Competing Exponential Clocks in a Superposed Stream

One Arrival, Several Independent Clocks

Suppose several independent sources can produce the next event. Each source has its own exponential waiting time, but only the earliest of those times is observed as the next arrival in the combined stream. The minimum determines how long we wait; the identity of the source attaining it determines who supplied that arrival. Both quantities can be calculated directly from the exponential distributions.

Work on a probability space \((\Omega,\mathcal{F},\mathbb{P})\). For a finite collection of \(m\ge1\) independent Poisson counting processes \((N_t^{(r)})_{t\ge0}\), let each process have state space \(\mathbb{Z}_{\ge0}\) and rate \(\lambda_r\), where \(r\in\{1,\ldots,m\}\). The time to the first arrival from source \(r\) is \(T_r=\inf\{t\ge0:N_t^{(r)}\ge1\}\). A positive-rate source has \(T_r\) exponentially distributed with rate \(\lambda_r\), meaning \(\mathbb{P}(T_r>t)=e^{-\lambda_r t}\) for \(t\ge0\). A source with rate zero has \(T_r=\infty\) almost surely.

We first treat the nondegenerate case in which every rate is positive. Then \(\Lambda>0\), and each waiting time has a continuous distribution. Consequently, two independent clocks tie with probability zero: for distinct \(i,j\), conditional on \(T_i=t\), the probability that \(T_j=t\) is zero for every \(t\), because \(T_j\) has no point masses. Thus there is almost surely a unique source of the next arrival.

The Minimum and the Winning Source

Proof. For \(t\ge0\), the event \(T>t\) is the event that every clock exceeds \(t\). Independence and the exponential survival formulas give

This is the survival function of an exponential random variable with rate \(\Lambda\), so \(T\) has that distribution and density \(f_T(t)=\Lambda e^{-\Lambda t}\) for \(t\ge0\). To calculate the probability that source \(i\) wins, integrate over the possible time \(t\) of its arrival. Its density at \(t\) is \(\lambda_i e^{-\lambda_i t}\), and every other clock must exceed \(t\), an event of probability \(\prod_{r\ne i}e^{-\lambda_r t}\) independent of \(T_i\). Therefore

The final equality follows by integrating the exponential: \(\int_0^\infty e^{-\Lambda t}\,dt=1/\Lambda\), since \(\Lambda>0\). Summing over all possible winners gives \(\sum_i\lambda_i/\Lambda=\Lambda/\Lambda=1\), as required. \(\square\)

The rate of the minimum is a sum, not an average: more independent sources give more opportunities for an early event. The winning probabilities are the individual rates divided by that sum. In particular, a source with twice another source’s rate is twice as likely to supply the next arrival, regardless of how long the clocks have already been running, provided the conditioning is that none has yet rung.

There is also a useful joint calculation. The probability that source \(i\) wins and the first arrival occurs by a deadline \(t\ge0\) is the integral of its winning density only up to that deadline:

This is not a new assumption about the sources; it follows from the same density-and-survival calculation used in the theorem. Summing over sources gives \(1-e^{-\Lambda t}=\mathbb{P}(T\le t)\), the probability that any source arrives by the deadline.

What Changes After a Quiet Interval?

Suppose no source has arrived by a fixed time \(s\ge0\). The waiting times still have an exponential race after \(s\). The conditioning event has positive probability, since \(\mathbb{P}(T>s)=e^{-\Lambda s}>0\). For \(x_1,\ldots,x_m\ge0\), independence gives the conditional joint survival probability

Proof. The conditioning event \(\{T_1>s,\ldots,T_m>s\}\) has probability \(\prod_r e^{-\lambda_r s}=e^{-\Lambda s}>0\), so the conditional probability is defined. The displayed calculation factors into one survival function for each \(r\). Each factor is \(e^{-\lambda_r x_r}\), the survival function of an exponential variable with rate \(\lambda_r\). Since the joint survival function factors into the product of these marginal survival functions for every nonnegative \(x_1,\ldots,x_m\), the residual waiting times are independent with the stated distributions. Applying the theorem to those residual times gives the claimed additional waiting-time distribution and winner probabilities. \(\square\)

This result is the race version of exponential memorylessness. A quiet interval does not favor one source over another: the residual clocks have the same rates as before. The conditioning here is specifically that all sources have remained quiet through time \(s\); conditioning on a different event need not have the same interpretation.

Zero Rates and a Common Mistake

The formulas also cover sources with rate zero if at least one source has positive rate. A zero-rate source has waiting time \(\infty\) almost surely and cannot supply the next arrival. The minimum is exponential with rate \(\Lambda\), the sum of the positive rates, and the winner probability for a zero-rate source is \(0/\Lambda=0\). If all rates are zero, then \(T=\infty\) almost surely and there is no next arrival; there is no finite-rate exponential distribution or winning source to calculate.

Common mistake: using the rates themselves as probabilities. With rates \(2\) and \(3\), it is wrong to say that the first source wins with probability \(2\) and the second with probability \(3\). Rates measure arrivals per unit time, not probabilities, and those numbers do not sum to \(1\). The correction is to divide each rate by the total rate: the probabilities are \(2/(2+3)=2/5\) and \(3/(2+3)=3/5\). Their sum is \(1\), and the combined waiting time has rate \(5\), not \(2\), \(3\), or their average.

The main distinction is worth keeping explicit: the minimum waiting time uses the sum of the rates, while the chance that a particular source wins uses that source’s rate divided by the sum. These calculations describe the first event in the race; they do not by themselves establish any further claims about the sequence of source labels.