Variance explained for binomial proportions

Question

In Rosenthal and Rubin (1979) ``A Note on Percent Variance Explained as A Measure of the Importance of Effects'', they give an example of where $r^2$ is deceptively low:

Suppose half the patients in a medical study are randomly assigned to a new medical treatment (X = 1) while the other half are assigned the standard medical treatment (X = 0). The dependent variable is “alive one year after treatment” (Y = 1) vs. “dead” (Y = 0). Suppose the study obtained the frequencies displayed in Table 1. Under the standard medical treatment, 30%of the patients live; while under the new treatment, 70% of the patients live. This certainly is a dramatic and important effect. Yet, for these data r2 = .16. The conclusion that the treatment is unimportant because it accounts for only 16%of the variance is simply wrong. Percent variance explainied can, in some cases, then, be a very deceptive measure.

The numbers are

      | Dead (Y=0) | Alive (Y=1) |
-------------------------------------
X = 0 | 35         | 15          | 50   
X = 1 | 15         | 35          | 50
-------------------------------------
      | 50         | 50          | 100

How do you actually calculate the $r^2$ here?

Answer 1

This is calculated directly from the formula for binomial variance, $np(1-p)$. First, for the null model, not using treatment as predictor, the estimated $p$ is $50/100=0.5$. The variance of the observation "total number alive" is then $np(1-p)=100\cdot 0.5 \cdot 0.5=25$. Under the alternative model, using the treatment predictor, we have two probability parameters, for each of the two groups defined by the treatment, which is $15/50=0.3$ and $35/50=0.7$, leading to a total binomial variance $50 \cdot 0.3 \cdot 0.7 + 50 \cdot 0.7 \cdot 0.3 = 21$. So the reduction in variance by the model is $25-21=4$, which is indeed $16\%$ of the total variance: $4/25=0.16$.