Principal Strata Defined by Competing Events

Why Define Strata From Both Treatment Conditions?

At a fixed follow-up horizon, a person may experience a target event, a competing event, or neither. Treatment can affect which event occurs first. A comparison of treatment-specific competing-event risks describes population outcomes under each treatment, but it does not by itself tell us which individuals would experience a competing event under both treatments, under neither, or under just one.

Principal strata classify individuals by jointly considering their counterfactual competing-event statuses under the two treatment conditions. This is a cross-world classification: each person belongs to one stratum, but only one treatment condition is observed for that person. The earlier material on Counterfactual Event Times and Causes supplies the potential-event-time framework used here. The focus in this tutorial is the definition and observability of the resulting strata, not a causal effect within a stratum.

Definition: Let $A\in\{0,1\}$ denote treatment, with 0 as control and 1 as treatment, and let $\tau$ be a fixed follow-up horizon. For treatment level $a$, let $T_C^a$ and $T_J^a$ be the potential times to the competing event and target event. Define the competing-event status $$ Z^a(\tau)=I(T_C^a\leq T_J^a,\ T_C^a\leq \tau). $$ A tie between event times is assigned to the competing event. If no competing event occurs by $\tau$, then $Z^a(\tau)=0$. The principal stratum is the ordered pair $(Z^0(\tau),Z^1(\tau))$.

The inequality $T_C^a\leq T_J^a$ makes the tie rule explicit: when the event times are equal, the competing event is counted as occurring first. If a study instead assigns ties to the target event, its definition must use $T_C^a<T_J^a$. The event-ordering convention must be fixed before statuses and strata are assigned. The horizon also matters: a competing event after $\tau$ does not make $Z^a(\tau)=1$.

This definition assumes well-defined treatment conditions and potential event times under each condition. For an observed-data interpretation, consistency means the observed status equals $Z^A(\tau)$ for the treatment actually received; no interference is also required for the usual individual-level potential-outcome interpretation. These conditions define the link between observed and counterfactual statuses. They do not make both statuses observable for any one person.

The Four Principal Strata

Because each status is binary, the joint classification has four possible values. The first entry refers to control and the second to treatment. Descriptive labels can be useful, but the ordered pair is the unambiguous definition.

Stratum	Counterfactual statuses	Interpretation at horizon $\tau$
(0, 0)	No competing event under either treatment	Competing-event status is absent under both conditions
(0, 1)	No competing event under control; competing event under treatment	Status changes from 0 to 1 under treatment
(1, 0)	Competing event under control; no competing event under treatment	Status changes from 1 to 0 under treatment
(1, 1)	Competing event under either treatment	Competing-event status is present under both conditions

These labels refer only to the competing-event status through $\tau$. For example, a person in stratum (0, 0) might experience the target event under one or both treatments, or might experience neither event by $\tau$. The stratum definition does not distinguish those possibilities. Similarly, “competing event under both” does not imply the event occurs at the same time under both treatments.

Worked Example: Applying the Horizon and Tie Rule

Suppose the study horizon is $\tau=5$ years and ties are assigned to the competing event, as in the definition above. For one hypothetical person, the potential event times are $T_C^0=3$, $T_J^0=3$, $T_C^1=6$, and $T_J^1=4$, measured in years. Determine the two statuses and the stratum.

Under control, $T_C^0=3\leq T_J^0=3$ and $T_C^0=3\leq5$, so $Z^0(5)=1$. The tie is counted as a competing event by the stated rule.

Under treatment, $T_C^1=6$ is after the horizon, so the condition $T_C^1\leq5$ fails. Therefore $Z^1(5)=0$, even though the target event occurs at year 4.

The hypothetical person’s stratum is $(1,0)$. If the horizon were instead 7 years, the treatment status would be 0 because the target event at year 4 precedes the competing event at year 6. The horizon is part of the estimand’s definition, not a detail that can be ignored.

What Is Observable?

For a person assigned treatment $A$, consistency gives the observed status $Z=Z^A(\tau)$. Thus a control-assigned person reveals the first coordinate, and a treatment-assigned person reveals the second. Neither reveals both coordinates. In particular, observing $Z=0$ under control does not show whether the person’s treatment status would be 0 or 1.

Randomization can identify the marginal distributions of $Z^0(\tau)$ and $Z^1(\tau)$ in the population, subject to ordinary sampling uncertainty and complete ascertainment of status by the horizon. It does not, by itself, identify the joint distribution of $(Z^0(\tau),Z^1(\tau))$. A randomized trial gives a comparison between groups, not two treatment assignments for the same person.

Key result: Write $\pi_{ij}=P(Z^0(\tau)=i,Z^1(\tau)=j)$ and $p_a=P(Z^a(\tau)=1)$. The marginals satisfy $$ p_0=\pi_{10}+\pi_{11},\qquad p_1=\pi_{01}+\pi_{11},\qquad \sum_{i,j\in\{0,1\}}\pi_{ij}=1. $$ Knowing $p_0$ and $p_1$ generally does not determine all four $\pi_{ij}$.

Worked Example: Bounds From Two Marginal Risks

In a hypothetical population, suppose the control and treatment probabilities of a competing event by $\tau$ are $p_0=0.30$ and $p_1=0.20$. Find the range of possible proportions in each stratum using only these two marginals.

Let $x=\pi_{11}$, the proportion in stratum (1, 1). The marginal equations require $\pi_{10}=0.30-x$ and $\pi_{01}=0.20-x$. The remaining proportion is $\pi_{00}=1-\pi_{10}-\pi_{01}-\pi_{11}=0.50+x$.

All four proportions must be nonnegative. In particular, $x\geq0$, $0.30-x\geq0$, and $0.20-x\geq0$. Hence $0\leq x\leq0.20$. As $x$ varies over that range, $\pi_{10}=0.30-x$ ranges from 0.30 down to 0.10, $\pi_{01}=0.20-x$ ranges from 0.20 down to 0, and $\pi_{00}=0.50+x$ ranges from 0.50 to 0.70.

For example, $x=0.10$ gives $(\pi_{00},\pi_{01},\pi_{10},\pi_{11})=(0.60,0.10,0.20,0.10)$, which reproduces the marginals: $p_0=0.20+0.10=0.30$ and $p_1=0.10+0.10=0.20$. The choice $x=0.20$ also reproduces both marginals, with proportions $(0.70,0,0.10,0.20)$. Since distinct joint distributions fit the same marginal risks, those risks alone do not identify the stratum proportions.

Deriving the General Bounds

The preceding calculation is an instance of the Fréchet bounds for two binary variables. The argument uses only the marginal probabilities and nonnegativity of the four joint probabilities; it does not assume independence or a causal model for how treatment changes status.

Parameterize the joint distribution.
Set $x=\pi_{11}$. The marginal equations imply $\pi_{10}=p_0-x$ and $\pi_{01}=p_1-x$.

Use the total probability.
Since the four cells sum to 1, $\pi_{00}=1-p_0-p_1+x$.

Impose nonnegativity.
The conditions $x\geq0$, $p_0-x\geq0$, $p_1-x\geq0$, and $1-p_0-p_1+x\geq0$ give the bounds below.

\max(0,p_0+p_1-1)\leq \pi_{11}\leq\min(p_0,p_1).

For any value of $\pi_{11}$ in these bounds, the other three cell probabilities follow from the equations in the flow. The bounds therefore describe the joint distributions compatible with the two marginals; they are not confidence intervals and do not express sampling uncertainty. In practice, $p_0$ and $p_1$ are estimated from data, so uncertainty in those estimates must also be considered if the bounds are used for statistical inference.

Worked Example: What Monotonicity Adds

Suppose the hypothetical marginal risks are $p_0=0.35$ and $p_1=0.20$. Consider the additional assumption of monotonicity $Z^1(\tau)\leq Z^0(\tau)$ for every person: treatment cannot change anyone’s competing-event status from 0 to 1. Calculate the stratum proportions implied by the marginals and this assumption.

The inequality rules out stratum (0, 1), so $\pi_{01}=0$. From $p_1=\pi_{01}+\pi_{11}=0.20$, it follows that $\pi_{11}=0.20$. Then $p_0=\pi_{10}+\pi_{11}=0.35$, giving $\pi_{10}=0.35-0.20=0.15$. Finally, $\pi_{00}=1-0-0.15-0.20=0.65$.

The resulting proportions are 0.65 in (0, 0), 0 in (0, 1), 0.15 in (1, 0), and 0.20 in (1, 1). They sum to 1, and their marginals are $p_0=0.15+0.20=0.35$ and $p_1=0+0.20=0.20$. This identification follows from the monotonicity assumption plus the marginals; the assumption itself is not verified by observing those marginals.

Under this monotonicity condition, a control-assigned person observed with $Z=0$ must be in (0, 0), because the only allowed stratum with $Z^0=0$ is (0, 0). A treatment-assigned person observed with $Z=1$ must be in (1, 1), because the only allowed stratum with $Z^1=1$ is (1, 1). Other observed treatment-status combinations do not determine an individual’s stratum.

Assumptions, Diagnostics, and Limitations

Monotonicity is a substantive restriction, not a consequence of the treatment risk being lower in a sample. A lower marginal risk under treatment is compatible with some people moving from status 0 to 1 and others moving from 1 to 0. The individual-level assertion $Z^1\leq Z^0$ says that the first kind of change never occurs. Its plausibility depends on the clinical setting, the treatment, the competing event, and the horizon.

A useful basic check is whether proposed stratum proportions are nonnegative, sum to 1, and reproduce the treatment-specific marginals. If monotonicity is asserted in the direction $Z^1\leq Z^0$, then $p_1\leq p_0$ must hold in the population, since $Z^1\leq Z^0$ for each person implies $E(Z^1)\leq E(Z^0)$. A clear contradiction between the assumed direction and credible population risks undermines that assumption. Conversely, $p_1\leq p_0$ does not establish monotonicity: offsetting individual changes can produce the same marginal ordering.

Incomplete follow-up creates another practical problem. If status by $\tau$ cannot be determined for censored individuals, the observed treatment-arm proportions are not automatically the marginal risks $p_a$. Valid estimation then requires appropriate assumptions and methods for censoring; a raw proportion among those with known status may be biased. The strata also depend on the chosen event definitions and horizon, so changing either can change stratum membership.

Compared with a treatment-specific cumulative incidence, principal strata answer a different question. A cumulative incidence is a marginal outcome under one treatment condition; a principal stratum is a joint classification using two counterfactual conditions. Compared with a hypothetical intervention that removes the competing event, principal stratification does not define an outcome under an event-elimination intervention. Its central limitation is that the joint counterfactual status is generally not observed, and assumptions that identify it must be defended rather than inferred from randomization alone.

Common Mistakes

Treating an observed status as the full stratum. Seeing $Z=0$ under control reveals $Z^0=0$, not $Z^1$. Without additional assumptions, the person could be in either (0, 0) or (0, 1).
Assuming randomization reveals the joint distribution. Randomization supports comparisons of marginal distributions. No participant is observed under both assignments, so randomization alone does not reveal the cross-world pairing.
Inferring monotonicity from a lower treatment risk. The condition $p_1<p_0$ is a marginal comparison; monotonicity rules out a particular individual-level change. The former does not prove the latter.
Leaving ties or the horizon unspecified. The definition must say how equal event times are ordered and which events count by $\tau$. Here ties are assigned to the competing event through $T_C^a\leq T_J^a$.

Key takeaway: Principal strata are defined by the pair $(Z^0(\tau),Z^1(\tau))$, but ordinary treatment assignment reveals only one coordinate per person. Marginal risks constrain the joint strata, while assumptions such as monotonicity can narrow or determine them only when their substantive content is credible.

Check Your Understanding

Answer each question using a fixed horizon and the tie convention specified in the definition above.

If $T_C^0=2$, $T_J^0=4$, $T_C^1=4$, $T_J^1=4$, and $\tau=5$, what is the person’s stratum?
Suppose $p_0=0.40$ and $p_1=0.30$. Give the bounds on $\pi_{11}$.
Under monotonicity $Z^1\leq Z^0$, which stratum is ruled out? If $p_0=0.40$ and $p_1=0.30$, calculate the other three proportions.
Does observing a treatment-assigned person with $Z=0$ identify that person’s stratum under $Z^1\leq Z^0$? Explain.

Solutions. (1) Control status is 1 because $2\leq4$ and $2\leq5$. Treatment status is 1 because the tie at year 4 is assigned to the competing event and occurs by year 5. The stratum is (1, 1). (2) The bounds are $\max(0,0.40+0.30-1)=0$ and $\min(0.40,0.30)=0.30$, so $0\leq\pi_{11}\leq0.30$. (3) Stratum (0, 1) is ruled out. Then $\pi_{11}=p_1=0.30$, $\pi_{10}=p_0-p_1=0.10$, and $\pi_{00}=1-0.30-0.10=0.60$. (4) No. The observation reveals $Z^1=0$, which is compatible with both (0, 0) and (1, 0) under the stated monotonicity condition.