᵈ₎ₙ) ∈ A, (X₍ᵈ₊₁₎ₙ,...,X₍ₙ₎ₙ) ∈ B | D = d) = P((X₍₁₎ₙ,...,X₍ᵈ₎ₙ) ∈ A | D = d) × P((X₍ᵈ₊₁₎ₙ,...,X₍ₙ₎ₙ) ∈ B | D = d)

The Answer:

The conditional independence follows because D=d creates a "natural partition" at threshold T. The order statistics below T depend only on the conditional distribution below T, while those above T depend only on the conditional distribution above T. Since the original variables are i.i.d. and the conditioning event D=d only specifies the count (not the specific assignment), the two groups become independent given this information.

Memory Tip:

Think of it as "sorting mail"! Once you know exactly how many letters go in the "urgent" pile (≤T) versus "regular" pile (>T), the way letters are arranged within each pile doesn't affect the other pile. The threshold T acts as a perfect separator once we condition on the count D=d.

Great question - this really showcases how conditioning can create independence in surprising ways! 🌟