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Heterogeneity across age cohorts

Heterogeneity across age cohorts. Danielle Posthuma & Hermine Maes Boulder – tc19. 0.5/1.0. 0.5/1.0. E 1. E 1. A 1. A 1. A 2. A 2. E 2. E 2. a female. a male. e female. e male. e female. e male. a female. a male. P 1. P 1. P 2. P 2. Sex differences – same heritability?.

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Heterogeneity across age cohorts

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  1. Heterogeneity across age cohorts Danielle Posthuma & Hermine Maes Boulder – tc19

  2. 0.5/1.0 0.5/1.0 E1 E1 A1 A1 A2 A2 E2 E2 afemale amale efemale emale efemale emale afemale amale P1 P1 P2 P2 Sex differences – same heritability? MZM DZM MZF DZF Boulder 2006 TC19

  3. 0.5/1.0 0.5/1.0 E1 E1 A1 A1 A2 A2 E2 E2 afemale amale efemale emale efemale emale afemale amale P1 P1 P2 P2 Sex differences – same genes? MZM DZM MZF DZF r E1 A1 A2 E2 amale emale efemale afemale P1 P2 Boulder 2006 TC19 DOS

  4. 0.5/1.0 0.5/1.0 E1 E1 A1 A1 A2 A2 E2 E2 ayoung aold eyoung eold eyound eold ayoung aold P1 P1 P2 P2 Differences across age cohorts – same heritability? Boulder 2006 TC19

  5. Exercise I: modifying the script to test for age heterogeneity • Open bmi_young.mx (in \\danielle\AGE_heterogeneity\) • This script: young males, 4 groups: 1 = calculation group – matrix declarations 2 = MZ data 3 = DZ data 4 = calculation group – standardized solution • ADE model, 1 grand mean, so 4 estimated parameters Boulder 2006 TC19

  6. Exercise I: modifying the script to test for age heterogeneity • Change this script so it will allow you to estimate ADE in the young and the older cohort by adding four groups for the older cohort • Then run it • If done correctly you should get -2ll =3756.552 and df = 1759 Boulder 2006 TC19

  7. Required modifications for Exercise I • Copy and paste all 4 groups • Change Select if agecat=2 in the two new data groups • Change matrices = group 5 in the two new data groups • Change #ngroups = 8 Boulder 2006 TC19

  8. Exercise II: Testing AE vs ADE for both cohorts • Test whether D is significant using option multiple • Check the unstandardized and standardized estimates of A and E in the young and old cohort under the AE model Boulder 2006 TC19

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  11. Since D is not significant, we will use the AE model to compare submodels: Drop W 1 1 1 W 5 1 1 Option issat End Save AE_cohort.mxs Boulder 2006 TC19

  12. Exercise III: Equality of variance components across age cohorts Add the option multiples with EQUATE command, use a get in between and test whether • ayoung = aold ? • eyoung = eold ? Boulder 2006 TC19

  13. Exercise III: Fit results Chi-squared 4.093 d.f. 1 Probability 0.043 Chi-squared 3.954 d.f. 1 Probability 0.047 • ayoung = aold ? EQ X 1 1 1 X 5 1 1 end get AE_cohort.mxs • eyoung = eold ? EQ Z 1 1 1 Z 5 1 1 End Boulder 2006 TC19

  14. Standardized estimates • Testing whether the standardized estimates of heritability and environment are equal across age. • As the standardized parameters are calculated we cannot change them in an option multiple, and cannot use the EQ statement. Instead we need to use a constraint group Boulder 2006 TC19

  15. Constraint group Title G9: Constraint Constraint Begin Matrices ; S comp 1 4 =S4 ; T comp 1 4 =S8 ; End matrices ; Constraint S=T; Option sat=3760.030, 1761 Option df = -3 End Boulder 2006 TC19

  16. Add up to 1 Fixed to 0 Adjusting df : Option df = -3 • MATRIX S [=P@V~] YOUNG 0.7924 0.0000 0.2076 0.0000 • MATRIX S [=P@V~] OLD 0.7046 0.0000 0.2954 0.0000 Boulder 2006 TC19

  17. Exercise IV: Testing homogeneity of standardized estimates • We want to test whether this constraint provides a worse fit than the unconstrained AE model, so fix D for both cohorts to zero • #ngroups #9 • Add constraint group • Fit results if done correctly: Chi-squared 6.030 d.f 1 Probability 0.014 Boulder 2006 TC19

  18. Testing whether the standardized estimates are equal is more significant, but now we cannot really distinguish anymore between A and E, as A+E = 1 • Danielle Dick will talk about interpretation of heterogeneity in unstandardized vs standardized parameters Boulder 2006 TC19

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