Discrete-Time Recursion For Competing-Risk Risk

From Interval Hazards to Absolute Risk

A set of cause-specific predictions is not yet a set of cumulative risks. A discrete cause-specific hazard describes risk during one interval, conditional on still being event-free at its start. To obtain risk by a later horizon, the calculation must account for the fact that fewer individuals remain event-free as time passes. The same event-free probability supplies the weight for every cause in each interval.

This tutorial derives that update and works through its arithmetic. The setting is a first-event competing-risks outcome observed over ordered intervals. It is not necessary to assume that competing causes are independent: the inputs are the conditional probabilities of each mutually exclusive first-event cause among those still event-free.

Definition: Let $0=t_0<t_1<\cdots<t_K$ be the interval boundaries. Let $T$ be time to the first event, and let $J\in\{1,\ldots,m\}$ identify its cause. For cause $j$ and interval $k$, the discrete cause-specific hazard is $$ h_{jk}=P(t_{k-1}<T\leq t_k,\ J=j\mid T>t_{k-1}). $$ The event-free probability at $t_k$ is $S_k=P(T>t_k)$, and the cause-$j$ cumulative incidence at $t_k$ is $F_{jk}=P(T\leq t_k,\ J=j)$. Set $S_0=1$ and $F_{j0}=0$.

The word hazard here refers to a conditional probability over a specified interval, not an instantaneous rate. For each interval, the cause-specific hazards must satisfy $h_{jk}\geq 0$ and $\sum_{j=1}^m h_{jk}\leq 1$. The remaining probability, $1-\sum_j h_{jk}$, is the conditional probability of staying event-free through that interval. These definitions concern the first event: after a first event occurs, the individual is no longer at risk of having a different first-event cause.

The Stepwise Recursion

At the start of interval $k$, a proportion $S_{k-1}$ of the population is event-free. Among that group, the conditional probability of cause $j$ during the interval is $h_{jk}$. Therefore the unconditional probability of experiencing cause $j$ in that interval is $S_{k-1}h_{jk}$. Adding this increment to the cumulative incidence already accrued gives the update. The corresponding event-free update multiplies the probability of being event-free at the start by the conditional probability of remaining event-free during the interval.

Theorem: Given valid discrete cause-specific hazards, update the event-free probability and each cause-specific cumulative incidence by $$ S_k=S_{k-1}\left(1-\sum_{j=1}^m h_{jk}\right), \qquad F_{jk}=F_{j,k-1}+S_{k-1}h_{jk},\quad j=1,\ldots,m. $$ Here $S_{k-1}$, not $S_k$, weights the event increments in interval $k$.

The placement of $S_{k-1}$ follows directly from conditioning on being event-free at the interval’s start. It also gives an algebraic check at every step: the cumulative probability of having experienced any first event plus the probability of remaining event-free must equal one.

\begin{aligned} S_k+\sum_{j=1}^m F_{jk} &=S_{k-1}\left(1-\sum_{j=1}^m h_{jk}\right) +\sum_{j=1}^m\left(F_{j,k-1}+S_{k-1}h_{jk}\right)\\ &=S_{k-1}+\sum_{j=1}^m F_{j,k-1}. \end{aligned}

Since $S_0+\sum_j F_{j0}=1$, repeated application of the update gives $S_k+\sum_j F_{jk}=1$ at every later interval. In particular, the total failure risk by $t_k$ is $1-S_k=\sum_j F_{jk}$. This identity checks the accounting; it does not establish that the supplied hazards are well estimated or appropriate for the prediction target.

Worked Calculations

Worked Example: Three Intervals and Two Causes

Suppose a hypothetical prediction supplies cause-specific interval probabilities for causes A and B. The interval hazards are $(0.10,0.05)$, $(0.20,0.10)$, and $(0.10,0.20)$, respectively. Each pair sums to at most one, so each interval leaves a nonnegative probability of remaining event-free. Calculate the cumulative incidences and event-free probability through interval 3.

Start with $S_0=1$, $F_{A0}=0$, and $F_{B0}=0$. In interval 1:

\begin{aligned} S_1&=1(1-0.10-0.05)=0.85,\\ F_{A1}&=0+1(0.10)=0.10,\\ F_{B1}&=0+1(0.05)=0.05. \end{aligned}

In interval 2, the starting event-free probability is $S_1=0.85$:

\begin{aligned} S_2&=0.85(1-0.20-0.10)=0.595,\\ F_{A2}&=0.10+0.85(0.20)=0.270,\\ F_{B2}&=0.05+0.85(0.10)=0.135. \end{aligned}

In interval 3, the starting event-free probability is $S_2=0.595$:

\begin{aligned} S_3&=0.595(1-0.10-0.20)=0.4165,\\ F_{A3}&=0.270+0.595(0.10)=0.3295,\\ F_{B3}&=0.135+0.595(0.20)=0.254. \end{aligned}

Thus by the end of interval 3, the predicted risks are 0.3295 for cause A and 0.2540 for cause B, while the event-free probability is 0.4165. The check is $0.3295+0.2540+0.4165=1.0000$. These are probabilities, or 32.95%, 25.40%, and 41.65%, respectively, by that horizon.

Worked Example: The Event-Free Weight Changes the Risk Increment

Consider a hypothetical two-interval prediction with hazards $(0.20,0.10)$ in interval 1 and $(0.05,0.15)$ in interval 2 for causes A and B. Find the cause-A cumulative incidence at the end of interval 2. Both interval hazard sums are valid: they are 0.30 and 0.20.

After interval 1, $S_1=1(1-0.20-0.10)=0.70$, and $F_{A1}=0.20$. The cause-A increment in interval 2 is not $0.05$ for the full original population; only the event-free fraction at the start of that interval is exposed to this conditional interval risk:

F_{A2}=F_{A1}+S_1h_{A2} =0.20+(0.70)(0.05)=0.235.

The cause-A cumulative incidence is therefore 0.235, or 23.5%, by the end of interval 2. For a full check, $S_2=0.70(1-0.05-0.15)=0.56$ and $F_{B2}=0.10+(0.70)(0.15)=0.205$. The probabilities sum to $0.235+0.205+0.56=1$.

Worked Example: Detecting an Invalid Interval Prediction

Suppose hypothetical predictions for two mutually exclusive first-event causes in one interval are $h_{1k}=0.65$ and $h_{2k}=0.45$. Each number lies between zero and one, but together they imply a conditional probability of failure of $0.65+0.45=1.10$. If $S_{k-1}=0.80$, the proposed update would give

$$ S_k=0.80(1-0.65-0.45)=0.80(-0.10)=-0.08. $$

A negative event-free probability is impossible. The predictions are not a valid set of conditional probabilities for mutually exclusive first-event causes on this interval. It is not enough to check each cause separately against one; the sum across causes must also be at most one. The inputs need to be reviewed or generated by a model that respects the joint probability constraint. Rescaling the hazards would alter the predictions and is not automatically justified.

Conditions, Checks, and Interpretation

The recursion is an accounting identity under its stated definitions, but using it for prediction requires care. The intervals must be ordered and the hazards must refer to the same interval boundaries and target population. Within an interval, causes must be mutually exclusive first-event categories, with any event-free outcome included as the remaining probability. The model supplying the hazards must produce conditional probabilities for individuals still event-free at each interval’s start. If the target is an individual with specified characteristics, the hazards must be appropriate for that individual; if the target is a population, they must be appropriate to the population-level prediction being made.

This calculation does not estimate the hazards or correct a misspecified prediction model. It takes the hazards as inputs. In data analysis, issues such as censoring and the method used to estimate conditional probabilities affect those inputs and require their own treatment. The recursion itself also does not assume independence between causes. It combines the probabilities of mutually exclusive causes conditional on the event-free state.

Useful diagnostics include checking that every $h_{jk}$ is between zero and one, that the cause-specific hazards sum to at most one in each interval, and that all resulting $S_k$ and $F_{jk}$ stay between zero and one. Check $S_k+\sum_jF_{jk}=1$ at each step, not only at the final horizon; an error in one interval can otherwise be hidden by later arithmetic. A failure of the identity usually signals an indexing, arithmetic, or update error. Passing it only verifies internal consistency.

The direct calculation $F_{jk}=\sum_{\ell=1}^k S_{\ell-1}h_{j\ell}$ is equivalent to applying the recursion and can be convenient for understanding each cause’s accumulated increments. The recursion is often easier to implement sequentially because it carries the current event-free probability and cumulative incidences forward. For continuous-time cause-specific hazards, cumulative incidence is derived using the event-free probability over continuous time; discrete interval probabilities should not be treated as instantaneous rates or inserted into a continuous-time formula without an appropriate conversion. The introductory material on deriving cumulative incidence from cause-specific hazards establishes the broader relationship; the present update is its discrete interval form. Aalen–Johansen estimation is another standard approach when the task is estimating state-occupation probabilities from event-history data, rather than simply combining supplied discrete hazard predictions.

Common pitfall: Adding the cause-specific hazards across intervals as if each applied to the original population over and over gives incorrect cumulative risk. In interval $k$, the increment for cause $j$ is $S_{k-1}h_{jk}$, not $h_{jk}$. The conditional hazard applies only to those who have remained event-free up to that interval.

Check Your Understanding

Use the recursion and show the event-free probability as well as the requested cause-specific risk where applicable.

With one cause and interval hazards 0.10 and 0.20, calculate the cumulative incidence and event-free probability after two intervals.
At the start of an interval, $S_{k-1}=0.60$. The hazards for causes A and B are 0.10 and 0.25. Find both interval risk increments and $S_k$.
Explain why hazards of 0.55 and 0.50 for two causes cannot be used together as conditional probabilities of mutually exclusive first events in the same interval.
For any number of causes, state the probability identity that should hold after each update and explain what it checks.

Solutions. (1) $S_1=1(1-0.10)=0.90$ and $F_1=0.10$. Then $S_2=0.90(1-0.20)=0.72$, and $F_2=0.10+0.90(0.20)=0.28$. (2) The increments are $0.60(0.10)=0.060$ for A and $0.60(0.25)=0.150$ for B. The event-free probability is $S_k=0.60(1-0.10-0.25)=0.390$. (3) Their sum is 1.05, so the implied event-free conditional probability is $1-1.05=-0.05$, which is impossible. (4) $S_k+\sum_jF_{jk}=1$. It checks that event-free probability and cumulative incidences account for all probability mass; it does not validate the model that supplied the hazards.

Discrete-Time Recursion For Competing-Risk Risk

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