Competing risks
Survival models are models for continuous risk, e.g. the force of
mortality, μ_{x}. We showed in an earlier post why this is more powerful
and efficient than modelling the rate of mortality,
q_{x}. Of course, if you have huge volumes of data,
you may not lose too much by modelling q_{x}.
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 singledecrement
model. If you have competing risks, however, then modelling
q_{x} 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 termassurance 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:
Simulation 
Lives 
Time lived 
Deaths 
1 
10000 
9954.5 
101 
2 
10000 
9953.5 
88 
3 
10000 
9940.9 
109 
4 
10000 
9939.0 
118 
5 
10000 
9952.8 
89 
Total 
50000 
49740.8 
505 
The estimate of the force of mortality is 505 / 49740.8 = 0.0102.
This can also be approximated by taking the estimate for
q_{x} (505 / 50000 = 0.0101) and converting it with
log(1q_{x}) = 0.0102. Either approach will yield a
close estimate of the underlying force of mortality, which we know to be
0.01.
For q_{x} 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:
Simulation 
Lives 
Time lived 
Deaths 
Lapses

1 
10000 
9497.0

94

936 
2 
10000 
9501.3

81

901 
3 
10000 
9465.3

102

909 
4 
10000 
9441.2

110

966 
5 
10000 
9484.0

84

925 
Total 
50000 
47388.7

471

4637

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 survivalmodel 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 q_{x} is far trickier. We first
note that the formula for the earlier singledecrement model cannot be used
for this doubledecrement 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(10.0094) = 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 exposedtorisk. The value of
q_{x} 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 q_{x} 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 qrate approach to competing risks demands complicated
adjustments which depend on the number of decrements. A failure to
make these adjustments will lead to underestimates of the risks. In
contrast, the μrate 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 μrate in multistate
models, which we also strongly support.
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