Univariate Model Fitting Sarah Medland QIMR – openMx workshop Brisbane 16/08/10.

Post on 14-Dec-2015

215 views 1 download

transcript

Univariate Model Fitting

Sarah Medland

QIMR – openMx workshop

Brisbane 16/08/10

Univariate Twin Models= using the twin design to assess

the aetiology of one trait (univariate)

1. Path Diagrams

2. Basic ACE R Script

1. Path Diagrams for Univariate Models

Basic Twin Model - MZ

Twin 1 Trait

A C E

c ea

Twin 2 Trait

E C A

c ae

1.0

1 1 1 111

1.0

Basic Twin Model - DZ

Twin 1 Trait

A C E

c ea

Twin 2 Trait

E C A

c ae

0.5

1 1 1 111

1.0

Basic Twin Model

Twin 1 Trait

A C E

c ea

Twin 2 Trait

E C A

c ae

rMZ = 1.0;

rDZ = 1.0

rMZ = 1.0;

rDZ = 0.5

1 1 1 111

The Variance

Since the variance of a variable is

the covariance of the variable

with itself, the expected variance

will be the sum of all paths from

the variable to itself, which follow

Wright’s rules

Variance for Twin 1 - A

Twin 1 Trait

A C E

c ea

1 1 1 a*1*a = a2

Variance for Twin 1 - C

a*1*a = a2

c*1*c = c2

Twin 1 Trait

A C E

c ea

1 1 1

Variance for Twin 1 - E

a*1*a = a2

c*1*c = c2

e*1*e = e2

Twin 1 Trait

A C E

c ea

1 1 1

Total Variance for Twin 1

a*1*a = a2

c*1*c = c2

e*1*e = e2

Twin 1 Trait

A C E

c ea

1 1 1

Total Variance = a2 + c2 + e2

Covariance - MZ

Twin 1 Trait

A C E

c ea

Twin 2 Trait

E C A

c ae

1.0

1 1 1 111

1.0 a2 + c2

Covariance - DZ

Twin 1 Trait

A C E

c ea

Twin 2 Trait

E C A

c ae

0.5

1 1 1 111

1.00.5a2 +

c2

Variance-Covariance Matrices

MZ Twin 1 Twin 2

Twin 1 a2 + c2 + e2

a2 + c2

Twin 2 a2 + c2 a2 + c2 + e2

Variance-Covariance Matrices

DZ Twin 1 Twin 2

Twin 1 a2 + c2 + e2

0.5a2 + c2

Twin 2 0.5a2 + c2 a2 + c2 + e2

Variance-Covariance Matrices

MZ Twin 1 Twin 2

Twin 1 a2 + c2 + e2 a2 + c2

Twin 2 a2 + c2 a2 + c2 + e2

DZ Twin 1 Twin 2

Twin 1 a2 + c2 + e2 0.5a2 + c2

Twin 2 0.5a2 + c2 a2 + c2 + e2

Why isn’t e2 included in the covariance?

Because, e2 refers to environmental influences UNIQUE to each twin. Therefore, this cannot explain why there is similarity between twins.

Why is a2 only .5 for DZs but not MZs?

Because DZ twins share on average half of their genes, whereas MZs share all of their genes.

2. Basic openMx ACE Script

OverviewOpenMx script

Running the script

Describing the output

ACE model

# Fit ACE Model with RawData and Matrices Input# -----------------------------------------------------------------------univACEModel <- mxModel("univACE", mxModel("ACE", # Matrices a, c, and e to store a, c, and e path coefficients mxMatrix( type="Lower", nrow=nv, ncol=nv, free=TRUE, values=10, label="a11", name="a" ), mxMatrix( type="Lower", nrow=nv, ncol=nv, free=TRUE, values=10, label="c11", name="c" ), mxMatrix( type="Lower", nrow=nv, ncol=nv, free=TRUE, values=10, label="e11", name="e" ), # Matrices A, C, and E compute variance components mxAlgebra( expression=a %*% t(a), name="A" ), mxAlgebra( expression=c %*% t(c), name="C" ), mxAlgebra( expression=e %*% t(e), name="E" ),# Algebra to compute total variances and standard deviations (diagonal only) mxAlgebra( expression=A+C+E, name="V" ),

Specify thematrices you

needTo build the

model

Twin 1 Trait

A C E

c ea

1 1 1

Do some algebra to get the variances

Start values?

a2 = additive genetic variance (A)c2 = Shared E variance (C)e2 = Non-shared E variance (E) Sum is modelled to be expected

Total VarianceStart Values for a, c, e:

(Total Variance / 3)

Twin 1 Trait

A C E

c ea

1 1 1

Standardize parameter estimates

# Algebra to compute total variances and standard deviations (diagonal only)

mxAlgebra( expression=A+C+E, name="V" ), mxMatrix( type="Iden", nrow=nv, ncol=nv, name="I"), mxAlgebra( expression=solve(sqrt(I*V)), name="iSD"), # Algebra to compute standardized path estimares and variance

components mxAlgebra( expression=a%*%iSD, name="sta"), mxAlgebra( expression=c%*%iSD, name="stc"), mxAlgebra( expression=e%*%iSD, name="ste"), mxAlgebra( expression=A/V, name="h2"), mxAlgebra( expression=C/V, name="c2"), mxAlgebra( expression=E/V, name="e2"),The regression coefficient a is standardized by: (a * SD(A)) / SD(Trait)

where SD(Trait) is the standard deviation of the dependent variable, and SD(A) is the standard deviation of the predictor, the latent factor ‘A’ (=1)

Twin 1 Trait

A

a

1

V1 * = V V

SDinv

= 1/SD a * 1/SD = a/SD

Standardize parameter estimates

# Algebra to compute total variances and standard deviations (diagonal only)

mxAlgebra( expression=A+C+E, name="V" ), mxMatrix( type="Iden", nrow=nv, ncol=nv, name="I"), mxAlgebra( expression=solve(sqrt(I*V)), name="iSD"), # Algebra to compute standardized path estimares and variance

components mxAlgebra( expression=a%*%iSD, name="sta"), mxAlgebra( expression=c%*%iSD, name="stc"), mxAlgebra( expression=e%*%iSD, name="ste"), mxAlgebra( expression=A/V, name="h2"), mxAlgebra( expression=C/V, name="c2"), mxAlgebra( expression=E/V, name="e2"),The heritability ‘h2’ is the proportion of the total variance due to A (additive

genetic effects; = A / V. Note: this will be “sta” squared. The standardized variance components for C and E are: C / V; E / V N

Twin 1 Trait

A

sta

1

VA / = “h2” “sta”2

Standardising dataV = A + C + E A/V = 73/233= .31

V = [73] + [90] + [70] C/V = 90/233= .39

V = [233] E/V = 70/233= .30

a = 8.55c = 9.49e = 8.35

SD = sqrt(V) = 15.26

sta = 8.55 / 15.26 = 0.56 squared = .31stc = 9.49 / 15.26 = 0.62 squared = .39ste = 8.35 / 15.26 = 0.55 squared = .30

# Algebra for expected variance/covariance matrix in MZ mxAlgebra( expression= rbind ( cbind(A+C+E , A+C), cbind(A+C , A+C+E)),

name="expCovMZ" ), # Algebra for expected variance/covariance matrix in DZ mxAlgebra( expression= rbind ( cbind(A+C+E , 0.5%x%A+C), cbind(0.5%x%A+C , A+C+E)),

name="expCovDZ" ) MZ Twin 1 Twin 2

Twin 1 a2 + c2 + e2 a2 + c2

Twin 2 a2 + c2 a2 + c2 + e2

DZ Twin 1 Twin 2

Twin 1 a2 + c2 + e2 0.5a2 + c2

Twin 2 0.5a2 + c2 a2 + c2 + e2

Twin 1 Trait

A C E

c ea

Twin 2 Trait

E C A

c ae

1/ 0.5

1 1 1 111

1.0

mxModel("MZ", mxData( observed=mzData, type="raw" ), mxFIMLObjective( covariance="ACE.expCovMZ",

means="ACE.expMean", dimnames=selVars ) ), mxModel("DZ", mxData( observed=dzData, type="raw" ), mxFIMLObjective( covariance="ACE.expCovDZ",

means="ACE.expMean", dimnames=selVars ) ), mxAlgebra( expression=MZ.objective + DZ.objective,

name="m2ACEsumll" ), mxAlgebraObjective("m2ACEsumll"), mxCI(c('ACE.A', 'ACE.C', 'ACE.E')))univACEFit <- mxRun(univACEModel, intervals=T)univACESumm <- summary(univACEFit)univACESumm

Sub-models# Fit AE model# -----------------------------------------------------------------------univAEModel <- mxModel(univACEFit, name="univAE", mxModel(univACEFit$ACE, mxMatrix( type="Lower", nrow=1, ncol=1, free=FALSE, values=0, label="c11",

name="c" ) ))univAEFit <- mxRun(univAEModel)univAESumm <- summary(univAEFit)univAESumm

You can fit sub-models to test the significance of your parameters

-you simply drop the parameter and see if the model fit is significantly worse than full

model

Twin 1 Trait

A E

Twin 2 Trait

E A

Sub-models# Fit CE model# -----------------------------------------------------------------------univCEModel <- mxModel(univACEFit, name="univCE", mxModel(univACEFit$ACE, mxMatrix( type="Lower", nrow=1, ncol=1, free=FALSE, values=0, label="a11", name="a" ) ))univCEFit <- mxRun(univCEModel)univCESumm <- summary(univCEFit)univCESumm

Twin 1 Trait

C E

Twin 2 Trait

E C

# Fit E model# -----------------------------------------------------------------------univEModel <- mxModel(univAEFit, name="univE", mxModel(univAEFit$ACE, mxMatrix( type="Lower", nrow=1, ncol=1, free=FALSE, values=0, label="a11",

name="a" ) ))univEFit <- mxRun(univEModel)univESumm <- summary(univEFit)univESumm

Twin 1 Trait

E

Twin 2 Trait

E

The E parameter can never not be dropped because it includes measurement error

OpenMx Output

univACESummfree parameters: name matrix row col Estimate 1 a11 ACE.a 1 1 8.5465042 c11 ACE.c 1 1 9.4884543 e11 ACE.e 1 1 8.3521974 mean ACE.Mean 1 1 94.147803

observed statistics: 2198 estimated parameters: 4 degrees of freedom: 2194 -2 log likelihood: 17639.91 saturated -2 log likelihood: NA number of observations: 1110 chi-square: NA p: NA AIC (Mx): 13251.91 BIC (Mx): 1127.663 adjusted BIC: RMSEA: NA

tableFitStatisticsmodels compared to saturated model

Name ep -2LL df AIC diffLL diffdf p

M1 : univTwinSat 10 17637.98 2188 13261.98 - - - M2 : univACE 4 17639.91 2194 13251.91 1.93 6 0.93M3 : univAE 3 17670.38 2195 13280.38 32.4 7 0 M4 : univCE 3 17665.27 2195 13275.27 27.3 7 0 M5 : univE 2 18213.28 2196 13821.28 575.3 8 0

Smaller -2LL means better fit. -2LL of sub-model is always higher (worse fit). The question is: is it significantly worse.Chi-sq test: dif in -2LL is chi-square distributed. Evaluate sig of chi-sq test. A non-sig p-value means that the model is consistent with the data.

Nested.fitmodels compared to ACE model

Name ep -2LL df diffLL diffdf p

univACE 4 17639.91 2194 NA NA NAunivAE 3 17670.38 2195 32.47 1 0 univCE 3 17665.27 2195 25.36 1 0 univE 2 18213.28 2196 573.37 2 0

Smaller -2LL means better fit. -2LL of sub-model is always higher (worse fit). The question is: is it significantly worse.Chi-sq test: dif in -2LL is chi-square distributed. Evaluate sig of chi-sq test. Critical Chi-sq value for 1 DF = 3.84A non-sig p-value means that the dropped parameter(s) are non-significant.

Estimates ACE model> univACEFit$ACE.h2 [,1][1,] 0.3137131

> univACEFit$ACE.c2 [,1][1,] 0.3866762

> univACEFit$ACE.e2 [,1][1,] 0.2996108