Competing Exponential Waiting Times

Why Competing Clocks Need Careful Probability Calculations

Suppose several independent processes are running at once: different sensors may trigger, separate components may fail, or distinct types of requests may arrive. Each process has its own waiting time. The time of the first event is the minimum of those waiting times, while the cause of that event identifies which clock rang first. Questions may also impose a deadline or ask what happens after the first event.

The key calculation combines the clocks’ rates, but it does not discard their individual rates. The combined rate determines when the first event occurs; each clock’s share of that combined rate determines its chance of being the cause. When events are considered in sequence, remove the clock that has already rung and calculate the next race among those still active.

The First Event: Time and Cause

The survival probability of an exponential waiting time with rate \(\lambda\) is \(P(X>t)=e^{-\lambda t}\) for \(t\geq 0\). For independent clocks, all clocks must survive past \(t\) for the first event to occur after \(t\). Independence therefore gives

Thus \(T\) is exponential with rate \(\Lambda\). Its density is \(f_T(t)=\Lambda e^{-\Lambda t}\) for \(t\geq0\). The cause probabilities are \(P(J=i)=\lambda_i/\Lambda\). These results combine in a useful joint formula.

The joint formula distinguishes two questions that are often confused. The probability that some clock rings by \(t\) is \(1-e^{-\Lambda t}\). The probability that a particular clock \(i\) rings first and does so by \(t\) includes the additional factor \(\lambda_i/\Lambda\). A deadline does not make every cause equally likely.

Why the Cause and First-Event Time Are Independent

A short density argument explains why the cause probability and the first-event time probability multiply. For clock \(i\) to be first at a time near \(t\), it must ring near \(t\), while every other clock must survive to \(t\). The density of this event, as a function of \(t\), is

The last expression factors into the probability \(P(J=i)=\lambda_i/\Lambda\) and the density \(f_T(t)=\Lambda e^{-\Lambda t}\). Integrating over any time interval shows that the cause and first-event time are independent. In particular, conditioning on the first event occurring before a deadline changes the distribution of \(T\), but it does not change the cause probabilities.

What Happens to the Remaining Clocks?

After the first event, its clock is no longer part of the race. The other clocks, conditional on having survived until the first event, retain exponential waiting-time distributions with their original rates. This follows from the memoryless property: for an exponential random variable \(X\) with rate \(\lambda\),

This says that, having waited \(s\) time units without a particular clock ringing, its additional wait has the same exponential distribution as its original wait. For independent clocks, the surviving clocks remain independent conditional on their survival to a fixed time. At the first event, the same calculation applies to each clock that did not ring: given the event time and the identity of the winner, each remaining clock has an exponential residual wait with its original rate. The next race therefore uses the sum of the rates of the clocks still active.

For example, if clock \(i\) rings first, the chance that clock \(k\) is next is \(\lambda_k/(\Lambda-\lambda_i)\), provided at least one clock remains. If the next event must also occur within \(u\) time units after the first, the probability of that next cause and deadline is

These are conditional probabilities for the race after the first event. To find the probability of a full sequence, multiply the first-stage probability by the conditional probability of the next stage, and continue in this way. This multiplication is valid because each later-stage factor is calculated conditional on the earlier specified events.

Conditioning on Survival to a Time

A problem may state that no event has occurred by a particular time and then ask about the next interval. The condition tells us that every clock survived to that time. By memorylessness, the residual waits have the same rates as before. The time origin can be shifted to the conditioning time, and the clocks can be treated as a fresh race.

A different question would ask for the probability that no event occurs by the conditioning time in the first place. That survival probability is \(e^{-\Lambda s}\). Include it only if the event of surviving to time \(s\) is part of the requested joint event. If the problem already conditions on survival, do not multiply by it again.

Common Exam Traps and a Reliable Process

Competing-clock problems are aligned with the SOA Exam P (Probability) syllabus because they combine exponential distributions, independence, conditional probability, and careful event definition. A useful sequence of steps is to identify the clocks still active at each stage, check that all rates use the same time unit, compute the relevant total rate, and then apply the cause share and deadline factor separately.

• Using only the deadline probability. \(1-e^{-\Lambda t}\) means some clock rings by \(t\). For a specified cause, multiply it by that clock’s rate share.
• Using the initial total rate at every stage. Once a clock has rung, remove it from the next race. The remaining total rate is the sum of the surviving clocks’ rates.
• Mixing time units. Rates per hour cannot be combined with a deadline in minutes without converting one of them. For example, a rate of \(0.5\) per hour is \(0.5/60\) per minute.
• Including survival twice. If survival to a time is given as a condition, memorylessness gives a fresh residual race. Do not multiply by the probability of that survival unless it is also part of the requested event.
• Multiplying cause shares without deadline factors. A rate share gives the probability of a cause being next with no time restriction. A deadline on that wait requires the corresponding factor \(1-e^{-\text{(remaining total rate)}\times\text{deadline}}\).