3110 Final Cheat Sheet

Change of variablesDistribution function techniqueFY(y)=P(Y≤y)=P(h(X1,…,Xn)≤y),fY(y)=∂FY(y)∂yTransformation techniquefy(y)=fX(h−1(y))⋅|∂h−1(y)∂y| if only one regionfY(y)=fX(h1−1(y))|dh1−1(y)dy|+fx(h2−1(y))||dh2−1(y)dy|fY1,...,Yn(y1,...,yn)=fX1,...,Xn(g1(y1,...,yn),...,gn(y1,...,yn))⋅|J||J|=det[∂X1∂Y1∂X1∂Y2∂X2∂Y1∂X2∂Y2] (integrate the dummy vars)Order StatisticsfX(1)=n[1−FX(x)]n−1fX(x)fX(n)=n[FX(x)]n−1fX(x)fX(r)(x)=n!(r−1)!(n−r)![FX(x)r−1]fX(x)[1−FX(x)]n−rfX~(x)=fX(n+1)(x)=(2n+1)!n!⋅n![FX(x)]nfX(x)[1−FX(x)]nUnbiasednessbias(θ^)=E(θ^−θ)=E(θ^)−θMSE(θ^)=E[(θ^−θ)2]=var(θ^)+[E(θ^)−θ]2Efficiency (compare unbiased estimators)efficiency(θ^)=CRLBVar(θ^)≤1 if Var(θ^)=CRLB→UMVUEvar(θ^)≥1nE[(∂ln⁡f(x;θ)∂θ)2]
Distribution PDF / PMF (E(X)) E(X²) (Var(X))
Uniform(a, b) 1b−a,a≤x≤b a+b2 a2+ab+b23 (b−a)212
Beta(α, β) Γ(α+β)Γ(α)Γ(β)xα−1(1−x)β−1,0<x<1 αα+β α(α+1)(α+β)(α+β+1) αβ(α+β)2(α+β+1)
Gamma(k, θ) 1Γ(k)θkxk−1e−x/θ,x>0 kθ k(k+1)θ2 kθ2
Poisson(λ) λxe−λx!,x=0,1,2,… λ λ(λ+1) λ
Binomial(n, p) (nx)px(1−p)n−x,x=0,1,…,n np np(1−p)+np2 np(1−p)
Geometric(p) (1−p)x−1p,x=1,2,3,… 1p 2−pp2 1−pp2
χ2(k) 12k/2Γ(k/2)xk/2−1e−x/2,x>0 k k(k+2) 2k
Expntl(λ) λe−λx,x>0 1λ 2λ2 1λ2
Normal(μ,σ2) 12πσe−12σ2(x−μ)2 μ μ2+σ2 σ2
limn→∞P(|θ^−θ|<ϵ)=P(−ϵ≤θ^−θ≤ϵ)=P(θ−ϵ≤θ^≤θ+ϵ)=1

If θ^ is an unbiased estimator of the parameter θ and the Var(θ^)→0 as n→∞, then θ^ is a consistent estimator of θ

 Chebyshev’s TheoremP(|X−μ|<kσ)≥1−1k2,orP(|X−μ|≥kσ)≤1k2

Sufficiency
If the conditional given θ^ depends on θ, θ^ is not sufficient

f(X1=x1,…,Xn=xn|θ^)=f(X1=x1,…,Xn=xn,θ^)g(θ^)← pdf of θ^=f(X1=x1,…,Xn=xn)g(θ^)

θ^ is a sufficient estimator iff the joint can be factorized (fact. theorem)

f(X1=x1,…,Xn=xn;θ)=g(θ^,θ)⋅h(x1,…,xn)Method of Moments (k moments for k parameters)E(X)=X¯,E(X2)=X¯2Method of maximum likelihood estimatorL(θ)=L(θ;x1,…,xn)=f(x1,…,xn;θ)=∏i=1nf(xi;θ)l(θ)=ln⁡L(θ)=ln⁡∏f(xi;θ)=∑i=1nln⁡f(xi;θ)dl(θ)dθ=0 to find crit point at θ^d2l(θ)dθ2|θ=θ^= to check maximum, should be <0 case 2+ parameters:if[∂2ℓ∂θ12∂2ℓ∂θ1∂θ2∂2ℓ∂θ2∂θ1∂2ℓ∂θ22]<0, then our θ1^,θ2^ are MLEs of θ1,θ2Bayesian Estimation(Prior distribution) g(θ)=Prior belief about θ(Likelihood) L(θ)=f(x;θ)=Data’s likelihood given θ(Posterior distribution) h(θ|x)=f(x,θ)f(x)=f(x;θ)g(θ)f(x)=L(θ)g(θ)f(x)L(θ)g(θ)=Unnormalized posteriorf(x)=∫all θL(θ)g(θ)dθ=Marginal likelihoodh(θ|x)=L(θ)g(θ)f(x)=Normalized posteriorθ^B=E(θ|x)=∫all θθh(θ|x)dθ=Bayesian estimate∑n=0xarn=a(1−rx+1)1−rConfidence Intervals:μ=X¯±z1−α/2⋅σn(known σ)μ=X¯±tn−1,α/2⋅Sn(unknown σ)μ=X¯±z1−α/2⋅Sn(CLT)μ1−μ2=(X¯1−X¯2)±z1−α2⋅σ12n2+σ22n2μ1−μ2=(X¯1−X¯2)±tν,α/2⋅S12n1+S22n2(n1,n2≥30)μ1−μ2=(X¯1−X¯2)±tn1+n2−2,α/2⋅Sp2(1n1+1n2)Sp2=(n1−1)S12+(n2−1)S22n1+n2−2σ2=(∑(Xi−μ)2χ1−α/2,n2,∑(Xi−μ)2χα/2,n2) (known μ)σ2=((n−1)S2χ1−α/2,n−12,(n−1)S2χα/2,n−12)p=p^±z1−α/2⋅p^(1−p^)np1−p2=(p^1−p^2)±z1−α/2⋅p^1(1−p^1)n1+p^2(1−p^2)n2σ12σ22=(S12S22⋅1F1−α/2,n1−1,n2−1,S12S22⋅1Fα/2,n1−1,n2−1)Test function: ϕ(x1,…,xn)={1if (x1,…,xn)∈C(reject H0)0otherwiseType I error=α=P(reject H0∣H0 true)Type II error=β=P(fail to reject H0∣H0 false)Power=1−β=P(reject H0∣H1 true)π(θ)=P(reject H0∣θ)(power function)Neyman–Pearson Lemma (for simple H0 vs simple H1):L0=∏i=1nf(xi;θ0)(likelihood under H0)L1=∏i=1nf(xi;θ1)(likelihood under H1)L0L1≤k⇒ reject H0Choose k to satisfy P(reject H0∣H0)=αCritical region: set of values making L0L1≤kLikelihood Ratio Test (general):L(θ)=∏i=1nf(xi;θ)(full likelihood)Λ=max L0max L=L(θ~)L(θ^)=∏i=1nf(xi,θ0)∏i=1nf(xi,θ^)Reject H0 if Λ≤k(or equivalently, −2ln⁡Λ≥c)Choose k (or c) so that test has level αAsymptotically: −2ln⁡Λ∼χdf2 under H0Z/T Tests for Mean(tn−1)C={(x1,…,xn);|x¯−μ0|≥Z1−α2σn} (2 sided test)θ<θ0:z≤−Z1−α,θ>θ0:z≥Z1−αDifference in Means (Unknown Variances)Zobs=x¯−y¯−δ0S12n1+S22n2n≥30t=x¯−y¯−δ0Sp2(1n1+1n2)∼tn1+n2−2n<30,pooled σ2Test for VariancesKnown μ:χ2=nσ^2σ02=∑(xi−μ)2σ02∼χn2Unknown μ:χ2=(n−1)s2σ02∼χn−12Reject H0:χobs2<χα/22 or χobs2>χ1−α/22F=S12S22∼Fn1−1,n2−1Reject H0:Fobs>F1−α/2, or Fobs<Fα/2Binomial Proportion Test (Exact)Two-sided:X≥K(α/2) or X≤K′(α/2), K smallest and K′ largest for α2One-sided:X≥K(α), where P(X≥K|H0)≤αBinomial Proportion (Normal Approx.)Z=X−nθ0nθ0(1−θ0)≈N(0,1)Continuity correction:Z=(x±12)−nθ0nθ0(1−θ0)Use +12:x≥nθ0,−12 if x<nθ0Chi-squared Test, Reject if χobs2≥χdf,1−α2∑i=1KZi2∼χk2=∑(xi−niθiniθi(1−θi))2=∑i∑j(fij−Eij)2Eij∼χdf2df=(k−1)(c−1), or k(c−1) if θj′s are given

To test for association (independence):

Eij=nπ^ij=nπ^i⋅π^⋅j,df=Ha−H0

Step 1. Estimate the parameter for the assumed distribution
Step 2. Compute the probability for each observation under the assumed distribution
Step 3. Compute the expected frequencies
Step 4. Test the goodness-of-fit of the assumed distribution to the observed data
df=k−1−# parameters estimated