Connecting Discrete Hazards to Survival Probabilities

From Conditional Hazards to Survival

A hazard at a particular event time is conditional: it describes the chance of an event among those who have remained event-free up to the relevant point. It is not, by itself, the probability of an event in the original population. Survival across several times therefore depends on multiplying successive conditional probabilities of avoiding the event. This connection is the basis for interpreting product-limit survival estimates and for understanding how discrete hazards accumulate.

Let $T$ be a nonnegative event time, and write $S_0(t)=P(T>t)$ for the probability of remaining event-free beyond time $t$. Choose ordered times $t_0<t_1<\cdots<t_m$. Define the interval hazard $h_j$ as the conditional probability of an event in the $j$th interval, given survival through its start. This definition allows the intervals to include all event times, even if the distribution is not concentrated at a single point in each interval.

Definition: For $j=1,\ldots,m$, the discrete interval hazard is $$ h_j=P(t_{j-1}<T\le t_j\mid T>t_{j-1}). $$ Its conditional survival factor is $1-h_j=P(T>t_j\mid T>t_{j-1})$. The accumulated discrete hazard through $t_m$ is $H_m=\sum_{j=1}^{m}h_j$. For event times with no probability between them, the intervals can be chosen so that $h_j=P(T=t_j\mid T>t_{j-1})$.

The starting probability matters. If the process begins at a time $t_0$ by which some events may already have occurred, then $S_0(t_0)$ need not equal 1. If instead the population is initially event-free, $S_0(t_0)=1$. The product formula retains this starting probability explicitly, so it remains valid in either case.

Why the Product Formula Holds

The conditional probability of surviving the first interval is $1-h_1$. For the next interval, the conditional probability of surviving it, given survival to $t_1$, is $1-h_2$. Multiplying these successive conditional probabilities gives the probability of surviving all intervals. In full, the chain rule gives:

\begin{aligned} S_0(t_m) &=S_0(t_0)\, P(T>t_1\mid T>t_0)\, P(T>t_2\mid T>t_1)\cdots P(T>t_m\mid T>t_{m-1})\\ &=S_0(t_0)\prod_{j=1}^{m}(1-h_j). \end{aligned}

This argument requires the hazards to cover the intervals from $t_0$ through $t_m$. Events before $t_0$ are accounted for in $S_0(t_0)$; events within each later interval are accounted for in its hazard. For a list of point event times, the corresponding intervals must contain all possible event times in the range. Otherwise, the product omits the chance of an event in a gap.

Theorem: If $S_0(t_0)=1$ and the intervals cover all possible event times through $t_m$, then $$ S_0(t_m)=\prod_{j=1}^{m}(1-h_j). $$ If $S_0(t_0)\ne 1$, multiply the product by $S_0(t_0)$. A hazard of 1 makes the survival probability zero from that interval onward.

Worked Example: Survival From Specified Interval Hazards

A hypothetical equipment-protection program tracks the time $T$ to a first qualifying breakdown. The population is event-free at time $t_0=0$. The conditional breakdown hazards in the intervals $(0,1]$, $(1,2]$, and $(2,3]$ years are $0.10$, $0.20$, and $0.25$, respectively. Find the probability of remaining breakdown-free beyond year 3 and the accumulated hazard. No interest discounting is involved, so $i=0$ for this probability calculation. No mortality assumption or mortality table is used; the event is equipment breakdown. The discrete distribution is specified by the three interval hazards and initial survival $S_0(0)=1$.

Method. Apply the product formula because the hazards cover all three consecutive intervals and the program begins with event-free equipment.

The interval survival factors are:

1-h_1=1-0.10=0.90,\qquad 1-h_2=1-0.20=0.80,\qquad 1-h_3=1-0.25=0.75.

Thus survival beyond year 3 is:

$$ S_0(3)=1(0.90)(0.80)(0.75)=0.5400. $$

The accumulated hazard is the sum of the interval hazards: $$ H_3=0.10+0.20+0.25=0.55. $$ The result means that the modeled probability of remaining breakdown-free beyond year 3 is $0.5400$, while the accumulated discrete hazard is $0.55$. These are different quantities.

Check. Starting with probability 1, the probability surviving the first interval is 0.90. Of that surviving group, 0.80 survive the second, leaving $0.90(0.80)=0.72$. Of those, 0.75 survive the third, leaving $0.72(0.75)=0.5400$. This agrees with the product. The common mistake avoided is adding the three hazards and treating $1-H_3=0.45$ as survival; hazards are conditional probabilities, so survival factors multiply.

Accumulated Hazard and the Exponential Approximation

The sum $H_m=\sum h_j$ is useful because it records the total of the interval hazard probabilities. But the survival probability is a product, not $1-H_m$, and in general it is not exactly $\exp(-H_m)$. To see the connection, take logarithms of the product when every $h_j<1$ and $S_0(t_0)=1$:

\log S_0(t_m)=\sum_{j=1}^{m}\log(1-h_j).

For small $h_j$, the expansion $-\log(1-h_j)=h_j+h_j^2/2+h_j^3/3+\cdots$ shows why the product is often approximated by $\exp(-H_m)$: keeping only the first-order terms gives $\log S_0(t_m)\approx-\sum h_j$. The neglected terms are positive, so when $S_0(t_0)=1$,

S_0(t_m) =\exp\left(\sum_{j=1}^{m}\log(1-h_j)\right) \le \exp\left(-\sum_{j=1}^{m}h_j\right) =\exp(-H_m).

This inequality holds for hazards between 0 and 1; if any hazard equals 1, survival is zero and the product formula applies directly. The approximation is generally closer when each hazard is small. If initial survival is less than 1, the product and inequality include that starting probability: $S_0(t_m)=S_0(t_0)\prod_j(1-h_j)$.

In an empirical product-limit calculation, $h_j$ is commonly estimated at an event time by $d_j/r_j$, where $d_j$ is the number of events and $r_j$ is the number at risk just before that time. The estimated survival factor is $1-d_j/r_j$. Interpreting this as an estimate for the target event-time distribution relies on appropriate handling of censoring, commonly independent censoring conditional on the information used to define the groups. The arithmetic product itself does not remove bias from informative censoring.

Worked Example: Building a Product-Limit Estimate

In a hypothetical study of 40 leased sensor units, all are operating at the start. At three successive inspection times, the risk-set sizes are $r_1=40$, $r_2=32$, and $r_3=24$; the numbers of first failures are $d_1=8$, $d_2=8$, and $d_3=6$. There is no censoring before these inspection times. Estimate survival just after the third time and calculate accumulated hazard. Use no discounting, $i=0$; no mortality assumption applies because the modeled event is sensor failure. The empirical hazards are $d_j/r_j$.

Method. Estimate each conditional event probability from its risk set, then multiply the corresponding conditional survival factors.

At the first time: $$ \widehat h_1=\frac{8}{40}=0.20,\qquad 1-\widehat h_1=0.80. $$ After 8 failures, $40-8=32$ units remain at risk, as specified. At the second time: $$ \widehat h_2=\frac{8}{32}=0.25,\qquad 1-\widehat h_2=0.75. $$ After a further 8 failures, $32-8=24$ remain at risk. At the third time: $$ \widehat h_3=\frac{6}{24}=0.25,\qquad 1-\widehat h_3=0.75. $$ Therefore: $$ \widehat S(t_3)=(0.80)(0.75)(0.75)=0.4500. $$ The accumulated estimated hazard is: $$ \widehat H_3=0.20+0.25+0.25=0.70. $$

Check. The first time leaves an estimated fraction $0.80$ functioning; the second leaves $0.80(0.75)=0.60$; the third leaves $0.60(0.75)=0.4500$. The risk-set sequence is consistent with the failure counts. The common mistake avoided is using the original 40 units as the denominator at every time: each hazard is conditional on being at risk at that time.

Worked Example: Accounting for Initial Survival

A hypothetical monitoring record begins at time $t_0=0$, but 10% of the original population had already experienced the event by then. Among the population surviving beyond time 0, the hazards in $(0,1]$ and $(1,2]$ are $h_1=2/9$ and $h_2=3/7$. Find survival beyond time 2 as a probability in the original population. No interest discounting is used ($i=0$); no mortality assumption is involved. The distribution is specified by $S_0(0)=0.90$ and these two conditional interval hazards.

Method. Include the probability of reaching the start of the hazard sequence, then multiply by both interval survival factors.

The first conditional survival factor is: $$ 1-h_1=1-\frac{2}{9}=\frac{7}{9}. $$ The second is: $$ 1-h_2=1-\frac{3}{7}=\frac{4}{7}. $$ Thus: $$ S_0(2)=0.90\left(\frac{7}{9}\right)\left(\frac{4}{7}\right) =0.90\left(\frac{4}{9}\right)=0.40. $$ The accumulated hazard over the two observed intervals is: $$ H_2=\frac{2}{9}+\frac{3}{7} =\frac{14+27}{63}=\frac{41}{63}\approx0.6508. $$

Check. Of the original population, $0.90$ reaches time 0 event-free. The first interval leaves $0.90(7/9)=0.70$, and the second leaves $0.70(4/7)=0.40$. This verifies the result. The common mistake avoided is omitting the initial factor and reporting $4/9$, which is survival conditional on having already survived to time 0, not survival in the original population.

Common Errors and Practical Interpretation

The product formula is simple, but its validity depends on defining the hazards and starting point carefully. In particular, a list of hazards at selected times cannot account for events that could occur before the first listed time or between listed times unless those possibilities are included in the initial survival or in the hazard intervals.

Common Exam Trap: Missing an Event-Time Interval

Suppose $S_0(0)=0.90$, and the conditional hazards for the complete intervals $(0,1]$ and $(1,2]$ are $2/9$ and $3/7$. A mistaken calculation omits the initial survival and reports: $$ \left(1-\frac{2}{9}\right)\left(1-\frac{3}{7}\right) =\frac{7}{9}\frac{4}{7}=\frac{4}{9}\approx0.4444. $$

That answer is conditional on survival to time 0. For survival in the original population, the correct calculation is: $$ S_0(0)\left(1-\frac{2}{9}\right)\left(1-\frac{3}{7}\right) =0.90\left(\frac{4}{9}\right)=0.4000. $$

A related error is to define hazards only at a few event times while silently omitting possible events in the gaps. The correction is to choose intervals that cover the time range, or explicitly include the probability of surviving to the start of the listed sequence. Always state what population the survival probability is measured from.

For interpretation, keep three objects distinct: $h_j$ is a conditional event probability in one interval; $H_m$ is the sum of those probabilities; and $S_0(t_m)$ is the probability of surviving all intervals, found by multiplying conditional survival factors and including initial survival when needed. The exponential expression $\exp(-H_m)$ can be a useful small-hazard approximation, but it does not replace the product when the interval hazards are appreciable.

Key takeaway: Survival is built by chaining conditional survival probabilities. The product is valid only when the starting survival is included and the hazard intervals cover the time range; accumulated hazard is a sum, not a substitute for that product.

Check Your Understanding

For each problem, identify the survival factors before combining them. Answers are given to four decimal places where appropriate.

A hypothetical device begins event-free and has interval hazards $0.05$, $0.10$, and $0.20$ over three consecutive intervals. Find survival through all three intervals and the accumulated hazard. (Answer: survival $=0.6840$; accumulated hazard $=0.35$.)
A study begins with $S_0(t_0)=0.80$. The hazards in the next two complete intervals are $0.25$ and $0.40$. Find survival in the original population at the end of the second interval. (Answer: $0.80(0.75)(0.60)=0.3600$.)
At two successive event times, 5 of 50 at-risk units fail and then 9 of the remaining 45 fail. Estimate survival just after the second time and give the sum of the estimated hazards. Assume no censoring before either time. (Answer: survival $=(45/50)(36/45)=0.7200$; accumulated hazard $=0.10+0.20=0.30$.)
An initially event-free population has interval hazards $0.20$, $0.10$, and $0.50$. Calculate survival after all three intervals. Is $1$ minus the accumulated hazard the survival probability? (Answer: survival $=(0.80)(0.90)(0.50)=0.3600$; no, $1-(0.20+0.10+0.50)=0.20$ is not the product-limit survival.)
A population has initial survival $S_0(t_0)=0.92$, followed by hazards $0.15$ and $0.25$ in two intervals. Find survival beyond the second interval. (Answer: $0.92(0.85)(0.75)=0.5865$.)

What You'll Learn

From Conditional Hazards to Survival

Why the Product Formula Holds

Worked Example: Survival From Specified Interval Hazards

Worked Example: Survival From Specified Interval Hazards

Accumulated Hazard and the Exponential Approximation

Worked Example: Building a Product-Limit Estimate

Worked Example: Building a Product-Limit Estimate

Worked Example: Accounting for Initial Survival

Worked Example: Accounting for Initial Survival

Common Errors and Practical Interpretation

Common Exam Trap: Missing an Event-Time Interval

Check Your Understanding

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