PATH ANALYSIS

(see Chapters 5,6 of Neale & Cardon)

- allows us to represent linear models for the relationships betweenvariables in diagrammatic form

e.g. a genetic model; a factor model; a regression model.

- makes it easy to derive expectations for the variances and covariancesof variables in terms of the parameters of the proposed linear model.

- permits easy translation into a matrix formulation as used by programssuch as MX.

dcae

P

E A C D

PHENOTYPE

P = eE + aA + cC + dD

e.g. UNIQUEENVIRONMENT

ADDITIVEGENETICEFFECT

SHAREDENVIRONMENT

DOMINANCEGENETICEFFECT

1 1 11

Figure 1. A PATH MODEL FOR MZ TWIN PAIRS

EXOGENOUSVARIABLES

ENDOGENOUSVARIABLES

1.0 (MZT) or 0.0 (MZA)

ET1 AT1 CT1 DT1

dcae dcae

1

PT2PT1

1

1 1 1 1 1 1ET2 AT2 CT2 DT2

1.0

1.0

TWIN 1 TWIN 2

CONVENTIONS OF PATH ANALYSIS

Squares or rectangles denote observed variables.

Circles or ellipses denote latent (unmeasured) variables.

Upper-case letters are used to denote variables.

Lower-case letters (or numeric values) are used to denote covariances or path coefficients.

Single-headed arrows or paths ( ) are used to represent hypothesized causal relationships between variables under a particular model -- where the variable at the tail of the arrow is hypothesized to have a direct causal influence on the variable at the head of the arrow (e.g., AT1 PT1).

Double-headed arrows ( ) are used to represent a covariance between two variables, which may arise through common causes not represented in the model. Double-headed arrows ( ) may also be used to represent the variance of a variable.

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 (A P) or from endogenous variables to other endogenous variables(P1 P2).

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.

ASSUMPTIONS OF OUR SIMPLE PATH MODEL FOR MZ TWIN PAIRS:

(i) linear additive effects -- no multiplicative (genotype x environment) interactions;

(ii) E, A, C, D are mutually uncorrelated -- no genotype-environment covariance;

(iii) path coefficients e, a, c, d are the same for first, second twins;

(iv) no reciprocal sibling environmental influences; i.e., no direct paths P1 P2

TRACING RULES OF PATH ANALYSIS

(Simplified rules assuming no ‘feedback loops’, i.e., no reciprocal causation -reciprocal sibling environmental influences - etc.)

(1) Trace backwards, change direction at a 2-headed arrow, then traceforwards.

implies that we can never trace through two 2-headed arrows inthe same chain.

(2) The expected covariance between two variables, or the expectedvariance of a variable, is computed by multiplying together all thecoefficients in a chain, and then summing over all possible chains.

CONTRIBUTION

e.g., for the MZ twin pair covariance we have:

a 1 aa2

c 1 c

d 1 d

CovMZ = a2 + c2 + d2

P1 AT1 AT2 P2

P1 CT1 CT2 P2 c2

P1 DT1 DT2 P2 d2

e.g., for the total phenotypic variance for MZ twins:

CONTRIBUTIONe 1 e

e2

a 1 a

c 1 c

d 1 d

VarMZ = e2 + a2 + c2 + d2

P1 ET1 ET1 P1

P1 AT1 AT1 P1 a2

P1 CT1 CT1 P1 c2

P1 DT1 DT1 P1 d2

PATH ANALYSIS: EXERCISES

Exercise 1. Using Figure 1, what is the expected covariance between MZtwin pairs reared apart (MZAs)?

Exercise 2. Using Figure 1, what is the expected variance of twins fromMZ pairs reared apart?

Figure 2. DZ TWIN PAIRS/FULL SIBLINGS

1 (DZT/FST) or 0 (DZA/FSA)

D E A C E A C D

ched dche

1 1 1 1 1 1 1 1

0.25

0.5 (1+x)

x = 0 under random mating.

PTWIN 2PTWIN 1

Exercise 3. Using Figure 2, and assuming random mating (x=0), what isthe expected covariance between DZ twin pairs reared together?

Exercise 4. Using Figure 2, what is the expected variance of DZ twins orfull siblings reared together?

Exercise 5. Using Figure 2, what is the expected covariance of DZ twinsor full sibling pairs reared apart?

Figure 3. BIOLOGICAL PARENT - OFFSPRING

E C A D E C A D

dacedace

0.25

PMOTHER PFATHER

0.50.5

0.5

0.5

E C A D D A C E

dhce echd

PCHILD 2PCHILD 1

RA

1

RA

1

0.5

0.5

1 1 1 1 1 1 1 1

1 1 11 11

1

Exercise 6. Using Figure 3, what is the expected covariance of parent andoffspring?

Exercise 7. Using Figure 3, what is the expected covariance of biologicalparent and adopted-away offspring?

Exercise 8. What 10 assumptions are implied by Figure 3?

Figure 4. BIOLOGICAL UNCLE / AUNT - NEPHEW / NIECE

1

0.5

E C A D E C A D

d

PAUNT/UNCLE

PMOTHER

ace dace

0.25

E C A D

0.5

0.5

PNEPHEW/NIECE

E C A D

dace

RA 0.5

PFATHER

dace

1

Exercise 9. Using Figure 4, what is the expected covariance betweenbiological uncle and nephew?

Figure 5. OFFSPRING OF MZ TWIN PAIRS

E D C A A C D E

d ace

PSPOUSE1

½

E D A C E D A C

cade

RA

0.5

1 1 1 1 1 1 1 1

1 1

1

1

11

PTWIN1 PSPOUSE2

POFFSPRING (Twin 1)POFFSPRING (Twin 2)

E D C A

½

1 1 1 1

c eda

PTWIN2

A C D E1 1 1 1

cade

A

0.5

R

11

1

d ace c eda

½ ½

Exercise 10. Using Figure 5, what is the expected covariance betweenMZ uncle/aunt and nephew/niece, i.e. between an MZ twin and the child of his or her cotwin?

Exercise 11. What is the expected covariance between first cousins related through MZ twins (“MZ half sibs”)?

Figure 6. DIRECTION-OF-CAUSATION MODEL: MZ PAIRS

E C1 1

CONDUCTPROBLEMS TWIN 1

e c

E C1 1

CONDUCTPROBLEMS TWIN 2

e c

1

E A1 1

ALCOHOLDEPENDENCE TWIN 1

e a

E A1 1

ALCOHOLDEPENDENCE TWIN 2

e a

1 ii

Exercise 12. Using Figure 6,

(1) What is the expected MZ covariance for(a) conduct problems;(b) alcohol dependence;(c) conduct problems in one twin and alcohol

dependence inthe cotwin?

(2) Redraw Figure 6 so that the direction of causation is now

ALCOHOL DEPENDENCE CONDUCT

PROBLEMS

What are the expected MZ covariances now for(a) conduct problems;(b) alcohol dependence;(c) conduct problems in one twin and alcohol

dependence inthe cotwin?

(3) What are the corresponding expected DZ covariances when

(a) CONDUCT PROBLEMS ALCOHOL DEPENDENCE

(b) ALCOHOL DEPENDENCE CONDUCT PROBLEMS

PART TWO:

PATH ANALYSIS: FROM PATH DIAGRAMS TO MATRICES

It is straightforward to convert from a path diagram to a matrix representation of a path model, and vice versa. Here we illustrate a computationally efficient approach -though other alternatives exist (see e.g. the MX manual).

Consider first a model with no feedback loops --for example Figure 1 for MZ twin pairs reared together, where all endogenous variables are directly observable, and there are no paths from endogenous variables to other endogenous variables.

Let X denote the covariance matrix for the exogenous variables. The i, j-th element of the matrix X (where i denotes Row, and j denotes Column) will be equal to the covariance of the i-th, j-th exogenous variables. X will be a square matrix.

XMZ =

1.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0

0.0 1.0 0.0 0.0 0.0 1.0 0.0 0.0

0.0 0.0 1.0 0.0 0.0 0.0 1.0 0.0

0.0 0.0 0.0 1.0 0.0 0.0 0.0 1.0

0.0 0.0 0.0 0.0 1.0 0.0 0.0 0.0

0.0 1.0 0.0 0.0 0.0 1.0 0.0 0.0

0.0 0.0 1.0 0.0 0.0 0.0 1.0 0.0

0.0 0.0 0.0 1.0 0.0 0.0 0.0 1.0

Let W denote the matrix of paths from the exogenous variables to the endogenous variables. The i, j-th element of the matrix W will be equal to the path from the j-th exogenous variable to the i-th endogenous variable. W will be a rectangular matrix, with number of rows equal to the number of endogenous variables, number ofcolumns equal to the number of exogenous variables.

W = e a c d 0 0 0 00 0 0 0 e a c d

For this simplified case, with no paths from endogenous variables to other endogenous variables, and all endogenous variables directly observable, we can derive the expected covariance matrix as a product of matrices,

MZ = WXMZ W

WX = e a c d 0 a c d

0 a c d e a c d

e a c d 0 a c d

0 a c d e a c dWXW = e 0

a 0

c 0

d 0

0 e

0 a

0 c

0 d

MZ = e2 + a2 + c2 + d2 a2 + c2 + d2

a2 + c2 + d2 e2 + a2 + c2 + d2

Exercise 13. Using Figure 2, write down the elements of the covariance matrix of the exogenous variables for DZ pairs, XDZ.

Derive DZ = WXDZ W

Figure 7 represents a more complicated model, with paths

from endogenous variables to other endogenous variables --

in this example, representing the reciprocal environmental

influences of the twins upon each other (or perhaps a rater

contrast effect, if twin data are ratings by the same parent).

Figure 7. MODEL WITH RECIPROCAL SIBLING EFFECTS

1 (MZT/DZT/FST) or 0 (MZA/DZA/FSA)

D E A C E A C D

caed dcae

1 (MZ) or 0.25 (DZ/FS)

1 (MZ) or 0.5 (DZ/FS)

PTWIN 2PTWIN 1

i - reciprocal sibling interaction effect

0 (MZA/DZA/FSA) or i (MZT/DZT/FST)

0 (MZA/DZA/FSA) or i (MZT/DZT/FST)

Y = 0 i

i 0

Let Y denote the matrix of paths from endogenous variables to other endogenous variables. Yi,j is the path from the j-th variable to the i-th variable, with Yi,i = 0 by definition.

It may be shown (see Neale and Cardon, C6), again assuming that all endogenous variables are directly observable, that

where Z = (I - Y)-1

and

Here I is the identity matrix, and -1 denotes the inverse.

MZ = ZWXMZ W Z

DZ = ZWXDZ W Z

Exercise 14. What are the expected covariance matrices for MZ and DZ twin pairs reared together, and for MZ and DZ twin pairs reared apart, under the model of Figure 7?

Exercise 15. Redraw Figure 6 for the case of reciprocal causation, i.e. a path i from ALCOHOL DEPENDENCE to CONDUCT PROBLEMS, in addition to the path i from CONDUCT PROBLEMS to ALCOHOL DEPENDENCE. What is the expected 4x4 covariance matrix for MZ pairs for this case?

THE END