Path Analysis Danielle Dick Boulder 2008

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Presentation transcript:

Path Analysis Danielle Dick Boulder 2008

Path Analysis Allows us to represent linear models for the relationships between variables in diagrammatic form Makes it easy to derive expectation for the variances and covariances of variables in terms of the parameters proposed by the model Is easily translated into matrix form for use in programs such as Mx

Example

Conventions of Path Analysis Squares or rectangles denote observed variables. Circles or ellipses denote latent (unmeasured) variables. (Triangle denote means, used when modeling raw data) Upper-case letters are used to denote variables. Lower-case letters (or numeric values) are used to denote covariances or path coefficients.

Conventions of Path Analysis Single-headed arrows or paths (–>) are used to represent causal relationships between variables under a particular model - where the variable at the tail is hypothesized to have a direct influence on the variable at the head. D –> A Double-headed arrows (<–>) are used to represent a covariance between two variables, which may arise through common causes not represented in the model. They may also be used to represent the variance of a variable. D <–> E

Conventions of Path Analysis Double-headed arrows may not be used for any variable which has one or more single-headed arrows pointing to it - these variables are called endogenous variables. Other variables are exogenous variables. Single-headed arrows may be drawn from exogenous to endogenous variables or from endogenous variables to other endogenous variables.

Conventions of Path Analysis Omission of a two-headed arrow between two exogenous variables implies the assumption that the covariance of those variables is zero (e.g., no genotype-environment correlation). Omission of a direct path from an exogenous (or endogenous) variable to an endogenous variable implies that there is no direct causal effect of the former on the latter variable.

Tracing Rules of Path Analysis Trace backwards, change direction at a double-headed arrow, then trace forwards. This implies that we can never trace through multiple double-headed arrows in the same chain. The expected covariance between two variables, or the expected variance of a variable, is computed by multiplying together all the coefficients in a chain, and then summing over all possible chains.

Example EXOGENOUS VARIABLES ENDOGENOUS VARIABLES

Exercises Cov AB = Cov BC = Cov AC = Var A = Var B = Var C = Var E

Covariance between A and B Cov AB = kl + mqn + mpl

Exercises Cov AB = Cov BC = Cov AC = Var A = Var B = Var C = Var E

Expectations Cov AB = kl + mqn + mpl Cov BC = no Cov AC = mqo Var A = k2 + m2 + 2 kpm Var B = l2 + n2 Var C = o2 Var E = 1

Quantitative Genetic Theory Observed behavioral differences stem from two primary sources: genetic and environmental

Quantitative Genetic Theory Observed behavioral differences stem from two primary sources: genetic and environmental 1 1 EXOGENOUS G E VARIABLES g e ENDOGENOUS P VARIABLES PHENOTYPE

Quantitative Genetic Theory There are two sources of genetic influences: Additive and Dominant 1 1 1 A D E a d e P PHENOTYPE

Quantitative Genetic Theory There are two sources of environmental influences: Common (shared) and Unique (nonshared) 1 1 1 1 A D C E a d c e P PHENOTYPE

In the preceding diagram… A, D, C, E are exogenous variables A = Additive genetic influences D = Non-additive genetic influences (i.e., dominance) C = Shared environmental influences E = Nonshared environmental influences A, D, C, E have variances of 1 Phenotype is an endogenous variable P = phenotype; the measured variable a, d, c, e are parameter estimates

Univariate Twin Path Model 1 1 1 1 A1 D1 C1 E1 a d c e P

Univariate Twin Path Model 1 1 1 1 1 1 1 1 A1 D1 C1 E1 E2 C2 D2 A2 a d c e e c d a PTwin1 PTwin2

Univariate Twin Path Model MZ=1.0 DZ=0.5 MZ=1.0 DZ=0.25 MZ & DZ = 1.0 1 1 1 1 1 1 1 1 A1 D1 C1 E1 E2 C2 D2 A2 a d c e e c d a PTwin1 PTwin2

Assumptions of this Model All effects are linear and additive (i.e., no genotype x environment or other multiplicative interactions) A, D, C, and E are mutually uncorrelated (i.e., there is no genotype-environment covariance/correlation) Path coefficients for Twin1 = Twin2 There are no reciprocal sibling effects (i.e., there are no direct paths between P1 and P2

Tracing Rules of Path Analysis Trace backwards, change direction at a double-headed arrow, then trace forwards. This implies that we can never trace through double-headed arrows in the same chain. The expected covariance between two variables, or the expected variance of a variable, is computed by multiplying together all the coefficients in a chain, and then summing over all possible chains.

Calculating the Variance of P1 MZ=1.0 DZ=0.5 MZ=1.0 DZ=0.25 MZ & DZ = 1.0 1 1 1 1 1 1 1 1 A1 D1 C1 E1 E2 C2 D2 A2 a d c e e c d a PTwin1 PTwin2

Calculating the Variance of P1 MZ=1.0 DZ=0.5 MZ=1.0 DZ=0.25 MZ & DZ = 1.0 1 1 1 1 1 1 1 1 A1 D1 C1 E1 E2 C2 D2 A2 a d c e e c d a PTwin1 PTwin2

Calculating the Variance of P1 MZ=1.0 DZ=0.5 MZ=1.0 DZ=0.25 MZ & DZ = 1.0 1 1 1 1 1 1 1 1 A1 D1 C1 E1 E2 C2 D2 A2 a d c e e c d a PTwin1 PTwin2 P=a(1)a

Calculating the Variance of P1 MZ=1.0 DZ=0.5 MZ=1.0 DZ=0.25 MZ & DZ = 1.0 1 1 1 1 1 1 1 1 A1 D1 C1 E1 E2 C2 D2 A2 a d c e e c d a PTwin1 PTwin2 P=a(1)a + d(1)d + c(1)c + e(1)e

Calculating the Variance of P1 = 1a2 P1 A1 A1 P1 d 1 d = 1d2 P1 D1 D1 P1 c 1 c = 1c2 P1 C1 C1 P1 e 1 e 1e2 = P1 E1 E1 P1 Var P1 = a2 + d2 + c2 + e2

Calculating the MZ Covariance MZ=1.0 DZ=0.5 MZ=1.0 DZ=0.25 MZ & DZ = 1.0 1 1 1 1 1 1 1 1 A1 D1 C1 E1 E2 C2 D2 A2 a d c e e c d a PTwin1 PTwin2

Calculating the MZ Covariance MZ=1.0 DZ=0.5 MZ=1.0 DZ=0.25 MZ & DZ = 1.0 1 1 1 1 1 1 1 1 A1 D1 C1 E1 E2 C2 D2 A2 a d c e e c d a PTwin1 PTwin2

Calculating the MZ Covariance 1 a = 1a2 P1 A1 A2 P2 d 1 d = 1d2 P1 D1 D2 P2 c 1 c = 1c2 P1 C1 C2 P2 e e = P1 E1 E2 P2 CovMZ = a2 + d2 + c2

Calculating the DZ Covariance MZ=1.0 DZ=0.5 MZ=1.0 DZ=0.25 MZ & DZ = 1.0 1 1 1 1 1 1 1 1 A1 D1 C1 E1 E2 C2 D2 A2 a d c e e c d a PTwin1 PTwin2

Calculating the DZ covariance CovDZ = ?

Calculating the DZ covariance 0.5 a = 0.5a2 P1 A1 A2 P2 d 0.25 d = 0.25d2 P1 D1 D2 P2 c 1 c = 1c2 P1 C1 C2 P2 e e = P1 E1 E2 P2 CovDZ = 0.5a2 + 0.25d2 + c2

Twin Variance/Covariance VarTwin1 Cov12 Cov21 VarTwin2 a2 + d2 + c2 + e2 a2 +d2 + c2 a2 + d2 + c2 a2 + d2 + c2 + e2 MZ = a2 + d2 + c2 + e2 0.5a2 + 0.25d2 + c2 0.5a2 + 0.25d2 + c2 a2 + d2 + c2 + e2 DZ =