Constant Hazard and Geometric Waiting Times

Why Look at Hazard Rather Than Waiting-Time Probability?

A waiting-time model describes when an event first occurs. In a hypothetical equipment-monitoring program, for example, the event might be the first week a fault is detected. The probability that detection occurs in a particular week depends both on the chance of detection in that week and on whether the equipment has remained undetected until then. A discrete-time hazard separates those two ideas: it measures the event probability for the current period among cases still at risk at its start.

For a geometric waiting time, the per-period success probability is the same in every period, provided no success has occurred yet. The hazard makes this feature precise. It is a conditional probability, not the unconditional probability that the first success occurs in a specified period.

Definition: Let $T$ be a waiting time taking values $1,2,3,\ldots$, where $T=t$ means the first event occurs in period $t$. The discrete-time hazard in period $t$ is $$ h_t=P(T=t\mid T\ge t) $$ provided $P(T\ge t)>0$. The condition $T\ge t$ means that the event did not occur in any of periods $1,\ldots,t-1$; the case is still at risk at the start of period $t$.

The denominator is the probability of reaching the start of period $t$ without the event. The numerator is the probability that the first event occurs in period $t$. Thus, if $T$ has probability mass function $f_T(t)=P(T=t)$,

h_t=\frac{P(T=t)}{P(T\ge t)}.

This ratio is defined only when the denominator is positive. Some references define a survival function using $P(T>t)$; that is the probability of surviving past period $t$. Here the risk set at the start of period $t$ is instead represented by $P(T\ge t)$, which includes cases whose event will occur during period $t$.

Hazard Under the Geometric Model

Suppose a sequence of trials has two outcomes, success and failure, and each trial is independent with the same success probability $p$, where $0<p\le 1$. Let $T$ be the number of the trial on which the first success occurs. Then $T$ has a geometric distribution on $1,2,3,\ldots$. For the event to occur first on trial $t$, the first $t-1$ trials must fail and trial $t$ must succeed.

P(T=t)=(1-p)^{t-1}p,\qquad t=1,2,3,\ldots

To reach the start of trial $t$, all of the first $t-1$ trials must have been failures. Therefore $P(T\ge t)=(1-p)^{t-1}$, and dividing the probability of first success on trial $t$ by the probability of reaching trial $t$ gives the hazard.

h_t=\frac{(1-p)^{t-1}p}{(1-p)^{t-1}}=p.

The cancellation shows why the hazard is constant: after conditioning on all earlier trials being failures, the probability of success on the current trial remains $p$. This follows from the independent, identical-trial assumptions. If trial probabilities vary over time, or outcomes are dependent, that conclusion need not hold.

Theorem: If $T$ is the trial number of the first success in independent Bernoulli trials with fixed success probability $p$, then $h_t=p$ for every $t$ with $P(T\ge t)>0$. Conversely, if a positive-integer waiting time has $h_t=p$ for every $t$, where $0<p\le1$, then it has the geometric probability mass function $P(T=t)=(1-p)^{t-1}p$.

Why Constant Hazard Gives a Geometric Distribution

The converse can be justified by building the waiting-time probabilities one period at a time. Write $q_t=P(T\ge t)$ for the probability of being at risk at the start of period $t$. Since the event either occurs in period $t$ or the case remains at risk for period $t+1$, the conditional hazard $h_t$ leaves a fraction $1-h_t$ of the period-$t$ risk set to continue.

P(T\ge t+1)=P(T\ge t)(1-h_t).

If $h_t=p$ in every period, start with $P(T\ge1)=1$, because the first event has not yet occurred before period 1. Repeatedly applying the relation gives $P(T\ge t)=(1-p)^{t-1}$. The event probability in period $t$ is the period-start risk probability multiplied by the hazard:

P(T=t)=P(T\ge t)h_t=(1-p)^{t-1}p.

This establishes the geometric mass function from the constant-hazard condition. For $0<p\le1$, the probability of never having the event is zero, since $(1-p)^t$ tends to zero as $t$ increases. If $p=0$, the hazard is zero in every period but the event never occurs; that is not a proper geometric waiting time on the positive integers.

Worked Examples

Worked Example: Hazard Versus Unconditional Probability

A hypothetical sensor has an independent probability $p=0.20$ of detecting a fault in each weekly test, provided it has not detected one earlier. Let $T$ be the week of the first detection. Find (a) the probability that the sensor is still at risk at the start of week 4, (b) the probability that first detection occurs in week 4, and (c) the hazard in week 4. State the method and identify a common mistake.

Method: use the geometric survival probability for the risk set, then multiply by the period hazard for the event probability. The probability of failure to detect in one week is $1-p=0.80$. To be at risk at the start of week 4, there must be no detection in weeks 1, 2, or 3:

P(T\ge4)=0.80^3=0.512.

For first detection in week 4, those three earlier tests must have no detection and the fourth test must detect the fault:

$$ P(T=4)=0.80^3(0.20)=0.512(0.20)=0.1024. $$

The week-4 hazard conditions on being at risk at the start of that week:

h_4=P(T=4\mid T\ge4)=\frac{0.1024}{0.512}=0.20.

Check: $P(T=4)$ equals the risk probability $0.512$ multiplied by the hazard $0.20$, giving $0.1024$, as required. The common mistake is to report $0.20$ as the unconditional probability of first detection in week 4. It is the conditional hazard; the unconditional probability is $0.1024$.

Worked Example: Verifying a Constant Hazard Over Several Periods

A hypothetical review process examines an application once per day. Each review independently resolves the application with probability $p=0.25$, provided it is unresolved at the start of that day. Let $T$ be the day of first resolution. Calculate the risk probability and event probability for day 3, then calculate the day-3 hazard. Explain what the result says about the waiting-time model.

Method: calculate the period-specific mass and divide by the period-start risk probability. The application reaches day 3 unresolved only if the first two reviews do not resolve it. Each has probability $1-0.25=0.75$ of not resolving it, so

P(T\ge3)=0.75^2=0.5625.

For first resolution on day 3, multiply that risk probability by the day-3 resolution probability:

$$ P(T=3)=0.75^2(0.25)=0.5625(0.25)=0.140625. $$

Consequently,

h_3=\frac{P(T=3)}{P(T\ge3)} =\frac{0.140625}{0.5625}=0.25.

The same calculation in any day $t$ gives $P(T=t)=0.75^{t-1}(0.25)$ and $P(T\ge t)=0.75^{t-1}$, so $h_t=0.25$ for every day. The waiting time is geometric on $1,2,\ldots$ with parameter $0.25$. Check: the day-3 mass $0.140625$ is smaller than the hazard $0.25$, because only $0.5625$ of applications reach the start of day 3. The mistake to avoid is treating the conditional daily probability as the unconditional probability of resolution on day 3.

Worked Example: Changing Trial Probabilities Change the Hazard

In a hypothetical three-stage screening process, the chance of detecting a condition is 0.10 in stage 1, 0.20 in stage 2, and 0.30 in stage 3, conditional on no earlier detection. Assume the stage outcomes follow these stated conditional probabilities. Calculate the hazards for the first three stages and the probability of first detection in stage 3. Is the waiting time geometric?

Method: calculate each hazard from its own conditional probability, and calculate a stage-specific event probability by multiplying by the probability of reaching that stage. By the way the model is specified, the hazards are

h_1=0.10,\qquad h_2=0.20,\qquad h_3=0.30.

To reach stage 3, the condition must not have been detected in stages 1 and 2. The corresponding probability is

P(T\ge3)=(1-0.10)(1-0.20)=0.90(0.80)=0.72.

The probability of first detection in stage 3 is the probability of reaching it multiplied by its hazard:

P(T=3)=P(T\ge3)h_3=0.72(0.30)=0.216.

The waiting time is not geometric, because the hazards are not constant: $0.10$, $0.20$, and $0.30$ differ. Check: conditioning the stage-3 event probability on reaching stage 3 gives $0.216/0.72=0.30$, matching $h_3$. The common mistake is to use a single geometric parameter despite the stated change in per-stage probabilities.

Interpreting Hazard Carefully

The hazard answers a conditional, period-specific question: among cases that have not yet experienced the event, what is the probability it occurs in the next period? It is neither the probability that the event has occurred by that period nor the probability that it first occurs in that period without conditioning. For a geometric model, all three quantities can be related, but they are not interchangeable.

Hazard: $P(T=t\mid T\ge t)$, the event probability for period $t$ among cases at risk then.
Period-specific mass: $P(T=t)=P(T\ge t)h_t$, the unconditional probability that the first event occurs in period $t$.
Risk at the start of period $t$: $P(T\ge t)$, the probability of no event in the first $t-1$ periods.

A constant hazard is a model property, not a guarantee that observed percentages in every finite sample will be identical. If a sample records different event proportions among the cases at risk in different periods, those observed proportions estimate the underlying hazards and can vary by chance. The model assumption is that the underlying conditional probability is fixed, not that every realized sample fraction must match it exactly.

The geometric interpretation also depends on the unit of time. If each trial is one day, $T=4$ means the first event occurs on day 4, after three event-free days. If the trials are weeks, the same index refers to weeks instead. A constant per-trial probability is not automatically a constant probability for a differently sized time interval.

Key takeaway: For a positive-integer waiting time, the hazard in period $t$ is $P(T=t\mid T\ge t)$. Under independent, identical Bernoulli trials with success probability $p$, this conditional probability is $p$ in every period, and the resulting waiting time is geometric.

Check Your Understanding

For each problem, identify the period-start risk probability before calculating an unconditional period-specific event probability.

A hypothetical test has an independent 0.12 probability of detecting a signal on each attempt if it has not been detected earlier. Find the hazard on attempt 5 and the probability of first detection on attempt 5. (Answer: $h_5=0.1200$; $P(T=5)=0.0720$.)
A geometric waiting time has success probability $p=0.30$. Find $P(T\ge4)$, $P(T=4)$, and $h_4$. (Answer: $0.3430,\ 0.1029,\ 0.3000$.)
A process has conditional event probabilities 0.10, 0.25, and 0.40 in periods 1, 2, and 3, respectively. Find $P(T\ge3)$, $P(T=3)$, and the period-3 hazard. Is the waiting time geometric? (Answer: $P(T\ge3)=0.6750$, $P(T=3)=0.2700$, $h_3=0.4000$; no, the hazards are not constant.)
A geometric waiting time has $P(T=1)=0.18$. Find the common hazard, $P(T\ge5)$, and $P(T=5)$. (Answer: $h_t=0.1800$ for every $t$; $P(T\ge5)=0.4521$; $P(T=5)=0.0814$.)
In a hypothetical cohort with no censoring, 100 cases are at risk at the start of period 1 and 20 events occur. The remaining 80 cases are at risk at period 2, where 16 events occur. The remaining 64 are at risk at period 3, where 12 events occur. Calculate the three observed hazards and state whether they are exactly constant. (Answer: $20/100=0.2000,\ 16/80=0.2000,\ 12/64=0.1875$; they are not exactly constant.)

What You'll Learn