Categorical Data

Frühling Rijsdijk1 & Caroline van Baal2

1IoP, London 2Vrije Universiteit, A’dam

Twin Workshop, Boulder Tuesday March 2, 2004

Aims

• Introduce Categorical Data

• Define liability and describe

assumptions of the liability model

• Show how heritability of liability can be

estimated from categorical twin data

• Practical exercises

Categorical data

Measuring instrument is able to only discriminate between two or a few ordered categories : e.g. absence or presence of a disease

Data therefore take the form of counts, i.e. the number of individuals within each category

Univariate Normal Distribution of Liability

Assumptions:

(1) Underlying normal

distribution of liability

(2) The liability distribution has

1 or more thresholds (cut-offs)

The standard Normal distribution

Liability is a latent variable, the scale is arbitrary,

distribution is, therefore, assumed to be a Standard Normal Distribution (SND) or z-distribution:

• mean () = 0 and SD () = 1• z-values are the number of SD away from the mean

• area under curve translates directly to probabilities > Normal Probability Density function ()

-3 3-1 0 1 2-2

68%

Standard Normal Cumulative Probability in right-hand tail(For negative z values, areas are found by symmetry)

Z0 Area

0 .50 50%.2 .42 42%.4 .35 35%.6 .27 27%.8 .21 21% 1 .16 16%1.2 .12 12%1.4 .08 8%1.6 .06 6%1.8 .036 3.6%2 .023 2.3%2.2 .014 1.4%2.4 .008 .8%2.6 .005 .5%2.8 .003 .3%2.9 .002 .2%

-3 3z0

Area=P(z z0)

When we have one variable it is possible to find a z-value (threshold) on the SND, so that the proportion exactly matches the observed proportion of the sample

• i.e if from a sample of 1000 individuals, 150 have met a criteria for a disorder (15%): the z-value is 1.04

-3 3

1.04

Two categorical traits

Trait2Trait1

0 1

0 00 01

1 10 11

When we have two categorical traits, the data are represented in a Contingency Table, containing cell counts that can be translated into proportions

0 = absent1 = present

Trait2Trait1

0 1

0 545 (.76)

75(.11)

1 56(.08)

40(.05)

Categorical Data for twins:

When the measured trait is dichotomous i.e. a disorder either present or absent in an unselected sample of twins:

cell a: number of pairs concordant for unaffected

cell d: number of pairs concordant for affected

cell b/c: number of pairs discordant for the disorder

0 = unaffected1 = affected

Twin2Twin1

0 1

0 00 a

01 b

1 10 c

11 d

Joint Liability Model for twin pairs

• Assumed to follow a bivariate normal distribution

• The shape of a bivariate normal distribution is determined by the correlation between the traits

• Expected proportions under the distribution can be calculated by numerical integration with mathematical subroutines

Bivariate Normal

R=.00 R=.90

Bivariate Normal (R=0.6) partitioned at threshold 1.4 (z-value) on both liabilities

Liab 2Liab 1

0 1

0 .87

.05

1 .05

.03

Expected Proportions of the BN, for R=0.6, Th1=1.4, Th2=1.4

Twin2Twin1 0 1

0 00 a 01 b

1 10 c 11 d

Correlated dimensions:

The correlation (shape) and the two thresholds determine the relative proportions of observations in the 4 cells of the CT.

Conversely, the sample proportions in the 4 cells can be used to estimate the correlation and the thresholds.

ad

bc

acbd

• A variance decomposition (A, C, E) can be applied to liability, where the correlations in liability are determined by path model

• This leads to an estimate of the heritability of the liability

Twin Models

ACE Liability Model

11

Twin 1

C EA

L

C AE

L

Twin 2

Unaf ¯Aff Unaf ¯Aff

1

1/.5

Summary

It is possible to estimate a correlation between categorical traits from simple counts (CT), because of the assumptions we make about their joint distributions

How can we fit ordinal data in Mx?

Summary statistics: CTMx has a built-in fit function for the maximum-likelihood analysis of 2-way Contingency Tables>analyses limited to only two variables

Raw data analyses- multivariate- handles missing data- moderator variables

Model-fitting to CT

Mx has a built in fit function for the maximum-likelihood analysis of 2-way Contingency Tables

The Fit Function is twice the log-likelihood of the observed frequency data calculated as:

ijc

jij

r

i

PnL ln2ln211

nij is the observed frequency in cell ijpij is the expected proportion in cell ij

Expected proportionsAre calculated by numerical integration of the

bivariate normal over two dimensions: the liabilities for twin1 and twin2

e.g. the probability that both twins are affected :

Φ is the bivariate normal probability density function, L1 and L2 are the liabilities of twin1 and twin2, with means 0, and is the correlation matrix of the two liabilities.

21T T

21 dLdL)Σ,0;L,L(

d

For example: for a correlation of .9 and thresholds (z-values) of 1, the probability that both twins are above threshold (proportion d) is around .12

The probability that both twins are are below threshold (proportion a) is given by another integral function with reversed boundaries :

B

BL2

L1

21

T T

21 dLdL)Σ,0;L,L(

a

and is around .80 in this example

log-likelihood of the data under the model subtracted from

log-likelihood of the observed frequencies themselves:

tot

ijc

1jij

r

1i n

nlnn2Lln2

χ² statistic:

The model’s failure to predict the observed data i.e. a bad fitting model,is reflected

in a significant χ²

Model-fitting to Raw Ordinal Data

ordinal ordinalZyg respons1 respons21 0 01 0 01 0 12 1 02 0 01 1 12 . 12 0 .2 0 1

Model-fitting to Raw Ordinal Data

The likelihood of a vector of ordinal responses is computed by the Expected Proportion in

the corresponding cell of the MN

Expected proportion are calculated by numerical integration of the MN normal over n dimensions. In this example it will be two, the liabilities for twin1 and twin2

2121 ),( dxdxxxT

T

2121 ),( dxdxxxT

T

2121 ),( dxdxxxT T

2121 ),( dxdxxx

T T

(0 1)

(1 0)

(0 0) (1 1)

2121 ),( dxdxxxT T

is the MN pdf, which is a function of, the correlation matrix of the variables

By maximizing the likelihood of the dataunder a MN distribution, the ML estimate of the

correlation matrix and the thresholds are obtained

Practical Exercise 1

Simulated data for 625 MZ and 625 DZ pairs (h2 = .40 c2 = .20 e2 = .40 > rmz=.60 rdz=.40) Dichotomized 0 = bottom 88%, 1 = top 12%This corresponds to threshold (z-value) of 1.18

Observed counts:MZ DZ

0 1 0 10 508 48 0 497 591 3534 1 49 20

Raw ORD File: bin.datScripts: tetracor.mx and ACEbin.mx

Practical Exercise 2Same simulated data Categorized 0 = bottom 22%, 1 = mid 66%,2 = top 12%This corresponds to thresholds (z-values) of -0.75 1.18

Observed counts:MZ DZ

0 1 2 0 1 20 80 58 1 0 63 74 21 68 302 47 1 71 289 572 1 34 34 2 4 45 20Raw ORD File: cat.datAdjust the correlation and ACE script

Threshold Specification in Mx

2 CategoriesThreshold Matrix : 1 x 2T(1,1) T(1,2) threshold twin1 & twin2

3 CategoriesThreshold Matrix : 2 x 2T(1,1) T(1,2) threshold 1 for twin1 & twin2T(2,1) T(2,2) threshold 2 for twin1 & twin2

Threshold Specification in Mx

3 Categoriesnthresh=2 nvar=2Matrix T: nthresh nvar (2 x 2)T(1,1) T(1,2) threshold 1 for twin1 & twin2T(2,1) T(2,2) increment

L LOW nthresh nthreshValue 1 T 1 1 to T nthresh nthresh

Threshold Model L*T /

1 01 1

t11 t12

t21 t22

* =t11 t12

t11 + t21 t12 + t22

Using Frequency Weights

ordinal ordinalZyg respons1 respons2

FREQ

1 0 0 5081 0 1 481 1 0 351 1 1 342 0 0 4972 0 1 592 1 0 492 1 1 20

The 1250 lines data file (bin.dat) can be summarized like this

Using Frequency WeightsG1: Data and model for MZ correlationDAta NGroups=2 NInput_vars=4 Missing=.Ordinal File=binF.datLabels zyg bin1 bin2 freqSELECT IF zyg = 1SELECT bin1 bin2 freq /DEFINITION freq /

Begin Matrices; R STAN 2 2 FREE !Correlation matrixT FULL nthresh nvar FREE !thresh tw1, thresh tw2L LOW nthresh nthreshF FULL 1 1End matrices;Value 1 L 1 1 to L nthresh nthresh ! initialize L

COV R / !Predicted Correlation matrix for MZ pairs

Thresholds L*T / !to ensure t1>t2>t3 etc.......

FREQ F /

The likelihood of this vector of observations is the Expected Proportion in

the corresponding cell of the MN :

2211

1 2

2211

1 2

1

),,,( dydxdydxyxyxtx ty

tx ty

ty

Example: 2 Variables measured in twins:

x has 2 cat > 0 below Tx1 , 1 above Tx1

y has 3 cat > 0 below Ty1 , 1 Ty1 - Ty2 , 2 above Ty2

Ordinal respons vector (x1, y1, x2, y2) For example (1 2 0 1)

Proband-Ascertained Samples

For rare disorders (e.g. Schizophrenia), selecting a random sample of twins will lead to the vast majority of pairs being unaffected.

A more efficient design is to ascertain twin pairs through a register of affected individuals.

When an affected twin (the proband) is identified, the cotwin is followed up to see if he or she is also affected.

There are several types of ascertainment

Twin 1Twin 2

0 1

0 00 a

01 b

1 10 c

11 d

Complete Ascertainment

Types of ascertainment

ProbandCo- twin

0 1

0 00 a

01 b

1 10 c

11 d

Single Ascertainment

Omission of certain classes from observation leads to an increase of the likelihood of observing the remaining individuals

Mx corrects for incomplete ascertainment by dividing the likelihood by the proportion of the population remaining after ascertainment

CT from ascertained data can be analysed in Mx by simply substituting a –1 for the missing cells

Ascertainment Correction

CTable 2 2-1 11-1 13

Summary

For a 2 x 2 CT: 3 observed statistics, 3 parameters (1 correlation, 2 threshold) df=0 any pattern of observed frequenciescan be accounted for, no goodness of fit of the normal distribution assumption.

This problem is solved when we have a CTwhich is at least 3 x 2: df>0A significant 2 reflects departure from normality.

SummaryPower to detect certain effects increaseswith increasing number of categories >continuous data most powerful

For raw ordinal data analyses, the first category must be coded 0!

Threshold specification when analyzing CT aredifferent