Minima of Independent Exponential Variables

Competing Exponential Variables

Suppose several independent events are waiting to occur, and each waiting time follows an exponential distribution. The first event to occur is determined by the minimum of those waiting times. Examples include the earliest failure among independent components or the first arrival among several independent sources. The key feature is that the minimum is itself exponential, but its rate is the sum of the individual rates.

Let \(X_1,\ldots,X_n\) be independent exponential variables, with \(X_i\) having rate \(\lambda_i>0\). In the rate parameterization, \(X_i\) has density \(\lambda_i e^{-\lambda_i t}\) for \(t\geq 0\), and survival probability \(P(X_i>t)=e^{-\lambda_i t}\). Define \(T\) to be the first waiting time and \(J\) to be the index of the variable that achieves it.

Definition: For independent exponential variables \(X_1,\ldots,X_n\) with respective rates \(\lambda_1,\ldots,\lambda_n>0\), define \(T=\min\{X_1,\ldots,X_n\}\) and \(J=\operatorname{argmin}_{1\leq i\leq n}X_i\). The total rate is \(\Lambda=\sum_{i=1}^n\lambda_i\). Since the variables have continuous distributions and are independent, ties occur with probability zero, so \(J\) is uniquely defined with probability one.

Distribution of the Minimum

The event that the minimum exceeds \(t\) occurs exactly when every waiting time exceeds \(t\). Independence therefore turns the probability of this joint event into a product.

Theorem: If \(X_1,\ldots,X_n\) are independent exponential variables with rates \(\lambda_i>0\), then \(T=\min_i X_i\) is exponential with rate \(\Lambda=\sum_i\lambda_i\). Its survival function, CDF, and density are

P(T>t)=e^{-\Lambda t},\quad F_T(t)= \begin{cases} 0,&t<0,\\ 1-e^{-\Lambda t},&t\geq 0, \end{cases} \quad f_T(t)=\Lambda e^{-\Lambda t},\quad t\geq 0.

Proof. For \(t\geq 0\), the event \(\{T>t\}\) is the intersection of the events \(\{X_i>t\}\). By independence and the exponential survival functions,

P(T>t) =P(X_1>t,\ldots,X_n>t) =\prod_{i=1}^n P(X_i>t) =\prod_{i=1}^n e^{-\lambda_i t} =e^{-t\sum_{i=1}^n\lambda_i} =e^{-\Lambda t}.

For \(t\geq 0\), \(F_T(t)=P(T\leq t)=1-P(T>t)=1-e^{-\Lambda t}\). For \(t<0\), all the \(X_i\) are nonnegative, so \(F_T(t)=0\). Differentiating the CDF for \(t>0\) gives \(f_T(t)=\Lambda e^{-\Lambda t}\). This density integrates to one because \(\int_0^\infty\Lambda e^{-\Lambda t}\,dt=1\). Thus \(T\) has an exponential distribution with rate \(\Lambda\). \(\square\)

The rates add because surviving until time \(t\) requires every individual clock to keep running until \(t\). A larger total rate means a shorter wait for the first event. For example, the mean waiting time of \(T\) is \(1/\Lambda\), using the mean of an exponential variable with rate \(\Lambda\).

Worked Example: Three Competing Failure Times

Three independent components have exponential failure times with rates \(2\), \(3\), and \(5\) per hour. Find the distribution of the time until the first failure and the probability that each component fails first.

The total rate is \(\Lambda=2+3+5=10\) per hour. Therefore the first-failure time \(T\) has density \(f_T(t)=10e^{-10t}\) for \(t\geq 0\), and \(P(T>t)=e^{-10t}\).

The probability that component \(i\) fails first is its rate divided by the total rate. Thus the probabilities are \(2/10=0.2\), \(3/10=0.3\), and \(5/10=0.5\). They sum to \(0.2+0.3+0.5=1\), as they must because exactly one component fails first with probability one.

For example, the probability all three components last more than \(0.2\) hours is \(P(T>0.2)=e^{-10(0.2)}=e^{-2}\), approximately \(0.1353\). The probability at least one fails within \(0.2\) hours is the complement, \(1-e^{-2}\), approximately \(0.8647\).

Which Variable Is Smallest?

The minimum's distribution gives the time of the first event, but not its identity. To calculate which variable is smallest, consider the possibility that \(X_i\) takes a value near \(t\) while all other variables exceed \(t\). The density for this to happen is the density of \(X_i\) at \(t\), multiplied by the other variables' survival probabilities.

Theorem: Under the same assumptions, for each \(i\),

P(J=i)=P(X_i=T)=\frac{\lambda_i}{\Lambda}.

Proof. For \(t\geq 0\), the joint density contribution for \(X_i\) to be the minimum at time \(t\) is its density at \(t\) times the probability that every other variable exceeds \(t\). Independence gives

\lambda_i e^{-\lambda_i t} \prod_{j\ne i}P(X_j>t) = \lambda_i e^{-\lambda_i t} \prod_{j\ne i}e^{-\lambda_jt} = \lambda_i e^{-\Lambda t}.

Integrating this density contribution over all possible first-event times gives

P(J=i)=\int_0^\infty \lambda_i e^{-\Lambda t}\,dt =\lambda_i\left[\frac{-1}{\Lambda}e^{-\Lambda t}\right]_{0}^{\infty} =\frac{\lambda_i}{\Lambda}.

The probabilities add to \(\sum_i\lambda_i/\Lambda=1\). Also, ties have probability zero: for any pair \(i\ne j\), independence and continuity imply \(P(X_i=X_j)=0\), and there are only finitely many pairs. Thus the events \(J=i\) partition the outcomes apart from a probability-zero set. \(\square\)

Worked Example: The First Arrival Before a Deadline

Two independent sources generate arrivals with exponential waiting times. Source A has rate \(0.4\) per day, and source B has rate \(0.6\) per day. Find the probability that B is the first source to arrive and that its arrival occurs within half a day.

The total rate is \(0.4+0.6=1\) per day. The density contribution for B to be first at time \(t\) is \(0.6e^{-t}\): B's density is \(0.6e^{-0.6t}\), and A's survival probability is \(e^{-0.4t}\), so their product is \(0.6e^{-t}\).

Integrating from time \(0\) to \(0.5\) gives

P(J=B,\ T\leq 0.5) =\int_0^{0.5}0.6e^{-t}\,dt =0.6(1-e^{-0.5}) \approx 0.2361.

As a check, the probability B is first at any time is \(0.6/(0.4+0.6)=0.6\). The deadline event is a subset of that event, and \(0.2361\leq 0.6\), as expected.

Independence of the First Time and Its Identity

In this setting, knowing which variable is smallest gives no information about the value of the minimum. This follows from a joint-density factorization, not merely from the separate formulas for \(T\) and \(J\).

Theorem: \(T\) and \(J\) are independent. More precisely, the joint density contribution for \(T=t\) and \(J=i\) is

f_{T,J}(t,i)=\lambda_i e^{-\Lambda t} =f_T(t)P(J=i),\quad t\geq 0.

Proof. The joint density contribution was obtained above: \(f_{T,J}(t,i)=\lambda_i e^{-\Lambda t}\). Since \(f_T(t)=\Lambda e^{-\Lambda t}\) and \(P(J=i)=\lambda_i/\Lambda\),

f_T(t)P(J=i) =\Lambda e^{-\Lambda t}\frac{\lambda_i}{\Lambda} =\lambda_i e^{-\Lambda t} =f_{T,J}(t,i).

Consequently, for any measurable set \(A\subseteq[0,\infty)\),

P(T\in A,\ J=i) =\int_A\lambda_i e^{-\Lambda t}\,dt =P(J=i)\int_A\Lambda e^{-\Lambda t}\,dt =P(J=i)P(T\in A).

Since \(J\) takes values in the finite set \(\{1,\ldots,n\}\), this factorization for every \(i\) and every measurable \(A\) establishes independence. \(\square\)

Worked Example: A First Event and Its Waiting Time

A device has two independent potential failure mechanisms. Mechanism A has rate \(1/3\) per hour and mechanism B has rate \(1/6\) per hour. Find the distribution of the time to the first failure, the probability A causes it, and the probability that A causes it within three hours.

The combined rate is \(\Lambda=1/3+1/6=1/2\) per hour. Thus \(T\) is exponential with rate \(1/2\), so \(P(T>t)=e^{-t/2}\) for \(t\geq0\). The probability A is first is \((1/3)/(1/2)=2/3\).

For the joint event that A is first and the first failure occurs by three hours, integrate its density contribution over that interval:

P(J=A,\ T\leq3) =\int_0^3\frac13e^{-t/2}\,dt =\frac23(1-e^{-3/2}) \approx 0.5179.

Independence provides a second calculation. The probability of any failure by three hours is \(1-e^{-3/2}\), so multiplying by \(P(J=A)=2/3\) gives the same result. The check confirms that the identity of the first mechanism and the time of the first failure separate into independent parts.

Common Pitfall and Why the Result Matters

A common mistake is to say that the smallest rate wins most often, or that the minimum has the smallest of the individual rates. Both conclusions reverse the role of rate: a larger exponential rate means a faster event, not a longer wait. If one rate is \(5\) and another is \(2\), the variable with rate \(5\) is more likely to be the minimum, with probability \(5/(5+2)\), and their minimum has rate \(7\).

The formula relies on independence. Without independence, the probability that all waiting times exceed \(t\) need not equal the product of their survival probabilities, so the sum-rate conclusion does not follow. When independence and exponential waiting times are appropriate, however, the results provide a compact way to analyze competing event types: total rate determines the time to the first event, and each rate's share of the total determines which event occurs first.

Key takeaway: For independent exponential waiting times, the minimum is exponential with rate equal to the sum of the rates. The probability that variable \(i\) is the minimum is its rate divided by the total rate, and the minimum time is independent of the identity of the minimizing variable.

Check Your Understanding

Try these questions using the definitions and results above.

Three independent exponential variables have rates \(1\), \(4\), and \(5\). What is the rate of their minimum, and what is the probability the second variable is smallest?
Explain why independence is needed to multiply the survival probabilities when deriving the distribution of a minimum.
Two independent exponential variables have rates \(0.2\) and \(0.8\) per day. What is the probability the first variable is the minimum?
For rates \(\lambda_1,\ldots,\lambda_n\), write the joint density contribution for the minimum to occur at time \(t\) and be attained by variable \(i\).
In the competing exponential setting, are the minimum time and the identity of the smallest variable independent? State the factorization that justifies your answer.