# 1 Regression as Moment Structure. 2 Regression Equation Y =  X + v Observable Variables Y z = X Moment matrix  YY  YX  =  YX  XX Moment structure.

## Presentation on theme: "1 Regression as Moment Structure. 2 Regression Equation Y =  X + v Observable Variables Y z = X Moment matrix  YY  YX  =  YX  XX Moment structure."— Presentation transcript:

1 Regression as Moment Structure

2 Regression Equation Y =  X + v Observable Variables Y z = X Moment matrix  YY  YX  =  YX  XX Moment structure  =   2  XX +  vv  XX  =  XX  XX Parameter vector  = ( ,  XX,  vv )’

3 Sample: z 1, z 2,..., z n n iid Sample Moments S = n -1  z i z i ’ s yy s yx S = s yx s xx Fitting S to  =  Estimator  S close to  3 moment equations s yy =  2  XX +  vv s yx =  XX s xx =  XX with 3 (unknown) parameters Parameter estimates  = (s yx /s xx, s XX, s yy - (  ) 2 s XX )’  is the same as the usual OLS estimate of  ^ ^ ^ ^ ^ ^

4 Regression Equation Y =  x + v X = x + u Observable Variables Y z = X Moment structure  =   2  XX +  vv  XX  =  XX  XX +  uu Parameter vector  = ( ,  XX,  vv,  uu )’ new parameter

5 Sample: z 1, z 2,..., z n n iid Sample Moments S := n -1  z i z i ’ s yy s yx S = s yx s xx Fitting S to  =  Estimator  = S close to  3 moment equations s yy =  2  xx +  vv s yx =  xx s xx =  xx +  uu with 4 (unknown) parameters Parameter estimates  = ??  is the same as the usual OLS estimate of  ^ ^^ ^ ^

6 The effect of measurement error in regression x Y X u v  Y =  (X -u)+ v =  X + (v -  u) =  X + w, where w = v -  u Note that w is correlated with X, unless u or  equals zero So, the classical LS estimate b of  is neither ubiased, neither consistent. In fact, b --->  YX /  XX =  xx /  XX )= k  k is the so called Fiability coefficient (reliability of X). Since 0  k  1 b suffers from downward bias

7 Regression Equation Y =   x 1 +   x 2...+  p x p + v X k = x k + u k Observable Variables b = S XX -1 S XY does not converge to  b* := (S XX -  uu ) -1 S XY In multiple regression Examples with EQS of regression with error in variables Using suplementary information to assessing the magnitude of variances of errors in variables.

8 Path analysis & covariance structure Example with ROS data

9 Sample covariance matrix ROS92 ROS93ROS94 ROS95 ROS9272.07 ROS9329.5636.21 ROS9430.2131.0946.51 ROS9527.6324.0435.19 46.62 Mean: 6.27 7.35 10.02 8.80 n = 70 ROS92 ROS93 ROS94 F b1b2b3 SEM: bj = ? It is a valid model ?

10 Calculations b 1 b 2 = 29.56 b 1 b 3 = 30.21 b 2 b 3 = 31.09 b 1 b 2 /b 1 b 3 = b 2 /b 3 = 29.56/30.21--> b 2 =.978b 3 31.09 = b 2 b 3 = b 3 (.978b 3 ) --> b 3 2 = 31.09/.978 b 3 = 5.64 In the same way, we obtain b 1 =5.34 b 2 =5.52 Model test in this case is CHI2 = 0, df = 0

11 Fitted Model R92R94R93 F 5.345.525.64 CHI2 = 0, df = 0 43.345.8014.74 1

12 /TITLE FACTOR ANALYSIS MODEL (EXAMPLE ROS) /SPECIFICATIONS CAS=70; VAR=4; /LABEL V1=ROS92; V2=ROS93; V3=ROS94; V4=ROS95; /EQUATIONS V1 = *F1 + E1; V2 = *F1 + E2; V3 = *F1 + E3; /VARIANCES F1 = 1.0; E1 TO E3 = *; /COVARIANCES /MATRIX 72.07 29.56 36.21 30.21 31.09 46.51 27.63 24.04 35.19 46.62 /END

13 ROS92 =V1 = 5.359*F1 + 1.000 E1.974 5.504 ROS93 =V2 = 5.516*F1 + 1.000 E2.650 8.482 ROS94 =V3 = 5.637*F1 + 1.000 E3.753 7.482 VARIANCES OF INDEPENDENT VARIABLES ---------------------------------- E D --- --- E1 -ROS92 43.347*I I 8.205 I I 5.283 I I I I E2 -ROS93 5.789*I I 3.924 I I 1.475 I I I I E3 -ROS94 14.736*I I 4.693 I I 3.140 I I I I

14 … with the help of EQS RESIDUAL COVARIANCE MATRIX (S-SIGMA) : ROS92 ROS93 ROS94 V 1 V 2 V 3 ROS92 V 1 0.000 ROS93 V 2 0.000 0.000 ROS94 V 3 0.000 0.000 0.000 CHI-SQUARE = 0.000 BASED ON 0 DEGREES OF FREEDOM STANDARDIZED SOLUTION: ROS92 =V1 =.631*F1 +.776 E1 ROS93 =V2 =.917*F1 +.400 E2 ROS94 =V3 =.827*F1 +.563 E3

15 one - factor four- indicators model R93R95R94 F ** * CHI2 = ?, df = ? p-value = ? R92 * ****

16 … with the help of EQS /TITLE FACTOR ANALYSIS MODEL (EXAMPLE ROS) ! This line is not read /SPECIFICATIONS CAS=70; VAR=4; /LABEL V1=ROS92; V2=ROS93; V3=ROS94; V4=ROS95; /EQUATIONS V1 = *F1 + E1; V2 = *F1 + E2; V3 = *F1 + E3; V4 = *F1 + E4; /VARIANCES F1 = 1.0; E1 TO E4 = *; /COVARIANCES /MATRIX 72.07 29.56 36.21 30.21 31.09 46.51 27.63 24.04 35.19 46.62 /END

17 ROS92 =V1 = 4.998*F1 + 1.000 E1.966 5.175 ROS93 =V2 = 4.837*F1 + 1.000 E2.622 7.779 ROS94 =V3 = 6.417*F1 + 1.000 E3.653 9.833 ROS95 =V4 = 5.393*F1 + 1.000 E4.710 7.590 VARIANCES OF INDEPENDENT VARIABLES ---------------------------------- E D --- --- E1 -ROS92 47.090*I I 8.437 I I 5.581 I I I I E2 -ROS93 12.810*I I 2.775 I I 4.616 I I I I E3 -ROS94 5.332*I I 3.017 I I 1.767 I I I I E4 -ROS95 17.536*I I 3.682 I I 4.763 I I … with the help of EQS

18 RESIDUAL COVARIANCE MATRIX (S-SIGMA) : RESIDUAL COVARIANCE MATRIX (S-SIGMA) : ROS92 ROS93 ROS94 ROS95 V 1 V 2 V 3 V 4 ROS92 V 1 0.000 ROS93 V 2 5.383 0.000 ROS94 V 3 -1.862 0.049 0.000 ROS95 V 4 0.676 -2.048 0.583 0.000 CHI-SQUARE = 6.271 BASED ON 2 DEGREES OF FREEDOM PROBABILITY VALUE FOR THE CHI-SQUARE STATISTIC IS 0.04347 STANDARDIZED SOLUTION: ROS92 =V1 =.631*F1 +.776 E1 ROS93 =V2 =.917*F1 +.400 E2 ROS94 =V3 =.827*F1 +.563 E3

19 Fitted Model R93R95R94 F 4.846.425.40 CHI2 = 6.27, df = 2 p-value =.043 R92 4.99 17.545.3312.8147.10

20 /TITLE FACTOR ANALYSIS MODEL (EXAMPLE ROS) /SPECIFICATIONS CAS=70; VAR=4; /LABEL V1=ROS92; V2=ROS93; V3=ROS94; V4=ROS95; /EQUATIONS V1 = *F1 + E1; V2 = *F1 + E2; V3 = *F1 + E3; V4 = *F1 + E4; /VARIANCES F1 = 1.0; E1 TO E4 = *; /COVARIANCES /CONSTRAINTS (V1,F1)=(V2,F1)=(V3,F1)=(V4,F1); /MATRIX 72.07 29.56 36.21 30.21 31.09 46.51 27.63 24.04 35.19 46.62 /END

21 … estimation results ROS92 =V1 = 5.521*F1 + 1.000 E1.528 10.450 ROS93 =V2 = 5.521*F1 + 1.000 E2.528 10.450 ROS94 =V3 = 5.521*F1 + 1.000 E3.528 10.450 ROS95 =V4 = 5.521*F1 + 1.000 E4.528 10.450 CHI-SQUARE = 12.425 BASED ON 5 DEGREES OF FREEDOM PROBABILITY VALUE FOR THE CHI-SQUARE STATISTIC IS 0.02941

22... EQS use an iterative optimization method ITERATIVE SUMMARY PARAMETER ITERATION ABS CHANGE ALPHA FUNCTION 1 21.878996 1.00000 1.39447 2 5.741889 1.00000 0.43985 3 2.309283 1.00000 0.19638 4 0.477505 1.00000 0.18079 5 0.147232 1.00000 0.18014 6 0.056361 1.00000 0.18008 7 0.014530 1.00000 0.18007 8 0.005784 1.00000 0.18007 9 0.001423 1.00000 0.18007 10 0.000598 1.00000 0.18007

23 Exercise: a)Write the covariance structure for the one - factor four- indicators model b) From the ML estimates of this model, shown in previous slides, compute the fitted covariance matrix. c) In relation with b), compute the residual covariance matrix Note: For c), use the following sample moments: ROS92 ROS93ROS94 ROS95 ROS9272.07 ROS9329.5636.21 ROS9430.2131.0946.51 ROS9527.6324.0435.19 46.62 Mean: 6.27 7.35 10.02 8.80 n = 70

24 one - factor four- indicators model with means R93R95R94 F ** * CHI2 = ?, df = ? p-value = ? R92 * **** 1 * * * *

25 /TITLE FACTOR ANALYSIS MODEL (EXAMPLE ROS data) /SPECIFICATIONS CAS=70; VAR=4; ANALYSIS = MOMENT; /LABEL V1=ROS92; V2=ROS93; V3=ROS94; V4=ROS95; /EQUATIONS V1 = *V999+ *F1 + E1; V2 = *V999+ *F1 + E2; V3 = *V999+ *F1 + E3; V4 = *V999+ *F1 + E4; /VARIANCES F1 = 1.0; E1 TO E4 = *; /COVARIANCES /CONSTRAINTS ! (V1,F1)=(V2,F1)=(V3,F1)=(V4,F1); /MATRIX 72.07 29.56 36.21 30.21 31.09 46.51 27.63 24.04 35.19 46.62 /MEANS 6.27 7.35 10.02 8.80 /END

26 ROS92 =V1 = 6.270*V999 + 4.998*F1 + 1.000 E1 1.022.966 6.135 5.175 ROS93 =V2 = 7.350*V999 + 4.837*F1 + 1.000 E2.724.622 10.146 7.779 ROS94 =V3 = 10.020*V999 + 6.417*F1 + 1.000 E3.821.653 12.204 9.833 ROS95 =V4 = 8.800*V999 + 5.393*F1 + 1.000 E4.822.710 10.706 7.591 VARIANCES OF INDEPENDENT VARIABLES ---------------------------------- E D --- --- E1 -ROS92 47.092*I I 8.437 I I 5.582 I I I I E2 -ROS93 12.810*I I 2.775 I I 4.616 I I I I E3 -ROS94 5.332*I I 3.017 I I 1.767 I I I I E4 -ROS95 17.535*I I 3.682 I I 4.763 I I

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