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# MUTIVARIATE ANALYSIS STAT 616

Texas A&M

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This 50 page Class Notes was uploaded by Darien Kutch on Wednesday October 21, 2015. The Class Notes belongs to STAT 616 at Texas A&M University taught by Michael Sherman in Fall. Since its upload, it has received 38 views. For similar materials see /class/225752/stat-616-texas-a-m-university in Statistics at Texas A&M University.

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Date Created: 10/21/15

More on Discrimination and Classi cation More on CART and Extensions General Regression Modell fj1 quot39a l 6339 Z 1 n In the GLM yj 90 913331 l 9103ij Ej This equation holds for every observation in the popula tion This is a tree With one node and one leaf CART uses regression modelling at each split but in the end makes the same estimate for every observation in the terminal node EG In the Heart Attack Example all obserations are classi ed as high risk or low risk given that they are in the same terminal node Ie the model is g do in each terminal node This leads to TB Treed Regression TB is a compromise between GLM and CART Al lows for regression ts at terminal nodes This enables less complex and more interpretable trees to be t EG Say the most important split variable is 5 Then if wk lt c we predict g 0 8533 and for wk gt c we predict Q 36 8103 This can be written as a linear model of the form 1 3m 2156 CAYTZT rzrajr wherezlzlifkltcandzr21ifajkgtc Not a true linear model as the indicator variables are estimated from the data One can have more than one split and then t separate regressions The number of indicator variables is equal to the number of terminal nodes How to choose between models of this type Consider an alternative hypothesis where the split of the parent node is given by L xik lt c and R 36 gt c Then SSEfull Zltyi i2 Zlyz39 l8lill2Zlyi r8rirl2l i1 L R SSEL SSER Under the null hypothesis the split is unecessary so that SSEreduced ap 81056110 SSEp 21 Then SSEP SSEL SSERdffull dfreduced I F SSE SSERVW dffulll We have dfreduced 2 The dif culty is determining df full It s at least four but likely more as the split was estimated from the data Simulations show that the df full is relatively constant for a xed number of co variates and sample size over different distributions of covariates Thus it can be estimated from ONE extensive sim ulation over 19 n EG Model the selling price of a home based on knowledge of year built square ft of the house number of baths in Iron county UT By considering each independent variable the most signi cant variable is the year built By considering all possible years the split point is determined to be 1969 This indicates that older homes have a different market than newer homes In this data set 71 538 and p 14 Simulations show that in this case dffull 2 16 The F statistic in this case is F 21533 4769 1284116 2 8 3 4769 12841538 16 39 Compare to F1452 pvalue lt 001 Model is g 3962 1282 nabaths 236 60442 livearea Where z 1 if year lt 1969 and 2 1 if year gt 1969 Can then test for further splits results show that living area is the variable in the nal regression at each terminal node Bumping Bootstrap Umbrella of Model Parameters Recall that Bagging produces an accurate classi er reduces error rates but leads to a hard to interpret predictor Bumping is an attempt to reduce error rates from a given procedure CART subset regression While retaining an interpretable predictor Setup Training sample 2 21 27 from a distribution F We have a model for the data that depends on param eters 6 From the training sample we minimize a target criterion d argmngz 6 The proposal is to estimate 6 using bootstrap sam B and estimate d from each bootstrap sam ples 21 z ple d argmngzb 6 and then choose d as the value that gives the smallest value of Rz 6 dB d where d argmmbRz Include the original sample 2 With the bootstrap samples EG Rz6 Qynxx6 Where Q is the loss function typically y 772 in regression and I y 7E 77 in classi cation Comments If using the original training sample we nd the global minimizer of Rz 6 eg linear least squares re gression then Bumping Will give this global minimizer Why The bootstrap sample includes the original sample and the global minimizer cannot be improved upon However adaptive procedures like subset regression and classi cation trees often nd local minimizers so there is potential for bumping to give a better local min imum Data sets 1 p 5 predictors each U 1 1 The responses are y 1 if 331 gt 0 and 32 gt 0 y 0 otherwise 2 glass data Determine which type of glass 6 types from 9 chemical measurements on 214 observa tions 3 Breast Cancer data 699 cases 458 benign and 241 malignant 9 variables cell characteristics Error Rate Comparison Data CART Bumped Bagged Simulate 049 029 025 Glass 035 030 025 Cancer 049 047 026 Results 1 Bagging is the best 2 Bumping reduces error in each case 3 So if only error rate is important bagging is bet ter if tted model structure is important bumping is bet ter Test for MVN assumption We know that for MVN observations Yz39 H21Yz39 0 N X120 LaW of Large Numbers and CLT imply Yz39 WIS 1Yi y N X1207 for n large Graphic based on X2 plot 1 Obtain ti y y S1yi y and the ordered 751 S 7571 2 Graph the pair 75122 Where x is the 1001 5 n percentile of the X120 distribution Plot should show a straight line for Normal observations More formal tests based on Skewness of variates ti More in Mardia Sections 18 57 and D Agostino and Stephens 1986 More Examples of the Multivariate General Linear Model We did 1 Tests for u with unknown 2 and 2 Tests for E with unknown 1 MANOVA Multivariate Analysis of Variance Why is it ANOVA MANOVA We re testing means not variability Should it be ANOME MANOME Denote I normal data matrices by ygl Y yg2 Kyl Where yij N NpLz39 Z 17 7 Lz397 Z 171 1 Main Hypothesis to Test Equality of Means HO M1 a given 21 E This is Multivariate ANOVA MANOVA How to Test a LR Under HO fl y and 30 S Under H1 Hi E 2771 yz39jnz39 and 231 Wn Where W 2211 niSi Let B 2 nS W Note that under HO W N Wp n I and B N WPEI 1 and B and W are independent These imply that VVB WI N Amen 5 we reject the null if this is small or if B is large b UIT Univariate ANOVA rejects equal means if I I Z T2 z391 z391 is large SO we reject HOa if I I Z niayi y2 Z nia Sia z391 z391 is large The maximum value over a is the largest eigenvalue of I I ZmSal 2mm mi y W 1B Compare LRT and UIT B WW IW 1B We see LRT and UIT not the same for general I They were the same when I 1 Hotelling T2 2 Special case of MANOVA I 2 LRT gives the TWO sample Hotelling T2 test UIT also reduces to TWO sample Hotelling T2 test SO LRT and UIT same for I 1 2 not in general 3 Test for equality of Variances Recall Bartlett s univariate test of 0 0 Reject the null if nlns2 2211 nilns is large Multivariate Extension H02 21 Z Z 2 Under H0 MLE of 2139 is pooled S 2 221 Under H1 MLE s are 2i 2 Si We have MYS 281 12SI I2 SO 21mm nlnlSl 221 nilnSi 221 nilnSflS Distribution For large samples X120p1112 Small sample adjustment BOX s M Test 4 a Linear Constraints on the Mean In 1 We considered the test of equal means H0 211 I L Often we want to test a relationship between means For example in the Iris data we saw that Versicolor 7E Setosa There is a 3rd species of Iris Virginica and gene structure suggests 311 2 13 212 In general Consider the hypothesis q H0 3 Z 51 0 11 2 unknown How to test this hypothesis Let S denote the unbiased estimator of E from sam ple z39 z39 1 I Then the test statistic is the Hotelling T2 based on 231 55 q q T2 CZ y HOS1Z iyi Mo z391 z391 vvhere S is the unbiased pooled estimator of E and 6 1 2 231 Distribution T2T2p7n q Critical values from the F distribution b Comparing components of the mean vector Eg Compare several competing treatments to a control or placebo This is testing HO ul u1u11 1 This is a special case of H0Alc Where A is a 7 X p matrix and c is a given r vector In this case the test statistic is the Hotelling T2 based on 71 7 7 lAV C ASA 1AV C N F0 71 7 5 Behren s Fisher Problem Univariate data how to test M1 2 L2 from indepen dent normal obs lVlLE of M1 n2 is E1 E2 Var l T2 0n1 03712 estimated by 8711 83712 Approximate T test tn 3 T1 E2Sin1 83712 2 For large samples tn N N 0 1 for small samples Welch 1949 Biometrika gives approximate df Multivariate K Sample Behren s Fisher Problem Test H0411 L given 21 7E 7E 2 H1411 7E 7E L given 21 7E 7E 2 Approach to test Under H1 MLE s are componentwise means and variances Under HO Use differentiation to get MLE for u then get Eli s as before Need to iterate to nd solutions Find the logLikelihoods under HO and H1 Distribution of 2lnY Large Samples Use X2 approximation Small samples Can use the bootstrap MANCOVA In ANOVA we ve considered continuous responses Y and discrete covariates X Now allow continous and discrete covariates General Setup Y X1B1 X2B2 E where E is an n X 19 data matrix of errors from Np0 Z We will need two stage least squares to t this model Lemma Assume X1 and X2 are full rank and their columns are linearly independent Then X 2RX2 is nonsingular Where R I X1X 1X1 1X1 Proof Need to show that if aX2RX2 0 then a 0 But a X 2R RX2a a X 2RX2a 0 gt RXga 0 So X23 X1X 1X11X 1X2a le say which implies a0 andb0 Two Stage Least Squares Rewrite the Model as YXBE Where X X1X2 and B 1B 2 Let B1 and B2 denote the estimators of B1 and B2 in the full model B1 denotes the estimator of B1 in the submodel B2 0 Using Geometry we have that the projection of X1B1I X21 2 by I R onto the span of X1 is Xl l In matri CBS X1X 1X11X 1X1 1 Xgi g X1B1 OI39 X11 1 X1X 1X1 1X 1X2f32 X1B1 Thus we have B1 2 B1 X 1X1 1X 1X2f32 Note X2B2 X1X 1X1 1X 1X2f32 Y XE Y XB1 Now premultiply by X2 note that Y XE is or thogonal to X2 so that Xg szu 2 Xg RY Now use the Lemma to obtain 32 X2 RX21X2 RY We have El 2 X 1X1 1X 1Y when B2 0 and B1 and 32 as given above in the full model Testing First univariate responses ANCOVA Setup 3 X151 X252 ea Where the columns of X1 and X2 are independent A Estimation Have 51 XiX11Xiy 62 Xz RX21 1X2 Ry 61 31 X1X11X1X252 RSS 3 y X131 X252ly X131 X232 2 y Ry X232 Ry B Tests of hypotheses H1520 Note RSSH y Ry y X131 y X131 and the LRT rejects H1 if RSSH RSS is large SO the LRT is the F test KRSSH RSSwRssn q m under H1 H22A510 Recall X X1X2 then put A 0 Aogb 3 X39X1Xy and A043 A0X X 1A6 1A043 N Fam q T under H2 For Multivariate Observations Replace y by Y A Parameter estimates B1 B1 32 as before and let E Y X1B1 X2i 2yY X1B1 X2B2 B Tests of hypotheses H12B20 E Y RY X2B2 RY H 232an E N Wp2n q 7 and H N Wp2r are independent As before F H N Apn q 737 under H1 and p Values come from the Chi squared or F dist H2 AB1 0 As before X X17X2a the pm A 0 AOCI i X X 1X Y and H A0lt AOX X1A61AOCiD N Wp2 a and again F NApn q ra under H2 An application of Multivariate to Spatial Statistics Observe Z Zij 139 1n j 1p Index 139 Locations 81 3n Index j Variables of interest Note The rows of Z are not independent A Main Goal Predict Estimate 20k variable k at new location Optimal Estimator minimizes Elek 20k2 Solution 20k EZOkZ Simpler Assume a linear estimator n P 20k 2 Z Z 11 j1 Restrictions on Aij7S Require Unbiasedness Taking Expectations M 21 231 Ari7 So 2171 Am 2 1 and 2171 Aij 0 for j 7E k NOW assume Lj 0 for j 1p EZOk 20k TL p TL p TL p Eng 2 Z Z AijZOkZzj l Z Z Z Z AiinjZijZijq SO z1 j1z391j1 I 014407 0 2 21 231 AijOjkwa 0 z1 j1 271 1 231 331 AiinjOjjW F ij7 say Minimize FW subject t0 2 1 and EM Oforj 7A k 11 21 HOW La Grange Multipliers De ne 0130 2M 1 and Gm ZAM forj 7A k z391 z391 G 01 pr and Z mG Take derivatives Wrt Aij7s and m Get up 1 equations have up p unknowns Unique Solution Simpler p 1 Minimize CO 0 2 Z AZC0 z Z Z AiAiOQ zquot z391 z391 11 Z 17 subject to Giz39 1n Zn 1 0 z 1 Let m be a Lagrange multiplier and let Taking derivatives dHdAz39 2C 0z 2231 AiCz39zquot m dHdm Z 1 Ai 1 Setting them to 0 we have CHLQ 3AyC U n 2 271 and z391 or letting Oij Cz39j O11 O12 CM 1 A1 001 O21 O22 O2n 1 A2 002 1 1 H 1 0 n 2 1 Compactly CA 2 00 We have A 3 100 These are the Kriging weights and 20 2 Z 11 is the Kriging Predictor of 20 Practical issue How to get the Covariance function needed for CM 19 variables situation is called Co Kriging Need to estimate and model all covariance func tions Factor Analysis Introduction Similar to PC goal is to nd a few factors that de scribe a large number of variables First famous factor analysis From a number of students taking 3 exams in Clas sics 1 French 2 and English Spearman in 1904 gave the following Correlation matrix 1 083 078 S 083 1 067 078 067 1 Spearman believed one factor general ability gen erates the three exams Ie 371 Alf 151 372 Z Agf 112 133 2 Agf U3 Where f is the underlying common factor A1 A2 A3 are the factor loadings and ui s are errors Two sources of errors 1 Ability in a subject not perfectly determined by any factor 2 Measurement on exam not a perfect measure of ability in subject Model Let X be a random p Vector with EX u and Varx Z A model with k factors is given by X uzAf l u Where A is a p x k matrix of parameters and f and u are random vectors The items in f are common factors and the items in u are unique factors Usual Assumptions All factors are uncorrelated and common factors stan dardized ComponentWise k 33139 Hi I ZAz jfj 151 j1 for i 1 p Thus k j1 Two components to Varx communality h Zia A12 unique variance 1 Using assumptions 1 5 have 2 2 AA II Factor Loadings A are not unique X u Af u implies that X u AGGf u for any orthogonal matrix G In practice choose ONE speci c rotation G Via a constraint This condition induces kk 1 2 constraints on parameters What is the simpli cation Via a factor model Recall Z AA I II E has 191 12 parameters A has pk parameters II has 19 parameters But Orthogonality constraint induces kk 1 2 con straints on parameters Have 8 pp 12 pkp kk 1V2 EG Test Score Data 195 fork1s5 for 22321 fork3s 2 In general 3 gt 0 gt no unique solution 3 0 gt one exact solution 8 lt 0 gt in nitely many solutions 3 gt 0 is reasonable factor model Estimation A Principal Factor Analysis B Maximum Likelihood Practical Issue x u AGgtltG fgt u How to choose the rotation matrix G Goal is to get disjoint sets of variables each set asso ciated With one factor Then the factor fj is a weighted average over the variables i for Which A is large One 7 such rotation is the Varimax rotation Main use of Factor Analysis is Psychological and Ed ucational testing Give an exam to patients With 71 items want to nd out What the exam is measuring Find k fac tors try to interpret the meaning of the variables in each Test for MVN assumption We know that for MVN observations Yz39 H21Yi 0 N X120 LaW of Large Numbers and CLT imply Yz39 WIS 16 y N X120 for 71 large Graphic based on X2 plot 1 Obtain t1 yz y S1yZ y and the ordered ta 3 1tn 2 Graph the pair ta x Where X12 is the 1001 5 71 percentile of the X120 distribution Plot should show a straight line for Normal observations More formal tests based on Skewness of variates ti More in Mardia Sections 18 57 and D Agostino and Stephens 1986 More Examples of the Multivariate General Linear Model We did 1 Tests for u With unknown 2 and 2 Tests for E with unknown u MANOVA Multivariate Analysis of Variance Why is it ANOVA MANOVA We re testing means not variability Should it be ANOME MANOME Denote I normal data matrices by 3 21 Y 3 22 Ky Where yij N NpLz39 Z 1 nZ Z 1 1 Main Hypothesis to Test Equality of Means H0 211 a given 21 Z This is Multivariate ANOVA MANOVA How to Test a LR Under H0 fl y and 30 S Under H1 Hi E 231 yijquotz39 and 231 W Where W 2 221 nilSi Let B 2 718 W Note that under H0 W N Wp2n I and B N WpZI 1 and B and W are independent These imply that VVB W N Am a b we reject the null if this is small or if B is large b UIT Univariate ANOVA rejects equal means if I I T2 i1 i1 is large So we reject HOa if I I Emam y2 niaSia 11 11 is large The maximum value over a is the largest eigenvalue of I I 27150 1 y3 z y W1B 11 11 Compare LRT and UIT B WW I W 1B We see LRT and UIT not the same for general I They were the same when I 1 Hotelling T2 2 Special case of MANOVA I 2 LRT gives the Two sample Hotelling T2 test UIT also reduces to Two sample Hotelling T2 test So LRT and UIT same for I 1 2 not in general 3 Test for equality of Variances Recall Bartlett s univariate test of of 0 Reject the null if nln32 221 71117183 is large Multivariate Extension H02 21 Z Z 2 Under H0 MLE of 21 is pooled S 2 221 Under H1 MLE s are 21 2 Si We have MYSquot281quot12SIquotI2 So 2lnY nlnS 211 mlnlSzl 2f1mzns13 Distribution For large samples xp1112 Small sample adjustment Box s M Test 4 a Linear Constraints on the Mean In 1 We considered the test of equal means H0 211 I 21 Often we want to test a relationship between means For example in the Iris data we saw that Versicolor 7E Setosa There is a 3rd species of Iris Virginica and gene structure suggests 311 2 13 212 In general Consider the hypothesis q H0 1 Z 521 M0 i1 2 unknown How to test this hypothesis Let S denote the unbiased estimator of E from sam ple i i 1 I Then the test statistic is the Hotelling T2 based on 231 67 q q T2 CZ 67 Mo S 1Z zE 0 11 11 Where S is the unbiased pooled estimator of Z and c 1 2 231 Distribution T2T2pn q Critical values from the F distribution b Comparing components of the mean vector Eg Compare several competing treatments to a control or placebo This is testing H0 Ll L LI1 MI This is a special case of HOZALZC Where A is a 7quot X 19 matrix and c is a given r vector In this case the test statistic is the Hotelling T2 based On 71 rrAy c ASA 1A7 c N F7quot 71 7quot 5 Behren s Fisher Problem Univariate data how to test M1 2 L2 from indepen dent normal obs MLE of M1 M2 is T1 E2 Var l T2 Ifm 03712 estimated by sfm 83712 Approximate T test tn 1 Ug l T28in1 83712 2 For large samples tn N N 0 1 for small samples Welch 1949 Biometrika gives approximate df Multivariate K Sample Behren s Fisher Problem Test H0 211 a given 21 7E 7E 21 H1 211 7E 7E a given 21 7E 7E 2 Approach to test Under H1 MLE s are componentwise means and variances Under H0 Use differentiation to get MLE for u then get Eli s as before Need to iterate to nd solutions Find the logLikelihoods under H0 and H1 Distribution of 2ZnY Large Samples Use X2 approximation Small samples Can use the bootstrap MANCOVA In ANOVA we ve considered continuous responses Y and discrete covariates X Now allow continous and discrete covariates General Setup Y X1B1 X2B2 E where E is an n X 19 data matrix of errors from Np0 Z We will need two stage least squares to t this model Lemma Assume X1 and X2 are full rank and their columns are linearly independent Then X 2RX2 is nonsingular where R I X1X 1X1 1X1 Proof Need to show that if aXglRX2 0 then a 0 But a XgR RXga a X 2RX2a 0 gt RXga 0 So X23 X1X 1X11X 1X2a le say which implies a0 andb0 Two Stage Least Squares Rewrite the Model as YXBE where X X1X2 and B 1B 2 Let B1 and B2 denote the estimators of B1 and B2 in the full model B1 denotes the estimator of B1 in the submodel B2 0 Using Geometry we have that the projection of X1B1 I X2B2 by I R onto the span of X1 is X1B1 In matri CGSZ X1X 1X11X 1X1 1 X2132 X1B1 01 X1131 X1X 1X1 1X 1X2f32 X1B1 Thus we have B1 2 B1 X 1X11X 1X2 2 Note X232 X1X 1X1X 1X2 2 Y X13 2 Y XB1 Now premultiply by X2 note that Y XE is or thogonal to X2 so that Xg RX2 2 Xg RY NOW use the Lemma to obtain 32 X2 RX21X2 RY We have B1 X 1X1 1X 1Y when B2 0 and B1 and 32 as given above in the full model Testing First univariate responses ANCOVA Setup 3 X131 X232 e Where the columns of X1 and X2 are independent A Estimation Have Bl XiX11Xiy 82 X2 RX21 1X2 Ry Bl 81 X1X11X1X2B2 R55 5 y X181 X282ly X181 X282 2 y Ry X252 Ry B Tests of hypotheses 13128220 Note RSSH y Ry y X181 y X181 and the LRT rejects H1 if RSSH RSS is large So the LRT is the F test HRSSH RssgtrRssltn q m under H1 H22A610 Recall X X1X2 then put A 0 Amt X X 1Xy and A0IlAOXX1A6llAo N Fan q r under H2 For Multivariate observations Replace y by Y A Parameter estimates B1 B1 32 as before and let E Y X1B1 X2132IY X1B1 X2132 B Tests of hypotheses H12B20 E Y RY X2B2 RY H X232 RY E N Wp2n q 7quot and H N Wp27 are independent As before F N Apn q 737quot under H1 and p Values come from the Chi squared or F dist H2 AB1 0 As beforeX X1X2a then pm A 0 AOCID i X X 1X Y and H AocigtyA0X X1Ag1Aoc N W142 a and again F Apn q ra under H2 An application of Multivariate to Spatial Statistics Observe Z Zij i 1 n j 1 p Index 139 Locations 31 3 Index j Variables of interest Note The rows of Z are not independent A Main Goal Predict Estimate 20k variable k at new location Optimal Estimator minimizes E 20k 20k2 Solution 20k EZOkZ Simpler Assume a linear estimator n P 20k 2 Z Z 11 j1 Restrictions on LijS Require Unbiasedness Taking Expectations W 21 231 ijLj So 21 Aik 1 and 221 A 0 forj 7E k NOW assume uj 0 forj 1 p EZOk 20k n p n P n P EZ k 2 Z Z AijZOkZZj l Z Z Z Z Aiinjzijzijl 11 j1 11 j1 171 371 Civic 0 2 21 231 AijCjkm 0 21 1 231 231 Aiin j iji 1quot Fz j say So Minimize FWj subject to 2M 1 and EM Oforj 7 11 21 HOW La Grange Multipliers De ne Gm 2M 1 and GM 2AM forj 7A k z391 z391 G 01 Gp and Z mG Take derivatives Wrt AZj s and m Get up 1 equations have up p unknowns Unique Solution Simpler p 1 Minimize CO 0 2 i now i i i AiAiCe zquot i1 i1zquot1 Z 1 subject to GOVJ 1 n 2 Zn 1 0 2391 Let m be a Lagrange multiplier and let Taking derivatives dHdAz ZC0 2221 AiCi Iquot m 21 l 1 Setting them to 0 we have Carn23AyC 6 m2 171 and i1 or letting Cij Ciajl5 C11 C12 C17 1 A1 001 C21 C22 C27 1 A2 002 I 5 E E 3 I I 5 I I 5 I 1 1 H 1 0 n 2 1 Compactly C A 2 CO We have A C1c0 These are the Kriging weights and 20 I Z 11 is the Kriging Predictor of 20 Practical issue How to get the Covariance function needed for Cij 19 variables situation is called Co Kriging Need to estimate and model all covariance func tions

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