2 Bivariate Normal Distribution of Liability

When we have only one variable, there is no goodness-of-fit test for the liability model because it always gives a perfect fit. However, this is not necessarily so when we move to the multivariate case. Consider first, the example where we have two variables, each measured as a simple `yes/no' binary response. Data collected from a sample of subjects could be summarized as a contingency table:

	Item 1
Item 2	No	Yes
Yes	13	55
No	32	15

It is at this point that we encounter the crucial statistical concept of degrees of freedom (df). Fortunately, though important, calculating the number of df for a model is usually very easy; it is simply the difference between the number of observed statistics and the number of parameters in the model. In the present case we have a $2\times 2$ contingency table in which there are four observed frequencies. However, if we take the total sample size as given and work with the proportion of the sample observed in each cell, we only need three proportions to describe the table completely, because the total of the cell proportions is 1 and the last cell proportion always can be obtained by subtraction. Thus in general for a table with $r$ rows and $c$ columns we can describe the data as $rc$ frequencies or as $rc-1$ proportions and the total sample size. The next question is, how many parameters does our model contain? The natural extension of the univariate normal liability model described above is to assume that there is a continuous, bivariate normal distribution underlying the distribution of our observations. Given this model, we can compute the expected proportions for the four cells of the contingency table. The model is illustrated graphically as contour and 3-D plots in Figure 2.2.

Figure 2.2: Contour and 3-D plots of the bivariate normal distribution with thresholds distinguishing two response categories. Contour plot in top left shows zero correlation in liability and plot in bottom left shows correlation of .9; the panels on the right shows the same data as 3-D plots.

The figures contrast the uncorrelated case ($r=0$) with a high correlation in liability ($r=.9$) and are dramatically similar to the scatterplots of data from unrelated persons and from MZ twins, shown in Figures 1.2 and 1.4 on pages  and . By adjusting the correlation in liability and the two thresholds, the model can predict any combination of proportions in the four cells. Because we use 3 parameters to predict the 3 observed proportions, there are no degrees of freedom to test the goodness of fit of the model. This can be seen when we consider an arbitrary non-normal distribution created by mixing two normal distributions, one with $r=+.9$ and the second with $r=-.9$, as shown in Figure 2.3. With thresholds imposed as shown, equal

Figure 2.3: Contour plots of a bivariate normal distribution with correlation .9 (top); and of a mixture of bivariate normal distributions (bottom), one with .9 correlation and the other with -.9 correlation. One threshold in each dimension is shown.

proportions are expected in each cell, corresponding to a zero correlation and zero thresholds, not an unreasonable result but with just two categories we have no knowledge at all that our distribution is such a bizarre non-normal example. The case of a $2\times 2$ contingency table is really a `worst case scenario' for no degrees of freedom associated with a model, since absolutely any pattern of observed frequencies could be accounted for with the liability model. Effectively, all the model does is to transform the data; it cannot be falsified.