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The lower median of an odd empirical sample is its middle order statistic

Statement

Let n=2r+1≥1 and x1,…,xn∈R. Let Pn=n−1∑iδxi be their empirical probability law and write x(1)≤⋯≤x(n) for the sorted sample. Then the lower median m(Pn) is x(r+1).

Facts & Assumptions

Given: An odd sample size n=2r+1≥1, real observations x1,…,xn, and their empirical law Pn.

[F1]

A finite empirical measure is a probability law: Pn(A)=n−1∑iδxi(A) is a probability law.

[F2]

The lower median of a real probability law: m(Pn)=inf⁡{t:Pn((−∞,t])≥1/2}.

Proof

technique · direct
1.1givenF1F2algebra

If t<x(r+1), at most r sample coordinates are at most t. Therefore Pn((−∞,t])≤r/(2r+1)<1/2, so no such t belongs to the defining set in [F2].

1.2givenF1F2algebra

At t=x(r+1), at least r+1 sample coordinates are at most t. Hence Pn((−∞,t])≥(r+1)/(2r+1)>1/2, so x(r+1) belongs to that defining set.

2.1step 1.1step 1.2∎

Steps 1.1–1.2 show that the defining set has no point below x(r+1) and contains x(r+1) itself. Its infimum is therefore x(r+1), as claimed.

Depends on

Used by

Dependency tree · two levels

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