样本方差无偏性证明.pdf
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样本方差无偏性证明.pdf

样本方差无偏性证明.pdf

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TheDerivationofSampleVariance'sunbiasednessAnarticleinLATEX2eMaiarJanuary23,2013Contents1Question11.1Variance................................11.2SampleVariance...........................11.3Unbiasedness.............................22Answer42.1Derivation...............................42.2Tips..................................5Chapter1Question1.1VarianceWell,andherebeginsmyfirstarticle(actuallyabook,that'snotajoke...)1writteninLATEX2e.It'sdefinedasfollows:pi=P(X=xi)(1.1)nXE(X)=xipi(1.2)i=1n2X2D(X)=EX−E(X)=xi−E(X)pi(1.3)i=11.2SampleVarianceWeintroduceanewvariableXasequation(1.4)onpage1shows:n1XX=X(1.4)nii=1Thennn1X1XE(X)=EX=E(X)=µ(1.5)ninii=1i=1nn1X1Xσ2D(X)=DX=D(X)=(1.6)n2in2ini=1i=11ù´·1˜‡5"2QuestionSampleVarianceisdefinedasequation(1.7):n1XS2=(X−X)2(1.7)n−1ii=11.3UnbiasednessTheunbiasednessofSampleMeanandSampleVarianceisasfollows:8<E(X)=E(X)(1.8):E(S2)=D(X)Nowwegettosolveit.1.3Unbiasedness3Thispageisintendedtobeblank!!Chapter2Answer2.1DerivationFirstly,weallknow:D(X)=E(X2)−(E(X))2E(X2)=D(X)+(E(X))2Andthemeansandvariances:E(Xi)=E(X)=µσ2D(X)=σ2;D(X)=inSoweget2222E(Xi)=D(Xi)+µ=σ+µ22σE(X)=D(X)+µ2=+µ2nXX+XX+···+X2+···+XXE(XX)=E1i2iiniinn1X=E(XX)+E(X2)nkiik=0k6=i1=((n−1)µ2+(σ2+µ2))nσ2=+µ2n2=E(X)2.2Tips5Nowlet'sbeginthederivation:n1XE(S2)=E(X−X)2n−1ii=1n1X2=EX2−2XX+Xn−1iii=1n1X2=EX2−2XX+Xn−1iii=1n1X=2nµ2+(n+1)σ2−2E(XX)n−1ii=11σ2=2nµ2+(n+1)σ2−2n(+µ2)n−1n=σ2=D(X)That'sall.Hopeyouenjoyit.2.2TipsWhat'sthedifferencebetweenxiandXi??Thinkaˆutit,whenyougetananswer,youcangobacktoclicktheblankpage3andcommentonmyarticle...andhereitends.