Estimates and Tests Without Covariates

The simplest question we might ask about the multiple event types is whether the type-specific hazard functions are the same for all event types, that is, h_j(t) = h(t) for all j. For the leader data, just looking at the frequencies of the three event types suggests that deaths due to natural causes (n=27) are much less likely to occur than the other two types, which have identical frequencies (n=165). In fact, we can easily obtain a formal test of the null hypothesis from these three frequencies. If the null hypothesis of equal hazards is correct, the expected frequencies of the three event types should be equal in any time interval. Thus, we have an expected frequency of 119=(472 - 115)/3 for each of the three types. Calculating Pearson’s chi-square test (by hand) yields a value of 88.9 with 2 d.f. So we can certainly conclude that the hazard for naturally occurring deaths is lower than for the other two exit modes.

Although the hazards are not equal, it’s still possible that they might be proportional (in a different sense than that of Chapter 5); that is, if the hazard for death changes with time, the hazards for constitutional and nonconstitutional exits may also change by a proportionate amount. We can write this hypothesis as

Equation 6.6

where the ω_js are constants of proportionality.

We can obtain a graphic examination of this hypothesis by using PROC LIFETEST to estimate log-log survivor functions for each of the three event types. If the hazards are proportional, the log-log survivor functions should be parallel, as explained in Chapter 5 (see the section Estimating Survivor Functions). Here’s the SAS code to do it:

proc lifetest data=leaders notable outsurv=a;
   time years*lost(0,2,3);
run;

proc lifetest data=leaders notable outsurv=b;
   time years*lost(0,1,3);
run;

proc lifetest data=leaders notable outsurv=c;
   time years*lost(0,1,2);
run;

data combined;
   set a b c;
   retain outcome 0;
   if years=0 and survival=1 then outcome=outcome+1;
   lls=log(-log(survival));
run;

proc gplot data=combined;
   where _censor_=0;
   symbol1 interpol=join color=black v=none line=1;
   symbol2 interpol=join color=black v=none line=2;
   symbol3 interpol=join color=black v=none line=8;
   plot lls*years=outcome;
run;

For each type of exit, we estimate the survivor curve by treating the other types as censoring. The DATA step concatenates the three data sets, creates a variable called OUTCOME (with values of 1, 2, or 3 corresponding to constitutional exits, natural deaths, and nonconstitutional exits) and computes the log-log survivor function. Finally the three curves are plotted on a single graph.

Output 6.1 shows the log-log survivor curves for the three event types. The curves for constitutional and nonconstitutional exits are virtually indistinguishable. Not surprisingly, the curve for natural deaths is much lower than the other two curves. There is also some tendency for the natural death curve to move closer to the other two in later years, which is evidence of nonproportionality.

We can also examine smoothed hazard plots using the SMOOTH macro described in Appendix 1, “Macro Programs.” In this case, we can invoke the macro with this statement, which produces the plot in Output 6.2:

%smooth(data=combined,time=years,width=4)

Here, we see more dramatic differences between the plots for constitutional and nonconstitutional exits. The hazard for constitutional exits drops much more rapidly at first, but then rises again after year 15. By contrast, the hazard for nonconstitutional exit declines rather slowly. In interpreting these curves, it’s important to keep in mind that for both constitutional and nonconstitutional exits, more than 80 percent of the exits occurred before the seventh year. Consequently, the hazard estimates for later years are based on a relatively small number of observations and may be unreliable.

Output 6.2. Smoothed Hazard Functions for Three Types of Exits

[View full size image]

Cox and Oakes (1984) proposed a parametric test of the proportional hazards hypothesis in equation (6.6). Consider the model

Equation 6.7

If β_j= β for all j, then the proportional hazards hypothesis is satisfied. Otherwise this model says that the log-hazards for any two event types diverge linearly with time. Cox and Oakes showed that if there are two event types, equation (6.7) implies a logistic regression model for type of event, with time of the event as an independent variable. Under the proportional hazards hypothesis, the coefficient for time will be 0.

For more than two event types, equation (6.7) implies a multinomial logit model for event type. Here’s how to estimate that model with PROC CATMOD:

proc catmod data=leaders;
   where lost ne 0;
   direct years;
   model lost=years / noprofile;
run;

Notice that leadership spells that are still in progress (LOST=0) when the observation is terminated are excluded from the analysis. The DIRECT statement tells CATMOD to treat YEARS as a quantitative rather than a categorical variable. Output 6.3 shows selected portions of the analysis.

Output 6.3. Test of Proportionality with PROC CATMOD

               MAXIMUM-LIKELIHOOD ANALYSIS-OF-VARIANCE TABLE
Source                    DF   Chi-Square      Prob
             --------------------------------------------------
             INTERCEPT                 2        69.33    0.0000
             YEARS                     2        17.89    0.0001

             LIKELIHOOD RATIO         42        68.45    0.0061



                  ANALYSIS OF MAXIMUM-LIKELIHOOD ESTIMATES

                                             Standard    Chi-
      Effect            Parameter  Estimate    Error    Square    Prob
      -----------------------------------------------------------------
      INTERCEPT                 1    0.0134    0.1432     0.01   0.9253
                                2   -2.5393    0.3140    65.40   0.0000
      YEARS                     3  -0.00391    0.0267     0.02   0.8834
                                4    0.1394    0.0359    15.10   0.0001

Looking first at the ANOVA table, we see that the effect of YEARS is highly significant, indicating a rejection of the proportionality hypothesis. The coefficients in the lower half of the output tell us which hazard functions are proportional and which are not. Parameter 3 is the β coefficient for the contrast between type 1 (constitutional exit) and type 3 (nonconstitutional exit). The chi-square statistic is minuscule, indicating that proportionality cannot be rejected for these two hazard types. Moreover, the fact that parameter 1 is also near 0 suggests that the hazard functions for these two event types are nearly identical. On the other hand, parameter 4 is highly significant. This coefficient is a contrast between the type 2 hazard (death from natural causes) and the type 3 hazard (nonconstitutional exit). Its value tells us that the hazard for natural death increases much more rapidly with time than the hazard for nonconstitutional exit; specifically, their ratio increases by about 15 percent each year (calculated as 100(e^.1394–1)).