Yonsei University
Chae-Yeon Xon
Yonsei University, School of Economics
Home/Resources/Inference, Selection, and Simulation/II-02
Conditional Probability, Design, and Partial Identification · Chapter II-02

Symmetry, Order Statistics, and Endogenous Selection

An odd statistic has zero expectation under central symmetry when its index is fixed. If the index is selected using the same shocks, the composed statistic may lose that symmetry. Order-statistic calculations and joint reflection conditions reveal the missing step.

Conceptual map

  1. II-02.01Symmetry of fixed transformations
  2. II-02.02Order statistics
  3. II-02.03Selected indices
  4. II-02.04Joint reflection conditions
  5. II-02.05Numbers and sizes of markets

1. Symmetry must act on the complete statistic

Central symmetry makes an integrable odd function mean zero. A data-dependent selection rule can turn individually odd candidates into a non-odd selected statistic.

Let \(X\in \mathbb{R}^{n}\) satisfy \(X\equiv d-X\). If an integrable statistic \(T\) obeys \(T(-x)=-T(x)\), then \(\mathbb{E}[T(X)]=0\).

Proof. Distributional symmetry gives \(\mathbb{E}[T(X)]=\mathbb{E}[T(-X)]=-\mathbb{E}[T(X)]\). Integrability makes both expectations finite. ∎

For order statistics, reflection reverses rank: if \(X_{(1)}\le\cdots\le X_{(n)}\), then \(X_{(k)}(-x)=-X_{(n+1-k)}(x)\). The median of an odd sample is therefore odd.

2. Endogenous selection produces a signed absolute median

Let \(X_{1},X_{2},X_{3}\) be independent \(\operatorname{Uniform}(-1,1)\) variables and let \(M=X_{(2)}\). Define two candidate statistics \(T_{1}=M\) and \(T_{2}=-M\). Each has mean zero. Select \(K=1\) when \(M\le0\) and \(K=2\) when \(M>0\). Then

\[T_{K}=-|M|.\](1)

The median density is \(f_{M}(m)=3(1-m^{2})/4\) on [\(-1\),1]. Hence \(\mathbb{E}|M|=3/8\) and \(\mathbb{E}[T_{K}]=-3/8\). Also E[\(M^{2}\)]=1/5, so \(\operatorname{Var}(T_{K})=1/5-9/64=19/320\).

Under joint reflection, \(K\) switches candidates while the selected value remains unchanged: \(T_{K}(-x)=T_{K}(x)\). The composed statistic is even and nonpositive.

3. A selected sign result needs equivariance of the rule

Proposition 1 · Joint reflection criterion

Let candidates \(T_{k}\) and selection rule \(K(x)\) be measurable. The selected statistic has mean zero under a centrally symmetric law whenever it is integrable and

\[T_{K(-x)}(-x)=-T_{K(x)}(x)\](2)

outside a probability-zero set.

Proof. Condition (2) states that the complete composition is odd. Apply the symmetry argument from Section 1. ∎

Candidate-wise oddness alone does not imply (2), because reflection can change the selected index.

4. Ties, atoms, and large-market limits

Failure case · Fixed-index symmetry applied after endogenous selection

In (1), both candidates are odd and centered. The selection rule chooses the candidate with the nonpositive realized value, producing expectation \(-3/8\).

The event \(M=0\) has probability zero under the continuous example. With atoms, its probability can be positive and the tie rule enters the distribution of \(K\). For even samples, the definition of median also matters. A law-of-large-numbers limit for a selected share must retain the joint selection map; pointwise convergence of each fixed candidate does not validate interchange with a data-dependent index without uniform control.

5. Implementation, exercises, and sources

Implement the reflection map on the full data vector and recompute the selection rule after reflection. Track ties explicitly. In Monte Carlo work, report the seed, draw count, sample standard deviation, and Monte Carlo standard error of each estimated mean; compare with any available exact integral.

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Exercises

  1. Derive the median density and all moments displayed in Section 2.
  2. Replace the selection rule by the candidate with the nonnegative value and compute its mean.
  3. Repeat the construction for a discrete symmetric law with an atom at zero under two tie rules.
Partial solutions

1. The second order-statistic density is \(6F(m)[1-F(m)]f(m)\), which simplifies to 3(\(1-m^{2}\))/4. Integrating on [0,1] gives 3/8 and 1/5. 2. The selected statistic becomes \(\lvert M\rvert\) and has mean 3/8.

  1. H. A. David and H. N. Nagaraja, Order Statistics, chapters on densities and moments of order statistics.Finite-sample median distribution.
  2. A. W. van der Vaart, Asymptotic Statistics, chapters on quantiles and stochastic convergence.Large-sample quantile arguments.
  3. Patrick Billingsley, Probability and Measure, sections on distributional symmetry and integration.Symmetry-based expectation identities.

6. Audit checkpoint

State the reflection map, candidate statistics, complete selection rule, integrability, rank convention, atom and tie behavior, sample parity, fixed versus selected index, limiting mode, and Monte Carlo error.

7. Scope boundary

The chapter treats finite-sample reflection and an elementary order-statistic example. Uniform empirical-process results for large candidate classes require additional tools.

Prerequisites