IetXbeannt-tkmatr’meffullmlumnralt. Partitienxasx=[.3t’1 Kg].trhere Xlxklanthgisnttkz1kl+kg=k [a] Shaw that X: Full eulltrnn real: and therein-re…

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3. Let X be an n×k matrix of full column rank. Partition X as X = [ X1 X1 isn×k1 andX2 isn×k2,k1 +k2 =k.X2 ], where(a) Show that X2 has full column rank and therefore (X2′ X2)−1 exists.(b) Define M2 = In −X2 (X2′ X2)−1 X2′ and X ̃1 = M2X1. Show that X ̃1 has full column
3. IetXbeannt-tkmatr’meffullmlumnralflt. Partitienxasx=[.3t’1 Kg].trhereXlfifixklanthgisnttkz1kl+kg=k [a] Shaw that X:- hm: Full eulltrnn real: and therein-re (3.“;er exist-i[h] Define M1 = rn—x: [X;X1}’1X5 and £1 = teem. Shaw that .t’. hasftfllmltnnn. .. —1ranhahd therefore (may) ={X;M1x1}’1 ex’HLs.4. (Schneider :1 pertitahued litter]: reggemten medel 1t" = I131 + 1133+ U, nag mulethat the meme-time ef the times] nerrnul linear regneaien bald- Let 31 he the {its eeeeetee et’ 31. Show that Feriflflthg} = ettxteex. 1-1. where .u: =I. — mete} ext. 5. Fer :- a ((1,1), let p.111.) 2 e [r — me e m} be the“ehedr“ fimetieu. see..- that msgigpeefl’ – c} = merit-35F – r1- 6. Censtder the feltewing simple regressien medel:fi=fiX.+Li’.. i=1…..rt. 1Where X. is a single regrmser. and fill. m NIDJI. Sana-use that the emmetn’tianeenstruets a seleetit’e sub-sample by ilteluding an]; the fltflf‘t‘fllm [rare the originalsample that satfifir I": v: e fer settle eenstant e.[a] Find the espresfim fer EEUeIXi} in the sub-sample in terms lien integral with respect te the standard nermal PDF. DD we have exegenfity {the mean indepen-denee ef the errer U.- frem the regrmer]: in the suhsampte? Explain. {b} Suppese that .5 :1- 0. Find hm: EEIJZIJL’a} changes with I. III daisies dE [HIE-‘1—rLt’. .