Supplementary material for: Efficient moment calculations for variance components in large unbalanced crossed random effects models

Transcript

1 Supplementay mateal fo: Effcent moment calculaton fo vaance component n lage unbalanced coed andom effect model Katelyn Gao Stanfod Unvety At B Owen Stanfod Unvety Augut 015 Abtact Th a upplementay document contanng poof fo ome eult n the man document The ecton numbe contnue whee that document left off Some contextual mateal epeated fo claty Alo a th a upplementay document, mateal that tadtonally left out a beng tedou algeba ncluded n full detal, makng the numeou tep eae to follow and check 10 Patally obeved andom effect model The andom effect model Y = µ + a + b + e,, N 100 d d d fo a Fa, b Fb and e Fe ndependent of each othe Thee andom vaable have mean 0, vaance σa, σ B, σ E and kutoe κ A, κ B, κ E, epectvely We wll not need the kewnee We ue lette,,, to ndex ow Lette,,, ae ued fo column In ntenet applcaton, the actual ndce may be people atng tem, tem beng ated, cooke, URL, IP addee, quey tng, mage dentfe and o on We mplfy the ndex et to N fo notatonal convenence One featue of thee vaable that we fully expect futue data to bng htheto uneen level That why a countable ndex et appopate We wll want to etmate σa, σ B, σ E and get a fomula fo the vaance of thoe etmate Many, pehap mot, of the Y value ae mng Hee we aume that the mngne not nfomatve Fo futhe dcuon ee Secton 71 of the man document The vaable Z {0, 1} take the value 1 f Y avalable and 0 othewe The total ample ze N = Z We aume that 1 N < We alo need N = Z and N = Z The numbe of unque obeved ow and column ae, epectvely, R 1 N>0, and C 1 N >0 In the um above, only fntely many ummand ae nonzeo When we um ove,,,, the um ove the et { N > 0} Smlaly um ove column ndce,,, ae ove the et { N > 0} Thee ange ae what one would natually get n a pa ove data log howng all ecod We fequently need the numbe of column ontly obeved n two ow uch a and Th Z Z = ZZ T Smlaly, column and ae ontly obeved n Z Z = Z T Z ow The matx Z encode eveal dffeent meauement egme a pecal cae Thee nclude coed degn, neted degn and IID amplng, a follow A coed degn wth an R C matx of completely obeved data can be epeented va Z = 1 1 R 1 1 C If max N = 1 and max N > 1 then the data have a neted tuctue, wth N dtnct ow n column and Z T Z = 0 fo Smlaly 1

2 max N = 1 wth max N > 1 yeld column neted n ow If max N = max N = 1 then we have N IID obevaton We note ome dentte: ZZ T = Z Z = N, and 101 ZZT = Z Z = Z N 10 We need ome notaton fo equalty among ndex et The notaton 1 = mean 1 = 1 = It dffeent fom 1 {,}={,} whch we alo ue Addtonally, 1 mean 1 1 = 11 Weghted U tattc We wll wok wth weghted U-tattc U a = 1 u Z Z Y Y U b = 1 v Z Z Y Y, and U e = 1 w Z Z Y Y, fo weght u, v and w choen below We can wte U a = u N N 1 whee an unbaed etmate of σ B + σ E fom wthn any ow wth N Unde ou model the value n ow ae IID wth mean µ + a and vaance σb + σ E, and o Va = σb + σe N 1 + κb + e N whee κb + e = κ B σ B + κ Eσ E /σ B + σ E the kuto of Y fo the gven and any Thu Va = σ B + σ E N 1 + κ Bσ B N + κ Eσ E N 103 Invee vaance weghtng then ugget that we weght popotonally to a value between N and N 1 Weghtng popotonal to N 1 ha the advantage of zeong out ow wth N = 1 Th condeaton motvate u to take u = 1/N, and mlaly v = 1/N If U e domnated by contbuton fom e then the obevaton ente ymmetcally and thee no eaon to not take w = 1 Even f the e do not domnate, the tattc U e compae moe data pa than the othe It unlkely to be the nfomaton lmtng tattc So w = 1 a eaonable default If the data ae IID then only U e above nonzeo Th appopate a only the um σa + σ B + σ E can be dentfed n that cae Fo data that ae neted but not IID, only two of the U-tattc above ae nonzeo and n that cae only one of σa and σ B can be dentfed epaately fom σ E The U-tattc we ue ae then U a = 1 U b = 1 U e = 1 Z Z Y Y Z Z Y Y, Z Z Y Y Becaue we only um ove wth N > 0 and wth N > 0, ou um neve nclude 0/0 and 10

3 111 Expected U-tattc Hee we fnd the expected value fo ou thee U-tattc Lemma 1 Unde the andom effect model 100, the U-tattc n 10 atfy EU a 0 N R N R σa EU b = N C 0 N C EU e N N N σ B 105 N N N Poof Ft we note that Ea a = σ A1 1 = Eb b = σ B1 1 =, Ee e = σ E1 1 = 1 = Now Y Y = b b + e e, and o EU a = 1 Z Z σ B 1 1 = + σe1 1 = 1 = = σ B + σ E Z Z 1 1 = = σ B + σ E Z 1 1 = and σ E = σ B + σ E N 1 = σ B + σ EN R The ame agument gve EU b = σa + σ E N C The matx n 105 Ou moment baed etmate ae 0 N R N R M N C 0 N C N N N 106 N N N ˆσ A ˆσ B ˆσ E = M 1 They ae only well defned when M nonngula The detemnant of M N R [ N CN N N CN N ] + N R [ N CN N ] U a U b U e 107 = N R [ N C N N ] + N R [ N CN N ] = N RN C[N N N + N] The ft facto potve o long a max N > 1, and the econd facto eque max N > 1 We aleady knew that we needed thee condton n ode to have all thee U-tattc depend on the Y It tll of nteet to know when the thd facto potve It uffcent that no ow o column ha ove half of the data 3

4 1 The vaance Fom equaton 107 we get Va ˆσ A ˆσ B ˆσ E = M 1 Va U a U b U e M 1 whee M gven at 106 So we need the vaance and covaance of the thee U tattc To fnd vaance, we wll wok out EU fo ou U-tattc Thoe nvolve EY Y Y Y = E a a + b b + e e a a + b b + e e [ a = E a + b b + e e + a a b b + a a e e + b b e e a a + b b + e e + a a b b + a a e e + b b e e ] Th expeon nvolve 8 ndce and t ha 36 tem Some of thoe tem mplfy due to ndependence and ome vanh due to zeo mean To hoten ome expeon we ue B A,, Ea a a a D A, Ea a, and, Q A,, Ea a a a wth mnemonc blnea, dagonal and quatc Thee ae mlaly defned tem fo component B Fo the eo tem we have B E,, Ee e e e D E,, Ee e, and, Q E,, Ee e e e The genec contbuton EY Y Y Y to the mean quae of a U-tattc equal Q A,, + Q B,, + Q E,, + D A, D B, + D A, D E,, + D B, D A, + D B, D E,, D E,, D A, + D E,, D B, + B A,, B B,, + B A,, B E,, + B B,, B E,, The othe tem ae zeo 11 Vaance pat Hee we collect expeon fo the quantte appeang n the genec tem of ou quaed U-tattc Lemma In the andom effect model 100, B A,, = A σ 1= 1 = 1 = + 1 =, B B,, = B σ 1= 1 = 1 = + 1 =, and B E,, = E σ 1= 1 = 1 = + 1 =

5 Poof The ft one follow by expandng and ung Ea a = σ A 1 =, et cetea The othe two ue the ame agument Lemma 3 In the andom effect model 100, Poof Take = and = n Lemma D A, = σ A1 1 = D B, = σ B1 1 = D E,, = σ E1 1 = Lemma In the andom effect model 100, Q A,, = 1 1 σ A + κ A + 1 {, } + 1 {, } + 1 {, }={, } Q B,, = 1 1 σ B + κ B + 1 {, } + 1 {, } + 1 {, }={, } Q E,, = 1 1 σ E + κ E + 1 {, } + 1 {, } + 1 {, }={, } Poof We pove the ft one; the othe ae mla Th quantty 0 f = o = When and, thee ae 3 cae to conde: {, } {, } = 0, {, } {, } = 1 and {, } {, } = The kuto defned va κ A = Ea /σ A 3, o Ea = κ A + 3σ A Fo no ovelap, we fnd Fo a ngle ovelap, Fo a double ovelap, Ea 1 a a 3 a = Ea 1 a = σ A Ea 1 a a 1 a 3 = Ea 1 a 1 a + a a 1 a 1 a 3 + a 3 = Ea 1 + 3σ A = σ Aκ A + 6 Ea 1 a = Ea 1 a 1 a 3 + 6a 1a a 3 1a + a = Ea 1 + 6σ A = σ Aκ A + 1 A a eult, σa, {, } {, } = 0, Ea a a a = σa κ A + 6, {, } {, } = 1, σa κ A + 1, {, } {, } =, and o Ea a a a equal 1 1 σa + κ A + 1 {, } + 1 {, } + 1 {, }={, } 1 Vaance of U a We wll wok out EU a and then ubtact EU a Ft we wte U a = 1 Fo EUa we ue the pecal cae = and = of 108, EUa = 1 [ Z Z Z Z Z Z Z Z Y Y Y Y 5

6 = 1 Q A,, + Q B,, + Q E,, + D A, D B, + D A, D E,, + D B, D A, + D B, D E,, + D E,, D A, + D E,, D B, + B A,, B B,, + B A,,B E,, + B B,, B E,, ] [ Z Z Z Z Q B,, } {{ } Tem 1 + Q E,, Tem + D B, D E,, + D E,, D B, + B B,, B E,, Tem 3 Tem Tem 5 afte elmnatng tem that ae alway 0 We handle thee fve um n the next ububecton 11 U a tem 1 1 = σ B Z Z Z Z Q B,, Z Z Z Z κ B + 1 {, } + 1 {, } + 1 {, }={, } = σ B Z Z Z Z 1 1 = 1 1 = + κ B + 1 {, } + 1 {, } + 1 {, }={, } 11 1 and 13 1 Tem 1 now a um of fou tem, 11 though 1 Tem 11 σ B tme 1 = + Z Z Z Z 1 1 = 1 = + 1 = 1 = Z Z Z Z Tem 1 σb κ B + / tme Z Z Z Z Z Z Z Z = N N N N + 1 = N R = Z Z Z Z 1 1 = 1 = + 1 = 1 = 1 {, } Z Z Z Z 1 = + 1 = 1 = 1 = ] 6

7 + = Z Z Z 1 = + 1 = 1 = 1 = ZZ T Z Z Z 1 = Z Z 1 = ZZ T ZZT + ZZ T ZZ T + ZZ T = ZZ T N The expeon ZZT mplfe to N, changng t fom a ow quantty to a column quantty But the othe pat of th expeon ae equvalent to um of tem uch a Z N makng the column veon le convenent to wok wth Tem 13 the ame a tem 1 by ymmety of ndce Tem 1 σb tme = = = = = Z Z Z Z 1 1 = 1 1 = 1 {, }={, } Z Z Z Z {, }={, } Z Z Z Z = 1 = Z Z Z Z 1 Z Z Z Z ZZ T ZZ T Z Z Summng tem 11 though 1 yeld σb N R + κ B + ZZ T N + ZZ T ZZ T 1 1 U a tem 1 = σ E = σ E Z Z Z Z Q E,, Z Z Z Z κ E + 1 {, } + 1 {, } + 1 {, }={, } Z Z Z Z 1 1 7

8 1 + κ E + 1 = 1 {, } + 1 {, } and = 1 {, }={, } Tem 1 σ E tme Z Z Z Z 1 1 = N R by the ame poce that evaluated tem 11 Tem σ E κ E + / tme = Z Z Z Z = 1 {, } N Z Z Z Z 1 {, } + N Z Z Z Z 1 = 1 {, } N Z Z Z Z 1 = 1 {, } N Z Z Z Z 1 = 1 = 1 {, } whch educe to N Z Z Z Z 1 {, } + = N Z Z Z Z 1 = 1 {, } N Z Z Z Z 1 = 1 = N Z Z Z Z 1 = 1 = 1 = N Z Z Z Z 1 = + 1 = 1 = 1 = + = N Z Z Z Z 1 = 1 = + 1 = 1 = 1 = N Z Z Z Z 1 = 1 = N Z Z Z Z 1 = 1 = 1 = N Z Z Z N Z Z N Z Z + N Z Z + N Z N Z = N R R + R + 8

9 = N R + = N 1 The lat expeon eemble the dagonal pat of tem 1 Tem 3 the ame the ame a tem Tem σ E tme Z Z Z Z 1 = {, }={, } Th the ame um a the coeffcent n tem 1 ha except that t ha the addtonal contant = Impong = on that quantty yeld N ZZT N ZZT = 1 Tem thu σ E N R + κ E + N U a tem 3 and Thee tem ae equal by ymmety We evaluate tem 3 = 1 1 Z Z Z Z D B, D E,, Z Z D B, Z Z D E,, Now and by the ame tep σb σ E N R Z Z D B, = σ B = σ B Z Z 1 1 = N 1 = σ BN R Z Z D E,, = σe Z Z 1 1 = = σen R Theefoe tem 3 of EU a equal σ B σ E N R and the um of tem 3 and 1 U a tem 5 The tem equal = Z Z Z Z B B,, B E,, Z Z B B,, Z Z B E,, 9

10 Now Z Z B E,, = σ E Z Z 1= 1 = 1 = + 1 = = σ E Z Z 1= 1 = 1 = + 1 = Tem 5 then σe = σ Eσ B = σ Eσ B σ Eσ B σ Eσ B + σ Eσ B N Z Z Z Z 1= 1 = 1 = + 1 = BB,, N Z Z Z Z 1= 1 = 1 = + 1 = N Z Z Z Z 1 = 1= 1 = 1 = + 1 = N Z Z Z Z 1 = 1= 1 = 1 = + 1 = N Z Z Z Z 1 = 1= 1 = 1 = + 1 = N Z Z Z Z 1 = 1= 1 = 1 = + 1 =, whch we call tem 51, 5, 53 and 5 Next we fnd the coeffcent of σb σ E n thee fou tem Fo tem 51, we get N Z Z Z Z 1 = 1= 1 = 1 = + 1 = = = N 1 = N R N Z Z Z 1 1= 1 = + 1 = Fo tem 5, we get = = N R N Z Z Z Z 1 = 1= 1 = 1 = + 1 = N Z Z Z 1= 1 1 = + 1 = a well Tem 53 and 5 ae alo N 1, by the tep ued fo tem 5 and 51 epectvely A a eult tem 5 equal σb σ E N R 13 Combnaton Combnng the eult of the pevou ecton, we have EUa = σb N R + κ B + ZZ T N + ZZ T ZZ T 1 + σ Bσ EN R + σ Bσ EN R 10

11 + σe N R + κ E + N Subtactng EU a = N R σb + σ E we fnd VaU a = σb κ B + ZZ T 1 + σ Bσ EN R + σ E κ E N + N 1 + ZZ T ZZ T Check We can check ome pecal cae of th fomula 11 Row neted n column If fo ntance ow ae neted wthn column, then N = R, and all N = N = 1 and n th cae U a = 0 The above fomula gve VaU a = 0 fo th cae 1 Column neted n ow If column ae neted n ow, then ZZ T = 1 = N and equaton 109 yeld VaU a = σb κ B + N 1 = N + = + σ Bσ EN R + σ E σ Bκ B + + σ Eκ E + = κ B σ B + κ E σ E κ E + N 1 + N 1 + σe N 1 + σe 1 = N N σ Bσ EN R N σ Bσ EN R = κ B σ B + κ E σ E = κ B σ B + κ E σ E N 1 + σe N 1 + σ Bσ EN R N 1 + N Rσ B + σ E 110 When column ae neted n ow, then U a = N 1 and becaue the ow ae then ndependent, U a ha vaance σb + σe N 1 N 1 + κb 1 + e 11 N The kuto of b + e σ B σ κ B+E = κ B + E σb + κe σ E σb + σ E Theefoe fo column neted n ow VaU a = σb + σe N 1 + N 1 σ B σ κ B + E N σb + κe σ E σb + σ E = σ B + σ E N 1 + κ B σ B + κ E σ E N 1 = N Rσ B + σ E + κ B σ B + κ E σ E N 1, matchng the expeon 110 that come fom equaton 109 fo VaU a 11

12 13 σ B = 0 If σb = 0 then VaU a hould be the ame a t fo column neted n ow In th cae equaton 109 educe to VaU a = σe κ E + N κ E = σ E = σ E κ E N 1 + N 1 + N R N If ntead we ft take the column neted n ow pecal cae fom equaton 110 and then ubttute σb = 0, we get the ame expeon 1 σ E = 0 and κ B = In th pecal cae we take σe = 0 and take b U±1 Then σb = 1 and κ B = Then VaU a = σb ZZ T ZZ T 1 = ZZ T ZZ T 1 In th cae U a = 1 = 1 Z Z b b Z Z b b b + b = Z Z 1 b b = N Z Z b b and o VaU a = VaŨa whee Ũa = Z Z b b We ealy fnd that EŨa = Z Z 1 = = 1 = R To get the vaance of Ũa we need Eb b b b = 1 = 1 = + 1 = 1 = + 1 = 1 = 1 = 1 = 1 = Now EŨ a = = + + Z Z Z Z Eb b b b Z Z Z Z 1 = 1 = Z Z Z Z 1 = 1 = Z Z Z Z 1 = 1 = Z Z Z Z 1 = 1 = 1 = 1

13 = = = R + Z Z + Z Z Z Z Z Z Z Z ZZ T ZZ T ZZ T 1 Z Z ZZ T In th cae we get VaŨa = R + ZZ T ZZ T 1 R matchng the eult fom Coed degn In a coed degn N = C fo all and ZZ T = C fo all and Hee the vaance VaU a = σb κ B + C1 C 1 + C 1 C 1 + σe κ E + C1 C C 1 σ EN R = σb κ B + C1 C C 1 R + σe κ E + C1 C C 1 R σ ER 1C Vaance of U b Th cae exactly ymmetc to the one above wth VaU a gven by 109 Theefoe VaU b = σb κ A + Z T Z N + N 1 Z T Z Z T Z 1 + σ E κ E + N Vaance of U e A befoe, we fnd EUe and then ubtact EU e Now U e = 1 Z Z Z Z Y Y Y Y Fom 108, EU e = 1 Z Z Z Z [ Q A,, Tem 1 + Q B,, Tem + Q E,, Tem 3 + D A, D B, + D A, D E,, + D B, D A, + D B, D E,, Tem Tem 5 Tem 6 Tem 7 + D E,, D A, + D E,, D B, Tem 8 Tem 9 13

14 ] + B A,, B B,, + B A,, B E,, + B B,, B E,, Tem 10 Tem 11 Tem 1 We handle the twelve um n the next ubecton 161 U e Tem 1 A befoe, we plt tem 1 nto fou pat 1 Z Z Z Z Q A,, = 1 Z Z Z Z 1 1 σ A + κ A + 1 {, } + 1 {, } + 1 {, }={, } = σ A Z Z Z Z 1 1 = 1 1 = + κ A + 1 {, } + 1 {, } + 1 {, }={, } 11 1 and 13 1 Fo tem 11, we have σa tme Z Z Z Z 1 1 = 1 = + 1 = 1 = = Z Z Z Z Z Z Z Z Z Z Z Z + Z Z Z Z = N N N N N + N N = N N Tem 1 σa κ A + / tme Z Z Z Z 1 1 = 1 = + 1 = 1 = 1 {, } = Z Z Z Z 1 = + 1 = 1 = 1 = Z Z Z Z 1 = + 1 = 1 = 1 = Z Z Z Z 1 = + Z Z Z Z 1 = = Z Z Z Z + Z Z Z Z Z Z Z Z Z Z Z Z Z Z Z Z + Z Z Z Z Z Z Z Z + Z Z Z Z = N N + N N N N 3 1

15 N N 3 N N 3 + N N N 3 + N = N N N N 3 + N By ymmety of ndce, tem 13 the ame a tem 1 Fo tem 1, we have σ A tme Z Z Z Z {, }={, } = Z Z Z Z =1 = = Z Z Z Z 1 = N N N N 1 1 = = Summng tem 11 to 1 gve σ A N N N N N + 3 N + σaκ A + N N N N 3 + N N 16 U e Tem We can ue the ymmety of the ole of A and B and the ndce Theefoe, tem equal to N N N + 3 N κ B + N N N N 3 + σ B N N 163 U e Tem 3 A befoe, we plt tem 3 nto fou pat 1 Z Z Z Z Q E,, = 1 Z Z Z Z 1 1 σ E = σ E Z Z Z Z 1 1 = 1 1 = + κ E + 1 {, } + 1 {, } 31 3 and 33 + κ E + 1 {, } + 1 {, } + 1 {, }={, } + 1 {, }={, } 3 15

16 Fo tem 31, we have σe tme Z Z Z Z 1 1 = 1 = + 1 = 1 = = Z Z Z Z Z Z Z Z Z Z + = N N 3 N 3 + N = N N 1 Z Z Tem 3 σ E κ E + / tme Z Z Z Z 1 1 = 1 = + 1 = 1 = 1 {, } = Z Z Z Z 1 = + 1 = 1 =1 = Z Z Z 1 = + 1 = 1 =1 = Z Z Z 1 = + Z Z 1 = = Z Z Z + Z Z Z Z Z Z Z Z Z + Z Z Z + Z = N 3 + N 3 N N N + N N + N = N 3 N + N = NN 1 By ymmety of ndce, tem 33 the ame a tem 3 Fo tem 3, we have σ E tme Z Z Z Z {, }={, } = Z Z Z Z =1 = = Z Z 1 = NN 1 Summng tem 31 to 3, we get σ ENN 1[NN 1 + ] + σ Eκ E + NN 1 16

17 16 U e Tem 1 = 1 Z Z Z Z D A, D B, Z Z D A, Z Z D B, The ft facto Z Z D A, = σ A Z Z 1 1 = = σan Z Z = σ AN N By the ame agument, the econd facto Z Z D B, = σ BN N, and o tem σ Aσ BN N N N 165 U e Tem 5 1 = 1 Z Z Z Z D A, D E,, Z Z D A, Z Z D E,, The ft facto computed n the pevou ecton The econd facto Z Z D E,, = σ E Z Z 1 1 = Thu, tem 5 = σen Z = σenn 1 σ Aσ ENN 1N N 166 U e Tem 6 By ymmety of ndce, th the ame a Tem : σ Aσ BN N N N 17

18 167 U e Tem 7 Th lke tem 5 wth facto A and B ntechanged Thu, tem 7 equal to σ Bσ ENN 1N N 168 U e Tem 8 By ymmety of ndce, th the ame a tem 5: σ Aσ ENN 1N N 169 U e Tem 9 By ymmety of ndce, th the ame a tem 7: σ Bσ ENN 1N N 1610 U e Tem 10 Z Z Z Z B A,, B B,, = σaσ B Z Z Z Z 1= 1 = 1 = + 1 = 1= 1 = 1 = + 1 = = σaσ B Z Z Z Z 1 =1 = 1 = 1 = 1 = 1 = + 1 = 1 = 1 = 1 = + 1 = 1 = + 1 = 1 = 1 = 1 = 1 =1 = + 1 =1 = + 1 =1 = 1 =1 = + 1 = 1 = 1 = 1 = 1 = 1 = + 1 = 1 = = σaσ B Z Z Z Z Z Z Z Z Z Z Z + Z Z Z Z Z Z Z Z + Z Z Z + Z Z Z Z Z Z Z Z Z Z Z Z + Z Z Z Z + Z Z Z Z Z Z Z + Z Z Z Z Z Z Z Z Z Z Z Z + Z Z Z = σaσ B N 3 N Z N N N N N Z + NN N Z N N + N 3 + N N N N N Z N N N Z + N N + N 3 N Z N N + N N N N N Z N Z N N + N 3 = σaσ B N 3 N Z N N + NN 18

19 1611 U e Tem 11 Z Z Z Z B A,, B E,, = σaσ E Z Z Z Z 1= 1 = 1 = + 1 = 1= 1 = 1 = + 1 = = σaσ E 1= 1 = 1 = 1 =1 = + 1 = 1 = Z Z Z Z 1 = 1 = + 1 = + 1 = 1 = 1 = 1 = 1 =1 = + 1 =1 = + 1 = 1 =1 = + 1 = 1 = 1 = 1 = 1 = 1 = + 1 = = σaσ E Z Z Z Z Z Z Z Z Z + Z Z Z Z Z Z + Z Z Z + Z Z Z Z Z Z Z Z Z + Z Z Z + Z Z Z Z Z Z + Z Z Z Z Z Z Z Z Z + Z Z Z = σaσ E Z Z Z Z Z Z = σaσ E N 3 N Z Z Z N Z Z Z Ue Tem 1 We can ue the ymmety wth tem 11, ntechangng ow column Thu, tem 1 σbσ E N 3 N N 17 Combnaton Summng up the eult of the pevou twelve ecton, we have EUe = σan σan N + 3σA N σ A N + σaκ A + N N N N 3 + N σbn N + 3σB N σ B N κ B + N N N N σe N N 3 + 3N N + σeκ E + NN 1 + σaσ BN NN N N N + σ Aσ ENN 1N N + σ Aσ BN N N N σ ENN 1N N + σaσ ENN 1N N σ ENN 1N N + σaσ B N 3 N Z N N + NN + σaσ E N 3 N N + σ B σ E N 3 N N Then, we have VaU e = EU e EU e 19

20 = EU e σ AN N σ BN N σ EN N σ Aσ BN N N N σ Aσ ENN 1N N σ Bσ ENN 1N N = σa N σ A N + σaκ A + N N N N 3 + N N σ B N κ B + N N N N σen N + σeκ E + NN 1 + σaσ B N 3 N Z N N + N N N + σaσ E N 3 N N + σ B σe N 3 N N Next, we mplfy the fom of th expeon The coeffcent of κ A + σ A N N N N 3 + N = N N N and mlaly fo that of κ B The coeffcent of κ E + σe NN 1 The emanng multple of σa N N and mlaly fo σb The emanng multple of σ E NN 1 The coeffcent of σ A σ B N 3 N Z N N + NN = NN NZ N N + N Z becaue N Z = N 3 Theefoe the coeffcent of σ A σ B N N NZ Applyng thee mplfcaton VaU e = σa N N N N + σ ENN 1 + κ A + σa NN N + κ B N N N + κ E + σenn 1 + σaσ B N N NZ + σaσ EN N N + σ B σ EN N The coeffcent of σa σ B a meaue of how cloe to a egula R C gd the data ae 18 Check 181 σ A = σ B = 0 If σ A = σ B = 0 then Y ae IID wth vaance σ E and kuto κ E Then U e = 1 Y Y = NN 1 e 0 N 113

21 whee e the uual ample tandad devaton appled to all N of the Y Thu VaU e = σen N 1 N 1 + κ E = σ N E N N 1 + κ E NN 1 Subttutng σ A = σ B = 0 n VaU e fom Secton 17 yeld σen N + σeκ E + NN 1 = N N 1σE + NN 1 κ E σe whch matche the fomula Equaton 113 become VaU e = σenn 1 + κ E + σenn 1 whch alo matche 18 IID amplng If max N = max N = 1, then the obevaton ae IID wth vaance σ Y = σ A + σ B + σ E and kuto κ Y = κ AσA + κ BσB + κ EσE σy Now U e NN 1 tme the ample tandad devaton of all N obevaton Thu VaU e = N N 1σ Y + NN 1 κ Y σ Y = N N 1σ A + σ B + σ E + σ Aσ B + σ Aσ E + σ Bσ E + NN 1 κ A σ A + κ B σ B + κ E σ E 11 In th cae, the fomula gve σ A + σ B N σ A N σ B N + σaκ A + N N κ B + N + σ EN N + σ Eκ E + NN 1 + σ Aσ B N N N 3 + N N N 3 N N N 3 + N Z N N + N N + σaσ E N 3 N N + σ B σe N 3 N N If we et all potve N = 1 and all potve N = 1 then N = N becaue thee ae now R = N ow n the data Smlaly N 3 and N um to N and thee powe of N alo um to N Next Z N N = Z = N The mot ubtle of thee um N N = 1 = N becaue the ndce un ove all wth N > 0 and all wth N > 0 Makng thee ubttuton we get VaU e = σ AN σ AN + σ Aκ A + N 3 N + N + σ BN σ BN + σ Bκ B + N 3 N + N + σ EN N + σ Eκ E + NN 1 + σ Aσ BN 3 N + N + σaσ EN 3 N σ EN 3 N = σa N N + NN 1 N N + NN 1 + σe NN 1 + NN 1 + σaσ BN N 1 1

22 whch matche equaton 11 Equaton 113 gve VaU e = σ A + σ Aσ EN N 1 + σ Bσ EN N 1 + NN 1 κ A σ A + κ B σ B + κ E σ E = N N 1σA + σe + N N 1 σaσ B + σaσ E + σaσ E + NN 1 κ A σa + κ B σb + κ E σe N N N N + σ ENN 1 + κ A + σa NN N + κ B N N N + κ E + σenn 1 + σaσ B N N NZ + σaσ EN N = σ AN N + σ BN N + σ ENN 1 N + σ B σ EN N + κ A + σ ANN 1 + κ B + σ BNN 1 + κ E + σ ENN 1 + σ Aσ BN N 1 + σ Aσ ENN N + σ Bσ ENN N 183 IID amplng agan If max N = 1 and σb = 0, then once agan the obevaton ae IID and The fomula gve VaU e = σ A VaU e = N N 1σ Y + NN 1 κ Y σ Y = N N 1σ A + σ B + σ E + σ Aσ B + σ Aσ E + σ Bσ E + σ B N + NN 1 κ A σ A + κ B σ B + κ E σ E N = N N 1σ A + σ E + σ Aσ E + NN 1 κ A σ A + κ E σ E 115 σ A N σ B N + σaκ A + N N κ B + N + σ EN N + σ Eκ E + NN 1 + σ Aσ B N N N N N 3 N N 3 + N N 3 + N Z N N + N N + σaσ E N 3 N N + σ B σe N 3 N N = σ A N σ A N + σaκ A + N N N + σ EN N + σ Eκ E + NN 1 + σ Aσ EN 3 N N 3 + N N = σan σan + σaκ A + N 3 N + N + σen N + σeκ E + NN 1 + σaσ EN 3 N = N N 1σA + σe + σaσ E + NN 1 κ A σa + κ E σe, matchng 115

23 13 Covaance of U a and U b We ue the fomula CovU a, U b = EU a U b EU a EU b, o we ut need to compute EU a U b Ung ou pefeed nomalzaton, Then, U a U b = 1 = 1 EU a U b = 1 We conde each tem epaately 131 U a U b Tem 1 Fo 11, we have 1 = σ E Z Z Y Y Z Z Y Y Z Z Z Z Y Y Y Y Z Z Z Z QE,, Tem 1 + D B, D A, + D B, D E,, + D E,, D A, Tem Tem 3 Tem Z Z Z Z Q E,, Z Z Z Z κ E + 1 {, } + 1 {, } + 1 {, }={, } = σ E N 1 Z Z Z Z κ E + 1 {, } + 1 {, } 11 1 and 13 σe = σe Z Z Z Z 1 1 = σe N = σ E N R N C Z Z 1 1 = Z Fo 1, we have σe κ E + / tme Z Z Z Z {, } = = Z Z Z Z = Z Z Z 1 1 N + 1 {, }={, } 1 Z Z 1 1 = Z 3

24 = Z Z Z 1 1 = 1 = + 1 = 1 = = Z Z = Z Z Z Z Z + Z Z Z + Z = Z N Tem 13 the ame a 1 by ymmety of ndce Fo 1, we have σe Z Z Z Z {, }={, } = 0, nce the lat ndcato mple = and = but the econd one 1 Summng up, tem 1 equal to σe N R N C + σ E κ E + Z N = σe N R N C + σ E κ E + N R C + Z 13 U a U b Tem U a U b Tem 3 = 1 = σaσ B Z Z Z Z D B, D A, Z Z Z Z σb1 1 = σa1 1 = = σaσ B N = σ Aσ B N R N C 1 ung the pevou ecton 13 U a U b Tem Z Z 1 1 = Z N Z Z Z Z D B, D E,, = 1 = σbσ E N R N C 1 Z Z 1 1 = Z Z Z Z Z σb1 1 = σe1 1 = Z Z Z Z D E,, D A,

25 ung the pevou ecton = 1 N 1 Z Z Z Z σe1 1 = σa1 1 = = σaσ E N R N C 135 Combnaton Addng up the fou tem, we have EU a U b = σe N R N C + σ E κ E + Z N + σ Aσ B N R N C + σ B σ E N R N C + σ A σ E N R N C, and o CovU a, U b = EU a U b EU a EU b = EU a U b σb + σeσ A + σen RN C = σ Eκ E + Z N Notce that CovU a, U b = 0 when σ E = 0 Th can be vefed by notng that when σ E = 0 then U a a functon only of a whle U b a functon only of b Theefoe U a and U b ae ndependent when σ E = 0 1 Covaance of U a and U e We ue the fomula CovU a, U e = EU a U e EU a EU e, o we ut need to compute EU a U e Ft, U a U e = 1 Z Z Y Y Z Z Y Y = 1 Z Z Z Z Y Y Y Y Then, EU a U e = 1 Z Z Z Z QB,, + Q E,, + D B, D A, + D B, D E,, Tem 1 Tem Tem 3 Tem + D E,, D A, + D E,, D B, + B B,, B E,, Tem 5 Tem 6 Tem 7 We conde each tem epaately 11 U a U e Tem 1 1 Z Z Z Z Q B,, = 1 Tem 11 equal to σb tme Z Z Z Z 1 1 σ B + κ B + 1 {, } + 1 {, } 11 1 and 13 Z Z Z Z {, }={, } 1

26 = Z Z 1 1 = Z Z 1 1 = = N Z N Z Z = N R N N Tem 1 equal to σb κ B + / tme Z Z Z Z {, } = Z Z Z Z = = = Z Z Z Z 1 1 = 1 1 = = N = N Z Z Z Z Z Z Z Z + Z Z Z N N Z Z Z + Z N N Z Z Z Z Z Z Z Z Z N N Z Z Z N + Z N = = Z NN N Z N N N 1 N N + N Tem 13 equal to tem 1 by ymmety of ndce Tem 1 equal to σb tme Z Z Z Z {, }={, } = Z Z Z Z =1 = = Z Z Z Z 1 1 = = Z Z N N = Z N Z Z Z Z N Summng the fou tem, we fnd that tem 1 equal to σb N R N N + σ B Z N Z N + σ Bκ B + Z N N N 1 6

27 1 U a U e Tem Fo 1, we get σ E tme 1 Z Z Z Z Q E,, = 1 Z Z Z Z 1 1 σ E + κ E + 1 {, } + 1 {, } 1 and 3 Fo, we get σe κ E + / tme Z Z Z Z 1 1 = NN 1 Z Z 1 1 = = NN 1 N = NN 1N R Z Z Z Z Z {, } = Z Z Z Z = = = = N Z Z Z 1 1 = 1 1 = + 1 {, }={, } Z Z Z 1 1 = 1 = + 1 = 1 = N N = N NR N + R = N RN 1 Z Z Z + Z Tem 3 the ame a by ymmety of ndce Fo tem, we get σe tme Z Z Z Z {, }={, } = Z Z Z Z =1 = = Z Z 1 = N Z = N R Addng up the fou tem, we fnd that tem equal σ ENN 1N R + σ EN R + σ Eκ E + N RN 1 7

28 13 U a U e Tem 3 1 Z Z Z Z D B, D A, = 1 = σ Aσ B Z Z Z Z σ B1 1 = σa1 1 = Z Z 1 1 = = σaσ B N = σaσ B N R N Z Z Z 1 1 = N Z Z N 1 U a U e Tem 1 Z Z Z Z D B, D E,, = 1 = σ Bσ E 15 U a U e Tem 5 Z Z Z Z σ B1 1 = σe1 1 = 1 = Z Z 1 1 = = σbσ E N R N Z = σbσ E N R N N 1 Z Z Z Z D E,, D A, = 1 = σ Aσ E Z Z 1 1 = 1 = Z Z Z Z σ E1 1 = σa1 1 = Z Z 1 1 = = σaσ E N R N N Z Z 1 1 = ung the eult fo tem 3 16 U a U e Tem 6 1 Z Z Z Z D E,, D B, = 1 = σ Bσ E Z Z Z Z σ E1 1 = σb1 1 = Z Z 1 1 = 8 Z Z 1 1 =

29 = σbσ E N R N Z Z = σbσ E N R N N 17 U a U e Tem 7 Z Z Z Z B B,, B E,, = Z Z Z Z σ B 1= 1 = 1 = + 1 = σe 1= 1 = 1 = + 1 = = σbσ E Z Z Z Z 1 = 1 = 1 = 1 =1 = + 1 = 1 = 1 = 1 = + 1 = + 1 = 1 = 1 = 1 = = σbσ E + = σbσ E N 1 =1 = + 1 =1 = + 1 = 1 =1 = + 1 = 1 = 1 = 1 = 1 = 1 = + 1 = Z Z Z Z Z Z Z Z + Z Z Z + Z Z Z + = σbσ EN N = σbσ ENN R 18 Combnaton Z Z + Z Z Z Z Z N 1 Z Z Z + Z Z Z Z Z Z Z Z Z Z Z Z + Z Z Z Z Z Z Z Z Z Z Z We add up the even tem, eplacng ome N and N expeon by equvalent ung N and N, gettng EU a U e = σb N R N + σ Bκ B + N + σ B Z N N N 1 Z N + σenn 1N R + σen R + σeκ E + N RN 1 + σaσ B N R N N + σ B σ E N R N N N + σaσ E N R N + σ Bσ ENN R N + σ B σe N R N Z N 9

30 Now EU a EU e = N RσB + σe σan N + σ BN N + σen N whch contan tem equallng eveal of thoe n EU a U e above Subtactng thoe tem fom EU a U e yeld CovU a, U e = σ B + σ Bκ B + Z N Z N Z N N N 1 + σ EN R + σ Eκ E + N RN 1 + σ Bσ ENN R 15 Covaance of U b and U e By ntechangng the ole of the ow and column n CovU a, U e, we fnd that CovU b, U e = σ A + σ Aκ A + Z N Z N Z N N N 1 + σ EN C + σ Eκ E + N CN 1 + σ Aσ ENN C 16 Aymptotc appoxmaton: poof of Theoem We uppoe that the followng nequalte all hold N ɛn, N ɛn, R ɛn, C ɛn, N ɛ N, N ɛ N, N ɛn, and N ɛn fo the ame mall ɛ > 0 The ft x nequalte ae aumed n the theoem tatement The lat two follow fom the ft two We alo aume that 0 < m κ A +, κ B +, κ E +, σ A, σ B, σ E m < Note that we can bound σa σ B, σ A σ E, and σ A σ B away fom 0 and unfomly wth thoe othe quantte afte eplacng m by mnm, m and m by maxm, m We alo uppoe that Z N ɛ N, and Z N ɛ N 116 The bound n 116 eem eaonable but t appea that they cannot be deved fom the ft eght bound above We begn wth the coeffcent of σ B κ B + n VaU a fom equaton 1 It ZZ T 1 + = N Z N + Z 30

31 = N 1 + Oɛ The thd, fouth and ffth tem n VaU a ae all Oɛ The econd tem contan ZZ T ZZ T 1 ZZT = Z Z = Z N = Oɛ It follow that VaU a = σb κ B + N 1 + Oɛ Smlaly VaU b = σa κ A + N 1 + Oɛ The expeon fo VaU e contan tem σa κ A +N N +σ B κ B +N N All othe tem ae Oɛ tme thee two, motly though N N, N N The coeffcent of σa σ B contan N Z N N ɛn Z N = ɛn N o t of malle ode than the lead tem, a well a N N ɛn A a eult VaU e = σ Aκ A + N N N + σ Bκ B + N N 1 + Oɛ Tunng to the covaance CovU a, U b = σ Eκ E + Z 1 + = σ Eκ E + N R C + OR = σ Eκ E + N1 + Oɛ Next CovU a, U e contan the tem σb κ B + N Z N = σb κ B + N N The tem appeang afte that one ae ON = OɛN N The laget tem pecedng t domnated by Z N ɛn Z N Z N = ɛn It follow that CovU a, U e = σb κ B + N N 1 + Oɛ and mlaly, CovU b, U e = σa κ A + N N 1 + Oɛ Next, ung 19 Vaˆσ A VaUe = N + VaU a N CovU a, U e N Oɛ = σaκ A + 1 N N1 + Oɛ, and mlaly Vaˆσ B = σbκ B + 1 N N 1 + Oɛ The lat vaance Vaˆσ E VaUa = N + VaU b N + VaU e N N 3 CovU a, U e N 3 CovU b, U e + N CovU a, U b 1 + Oɛ 31

32 = σeκ E Oɛ N Next we vefy that thee vaance etmate ae aymptotcally uncoelated facto we have Covˆσ A, ˆσ B = 1 N VaU e 1 N 3 CovU b, U e 1 N 3 CovU a, U e + 1 N CovU a, U b = 1 N σaκ A + N κ B + N 1 N σ Aκ A + = 1 N σ Eκ E + whch Oɛ tme Vaˆσ A and Vaˆσ B Lkewe N 1 N σ Bκ B + N + 1 N σ Eκ E + Ignong the 1 + Oɛ Covˆσ A, ˆσ E = 1 N 3 CovU a, U e + 1 N 3 CovU b, U e 1 N VaU e 1 N VaU a 1 N CovU a, U b + 1 N 3 CovU a, U e = σbκ B + N N + σaκ A + 1 N N σaκ A + N κ B + 1 N N σbκ B + 1 N N σeκ E + 1 N = σ Eκ E + 1 N whch much malle than Vaˆσ A Smlaly Covˆσ B, ˆσ E = σe κ E + /N, much malle than Vaˆσ B 17 Etmatng Kutoe To etmate the kutoe κ A, κ B and κ E n the above vaance expeon, t uffce to etmate fouth cental moment uch a µ A, = σa κ A + 3 and mlaly defned µ B, and µ E, Gven ˆσ A, ˆσ B, and ˆσ E, we can do th va GMM Conde the followng etmatng equaton and the expectaton, W a = 1 1 Z Z Y Y N W b = 1 1 Z Z Y Y N W e = 1 Z Z Y Y Ung pevou eult, EW a = 1 1 Z Z E[Y Y N = 1 1 Z Z E b b + e e N = 1 = 1 1 N Z Z E b b + 6b b e e + e e 1 N Z Z µb, + 6σB σ E + µ E, + 6σE 1 1= 3

33 = 1 Z Z µb, + 3σB + 1σ N Bσ E + µ E, + 3σ E 1 1= = κb + 3σB + 1σBσ E + κ E + σe + 3σE N 1 = N R µ B, + 3σB + 1σBσ E + µ E, + 3σE By ymmety, EW b = N C µ A, + 3σA + 1σAσ E + µ E, + 3σE Next EW e = 1 Z Z E Y Y = 1 Z Z E a a + b b + e e = 1 Z Z E a a + 6a a b b + 6a a e e + b b + 6b b e e + e e = 1 Z Z µa, + 6σA1 1 = + σaσ B1 1 = 1 1 = + σaσ E1 1 = + µ B, + 6σB1 1 = σ E1 1 = + µ E, + 6σE1 1 = 1 = = µ A, + 3σA + 1σAσ E Z Z 1 1 = + µ B, + 3σB + 1σBσ E Z Z 1 1 = + µ E, + 3σE Z Z 1 1 = 1 = + 1σ AσB Z Z 1 1 = 1 = + 1 = 1 = = µ A, + 3σA + 1σAσ EN Z Z + µ B, + 3σB + 1σBσ EN Z Z + µ E, + 3σ ENN 1 + 1σ Aσ BN Z Z Z Z + N = µ A, + 3σ A + 1σ Aσ EN N + µ B, + 3σ B + 1σ Bσ EN N + µ E, + 3σ ENN 1 + 1σ Aσ BN N N + N Thee expectaton ae all lnea n the fouth moment Theefoe, gven etmate of σa, σ B, and σ E, we can olve anothe thee-by-thee ytem of equaton to get etmate of the fouth moment Lettng M be the matx n equaton 106 we fnd that whee EW a EW b = M EW e µ A, µ B, µ E, 3N RσB + 1N Rσ B σ E + 3N Rσ E + 3N CσA + 1N Cσ A σ E + 3N Cσ E H H = 3σ A + 1σ Aσ EN N + 3σ B + 1σ Bσ EN N + 3σ ENN 1 + 1σ Aσ BN N N + N Fo plug-n method of moment etmato we eplace expected W -tattc by the ample quantte, eplace the vaance component by the etmate and olve the matx equaton gettng ˆµ A, et cetea Then ˆκ A = ˆµ A, /ˆσ A 3 and o on 33

34 18 Bet lnea pedcto Hee we pedct conde lnea pedcton of Y We begn wth pedcton of the fom Ŷ = Ŷλ = λ Z Y Then we conde pedcton of a educed fom that conde only the total n ow, n ow and n the whole data et 181 Poof of Lemma 51 Let Ŷ = Z λ Y and L = EY Ŷ Then L = µ 1 λ Z + VaY + VaŶ CovY, Ŷ Ft VaY = σa + σ B + σ E Next CovY, Ŷ = λ Z σ A 1 = 1 = + σ E1 = 1 = = σa λ Z λ Z + σeλ Z, and fnally VaŶ = λ λ Z Z σ A 1 = 1 = + σe1 = 1 = = σa λ λ Z Z λ λ Z Z + σe λ Z Thu L = µ 1 λ Z + σ A + σe + σa λ λ Z Z λ λ Z Z + σe σa λ Z λ Z + σeλ Z λ Z Now uppoe that we conde the lo L = Eµ + a + b Ŷ To do o we eplace VaŶ and CovY, Ŷ above by Vaa + b and Cova + b, Ŷ epectvely, yeldng 18 Statonay condton L = µ 1 λ Z + σ A + σa λ λ Z Z λ λ Z Z + σe σa λ Z λ Z λ Z The patal devatve of L wth epect to λ µ 1 λ Z Z + σ Eλ Z + σa Z Z λ 1 = + λ 1 = 3

35 Z Z λ 1 = + λ 1 = σa Z 1 = σ B Z 1 = Afte takng account of the ndcato functon we get Z µ 1 λ Z 1 + σeλ + σ A σaz 1 = σbz 1 = Z λ + σ B Z λ We can eplace Z 1 = by 1 = becaue of the leadng facto Z Th and a coepondng change to the coeffcent of σb yeld Z µ 1 λ Z 1 + σeλ + σ A Z λ + σ B Z λ σa1 = σb1 = The mplfed expeon no longe eque the double pme and o we fnd that the patal devatve of L wth epect to λ Z µ λ Z 1 + σeλ + σa Z λ Z λ σa1 = σb1 = 183 Poof of Lemma 5 Hee we conde Ŷ = λ 0 Y + λ a Y + λ b Y whee Y = Z Y, Y = Z Y, and Y = Z Y The mean quaed eo L = EY Ŷ Expandng t we get L = µ 1 λ 0 N + λ a N + λ b N + VaY + λ 0VaY + λ avay + λ bvay λ 0 CovY, Y λ a CovY, Y λ b CovY, Y + λ 0 λ a CovY, Y + λ 0 λ b CovY, Y + λ a λ b CovY, Y A befoe VaY = σa + σ B + σ E We et about fndng the othe tem Ft VaY = σa N N + σen, Second VaY = σ AN + σ BN + σ EN, VaY = σ AN + σ BN + σ EN and CovY, Y = σ AN + σ BN + σ EZ, CovY, Y = σ AN + σ BZ + σ EZ, CovY, Y = σ AZ + σ BN + σ EZ and The emanng tem ue omewhat longe agument CovY, Y = Z Z CovY, Y 35

36 = Z Z 1= σa + 1 = σb + 1 = 1 = σ E = σan Z N + σen, CovY, Y = σa Z N N + σen and then by ymmety Fnally CovY, Y = Combnng thee pece we fnd that Z Z CovY, Y = Z Z σ A 1 = 1 = + σ E1 = 1 = = σa Z Z Z Z + σez = Z σ A N + σ BN + σ E L = µ 1 λ 0 N λ a N λ b N + σ A + σ B + σ E + λ 0 σ A N N + σen + λ a σan N + σen + λ b σan N + σen λ 0 σan N + σez λ a σan Z + σez λ b σaz N + σez + λ 0 λ a σ AN + σ B + λ a λ b Z σ AN + σ BN + σ E Z N + σ EN + λ 0 λ b σ A Z N N + σen Now uppoe we conde ntead L = Eµ + a + b Ŷ Then we mut eplace VaŶ by Vaµ + a + b = σa + σ B and emove the σ E Z tem fom the covaance wth Y The eult L = µ 1 λ 0 N λ a N λ b N + σ A + λ0 σa N N + σen + λ a σan N + σen + λ b σan N + σen λ 0 σan N λ a σan Z λ b σaz N + λ 0 λ a σan Z N + σen + λ 0 λ b σa Z N N + σen + λ a λ b Z σ AN + σ BN + σ E 18 Poof of Theoem 51 Fom the eult of Lemma 5, we ee that L quadatc n λ Snce L bounded below by 0, t follow that L attan t mnmum on R 3, whch would be any oluton of the tatonaty condton λ L = 0 We fnd the component of th gadent 1 L = Nµ λ 0 N + λ a N + λ b N 1 + λ 0 σa N N + σ λ EN σan N 0 + λ a σan Z N + σen + λ b σa Z N N + σen 36

37 1 L = N µ λ 0 N + λ a N + λ b N 1 + λ a λ a + λ 0 σan σan N + σen σan Z Z N + σ EN + λ b Z σ AN + σ BN + σ E, and 1 L = N µ λ 0 N + λ a N + λ b N 1 + λ b σ λ AN N + σen σaz N b + λ 0 σa Z N N + σen + λ a Z σan N + σe whee We wte th a Hλ = c Nµ + σa N N N N N c = N µ + σa N Z = N N Z N µ + σa Z N N Z N and H a ymmetc matx wth uppe tangle H = H 11 H 1 H 13 H H 3 H 33 µ σa σb wth element H 11 = µ N + σa N N + σen H 1 = µ NN + σan Z N + σen H 13 = µ NN + σa Z N N + σen H = µ N + σ AN + σ BN + σ EN H 3 = µ N N + σ AZ N + σ BZ N + σ EZ, H 33 = µ N + σ AN + σ BN + σ EN and Ung T Z N and T Z N ome of thee mplfy: H 1 = µ NN + σ AN + σ BT + σ EN, H 13 = µ NN + σ AT + σ BN + σ EN and 185 Poof of Theoem 5 To begn wth, we note that N = Z N Z ɛn We wte λ 0 = 1 H33 H 13 c1 det H H 31 H 11 c 3 Then λ b det Hλ 0 = H 33 c 1 H 13 c 3 = N µ N + σa N + σenµ + N σb µ NN + σa Z N N + σen N µ = µ µ NN + σann NN + σenn µ NN σan Z N σbn 3 σen 37

38 µ N 3 + σan N 3 + σen µ NN σan Z N σbn 3 σen = µ σann NN + σenn σan Z N σbn 3 σen + σ B µ N 3 + σ AN µ NN σ AN Z N = µ σann + σenn σan Z N σen + σ B σ AN σ AN Z N = µ σ A + σ ENN 1 + Oɛ and det Hλ b = H 11 c 3 H 31 c 1 = µ N + σa N N + σenn µ µ NN + σa Z N N + σen Nµ + N σb = µ µ N N + σan N N N + σenn µ N N σan Z N σbnn σenn µ N N + σan N N N + σenn µ NN σan Z N σbn 3 σen = µ σan N N N σan Z N σbnn µ N N + σan N N N + σenn µ NN σan Z N σbn 3 σen = µ σ BN N 1 + Oɛ Thu Next λ 0 λ b = σ A + σ E σb N 1 + Oɛ det H = H 11 H 33 H13 = µ N + σa N N + σen µ N N + σan + σen µ NN + σa µ N N µ + σ B µ NN = µ N N σ B N Z N + σen A a eult the pedcton fo a new ow n a lage column eentally that column aveage plu O1/N tme the global aveage 38

39 186 Specal cae N = 0 and N = 1 Now uppoe that we have no data n the taget ow and exactly one olde obevaton n the taget column Let be the ngle ow wth Z = 1 Thee ae enough lage ow and column that the uual condton N N N hold but thee ae alo ome lghtly obeved ow and column Then c = N µ σa σb = Nµ + σ B µ + σ B, and H 11 = µ N + σa N N + σen H 13 = µ NN + σa Z N N + σen = µ N + σ AN + σ B + σ E, and H 33 = µ N + σ AN + σ BN + σ EN = µ + σ A + σ B + σ E Then λ 0 1 H33 H λ = 13 Nµ + σ b H 11 H 33 H13 B H 13 H 11 µ The detemnant µ N + σa N N + σen µ + σa + σe µ N + σan + σe µ N µ + σ A + σ B + σ E = µ N σ A + σ B + σ E The numeato fo λ 0 µ N H 33 Nµ + σ B H 13 µ + σ B µ + σ A + σ B + σ ENµ µ Nµ + σ B = σ A + σ ENµ, and o Smlaly, the numeato fo λ b λ 0 1 σa + σ E N σa + σ B + σ E H 13 Nµ + σ B + H 11 µ + σ B µ NNµ + µ N µ + σ B = µ σ BN and o In th cae, the pedcton fo Y σ B λ b σa + σ B + σ E σb Y + σa + σ E Y σa + σ B + σ E 39

40 19 Aymptotc weght: poof of Theoem 53 Hee we have 1 N ɛn, 1 N ɛn, N ɛn, N ɛn, N ɛn, N ɛn, N ɛn, N Z ɛnn, and N Z ɛnn The ft fve follow ealy fom 1 < 1/ɛ N, N ɛn The lat fou follow fom the othe Fo ntance N N ɛn = ɛn, and N Z Z ɛn = ɛn N We alo have 0 < m µ, σa, σ B, σ E M < Then µ N µ NN µ NN H = µ NN µ + σa N µ N N 1 + Oɛ µ NN µ N N µ N and ung ymbolc computaton va Wolfam Alpha, Septembe 6, 015 µ σa + σ B + σ A σ B 1 1 σa σ B µ N σa N N σb N N H 1 = 1 1 σa N N σa N Oɛ 1 1 σb N 0 N σb N The detemnant of H 1 σa σ B µ N N N Oɛ, o we need N 1 and N 1 to make matx nveon a contnuou opeaton Smlaly Nµ c = N µ + σa 1 + Oɛ N µ Thu gnong the Oɛ tem λ 0 = µ σa + σ B + σ A σ B σa σ B µ N Nµ 1 σ A N N N µ + σa 1 σ B N N N µ + σ B = µ σ A + σ B + σ A σ B σ A σ B N µ + σ A σ A N µ + σ B σ B N = µ σ A + σ B + σ A σ B σ A σ B N = 1 N µ σ B + σ A σ B σ A σ B N µ σ A + σ B σ B σ A σ B N The end eult 1/N of the ame ode of magntude a the ognal tem Theefoe λ 0 = 1/N1 + Oɛ Smlaly λ a 1 = σa N N Nµ + 1 σa N N µ + σa = µ σ A N + µ + σa σa N = 1 N and λ b = 1, N and both of thee appoxmaton nvolve multplcaton by 1 + Oɛ In th lmt then whch make ntutve ene a ˆµ + â + ˆµ + ˆb ˆµ Ŷ = Y 1 + Oɛ + Y 1 + Oɛ Y 1 + Oɛ 0

41 0 Smoothng pedcto In ome cae we may want a bette etmate of EY than Y telf Such a pedcto could take the fom Ŷ = Ŷλ = λ 0 Z Y + λ a Z Y + λ b Z Y + λ ab Z Y 117 It put ethe exta o educed weght on Y telf, dependng on the gn of λ ab Th pedcto only ueful when Z = 1, o t doe not apply n the new ow o new column cae ethe It only nontval when ou goal to etmate µ + a + b, not Y telf So we only conde L = EŶ µ a b hee Lemma 5 The MSE fo the lnea pedcto 117 L = µ 1 λ 0 N λ a N λ b N + σ A + λ0 σa N N + σen + λ a σan N + σen + λ b σan N + σen λ 0 σan N λ a σan Z λ b σaz N + λ 0 λ a σan Z N + σen + λ 0 λ b σa Z N N + σen + λ a λ b Z σ AN + σ BN + σ E Poof Th poblem only ae when Z = 1, whch we aume fo the et of th ecton Then EỸ µ + a + b = EŶ + λ ab Y µ + a + b = L + λ abey + λ ab EY Ŷ λ ab EY µ + a + b Now EY µ + a + b = µ + σ A + σ B and EY Ŷ = µ Nλ 0 + N λ a + N λ b + CovY, Ŷ fo CovY, Ŷ = CovY, λ 0 Y + λ a Y + λ b Y = λ 0 σ A N + σ BN + σ EZ + λa σ A N + σ BZ + σ EZ + λb σ A Z + σ BN + σ EZ = λ 0 σ A N + σ BN + σ E + λa σ A N + σ BZ + σ E + λb σ A Z + σ BN + σ E, nce we aume that Z = 1 Theefoe EỸ µ a b equal µ 1 λ 0 N λ a N λ b N + σ A + λ 0 σa N N + σen + λ a σan N + σen + λ b σan N + σen λ 0 σan N λ a σan λ b σa N + λ 0 λ a σan Z N + σen + λ 0 λ b σa Z N N + σen + λ a λ b σan N + σe + λ abµ + σa + σe + λ ab λ 0 σ A N N + σe + λa σ A N + σe + λb σ A N + σe λ ab µ + σ A + σ B + µ λ ab Nλ 0 + N λ a + N λ b Gatheng up the coeffcent of µ we get µ 1 λ 0 N λ a N λ b N λ ab + σ A + σ B + λ 0 1 σ A N N + σen

42 + λ a σan N + σen + λ b σan N + σen λ 0 σan N λ a σan λ b σa N + λ 0 λ a σan Z N + σen + λ 0 λ b σa Z N N + σen + λ a λ b σan N + σe + λ abσa + σe + λ ab λ 0 σ A N N + σe + λa σ A N + σe + λb σ A N + σe λ ab σ A + σ B Half of the devatve of th quaed eo wth epect to λ ab λ ab µ + σa + σe + λ 0 σ A N N + σe + λa σ A N + σe + λb σ A N + σe µ σ A σ B + µ Nλ 0 + N λ a + N λ b We ee that gven the othe λ choce, th devatve deceang at 0 hence we favo potve elf-weght f µ +σ A+σ B > λ 0 Nµ +σ AN +σ BN +σ E +λa N µ +σ AN +σ B +σ E +λb N µ +σ A+σ BN +σ E Futhemoe, the optmal elf-weght, gven the othe λ 1 µ + σ A + σ B + σ E µ + σa λ 0 Nµ + σan N + σe λ a N µ + σ AN + σ B + σ E λb N µ + σ A + σ BN + σ E The pont of th pedcto that we mght expect anothe obevaton to be made late n ow and column Then etmatng µ + a + b a bette way to pedct than epeatng the eale Y To ue Lemma 5 afte a econd pa, one can compute L a the gven quadatc functon n the fou vaable λ 0, λ a, λ b and λ ab The mnmze of that quadatc gve weght to apply n pedcton When σe vey mall then Y aleady cloe to µ + a + b and placng pecal weght on Y wll be advantageou