Survival models are models for continuous risk, e.g. the force of mortality, μx. We showed in an earlier post why this is more powerful and efficient than modelling the rate of mortality, qx. Of course, if you have huge volumes of data, you may not lose too much by modelling qx.
A crucial caveat here is that the risk you are modelling is the only means of exit from the population, i.e. what is known as a single-decrement model. If you have competing risks, however, then modelling qx becomes a lot more complicated. In this situation, having large volumes of data will not help you at all. In contrast, survival models can be fitted to competing risks without any additional assumptions.
By way of example, let's consider a term-assurance portfolio where each policyholder has the identical continuous risk of mortality at rate 0.01 per year. We will start with the simplifying assumption that this is the only means of exit from a policy. Five simulations of a portfolio of 10,000 lives are shown below:
The estimate of the force of mortality is 505 / 49740.8 = 0.0102. This can also be approximated by taking the estimate for qx (505 / 50000 = 0.0101) and converting it with -log(1-qx) = 0.0102. Either approach will yield a close estimate of the underlying force of mortality, which we know to be 0.01.
For qx the problems start when we make the model more realistic and include a competing risk of lapse at rate 0.1 per year. Since the lapse rate is ten times the mortality rate, we obviously expect around ten times more lapses than mortality claims in any portfolio. The five simulations shown below include the two competing risks of mortality and lapse:
This time the estimate of the force of mortality is 471 / 47388.7 = 0.0099, which is little changed from the figure in the previous table, and still close to the known underlying value of 0.01 in the simulations. We note that the number of deaths has gone down because some deaths are occurring after lapse and are hence unknown to us. However, this does not affect the survival-model approach because the time lived has dropped accordingly. In a survival model, the same basic approach applies regardless of the number of risks.
The situation for qx is far trickier. We first note that the formula for the earlier single-decrement model cannot be used for this double-decrement situation. If we simply divide the number of events by the number of lives we get 0.0094, which would convert to a force of mortality of -log(1-0.0094) = 0.0095. This is below the value of 0.01 which we know to underlie the simulations. The problem is that the number of lives on the denominator is a misleadingly high measure of the actual exposed-to-risk. The value of qx we have calculated here is a dependent rate, which is so named because it depends on the other decrements and is not a true, independent estimate of mortality. In order to get a proper answer we would have to modify the formula for qx to take account of the existence of a second risk. To make matters more complicated still, a different correction formula is required for different numbers of competing risks.
Thus, a q-rate approach to competing risks demands complicated adjustments which depend on the number of decrements. A failure to make these adjustments will lead to under-estimates of the risks. In contrast, the μ-rate approach is the same, regardless of how many competing risks there are. This is why the CMI Technical Standards Working Party expressed a preference for the μ-rate in multi-state models, which we also strongly support.