Why Parameterization Matters
An exponential model is often used for a nonnegative waiting time, such as the time until a hypothetical device fails or a service request is completed. The same model can be described by its rate or by its mean waiting time. These parameters are reciprocals, so confusing them can produce an estimate with the right numerical ingredients but the wrong interpretation or units.
This tutorial derives the maximum likelihood estimates (MLEs) from independent waiting times. The derivation also makes clear an important boundary case: if every recorded waiting time is zero, the likelihood has no finite maximizer.
Here \(\lambda\) is a rate per unit of time, while \(\theta\) is an average waiting time in units of time. For example, if waiting times are measured in hours, \(\lambda\) is measured per hour and \(\theta\) in hours. Write the observed values as \(x_1,\ldots,x_n\), and define their total and sample mean by
The likelihood uses the joint density of the independent observations, viewed as a function of the unknown parameter with the observed data held fixed. Since all observations share the same exponential distribution, multiplying their densities gives the rate likelihood
This expression applies to observations in the nonnegative support. If the recorded values are censored, rounded, or otherwise not exact waiting times, their likelihood contributions are different and this formula does not directly apply.
Deriving the Rate Estimate
For \(T>0\), maximize the log-likelihood rather than the likelihood itself. The logarithm is strictly increasing, so it preserves which parameter value maximizes the likelihood, while turning products into sums:
Differentiate with respect to the positive rate parameter and set the score equal to zero:
The score equation identifies a candidate, so it remains to check that it is a maximum. The second derivative is negative for every \(\lambda>0\), and hence the log-likelihood is strictly concave throughout its parameter space. In addition, when \(T>0\), the likelihood tends to zero as \(\lambda\) approaches either zero or infinity. Thus the candidate is the unique finite maximum.
The condition \(T>0\) is essential. Because every observation is nonnegative, \(T=0\) means all observations are zero. In that case \(L(\lambda)=\lambda^n\), which has no finite maximum over \(\lambda>0\). It is not valid to substitute \(T=0\) into \(n/T\).
Worked Example: Estimating a Rate From Waiting Times in Hours
Model and assumptions. Suppose four independent, completely observed waiting times are modeled as exponential with unknown rate \(\lambda>0\), and the times are recorded in hours. The observed times are 2, 3, 5, and 6 hours. Estimate the rate and mean waiting time.
Method. Use the rate likelihood and its MLE, then express the reciprocal estimate as the mean waiting time. The common mistake to avoid is taking the reciprocal of the total without accounting for the sample size.
First calculate the total and sample mean:
Since \(T=16>0\), the finite rate MLE is
The fitted mean waiting time is the reciprocal:
Check. The rate and mean estimates are reciprocals, and the mean estimate equals the sample mean, \(4\) hours. Using \(1/T=1/16\) as the rate would be wrong: it would omit the factor \(n=4\).
The Mean or Scale Parameterization
The exponential density can instead be written using the positive mean/scale parameter \(\theta\). Its likelihood for the same observations is
For \(T>0\), its log-likelihood and derivative are
Setting the derivative to zero gives \(-n\theta+T=0\), and therefore \(\theta=T/n=\bar{x}\). At this candidate, the second derivative is \(-n/\theta^2<0\), confirming a local maximum. The log-likelihood tends to negative infinity at both ends of the positive parameter space when \(T>0\), so this is also the unique global maximum.
The rate and mean results are two expressions for the same maximization. To see why, substitute \(\theta=1/\lambda\) into the scale likelihood:
Thus the parameterizations describe the same family of distributions and the same fitted model. Their numerical MLEs differ because they estimate reciprocal parameters: the rate estimate is \(n/T\), and the mean/scale estimate is \(T/n\). This is a direct result of maximizing the likelihood under each parameterization; it does not require treating a rate as though it were a mean.
Worked Example: Converting a Rate Between Minutes and Hours
Model and assumptions. A hypothetical service system produces five independent, exactly observed exponential waiting times of 3, 5, 8, 4, and 10 minutes. Assume a common rate and no censoring. Estimate the rate per minute and per hour, and estimate the mean waiting time.
Method. Find the total in the data's original unit, calculate the rate and mean, and then convert the rate to a different time unit. The mistake to avoid is reporting a per-minute rate as though it were per hour.
The sample size, total, and mean are
The total is positive, so the rate estimate is
There are 60 minutes in an hour. A rate expressed per hour is 60 times the rate expressed per minute:
The mean/scale estimate is \(1/\widehat{\lambda}=6\) minutes. In hours, that mean is \(6/60=0.1\) hour. Its reciprocal is \(1/0.1=10\) per hour, agreeing with the converted rate.
Check. The per-minute and per-hour rates describe the same waiting-time model: \(60(1/6)=10\). The mean is 6 minutes, not 10 minutes; 10 is the numerical rate when time is measured in hours.
Zeros in the Data and the Boundary Case
A zero among otherwise positive observations does not by itself prevent a finite MLE. The condition is on the total \(T\), not on whether every observation is strictly positive. The exponential density is defined at zero, with \(f(0;\lambda)=\lambda\), so an exact zero contributes a factor \(\lambda\) to the likelihood.
Worked Example: One Zero but a Finite MLE
Model and assumptions. Four independent, exactly recorded waiting times are assumed to follow one exponential distribution with unknown rate \(\lambda>0\). The times, in hours, are 0, 1, 3, and 4. Estimate the rate and scale, and compare the likelihoods at the estimates.
Method. Check the sum before applying the MLE formulas. This avoids the mistaken rule that any zero observation makes the MLE infinite.
The sample size and sum are
Therefore the rate estimate is
and the mean/scale estimate is
At the rate estimate, the likelihood is
At the scale estimate, the likelihood is
Check. The two likelihood values agree, and the rate and scale estimates are reciprocals. The zero observation is allowed; the positive total ensures a finite maximizer.
Common Mistakes and When the Formula Does Not Apply
The formulas are short, but their conditions and units matter. Before calculating, confirm that the sample consists of independent observations from one exponential distribution, that the values are complete and measured in a common unit, and that \(T>0\) if a finite estimate is required.
- Dropping the sample size. The rate estimate is \(n/T\), not \(1/T\). The reciprocal of the total is not the rate MLE unless the sample size happens to be one.
- Switching rate and mean. The estimate \(T/n\) is a time, so it estimates the mean/scale \(\theta\); its reciprocal estimates the rate \(\lambda\).
- Ignoring time units. A rate per minute and a rate per hour have different numerical values. If the unit of time is multiplied by 60, the numerical rate is also multiplied by 60.
- Claiming a finite estimate when \(T=0\). In that case \(n/T\) is undefined, and the likelihood is unbounded rather than maximized at a finite parameter value.
- Using the complete-data likelihood for censored or rounded times. A censored observation does not contribute an ordinary density value at an observed completion time. Rounding also means the data describe an interval rather than an exact point. Those observations require likelihood contributions appropriate to how they were recorded.
An exponential model is also inappropriate if the assumed common distribution or independence is not justified by the problem. Maximum likelihood does not make those assumptions true; it identifies the parameter value that maximizes the likelihood under the stated model. This estimation method is aligned with the SOA VEE Mathematical Statistics syllabus.
Check Your Understanding
For each problem, state the parameterization and keep the estimate's time unit explicit. Assume independent, completely observed exponential waiting times unless the problem says otherwise.
- Three waiting times are 1, 4, and 7 seconds. Find the MLE of the rate per second and the mean waiting time. (Answer: \(\widehat{\lambda}=0.25\) per second; \(\widehat{\theta}=4\) seconds.)
- Six independent waiting times, measured in hours, have total \(42\) hours. Find the rate per hour and the mean estimate. (Answer: \(\widehat{\lambda}=1/7\) per hour; \(\widehat{\theta}=7\) hours.)
- Four waiting times in days are 0, 1.2, 2.8, and 4.0. Find the rate and scale MLEs, and say whether a finite MLE exists. (Answer: \(T=8\) days, so \(\widehat{\lambda}=0.5\) per day and \(\widehat{\theta}=2\) days; a finite MLE exists.)
- A sample has \(n=8\) observations and a mean of \(3.5\) hours. Find the total and the MLE of the rate per hour. (Answer: \(T=8(3.5)=28\) hours; \(\widehat{\lambda}=8/28=2/7\) per hour.)
- Five exactly recorded waiting times are all zero. What happens to the likelihood in the rate parameterization, and is there a finite rate MLE? (Answer: \(L(\lambda)=\lambda^5\), which is unbounded as \(\lambda\to\infty\); there is no finite rate MLE.)