Identifying Separable Effects

From Component-Specific Hazards to a Mixed-Regime Risk

A separable effect compares outcomes under interventions on treatment components, including combinations that may not have been assigned as complete treatments. The central identification task is to determine the event risk under such a mixed combination using information from the treatment arms that were observed. The key is to combine the components at the level of cause-specific hazards, then derive the risk from those hazards. Finite-interval event probabilities cannot generally be mixed in the same way.

Consider two mutually exclusive first events: event 1, the event of interest, and event 2, the competing event. Let treatment be decomposed into components $A_Y$ and $A_D$, each taking values 0 or 1. The component $A_Y$ is the component allowed to affect the event-1 hazard; $A_D$ is the component allowed to affect the event-2 hazard. A regimen $(a_Y,a_D)$ sets both components to specified values. Write $T^{a_Y,a_D}$ for the time to the first event under that regimen and $J^{a_Y,a_D}$ for its type.

Definition: For event type $j$, the cause-specific hazard under regimen $(a_Y,a_D)$ is the instantaneous event rate at time $t$, conditional on having experienced neither event before $t$: $$ \lambda_j^{a_Y,a_D}(t) = \lim_{\Delta t\downarrow 0} \frac{ P(t\leq T^{a_Y,a_D}<t+\Delta t,\ J^{a_Y,a_D}=j \mid T^{a_Y,a_D}\geq t) }{\Delta t}. $$ It has units of inverse time. The cumulative incidence $F_1^{a_Y,a_D}(t)$ is the probability of experiencing event 1 by time $t$ before event 2.

The dismissible component conditions (DCC) assert that, among those still event-free under each regimen, the event-1 hazard does not depend on $A_D$, and the event-2 hazard does not depend on $A_Y$:

\lambda_1^{a_Y,a_D}(t)=\lambda_1^{a_Y,a_D'}(t), \qquad \lambda_2^{a_Y,a_D}(t)=\lambda_2^{a_Y',a_D}(t).

These are statements about continuous-time hazards, not about probabilities of an event over a fixed interval. In particular, they do not say that the interval probability of event 1 under a mixed regimen equals that probability in one observed treatment arm.

The Identifying Formula

Suppose the observed treatment is the two-component treatment with $A_Y=A_D=A$, so the observed regimens are $(0,0)$ and $(1,1)$. Let $\lambda_j^a(t)$ denote the cause-specific hazard for event $j$ in observed treatment arm $A=a$. Consistency links the observed diagonal regimens to their potential outcomes. The DCC then imply that the event-1 hazard under any regimen with $A_Y=a_Y$ is $\lambda_1^{a_Y}(t)$, and the event-2 hazard under any regimen with $A_D=a_D$ is $\lambda_2^{a_D}(t)$.

For this identification from observed arms, assume treatment assignment is randomized, or that conditional exchangeability holds given measured baseline covariates; each treatment value has positive probability in the relevant population; treatment versions are consistent with the component interventions; and censoring is independent of the event process conditional on the variables used in the analysis. For clarity, the formula below is written for a randomized population without covariates. The DCC and the existence of meaningful component interventions are additional causal assumptions; they are not established by fitting a hazard model.

Theorem: Under the DCC and the identification conditions just stated, with absolutely continuous event-time distributions, the cumulative incidence of event 1 under regimen $(a_Y,a_D)$ is identified by $$ F_1^{a_Y,a_D}(t) = \int_0^t \exp\!\left\{ -\int_0^u \big[\lambda_1^{a_Y}(v)+\lambda_2^{a_D}(v)\big]\,dv \right\} \lambda_1^{a_Y}(u)\,du. $$ The corresponding survival probability in the event-free state at time $u$ is $$ S^{a_Y,a_D}(u) = \exp\!\left\{ -\int_0^u \big[\lambda_1^{a_Y}(v)+\lambda_2^{a_D}(v)\big]\,dv \right\}. $$

Here the observed arm-specific hazards supply the component-specific hazards: the event-1 hazard comes from arm $a_Y$, and the event-2 hazard comes from arm $a_D$. The exponent combines their cumulative hazards to produce survival under the mixed regimen. This is the identifying step: DCC supplies the bridge from diagonal treatment arms to the mixed regimen, and the competing-risk event process converts those hazards to a risk.

Why the Formula Holds

Let $S(u)$ be the probability of remaining free of both events through time $u$ under a fixed regimen. Over a short interval of length $\Delta t$, the conditional probability of either event is approximately $(\lambda_1(u)+\lambda_2(u))\Delta t$. Thus the survival function obeys the differential equation below. The event-1 cumulative incidence increases at rate $S(u)\lambda_1(u)$, because event 1 can occur at time $u$ only for someone still event-free immediately beforehand.

\frac{dS(u)}{du} = -S(u)\{\lambda_1(u)+\lambda_2(u)\}, \qquad S(0)=1.

Divide by $S(u)>0$ and integrate from 0 to $u$. This gives $\log S(u)=-\int_0^u\{\lambda_1(v)+\lambda_2(v)\}\,dv$, hence the exponential survival expression. Since $dF_1(u)/du=S(u)\lambda_1(u)$ and $F_1(0)=0$, integration gives $F_1(t)=\int_0^t S(u)\lambda_1(u)\,du$. Substituting the DCC-implied mixed-regime hazards yields the theorem's formula. This derivation assumes ordinary continuous-time hazards; with discrete time intervals or event-time atoms, the corresponding survival and incidence calculation must use an appropriate discrete-time or product-integral formulation.

Worked Example: Constant Hazards and a Mixed-Regime Risk

Suppose a hypothetical randomized study estimates constant cause-specific hazards per year of $\lambda_1^0=0.10$, $\lambda_2^0=0.20$, $\lambda_1^1=0.15$, and $\lambda_2^1=0.05$. Assume the DCC holds. Calculate the event-1 risk by 2 years under $(A_Y,A_D)=(1,0)$, and compare it with the risk under $(0,0)$.

For $(1,0)$, DCC assigns the event-1 hazard from arm 1 and the event-2 hazard from arm 0. The total hazard is $0.15+0.20=0.35$ per year. With constant hazards, the integral simplifies to $F_1(t)=\{\lambda_1/(\lambda_1+\lambda_2)\}\{1-\exp[-(\lambda_1+\lambda_2)t]\}$. Therefore:

F_1^{1,0}(2) = \frac{0.15}{0.35}\{1-\exp[-(0.35)(2)]\} = 0.4286(1-0.4966) = 0.2158.

The identified event-1 risk is 0.2158, or about 21.6% by 2 years. Under $(0,0)$, both hazards come from arm 0:

F_1^{0,0}(2) = \frac{0.10}{0.30}\{1-\exp[-(0.30)(2)]\} = 0.3333(1-0.5488) = 0.1504.

The separable risk difference comparing $A_Y=1$ with $A_Y=0$, while holding $A_D=0$, is $0.2158-0.1504=0.0654$, or 6.54 percentage points by 2 years. It is a causal contrast only under the assumptions stated above; it is not automatically the effect of changing the complete treatment.

Worked Example: Changing the Competing-Event Component

Use the same hypothetical hazards to compare $(0,1)$ with $(0,0)$ at 3 years. Here $A_Y=0$ is held fixed, while the component assigned to the competing-event hazard changes from 0 to 1.

Under $(0,1)$, the event-1 hazard is $0.10$ per year and the competing-event hazard is $0.05$ per year. Thus:

F_1^{0,1}(3) = \frac{0.10}{0.15}\{1-\exp[-(0.15)(3)]\} = 0.6667(1-0.6376) = 0.2416.

For $(0,0)$, the corresponding calculation is $F_1^{0,0}(3)=\frac{0.10}{0.30}\{1-\exp[-(0.30)(3)]\} =0.3333(1-0.4066)=0.1978$. The risk difference is $0.2416-0.1978=0.0438$, or 4.38 percentage points by 3 years. The event-1 hazard is unchanged, yet the event-1 cumulative incidence is higher under $(0,1)$, because the smaller competing-event hazard leaves more individuals event-free and able to experience event 1. This is a risk contrast, not evidence that the component directly changes the event-1 hazard.

Worked Example: Piecewise Hazards and the Interval-Probability Pitfall

Consider a hypothetical setting with two one-year intervals. Suppose the observed arm-1 event-1 hazards are 0.12 in year 1 and 0.08 in year 2, while the observed arm-0 event-2 hazards are 0.18 and 0.10, respectively. Under the DCC, these are the component-specific hazards for regimen $(1,0)$. Calculate its event-1 risk by year 2.

In year 1, the mixed-regime total hazard is $0.12+0.18=0.30$. Its event-1 incidence contribution is $\frac{0.12}{0.30}(1-\exp[-0.30])=0.1037$. Event-free survival at year 1 is $\exp(-0.30)=0.7408$. In year 2, the total hazard is $0.08+0.10=0.18$, so the additional event-1 incidence is:

0.7408\frac{0.08}{0.18}\{1-\exp(-0.18)\} = 0.7408(0.4444)(0.1647) = 0.0542.

The total event-1 risk by year 2 is $0.1037+0.0542=0.1579$, or 15.8%. For comparison, in the observed arm-1 first year, the conditional probability of event 1 during that year is $\frac{0.12}{0.12+0.05}\{1-\exp[-(0.17)]\}=0.1104$, using the specified arm-1 event-2 hazard of 0.05. That is not the mixed-regime first-year probability, which is 0.1037. The competing hazard differs, changing both the total event rate and the chance of remaining event-free. Mixing the observed arms' interval probabilities would therefore not implement the DCC formula.

Assumptions, Diagnostics, and Alternatives

The formula is useful only when its causal and statistical conditions are credible. In a randomized study, randomization supports exchangeability of treatment assignment; in an observational study, exchangeability requires adequate adjustment for common causes of treatment and outcome. Positivity requires observations at each treatment value needed to estimate the component-specific hazards. Consistency requires that the observed treatment versions correspond to the component interventions being interpreted. Independent censoring is needed for the observed hazards to represent the event process rather than a selected subset. These conditions complement, but do not replace, the DCC.

Check whether estimated arm-specific hazards are supported by enough individuals at risk throughout the time range of interest. Sparse late risk sets, heavy censoring, or unstable hazard estimates can make the resulting integral unreliable. Check the event definitions and timing as well: the formula assumes mutually exclusive first events and a well-defined time origin. A proportional-hazards model is not required; hazards may vary with time. If a parametric model imposes proportional hazards or constant hazards, however, its fit and time-pattern assumptions need separate scrutiny.

The DCC is generally not testable from the observed diagonal treatment arms alone: the mixed regimens needed to check its cross-component equalities are not observed. It is a substantive assumption about how components act on each event process. An estimated hazard pattern cannot, by itself, establish that assumption. Sensitivity analysis can show how results change under alternative hazard relationships, but cannot turn the DCC into an empirically verified fact.

A total treatment-policy effect compares risks under the observed complete regimens $(1,1)$ and $(0,0)$; it does not require attributing their difference to separate components. A separable effect instead holds one component fixed and contrasts the other, relying on the component decomposition and DCC. Cause-specific hazard comparisons alone do not give cumulative-incidence effects, because the risk depends on survival from both event types. A hypothetical effect under elimination of the competing event asks a different question and should not be substituted for a mixed-component risk.

Common mistake: “The DCC says event 1 is governed by $A_Y$, so use the event-1 interval probability from arm $A_Y$ and combine it with the event-2 interval probability from arm $A_D$.” This is incorrect: DCC equates conditional cause-specific hazards, not finite-interval probabilities. The correction is to take the component-specific hazards, construct mixed-regime survival from their sum, and integrate survival times the event-1 hazard.

Check Your Understanding

Answer each question, then check the brief solutions below. Rates are per year unless otherwise stated.

Under the DCC, which observed treatment arm supplies the event-2 hazard for regimen $(1,0)$?
If $\lambda_1=0.20$ and $\lambda_2=0.10$ are constant under a specified regimen, calculate $F_1(2)$.
Why can reducing the event-2 hazard increase event-1 cumulative incidence even when the event-1 hazard stays fixed?
Name two assumptions, apart from the DCC, needed to identify component-specific hazards from observed treatment arms.

Solutions. (1) Arm 0 supplies $\lambda_2^0$, because the second component is set to 0. (2) $F_1(2)=\frac{0.20}{0.30}\{1-\exp[-0.60]\}=0.6667(0.4512)=0.3008$, or 30.1%. (3) A lower competing hazard increases event-free survival, leaving more individuals at risk of event 1; the incidence integral $ \int S(u)\lambda_1(u)\,du $ can therefore increase. (4) Examples include exchangeability or randomized treatment assignment, positivity, consistency, and independent censoring.

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