應(yīng)用回歸分析證明題

1、應(yīng)用回歸剖析證明題及答案一.證明殘差知足的拘束條件：證明：由偏導(dǎo)方程即得該結(jié)論：Q?00Q?111nnei0，xiei0。i1i12n?x)0(yii01i1n?x)x02(yii01ii1證畢.證明平方和分解式：SSTSSRSSE。證明：n2n?2SST(yiy)(yiy)yiyii1i1n2n?2n?y)(yi)2(yi)(y)(yiyiyiyii1i1i1n?nn?上式第三項(xiàng)2eiy2ei(eiyi01xi)0i1i1i1?nn2ei?xiei)001i1i1n)2n?)2即SST(?y(yiyiyii1i1SSRSSE證畢.三.證明三種查驗(yàn)的關(guān)系：?1Lxx=rn2；(2)F=SSR/

2、1?12Lxx=t2（1）t=?1r2SSE/(n2)=2?證明：因?yàn)閞Lxy?1LxxSSR,SSR?1Lxx2r2SST,LxxLyyLyySST2ei2SSTSSR?n2n2?LrLyyn2rn2所以t?Lyy;SSR1r2FSSR/1?12Lxx.SSE/(n2)2?證畢.四證明：Var(ei)11(xix)22。n(xix)2證明：因?yàn)閑iyi?yi(?yi01xi)yiy?1(xix)yi1n(xix)yi(xix)niyiLxx1于是Var(ei)Varyi1n(xix)yi(xix)niyiLxx11n(xix)yiVaryiVaryiVar(xix)n2Lxxi12Covyi

3、,1nyi2Covyi,(xix)yi(xix)ni1Lxx2Cov1ny,(xix)yi(xx)ni1iLxxi212(xix)22122(xix)22nLxx2Lxxn11(xix)22nLxx證畢.五證明：在一元回歸中，Cov(?0,?1)x2。Lxx證明：Cov(?0,?1)Cov1nyi(xix)yix,(xix)yini1LxxLxxnCovi1nCovi11x(xix)yi,n(xix)yinLxxi1Lxx1x(xix)yi,n(xix)yinLxxi1Lxxn1x(xix)(xix)2i1nLxxLxxx2Lxx證畢.六證明：?21SSE是偏差項(xiàng)方差2的無偏預(yù)計(jì)。np1證明：

4、因?yàn)镈(ei)1(xix)221(xix)2n而E(ei2)D(ei)E(ei)2D(ei)所以21E(?)ESSE1nD(ei)np1i1n1E(ei2)np1i11n(1hii)2np1i11(np1)22np1證畢.?2(XX)1。七證明：E()；D()證明：E(?)11E(XX)Xy(XX)XEy(XX)1XXD(?)?Cov(XX)1Xy,(XX)1XyCov,(XX)1XCovy,yX(XX)1(XX)1X2IX(XX)1(XX)1證畢.八證明：在多元線性回歸中，假定N(0,2In)，則隨機(jī)向量yN(X,2In)。九證明：當(dāng)yN(X,2In)時(shí)，則：?2(XX)1)；（）2。（1）

5、N(,SSE/(np1)2證明：?1Xy，X是固定的設(shè)計(jì)矩陣，所以，?（1）因?yàn)?XX)是y的線性變換。又當(dāng)N(0,2時(shí)，有隨機(jī)向量yN(X,2?In)In)，所以聽從正態(tài)散布，且?2(XX)1?N(,2(XX)1)。E(),D()，即有（2）：因?yàn)镾SEee?(y-y)(y-y)(I-H)y(I-H)yy(I-H)yyNy(X)N(X)NX0N借助于定理：設(shè)XN(0,In)，A為nn對(duì)稱陣，秩為r，則當(dāng)A知足：A2A，二次型XA2X:2r，只要證明：rk(N)np1即可。因?yàn)镹是冪等陣，所以有rk(N)tr(N)，故rk(N)trInX(XX)1XntrX(XX)1Xtr(XX)1XXnp1證畢.十證明：在多元線性回歸中，最小二乘預(yù)計(jì)?e不有關(guān)，即與殘差向量?。Cov(,e)0證明：Cov(?,e)Cov(XX)1Xy,(IH)y(XX)1XCovy,y(IH)(XX)1X2I(IH)(XX)1X2I(IX(XX)1X)0證畢.n?)，此中?etet1十一證明：DW2(1t2。nnet2et21t2t2證明：因?yàn)閚nnn(etet1)2et2et212etet1DWt2t2t2t2nnet2et2t2t2nnnetet1假如以為22，則有?t2，所以etet1nt2t22et2neett1DW21t2?.n2(1)et2t2