A Fixed Focal Rate With Changing Competing Rates

Why a Fixed Focal Rate Does Not Fix the Event Probability

Suppose two populations have the same cause-specific hazard for a focal event, such as hospitalization, but different hazards for a competing first event, such as death before hospitalization. The focal rate is the same in both populations, yet the proportion experiencing hospitalization by a specified time can differ. The reason is that the competing event changes how many people remain event-free and able to experience the focal event.

This distinction is important when comparing populations: a cause-specific hazard describes the instantaneous focal-event rate among people who are still free of every event, whereas focal-event incidence describes the probability of having experienced the focal event by a given time. These are different quantities. The earlier material on cause-specific hazards and subdistribution hazards develops their interpretations; here the focus is the specific implication of holding the focal rate fixed while changing competing rates.

Definition: Consider mutually exclusive types of first event, with one designated as the focal event. Let \(\lambda_1(t)\) be its cause-specific hazard at time \(t\), and let \(\lambda_2(t),\ldots,\lambda_K(t)\) be the hazards for the other first-event types. The focal cumulative incidence function \(F_1(t)\) is the probability of experiencing the focal event by time \(t\), before any competing first event. The event-free survival probability is \(S(t)\), the probability of experiencing no event of any type by time \(t\).

For continuous event times, the basic relationship is that focal incidence accumulates at the focal hazard multiplied by the probability of still being event-free. In particular,

F_1(t)=\int_0^t S(u)\lambda_1(u)\,du, \qquad S(t)=\exp\left\{-\int_0^t \sum_{j=1}^K\lambda_j(u)\,du\right\}.

These expressions assume the stated cause-specific hazards describe the event process for the population under consideration, and that the competing first-event types are defined consistently. The integral shows why the focal hazard alone is not enough to determine the focal incidence: competing hazards affect \(S(u)\), the event-free probability that weights the focal rate at every time.

Constant Rates: The Direct Comparison

First suppose all cause-specific hazards are constant over the time interval of interest. Write the focal rate as \(a>0\), and let \(b\geq 0\) be the sum of the constant rates for all competing first-event types. Then the total first-event hazard is \(a+b\), and the event-free probability at time \(u\) is \(\exp\{-(a+b)u\}\). Substituting this into the incidence formula gives

F_1(t)=\int_0^t a e^{-(a+b)u}\,du =\frac{a}{a+b}\left(1-e^{-(a+b)t}\right).

The model assumes constant hazards over the chosen interval; it does not say that every individual has the same observed event time. The rates describe the event process for the population. The units of \(a\) and \(b\) are inverse time, so the time unit used for \(t\) must match. The resulting \(F_1(t)\) is a probability and has no units.

Worked Example: Same Hospitalization Rate, Different Death Rates

Consider two hypothetical populations followed for five years. In both, the constant hospitalization cause-specific hazard is \(a=0.08\) per year. In Population A, the competing-event hazard is \(b=0.04\) per year; in Population B, it is \(b=0.16\) per year. Assume these are the only event types, the rates are constant, and the event types are recorded consistently.

For Population A, the total first-event rate is \(0.08+0.04=0.12\) per year. Thus,

F_{1,A}(5)=\frac{0.08}{0.12}\left(1-e^{-0.12(5)}\right) =0.6667(1-e^{-0.6}) =0.6667(0.4512) \approx 0.3008.

For Population B, the total rate is \(0.08+0.16=0.24\) per year:

F_{1,B}(5)=\frac{0.08}{0.24}\left(1-e^{-0.24(5)}\right) =0.3333(1-e^{-1.2}) =0.3333(0.6988) \approx 0.2329.

Under this model, the estimated five-year probabilities are about 30.1% in A and 23.3% in B, despite the same hospitalization cause-specific hazard. This is a model-based comparison, not evidence that the competing-event rate caused a difference between real populations.

Why Increasing a Constant Competing Rate Lowers Focal Incidence

The direction of the constant-rate comparison can be shown directly. Hold \(a\) and \(t>0\) fixed, and write \(x=a+b\). Differentiating the focal incidence formula with respect to the competing rate \(b\) gives

\frac{\partial F_1(t)}{\partial b} = a\frac{(1+xt)e^{-xt}-1}{x^2}, \qquad x=a+b.

For \(y=xt>0\), \(e^y>1+y\), so \((1+y)e^{-y}<1\). Therefore the numerator \((1+xt)e^{-xt}-1\) is negative, and \(\partial F_1(t)/\partial b<0\). At a fixed positive follow-up time, increasing the constant competing rate strictly reduces focal incidence when the focal rate remains fixed. The result concerns this specified constant-rate model; it is not a universal statement about every comparison involving time-varying or changing population characteristics.

Worked Example: Several Competing Rates

Suppose a hypothetical population has a constant focal cause-specific hazard of \(0.20\) per year, and consider two years of follow-up. Compare three settings: no competing events (\(b=0\)), a competing rate of \(0.10\) per year, and a competing rate of \(0.30\) per year. Assume the focal and competing rates are constant in each setting.

With \(b=0\), the focal incidence is

F_1(2)=\frac{0.20}{0.20}(1-e^{-0.20(2)}) =1-e^{-0.4} \approx 0.3297.

With \(b=0.10\), the total rate is \(0.30\) per year:

F_1(2)=\frac{0.20}{0.30}(1-e^{-0.30(2)}) =0.6667(1-e^{-0.6}) \approx 0.3008.

With \(b=0.30\), the total rate is \(0.50\) per year:

F_1(2)=\frac{0.20}{0.50}(1-e^{-0.50(2)}) =0.4000(1-e^{-1}) \approx 0.2528.

The focal probabilities decrease from about 33.0% to 30.1% to 25.3% as the constant competing rate rises. In the no-competing-event setting, the focal incidence equals the probability of any event by two years. In the other settings, some people experience a competing first event instead.

Changing Competing Rates: Timing Also Matters

When hazards vary over time, the constant-rate formula no longer applies. The integral expression still applies, however, if the cause-specific hazards are specified as functions of time. It accounts for the event-free probability at each moment, so a competing rate that is high early can reduce the number of people still at risk of the focal event later.

Worked Example: The Same Overall Competing-Rate Pattern in a Different Order

Compare two hypothetical populations over four years. Both have a constant focal hazard of \(0.10\) per year. Population A has competing hazard \(0.10\) per year during years 0 to 2, then \(0.30\) per year during years 2 to 4. Population B has these competing rates in the reverse order. Assume the piecewise-constant hazards are known, the event types are mutually exclusive first events, and there are no other event types.

For A, during the first two years the total rate is \(0.20\) per year. The focal incidence accumulated during that interval is

F_{1,A}(2)=\frac{0.10}{0.20}(1-e^{-0.20(2)}) =0.5000(1-e^{-0.4}) \approx 0.1648.

The probability of remaining event-free at year 2 is \(e^{-0.4}\approx 0.6703\). During years 2 to 4, the total rate is \(0.40\) per year. The focal incidence accumulated in that interval, conditional on being event-free at year 2, is

\frac{0.10}{0.40}(1-e^{-0.40(2)}) =0.2500(1-e^{-0.8}) \approx 0.1377.

Weighting this by the probability of reaching year 2 event-free gives

F_{1,A}(4)=0.1648+e^{-0.4}(0.1377) \approx 0.1648+0.0923 =0.2571.

For B, the first two years have total rate \(0.40\) per year, so

F_{1,B}(2)=\frac{0.10}{0.40}(1-e^{-0.40(2)}) =0.2500(1-e^{-0.8}) \approx 0.1377,

and the event-free probability at year 2 is \(e^{-0.8}\approx 0.4493\). In years 2 to 4, the total rate is \(0.20\) per year. Therefore,

F_{1,B}(4)=0.1377+e^{-0.8}\left\{\frac{0.10}{0.20}(1-e^{-0.20(2)})\right\} =0.1377+0.4493(0.1648) \approx 0.2117.

Both populations have the same focal hazard throughout, and both have the same integrated competing hazard over four years: \(0.10(2)+0.30(2)=0.80\). Yet their four-year focal incidences differ. The higher competing rate occurs earlier in B, leaving fewer people event-free for the later period in which its competing rate is lower. Thus, when rates change over time, their timing—not only their overall amount—can matter.

Interpreting Comparisons and Checking the Model

A useful comparison states both the rate condition and the probability result. For example: “Under the constant-rate model, the populations have the same focal cause-specific hazard, but the population with the higher competing-event hazard has lower focal cumulative incidence by five years.” This does not say that the focal hazard is a probability, or that the focal incidence is determined by that hazard alone.

The constant-rate calculation is useful for isolating the mechanism, but its assumptions need scrutiny before it is used to describe data. Check whether rates plausibly remain constant over the selected period, whether event types are defined and recorded consistently, and whether a single rate adequately summarizes each population. If observed rates change sharply over time, apply the time-varying incidence relationship rather than treating one average rate as constant. If the populations differ in their composition, a difference in their overall incidence can also reflect that composition; the simple calculation does not adjust for it.

The main alternatives answer different questions. A cause-specific hazard comparison focuses on the instantaneous rate among people still free of every event type. A cumulative-incidence comparison focuses on the probability of the focal event by a specified time, accounting for competing events. A subdistribution-hazard comparison uses a different risk-set construction and targets a different rate summary. These quantities should not be substituted for one another: choose the one that matches the question and state its interpretation.

Common Mistakes and Limitations

Assuming equal focal hazards mean equal focal probabilities. They do not when competing hazards differ; the event-free probability also contributes to focal incidence.
Reading a hazard as a probability. A rate such as \(0.08\) per year is not an 8% probability of the focal event in a year. The probability depends on the event process and time interval.
Using the constant-rate formula for changing rates. If rates vary, the event-free probability must be evaluated over time and used in the incidence integral.
Interpreting a population contrast as causal. A numerical contrast under specified rates does not establish that changing a competing rate would cause the focal incidence to change in a real population.
Ignoring the time scale or units. Rates per year must be combined with time in years; mixing months and years changes the calculation.

Finally, these formulas describe incidence from the event hazards being specified. In applied data, estimation also depends on issues such as how outcomes and follow-up are observed. A close numerical fit to a constant-rate calculation does not by itself establish that its assumptions are appropriate.

Key takeaway: With a fixed focal cause-specific hazard, changing the competing hazards can change focal-event incidence by changing the probability of remaining event-free. Under constant rates, a higher competing rate lowers focal incidence at a fixed positive time; under time-varying rates, both the level and timing of competing hazards matter.

Check Your Understanding

Try each question, then compare your reasoning with the solutions.

A population has constant focal hazard \(0.10\) per year and competing hazard \(0.10\) per year. Under the constant-rate model, calculate focal incidence by three years.
For a focal hazard of \(0.12\) per year, competing hazard of \(0.08\) per year, and two years of follow-up, calculate focal incidence.
Two populations have equal constant focal hazards, but Population B has a larger constant competing hazard. Under the assumptions in this tutorial, which has the larger focal incidence at a fixed positive time? Explain.
Why can two populations with the same integrated competing hazard over follow-up still have different focal incidences?
Does a lower focal incidence in a population with a higher competing hazard establish that the competing events caused the difference? Explain.

Solutions. (1) The total rate is \(0.10+0.10=0.20\) per year, so \(F_1(3)=\frac{0.10}{0.20}(1-e^{-0.20(3)})=0.5000(1-e^{-0.6})\approx 0.2256\). The three-year focal incidence is about 22.6%. (2) The total rate is \(0.20\) per year, so \(F_1(2)=\frac{0.12}{0.20}(1-e^{-0.20(2)})=0.6000(1-e^{-0.4})\approx 0.1978\), or about 19.8%. (3) Population A, with the smaller competing rate, has larger focal incidence. The derivative of \(F_1(t)\) with respect to the constant competing rate is negative for \(t>0\), with the focal rate held fixed. (4) The event-free probability depends on when competing hazards occur. An early high competing rate can remove more people from the event-free group before later focal events could occur, even if the integrated competing rate is unchanged. (5) No. The calculation describes a model-based association between specified hazards and incidence. It does not establish a causal effect in real populations, which may differ in other ways or violate the model assumptions.