An Exponential-Rate Interval From the Sample Minimum

Why the Sample Minimum Can Estimate an Exponential Rate

Suppose a study records independent waiting times until a specified event occurs. If the waiting times follow an exponential distribution, the rate parameter describes how quickly events tend to occur: a larger rate corresponds to shorter waits. The smallest observed wait contains information about that rate because, among several observations, a rapid event becomes more likely as the rate increases.

This tutorial uses the sample minimum to construct an exact confidence interval. The interval is not based on a normal approximation: its coverage follows directly from the distribution of the minimum. The method is aligned with the SOA VEE Mathematical Statistics syllabus.

Model and assumptions: Let \(X_1,\ldots,X_n\) be independent and identically distributed waiting times, each with an exponential distribution of unknown rate \(\lambda>0\). Thus \(X_i\sim\operatorname{Exponential}(\lambda)\), with density \(f_X(x)=\lambda e^{-\lambda x}\) for \(x\geq 0\), and survival function \(S_X(x)=e^{-\lambda x}\). The observations are complete, uncensored waiting times measured in a common time unit. Define the sample minimum by \(M=\min(X_1,\ldots,X_n)\).

The rate has units reciprocal to the time unit used for \(X_i\). For example, if waiting times are measured in hours, \(\lambda\) is reported per hour. This model does not describe a sample with right-censored observations unless the sampling and censoring mechanism are handled separately.

Distribution of the Sample Minimum

The event that the minimum exceeds \(m\) occurs exactly when every waiting time exceeds \(m\). Independence lets us multiply those probabilities:

\begin{aligned} P(M>m) &=P(X_1>m,\ldots,X_n>m)\\ &=\prod_{i=1}^{n}P(X_i>m)\\ &=(e^{-\lambda m})^n\\ &=e^{-n\lambda m},\qquad m\geq 0. \end{aligned}

Therefore \(M\) itself has an exponential distribution with rate \(n\lambda\). In particular, multiplying the minimum by \(n\lambda\) removes the unknown rate from its distribution.

Theorem: Under the stated model, \(M\sim\operatorname{Exponential}(n\lambda)\), so the pivotal quantity \(Q=n\lambda M\) has the standard exponential distribution, with rate 1. Its cumulative distribution function is \(P(Q\leq q)=1-e^{-q}\) for \(q\geq 0\).

A pivotal quantity is a statistic whose distribution does not depend on the unknown parameter. Here \(M\) depends on the unknown rate, but the scaled quantity \(n\lambda M\) always follows the same standard exponential distribution. This gives probability statements that can be inverted to obtain bounds for \(\lambda\).

Deriving an Equal-Tailed Confidence Interval

Let the desired confidence level be \(1-\alpha\), where \(0<\alpha<1\). For an equal-tailed interval, assign probability \(\alpha/2\) to each tail of the standard exponential pivot. Its lower and upper cutoffs are

a=-\ln(1-\alpha/2), \qquad b=-\ln(\alpha/2).

These values satisfy \(P(Q<a)=\alpha/2\) and \(P(Q>b)=\alpha/2\). Consequently, \(P(a\leq Q\leq b)=1-\alpha\). Substituting \(Q=n\lambda M\), and solving the two inequalities for the positive parameter \(\lambda\), gives the interval.

Confidence interval: If \(M>0\), an exact equal-tailed \(100(1-\alpha)\%\) confidence interval for the exponential rate \(\lambda\) is

\left[ \frac{-\ln(1-\alpha/2)}{nM}, \frac{-\ln(\alpha/2)}{nM} \right].

The endpoints are positive and ordered because \(-\ln(1-\alpha/2)<-\ln(\alpha/2)\) for \(0<\alpha<1\). The interval is often asymmetric: the exponential distribution is right-skewed, and equal probabilities in its two tails do not generally produce symmetric numerical cutoffs.

Find the observed minimum.
Set \(M\) equal to the smallest of the \(n\) recorded waiting times.

Choose the confidence level.
For a \(100(1-\alpha)\%\) equal-tailed interval, calculate \(-\ln(1-\alpha/2)\) and \(-\ln(\alpha/2)\).

Divide by \(nM\).
Use the smaller pivot cutoff for the lower rate limit and the larger cutoff for the upper rate limit.

Report the rate unit.
If \(M\) is in days, both limits are rates per day. Keep the time unit consistent throughout the calculation.

Why the Coverage Is Exact

The confidence claim can be checked by translating the event that the interval contains the fixed true rate. Write \(L=a/(nM)\) and \(U=b/(nM)\), where \(a\) and \(b\) are the equal-tailed cutoffs. Since \(n\), \(M\), and \(\lambda\) are positive, the inequalities can be rearranged without changing their directions:

\begin{aligned} L\leq\lambda\leq U &\iff \frac{a}{nM}\leq\lambda\leq\frac{b}{nM}\\ &\iff a\leq n\lambda M\leq b. \end{aligned}

The pivot \(n\lambda M\) has a standard exponential distribution for every \(\lambda>0\). Thus

\begin{aligned} P_\lambda(L\leq\lambda\leq U) &=P(a\leq n\lambda M\leq b)\\ &=(1-e^{-b})-(1-e^{-a})\\ &=e^{-a}-e^{-b}\\ &=(1-\alpha/2)-\alpha/2\\ &=1-\alpha. \end{aligned}

This is exact coverage under the model, not merely coverage that approaches the target as the sample size grows. The probability here is over repeated samples: the parameter \(\lambda\) is fixed, and the interval varies from sample to sample.

Worked Examples

Worked Example: A 95% Interval From Waiting Times in Days

In a hypothetical reliability study, six independent, uncensored components are observed from activation until a specified signal occurs. Assume each waiting time has an exponential distribution with unknown rate \(\lambda\) per day. The observed waiting times, in days, are \(0.7, 1.2, 0.3, 2.1, 0.9,\) and \(1.5\). Construct an equal-tailed 95% confidence interval for \(\lambda\).

Method. Use the sample-minimum pivot. The sample size is \(n=6\), and the smallest observation is \(M=0.3\) days. Thus \(nM=6(0.3)=1.8\) days.

For 95% confidence, \(\alpha=0.05\), so each tail has probability \(\alpha/2=0.025\). The pivot cutoffs are

a=-\ln(0.975)=0.0253178,\qquad b=-\ln(0.025)=3.688879.

Divide each cutoff by \(nM=1.8\):

L=\frac{0.0253178}{1.8}=0.0140654,\qquad U=\frac{3.688879}{1.8}=2.049377.

The interval is \([0.0141,\,2.05]\) per day, rounded to three significant figures. It is an interval for the rate, not for the mean waiting time. Check: the lower limit is positive and below the upper limit; its units are reciprocal days. The common mistake this method avoids is using the sample average in place of the minimum, which would not produce this pivot or this interval.

Worked Example: Converting Minutes Before Reporting a Rate

A hypothetical monitoring system records ten independent, complete waiting times until a sensor activates. Assume each is exponentially distributed with unknown rate \(\lambda\). The minimum observed wait is 18 minutes. Find a 90% equal-tailed interval for \(\lambda\) per hour.

Method. First express the minimum in hours, because the requested rate is per hour. Since \(60\) minutes equal one hour,

M=\frac{18}{60}=0.3\text{ hours},\qquad nM=10(0.3)=3\text{ hours}.

For 90% confidence, \(\alpha=0.10\), so each tail has probability \(0.05\). The cutoffs are

a=-\ln(0.95)=0.0512933,\qquad b=-\ln(0.05)=2.995732.

The rate limits are

L=\frac{0.0512933}{3}=0.0170978,\qquad U=\frac{2.995732}{3}=0.998577.

Therefore the 90% interval is \([0.0171,\,0.999]\) per hour, rounded to three significant figures. Check: both endpoints have units of inverse hours. The common mistake is to divide by \(10(18)\) while treating 18 as hours; that would mix minutes and hours and give a rate with the wrong scale.

Worked Example: A 99% Interval From a Small Sample

A hypothetical laboratory records five independent, uncensored waiting times until a reaction reaches a specified endpoint. Assume each time has an exponential distribution with unknown rate \(\lambda\) per month. The observations, in months, are \(2.4, 1.1, 3.0, 0.7,\) and \(1.9\). Calculate a 99% equal-tailed confidence interval.

Method. The sample size is \(n=5\), and the minimum is \(M=0.7\) months, so \(nM=5(0.7)=3.5\) months. For 99% confidence, \(\alpha=0.01\), and each tail has probability \(0.005\). Therefore,

a=-\ln(0.995)=0.00501254,\qquad b=-\ln(0.005)=5.298317.

Divide both cutoffs by \(3.5\):

L=\frac{0.00501254}{3.5}=0.00143215,\qquad U=\frac{5.298317}{3.5}=1.513805.

The 99% interval is \([0.001432,\,1.514]\) per month, rounded to four significant figures. Check: the interval is positive and ordered. Its large width is not a calculation error: the high confidence level requires more of the pivot’s upper tail to be included, and the sample size is small. The common mistake is to place the small cutoff in the upper-limit formula; the upper rate limit uses the larger pivot cutoff \(b\).

Common Mistakes and Scope

Reversing the endpoints. The lower rate limit uses \(-\ln(1-\alpha/2)\), while the upper limit uses \(-\ln(\alpha/2)\). The upper pivot cutoff is larger, so its rate limit is larger.
Reversing an inequality incorrectly. When isolating \(\lambda\), all quantities divided by are positive. The interval event becomes \(a\leq n\lambda M\leq b\), with the same order of inequalities.
Using the wrong observation. This interval is derived from the minimum, not the maximum or the sample mean. Substituting another statistic does not preserve the stated pivot distribution.
Ignoring censoring or dependence. The derivation uses complete independent observations from the same exponential distribution. Censoring, dependence, or differing rates can change the distribution of the minimum and invalidate the formula.
Misreading the rate parameter. The exponential rate is not the mean waiting time. Under this model, the mean is \(1/\lambda\), so a rate interval cannot be reported as a mean interval without transforming its endpoints appropriately.

Key takeaway: For independent, identically distributed exponential waiting times, the minimum has rate \(n\lambda\), making \(n\lambda M\) a standard exponential pivot. Inverting its equal-tailed probability limits gives an exact interval for \(\lambda\), provided the observations are complete and the time units are consistent.

Check Your Understanding

For each problem, assume the waiting times are independent, identically distributed, complete observations from the stated exponential model.

Four hypothetical machine cycles have waiting times \(0.8, 1.5, 0.6,\) and \(2.0\) hours until a diagnostic event. Find the 90% equal-tailed interval for the rate per hour. (Answer: \([0.0214,\,1.248]\) per hour, to four significant figures.)
Eight hypothetical patient follow-up waits have a sample minimum of 12 days. Find a 95% equal-tailed interval for the event rate per day. (Answer: \([0.000264,\,0.0384]\) per day, to three significant figures.)
Six independent waiting times have a minimum of 24 minutes. Find a 90% interval for the rate per hour, showing the time conversion. (Answer: \([0.0214,\,1.248]\) per hour, to four significant figures.)
A sample of five hypothetical network response waits has minimum \(0.4\) seconds. Construct a 99% equal-tailed interval for the rate per second. (Answer: \([0.002506,\,2.649]\) per second, to four significant figures.)
For a 95% interval, a sample of \(n=12\) has minimum \(M=0.25\) days. State the pivot cutoffs, rate interval, and exact coverage under the model. (Answer: \(a=0.0253178\), \(b=3.688879\), interval \([0.008439,\,1.2296]\) per day to four significant figures, and exact coverage \(0.95\).)

What You'll Learn