Exponential Order Statistics And Spacings

Why Exponential Spacings Are Useful

When several independent lifetimes or waiting times are ordered from earliest to latest, the gaps between consecutive order statistics are often easier to analyze than the order statistics themselves. For an iid exponential sample, these gaps have a particularly simple structure: they are independent exponential random variables, although their rates change as observations are removed from the running sample.

The key is to distinguish the first observed time, the second observed time, and the gap between them. If the first event occurs at time \(x_{(1)}\) and the second at \(x_{(2)}\), then the second gap is \(x_{(2)}-x_{(1)}\), not \(x_{(2)}\). The rate of that gap also differs from the rate of the first waiting time.

Model and the Minimum

Assume \(X_1,\ldots,X_n\) are independent and identically distributed exponential random variables with rate \(\lambda>0\). The rate has units reciprocal to those of \(X_i\); for example, if time is measured in months, \(\lambda\) is per month. Under this parameterization, the density is \(\lambda e^{-\lambda x}\) for \(x\geq 0\), and the mean is \(1/\lambda\).

Definition: Write \(X_{(1)}\leq X_{(2)}\leq\cdots\leq X_{(n)}\) for the sample values arranged in increasing order. The sample minimum is \(X_{(1)}=\min(X_1,\ldots,X_n)\). With probability one, there are no ties because the exponential distribution is continuous.

For \(t\geq 0\), the event \(X_{(1)}>t\) occurs exactly when every sample value exceeds \(t\). Independence therefore gives

P(X_{(1)}>t) =P(X_1>t,\ldots,X_n>t) =\prod_{i=1}^{n}P(X_i>t) =(e^{-\lambda t})^n =e^{-n\lambda t}.

Thus the minimum is exponential with rate \(n\lambda\). Its density is \(n\lambda e^{-n\lambda t}\) for \(t\geq0\), and its mean is \(1/(n\lambda)\). This result applies to independent, identically distributed exponential variables with the same rate. If the observations have different rates, the minimum is still exponential, but its rate is the sum of their rates; the equal-rate spacing result below does not apply unchanged.

Worked Example: Earliest Failure Among Identical Components

A hypothetical system contains five independent components. Each component’s lifetime is exponential with mean 10 months. Find the probability that the first component failure occurs after 4 months, the density of the first failure time at 4 months, and the expected time to the first failure.

Method and model. The common rate is \(\lambda=1/10=0.1\) per month, and \(n=5\). The minimum has rate \(n\lambda=5(0.1)=0.5\) per month.

Probability. The survival probability of the minimum at 4 months is \(P(X_{(1)}>4)=e^{-(0.5)(4)}=e^{-2}=0.1353\), to four decimal places.

Density. The density at 4 months is \(f_{X_{(1)}}(4)=0.5e^{-(0.5)(4)}=0.5e^{-2}=0.0677\) per month, to four decimal places.

Expected time. The mean of an exponential variable is the reciprocal of its rate, so \(E[X_{(1)}]=1/0.5=2\) months.

Check and trap avoided. The probability and density use the same minimum rate, \(0.5\) per month, and the mean is its reciprocal. Using the individual-component rate \(0.1\) per month would incorrectly give the lifetime of one component rather than the earliest failure among five.

Successive Spacings

Set \(X_{(0)}=0\), and define the \(j\)th spacing as the distance from the \((j-1)\)st order statistic to the \(j\)th:

D_j=X_{(j)}-X_{(j-1)},\qquad j=1,\ldots,n.

In particular, \(D_1=X_{(1)}\). For \(j\geq2\), \(D_j\) is a gap between consecutive ordered observations. The order statistics can be recovered by adding gaps: \(X_{(k)}=D_1+\cdots+D_k\).

Theorem: If \(X_1,\ldots,X_n\) are iid exponential with rate \(\lambda\), then the spacings \(D_1,\ldots,D_n\) are mutually independent, and \(D_j\) is exponential with rate \((n-j+1)\lambda\). The rates are \(n\lambda,(n-1)\lambda,\ldots,\lambda\), in that order.

The first gap has the largest rate because all \(n\) observations are initially still at risk of being the first. After the first observation has occurred, \(n-1\) values remain to determine the next gap; after the second, \(n-2\) remain, and so on. The joint-density derivation makes both the rates and independence precise.

Why the Spacings Are Independent

The joint density of the ordered sample is \(n!\lambda^n e^{-\lambda(x_1+\cdots+x_n)}\) on \(0<x_1<\cdots<x_n\). The factor \(n!\) accounts for the \(n!\) possible arrangements of distinct sample values before ordering. Transform from the ordered values to the spacings using \(x_k=d_1+\cdots+d_k\). This triangular transformation has Jacobian determinant 1, and its support becomes \(d_1,\ldots,d_n>0\).

Each \(d_j\) occurs in \(x_j,x_{j+1},\ldots,x_n\), a total of \(n-j+1\) terms. Consequently,

x_1+\cdots+x_n =\sum_{j=1}^{n}(n-j+1)d_j.

Substituting into the ordered-sample density gives

n!\lambda^n \exp\!\left(-\lambda\sum_{j=1}^{n}(n-j+1)d_j\right) = \prod_{j=1}^{n} \left[(n-j+1)\lambda e^{-(n-j+1)\lambda d_j}\right].

For the equality, the product of the rates is \(\lambda^n n(n-1)\cdots1=n!\lambda^n\), and the exponential factors multiply to the required exponent. Each bracket is an exponential density on \(d_j>0\), and the joint density factors into one density per spacing. A joint density that factors this way establishes mutual independence and gives the stated rate of each spacing.

Worked Example: A Probability Involving the First Two Gaps

Four independent hypothetical device lifetimes are exponential with rate \(0.25\) per day. Find the probability that the first failure occurs after 1 day and that the gap from the first to the second failure is at most 2 days.

Method and rates. Here \(n=4\) and \(\lambda=0.25\) per day. The first spacing \(D_1\) has rate \(4(0.25)=1\) per day. The second spacing \(D_2\) has rate \(3(0.25)=0.75\) per day. The spacings are independent.

The requested event is \(\{D_1>1,\ D_2\leq2\}\). By independence, its probability is the product of the two spacing probabilities:

\begin{aligned} P(D_1>1,\ D_2\leq2) &=P(D_1>1)P(D_2\leq2)\\ &=e^{-(1)(1)}\left(1-e^{-(0.75)(2)}\right)\\ &=e^{-1}(1-e^{-1.5})\\ &=0.3679(1-0.2231)\\ &=0.3679(0.7769)\\ &=0.2858. \end{aligned}

The answer is \(0.2858\), to four decimal places. The rates have units per day and the time values are in days, so each exponent is dimensionless. As a check on the common trap, using the individual rate \(0.25\) for \(D_1\) would give \(P(D_1>1)=e^{-0.25}\); the correct rate for the minimum is \(4(0.25)=1\).

Worked Example: A Gap Is Not the Second Order Statistic

Three independent hypothetical inspection times are exponential with rate \(0.4\) per hour. Find the probability that the first inspection occurs between 0.5 and 1.5 hours and the second inspection occurs more than 1 hour after the first.

Method and model. The event specifies \(D_1\) between 0.5 and 1.5 hours and \(D_2>1\) hour. With \(n=3\), the rates are \(3(0.4)=1.2\) per hour for \(D_1\) and \(2(0.4)=0.8\) per hour for \(D_2\). The third spacing is unrestricted, so it contributes probability 1. Independence lets us multiply the first two probabilities.

\begin{aligned} P(0.5<D_1<1.5,\ D_2>1) &=\left(e^{-(1.2)(0.5)}-e^{-(1.2)(1.5)}\right)e^{-(0.8)(1)}\\ &=(e^{-0.6}-e^{-1.8})e^{-0.8}\\ &=(0.5488-0.1653)(0.4493)\\ &=0.3835(0.4493)\\ &=0.1723. \end{aligned}

The probability is \(0.1723\), to four decimal places. The second condition is about the gap \(D_2=X_{(2)}-X_{(1)}\), so its rate is \(0.8\) per hour. Treating it as a condition on \(X_{(2)}\) would change the event and would not be justified by this calculation.

Reading Problems Carefully

A reliable solution begins by translating each phrase into a spacing or an order statistic. “The first event occurs after \(a\)” means \(D_1>a\). “The next event is more than \(b\) time units after the first” means \(D_2>b\). “The second event occurs after time \(c\)” instead means \(X_{(2)}>c\), and since \(X_{(2)}=D_1+D_2\), that is a condition on the sum of two spacings, not just on \(D_2\).

That distinction matters because independence applies directly to the spacings, not to the order statistics. For example, \(X_{(1)}\) and \(X_{(2)}\) are not independent: the second ordered time includes the first. Do not multiply their marginal probabilities as if they were independent. If a problem involves a sum of spacings or an event defined by several order statistics, first express the event accurately; a spacing decomposition does not by itself make every resulting calculation a product.

Key takeaway: For an iid exponential sample with rate \(\lambda\), the successive gaps are independent exponentials with rates \(n\lambda,(n-1)\lambda,\ldots,\lambda\). Identify whether each condition concerns an ordered value \(X_{(j)}\) or a gap \(D_j\) before applying those rates.

Check Your Understanding

For each problem, identify the relevant spacing rate before calculating.

Six independent lifetimes are exponential with rate \(0.3\) per week. Find the probability that the earliest lifetime exceeds 2 weeks and the density of the earliest lifetime at 1 week. (Answer: probability \(0.0273\); density \(0.2975\) per week.)
Four iid exponential observations have rate \(0.5\) per day. Find the probability that the second spacing exceeds 1 day, and give the mean of that spacing. (Answer: probability \(e^{-1.5}=0.2231\); mean \(2/3\) day.)
Three independent waiting times are exponential with rate \(0.2\) per hour. Find the probability that \(D_1>1\) hour and \(D_2\leq2\) hours. (Answer: \(e^{-0.6}(1-e^{-0.8})=0.3022\).)
Five independent hypothetical sensor lifetimes are exponential with mean 4 hours. Find the probability that the first failure is after 0.5 hour and the second gap is between 0.5 and 1.5 hours. (Answer: \(e^{-0.625}(e^{-0.5}-e^{-1.5})=0.2052\).)
Two iid exponential observations have rate \(0.4\) per day. Find the probability that the first observation exceeds 1 day and the gap between the two observations exceeds 2 days. (Answer: \(e^{-0.8}e^{-0.8}=e^{-1.6}=0.2019\).)

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