Mehmet Emre Tiryaki
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Transformations and Order Statistics

Change of Variables

For U=g1(X,Y)U = g_1(X, Y), V=g2(X,Y)V = g_2(X, Y) one-to-one, with inverse x=h1(u,v)x = h_1(u, v), y=h2(u,v)y = h_2(u, v),

fU,V(u,v)=fX,Y(h1(u,v),h2(u,v))⋅∣J∣,J=det⁡[∂x∂u∂x∂v∂y∂u∂y∂v]f_{U,V}(u, v) = f_{X,Y}\big(h_1(u, v), h_2(u, v)\big) \cdot |J|, \qquad J = \det \begin{bmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{bmatrix}

Order Statistics

For X1,…,XnX_1, \ldots, X_n IID with CDF FF and density ff, and X(1)≤⋯≤X(n)X_{(1)} \le \cdots \le X_{(n)} the sorted values,

FX(n)(x)=F(x)n,FX(1)(x)=1−[1−F(x)]nF_{X_{(n)}}(x) = F(x)^n, \qquad F_{X_{(1)}}(x) = 1 - [1 - F(x)]^n fX(n)(x)=nF(x)n−1f(x),fX(1)(x)=n[1−F(x)]n−1f(x)f_{X_{(n)}}(x) = nF(x)^{n-1}f(x), \qquad f_{X_{(1)}}(x) = n[1 - F(x)]^{n-1}f(x) fX(i)(x)=(ni−1,1,n−i)F(x)i−1f(x)[1−F(x)]n−if_{X_{(i)}}(x) = \binom{n}{i - 1, 1, n - i} F(x)^{i-1} f(x) [1 - F(x)]^{n-i}

For i<ji < j,

fX(i),X(j)(xi,xj)=(ni−1,1,j−i−1,1,n−j)F(xi)i−1f(xi)[F(xj)−F(xi)]j−i−1f(xj)[1−F(xj)]n−j Ixi<xjf_{X_{(i)}, X_{(j)}}(x_i, x_j) = \binom{n}{i - 1, 1, j - i - 1, 1, n - j} F(x_i)^{i-1} f(x_i) [F(x_j) - F(x_i)]^{j-i-1} f(x_j) [1 - F(x_j)]^{n-j} \, I_{x_i < x_j}

and the joint density of all of them is

f(x1,…,xn)=n! f(x1)f(x2)⋯f(xn) Ix1<x2<⋯<xnf(x_1, \ldots, x_n) = n! \, f(x_1) f(x_2) \cdots f(x_n) \, I_{x_1 < x_2 < \cdots < x_n}

Exchangeability

X1,…,XnX_1, \ldots, X_n are exchangeable if their joint distribution is the same under every permutation of (1,…,n)(1, \ldots, n), i.e. no variable is any more likely to be at a certain place. IID implies exchangeable. Sampling without replacement from a finite population gives exchangeable but not independent samples (see survey sampling).

Bivariate Normal

With Z1,Z2Z_1, Z_2 IID N(0,1)\mathcal{N}(0, 1),

X=μ1+σ1Z1,Y=μ2+σ2(ρZ1+1−ρ2 Z2)X = \mu_1 + \sigma_1 Z_1, \qquad Y = \mu_2 + \sigma_2\big(\rho Z_1 + \sqrt{1 - \rho^2}\, Z_2\big)

gives (X,Y)∼BVN(μ1,μ2,σ12,σ22,ρ)(X, Y) \sim \mathrm{BVN}(\mu_1, \mu_2, \sigma_1^2, \sigma_2^2, \rho), with ρ(X,Y)=Cov(X,Y)σXσY=ρ\rho(X, Y) = \frac{\mathrm{Cov}(X, Y)}{\sigma_X \sigma_Y} = \rho.

Variance and Conditioning

Var(X)=E[Var(X∣Y)]+Var(E[X∣Y])\mathrm{Var}(X) = E[\mathrm{Var}(X | Y)] + \mathrm{Var}(E[X | Y])

By Cauchy-Schwarz ∣E[XY]∣≤(E[X2]E[Y2])1/2|E[XY]| \le (E[X^2]E[Y^2])^{1/2}, so ∣ρ(X,Y)∣≤1|\rho(X, Y)| \le 1, and ∣ρ(X,Y)∣=1|\rho(X, Y)| = 1 implies Y=aX+bY = aX + b.