14 Order Statistics

1 Beta Distribution

In the last note we introduced Beta distribution in this example. Now we give a formal definiton.

Beta Distribution

For α,β,λ>0, X∼Gamma(α,λ),Y∼Gamma(β,λ), X⊥⊥Y. (Note: Gamma(1,λ)=Exp(λ).) Denote Z=XX+Y⊥⊥X+Y, then Z∼Beta(α,β). The p.d.f. of Z isfZ(z)=zα−1(1−z)β−1Γ(α+β)Γ(α)Γ(β)1{z∈(0,1)}.

Here Gamma function isΓ(α)=∫0∞tα−1e−tdt.For n∈N, Γ(n)=(n−1)!. E(Z)=αα+β.

fZ(z) can take quite different shapes for different α,β value.

Generalization: Xi∼Gamma(αi,λ),i=1,⋯,n. X1,⋯,Xn are independent. Then X1+⋯+Xn⊥⊥(X1,⋯,Xn)X1+⋯+Xn∼Dirichlet(α1,⋯,αn), with p.d.f f(x1,⋯,xn)=Γ(α1+⋯+αn)Γ(α1)⋯Γ(αn)1{∑i=1nxi=1}∏i=1n1{xi>0}xiαi−1.

2 Order Statistics

Order Statistics

(x1,⋯,xn) is a list of real numbers. The order statistics of (x1,⋯,xn) is a permutation of the list s.t. x(1)≤⋯≤x(n).
If x1,⋯,xn are distinct, x(j) is the j−th smallest element in {x1,⋯,xn}.

2.1 Distribution of Order Statistics

Suppose X1,⋯,Xn∼i.i.dF.

Claim

The p.d.f of X(1) is given byfX(1)(x)=nf(x)[1−F(x)]n−1.

Take derivative:fX(1)(x)=ddxFX(1)(x)=nf(x)[1−F(x)]n−1.

Similarly,

Claim

The p.d.f of X(n) is given byfX(n)(x)=nf(x)[F(x)]n−1.

Claim

The p.d.f of X(j) isfX(j)(x)=n(n−1j−1)f(x)[F(x)]j−1[1−F(x)]n−j.

Corollary: Uniform Order Statistics

Suppose U1,⋯,Un∼i.i.dUniform(0,1). Then plug in the formula: U(j)∼Beta(j,n−j+1).

2.2 General Order Statistics

Recall quantile function:

Quantile Function

Given a random variable X with c.d.f FX. Define qX(u)=inf{x∈R|FX(x)≥u},u∈(0,1).
Furthermore, for U∼Uniform(0,1),qX(U)=dX.

Then, for general X1,⋯,Xn∼i.i.dFX, X(j)=dqX(U(j)), where U(j)∼Beta(j,n−j+1) for j=1,⋯,n.

2.3 Joint Distribution of Order Statistics

Theorem

(X1,⋯,Xn) is exchangeable with joint density f(x1,⋯,xn). Then fX(1),⋯,X(n)(x1,⋯,xn)=n!f(x1,⋯,xn)1{X(1)<⋯<X(n)}.

2.4 Discussion on Gaps

Denote U1=U(1),Lj=U(j)−U(j−1),j=2,⋯,n, and Ln+1=1−U(n). Lj are the gaps between order statistics. Then fL1,⋯,Ln(l1,⋯,ln)=n!1{l1+⋯+ln<1}∏i=1n1{li>0}.
RHS is the joint density of (W1,⋯,Wn+1)W1+⋯+Wn+1, where W1,⋯,Wn+1∼i.i.dExp(λ)=Gamma(1,λ). So (L1,⋯,Ln+1)=d(W1,⋯,Wn+1)W1+⋯+Wn+1,(L1,⋯,Ln+1)∼Dirichlet(1,⋯,1),f(l1,⋯,ln+1)=n!1{∑i=1n+1li=1}∑i=1n1{li>0}.
The gaps are not independent, but (L1,⋯,Ln+1) is exchangeable. So L1=d⋯=dLn+1,L1=X(1)∼Beta(1,n).
∀i=1,⋯,n+1, E[Li]=11+n.