The Constant Hazard of the Exponential Distribution

Why Hazard Is More Than a Failure Probability

A waiting time can be described by the probability that the event has not happened yet. A different question is: among cases that have lasted until time $t$, how quickly is the event occurring just after $t$? The hazard rate answers that question. It is a conditional rate at a point in time, not the probability of an event over an interval of positive length.

For an exponential waiting time, the hazard is constant: the instantaneous event rate is the same whether little time or much time has elapsed. This gives a precise connection between the exponential model and memoryless waiting. The distinction between a rate and a probability matters in actuarial work, where a rate has units such as “per month,” but a probability is dimensionless.

Definition: Let $T$ be a nonnegative, absolutely continuous waiting time, with cumulative distribution function $F_T(t)$, density $f_T(t)$, and survival function $S_T(t)=P(T>t)$. At a time $t$ with $S_T(t)>0$, the hazard, when the limit exists, is $$ h_T(t)=\lim_{\Delta t\downarrow 0} \frac{P(t<T\le t+\Delta t\mid T>t)}{\Delta t}. $$ If the right derivative of $F_T$ at $t$ equals the selected density value $f_T(t)$, then $$ h_T(t)=\frac{f_T(t)}{S_T(t)}. $$ For an absolutely continuous distribution, this ratio agrees with the hazard defined through the limit almost everywhere. To assert the ratio at a particular time, the stated pointwise regularity condition is needed; a density can otherwise be changed at individual points without changing the distribution.

Deriving the Exponential Hazard

Use the rate parameterization $T\sim\operatorname{Exponential}(\lambda)$, where $\lambda>0$ is the rate per unit of time. The model has support $t\ge 0$, density $f_T(t)=\lambda e^{-\lambda t}$, and survival function $S_T(t)=e^{-\lambda t}$. Both functions are continuous, so the pointwise density-to-survival formula applies for every $t\ge 0$.

h_T(t)=\frac{f_T(t)}{S_T(t)} =\frac{\lambda e^{-\lambda t}}{e^{-\lambda t}} =\lambda,\qquad t\ge 0.

The exponential factor cancels. Thus the hazard does not depend on elapsed time. Its units are the inverse of the units used for $T$: if time is measured in hours, $\lambda$ and $h_T(t)$ are measured per hour. A hazard is not itself a probability. For example, a hazard of $0.2$ per hour does not mean that the event has probability $0.2$ in every hour; the probability over a full hour is $1-e^{-0.2}$, provided the exponential model applies.

Worked Example: Evaluating an Exponential Hazard

A hypothetical sensor has an operating lifetime $T\sim\operatorname{Exponential}(0.25)$, with time measured in days and rate $0.25$ per day. Find its hazard after two days of operation and the probability it fails during the next half-day, given that it has survived two days. State why these are different quantities.

Method. Use the density-to-survival ratio for the instantaneous hazard, then use conditional survival to obtain the probability over the specified interval.

The density and survival function are $$ f_T(t)=0.25e^{-0.25t},\qquad S_T(t)=e^{-0.25t}. $$ At $t=2$ days, $$ h_T(2)=\frac{0.25e^{-0.25(2)}}{e^{-0.25(2)}}=0.25\text{ per day}. $$

For the next half-day, conditional on survival to day 2, $$ P(2<T\le 2.5\mid T>2) =1-\frac{S_T(2.5)}{S_T(2)} =1-e^{-0.25(0.5)} =1-e^{-0.125}. $$ Since $e^{-0.125}\approx 0.8825$, the probability is $0.1175$, to four decimal places.

Check. The hazard has units of inverse days; the interval probability has no units and is between 0 and 1. The probability is not $0.25$, because $0.25$ is a rate per day, not a one-day failure probability. The calculation uses the same rate for the sensor’s entire lifetime, as required by the exponential model.

Common mistake avoided: treating the hazard as the probability of failure over a specified interval. The probability here is $0.1175$ over half a day, while the instantaneous hazard is $0.25$ per day.

Why Constant Hazard Gives Memoryless Waiting

The connection can be seen directly from conditional survival. For any $s,t\ge 0$ with $S_T(s)>0$, the probability of waiting at least another $t$ time units after already waiting $s$ is $S_T(s+t)/S_T(s)$. For the exponential distribution, this ratio depends on $t$, but not on $s$.

P(T>s+t\mid T>s) =\frac{S_T(s+t)}{S_T(s)} =\frac{e^{-\lambda(s+t)}}{e^{-\lambda s}} =e^{-\lambda t} =P(T>t).

So, conditional on having lasted $s$ time units, the remaining wait has the same survival probabilities as a fresh exponential waiting time. This is memorylessness. It does not say the event is guaranteed to happen soon; it says that the elapsed time does not change the model’s conditional distribution of the remaining wait.

There is also a converse under standard smoothness conditions. Suppose a nonnegative continuous waiting-time model has $S_T(0)=1$, positive differentiable survival on the time interval of interest, and constant hazard $h_T(t)=\lambda$. Since $f_T(t)=-S_T'(t)$, the density-to-survival formula gives the differential equation below. Solving it shows that the survival function must be exponential.

Write the hazard equation.
From $h_T(t)=f_T(t)/S_T(t)=\lambda$ and $f_T(t)=-S_T'(t)$, obtain $-S_T'(t)/S_T(t)=\lambda$.

Separate and integrate.
Because $S_T(t)>0$, divide by $S_T(t)$ and integrate from $0$ to $t$: $\int_1^{S_T(t)}du/u=-\lambda\int_0^t dr$, using $S_T(0)=1$.

Solve for survival.
The integrated equation is $\log S_T(t)=-\lambda t$, so $S_T(t)=e^{-\lambda t}$. This is the exponential survival function and yields the memoryless conditional-survival ratio above.

The converse also clarifies the model’s assumptions. A constant hazard $\lambda>0$ over all $t\ge 0$, together with the stated survival regularity and $S_T(0)=1$, produces an exponential distribution. If a waiting-time model has a changing hazard, it is not exponential under this definition, even if its hazard happens to equal a constant at one particular time.

Worked Example: Hazard Versus a Finite-Interval Probability

A hypothetical component has lifetime $T\sim\operatorname{Exponential}(0.08)$, measured in years, with rate $0.08$ per year. Given that it is operating at year 7, calculate its hazard at that time and the probability it fails in the following 3 years.

Method. Evaluate the constant hazard from the density-to-survival ratio, then calculate the conditional probability over the three-year interval from the survival ratio.

The density and survival function are $f_T(t)=0.08e^{-0.08t}$ and $S_T(t)=e^{-0.08t}$. Therefore, $$ h_T(7)=\frac{0.08e^{-0.08(7)}}{e^{-0.08(7)}}=0.08\text{ per year}. $$ The probability of failure from year 7 through year 10, conditional on operating at year 7, is $$ P(7<T\le 10\mid T>7) =1-\frac{S_T(10)}{S_T(7)} =1-e^{-0.08(3)} =1-e^{-0.24}. $$ Since $e^{-0.24}\approx 0.7866$, this probability is $0.2134$, to four decimal places.

Check. The answer is a probability for a three-year interval, not a rate. It is less than 1 and exceeds the probability for a shorter interval under the same model. The fact that the component already operated for 7 years does not alter the conditional calculation.

Common mistake avoided: multiplying the hazard by 3 and reporting $0.24$ as the probability. The product $0.08\times3=0.24$ is the exponent in the survival calculation; the probability is $1-e^{-0.24}=0.2134$.

How Hazard Changes Outside the Exponential Model

A density-to-survival ratio can also describe a non-exponential waiting-time model. In that case the ratio generally changes with time, so it is important not to assume memorylessness merely because a hazard is defined. The following comparison uses a uniform waiting time, whose density is continuous on the interior of its support.

Worked Example: A Uniform Waiting Time Has a Changing Hazard

A hypothetical inspection delay $T$ is uniformly distributed from 0 to 10 hours. Find the hazard at hour 4 and the probability the delay ends in the next hour, conditional on it having lasted more than 4 hours. Explain why this model is not memoryless.

Method. Derive the density and survival function on the interior of the support, take their ratio, and compare the conditional interval probability with the memoryless property.

For $0\le t\le 10$, the density is $f_T(t)=1/10$. For $0\le t\le10$, the survival function is $S_T(t)=(10-t)/10$. At $t=4$, survival is positive and both functions are continuous, so $$ h_T(4)=\frac{1/10}{(10-4)/10} =\frac{1}{6}\text{ per hour} \approx 0.1667\text{ per hour}. $$

The conditional probability that the delay ends during the next hour is $$ P(4<T\le5\mid T>4) =\frac{P(4<T\le5)}{P(T>4)} =\frac{1/10}{6/10} =\frac{1}{6} \approx 0.1667. $$ Here the interval is one hour long, so the numerical equality with the hazard at hour 4 is not a general rule that a hazard is an interval probability.

The hazard is not constant: for $0\le t<10$, $h_T(t)=1/(10-t)$, which increases as $t$ approaches 10. Also, the conditional survival probability after 4 hours of elapsed time is not the same as survival from a fresh start for every additional duration. Thus this uniform delay is not memoryless.

Check. At hour 4, $S_T(4)=0.6>0$, as required to condition on survival. At the endpoint $t=10$, survival is zero, so the conditional hazard given $T>10$ is not defined.

Common mistake avoided: calling every density-to-survival ratio a constant hazard. The ratio must be evaluated as a function of time; in this example it increases rather than remaining fixed.

Interpreting and Using the Hazard Correctly

The hazard is useful because it describes the local event rate among items, people, or processes that have survived to the present time. For an exponential model, it is constant and the conditional distribution of remaining waiting time is unchanged by elapsed time. That is a substantive model assumption, not a consequence of merely having a continuous lifetime.

For a small positive interval $\Delta t$, the definition implies that the conditional probability of an event in the next interval is approximately $h_T(t)\Delta t$, when the hazard is well behaved near $t$. For a finite interval, use the survival ratio or integrate the hazard; do not simply substitute a finite interval length into the instantaneous definition. Under a differentiable survival model, $$ h_T(t)=-\frac{d}{dt}\log S_T(t), \qquad S_T(t)=\exp\left(-\int_0^t h_T(u)\,du\right), $$ where the second identity uses $S_T(0)=1$. If $h_T(u)=\lambda$, this reduces to $S_T(t)=e^{-\lambda t}$.

Exam takeaway: For $T\sim\operatorname{Exponential}(\lambda)$, the density-to-survival ratio is $\lambda$ at every $t\ge0$, and the waiting time is memoryless. Keep the units attached to the hazard, check that survival at the conditioning time is positive, and use a survival ratio to obtain probabilities over finite intervals. The pointwise density-to-survival formula requires the appropriate regularity at the time being evaluated.

Check Your Understanding

For each problem, identify whether the requested result is a hazard or a finite-interval probability. Give probabilities to four decimal places.

A hypothetical cooling system has lifetime $T\sim\operatorname{Exponential}(0.5)$, with time measured in minutes. Find its hazard at 3 minutes and the probability it stops during the next 0.2 minute, given that it is still operating at 3 minutes. (Answer: hazard $0.5$ per minute; probability $1-e^{-0.1}=0.0952$.)
A hypothetical device has an exponential lifetime with rate $0.12$ per year. Given survival through 4 years, find the probability it fails in the next 2 years. (Answer: $1-e^{-0.24}=0.2134$.)
A delay is uniform on $[0,8]$ hours. Find its hazard at hour 2 and the probability it ends in the next hour given that it has lasted 2 hours. (Answer: hazard $1/6$ per hour; probability $1/6=0.1667$.)
A hypothetical process has a constant hazard of $0.03$ per month and positive differentiable survival with $S_T(0)=1$. Find the probability it ends within the next 5 months, given that it has lasted 20 months. (Answer: $1-e^{-0.15}=0.1393$.)
A hypothetical lifetime model has constant hazard $0.4$ per day and positive differentiable survival with $S_T(0)=1$. Find $S_T(2)$, then find the conditional probability of ending between days 2 and 5 given survival through day 2. (Answer: $S_T(2)=e^{-0.8}\approx0.4493$; conditional probability $1-e^{-1.2}=0.6988$.)

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