Exponential MLE With Right-Censored Losses

Why Right-Censored Exponential Data Need a Different Count

A right-censored record tells us how long an item was observed without an event, but not the eventual event time. For example, a hypothetical insurer might observe a loss amount only up to a policy limit: a loss below the limit is exact, while a loss above it is known only to exceed the limit. The limit contributes information about the loss distribution, but it is not an exact loss.

For an exponential model, the maximum likelihood estimate has a particularly simple form. The numerator counts observed events; the denominator adds the observed time or amount for every record, including censored records. Understanding why both pieces appear prevents the main calculation error.

Definition: Suppose the latent quantity \(X\) has an exponential distribution with rate \(\lambda>0\), so \(f_X(x)=\lambda e^{-\lambda x}\) and \(S_X(x)=P(X>x)=e^{-\lambda x}\) for \(x\geq 0\). For record \(i\), let \(t_i\) be its observed duration or threshold, and let \(z_i=1\) if an exact event is observed and \(z_i=0\) if the record is right-censored at \(t_i\). Define \(r=\sum_{i=1}^n z_i\), the number of exact events, and \(T=\sum_{i=1}^n t_i\), the total observed time or amount.

The model assumes that the observations share the same exponential rate and are independent. The censoring mechanism must be noninformative for estimating \(\lambda\): in particular, the censoring time must not carry additional information about the latent event time beyond the recorded censoring status and threshold. An independent censoring mechanism, or a known observation design whose likelihood factor does not involve \(\lambda\), supports the likelihood below. If censoring is informative and its dependence on \(X\) is not modeled, this calculation need not be valid.

This is censoring, not truncation. A censored record remains in the sample and contributes its observed duration; a truncated record is absent unless it meets a selection condition. The likelihood therefore uses the survival probability for a censored record rather than conditioning on inclusion.

Likelihood and the Event-Count Score Equation

For an exact observation at \(t_i\), the contribution involving \(\lambda\) is its density, \(f_X(t_i)\). For a record censored at \(t_i\), the information is \(X>t_i\), so its contribution is the survival probability \(S_X(t_i)\). Under the stated model, both contributions can be written together as

L(\lambda)=\prod_{i=1}^n f_X(t_i)^{z_i}S_X(t_i)^{1-z_i} =\prod_{i=1}^n \left(\lambda e^{-\lambda t_i}\right)^{z_i} \left(e^{-\lambda t_i}\right)^{1-z_i} =\lambda^r e^{-\lambda T}.

If the censoring mechanism has a likelihood factor, it can be omitted from this expression when that factor does not involve \(\lambda\). Taking logarithms gives the log-likelihood, score, and second derivative:

\ell(\lambda)=r\log\lambda-\lambda T,\qquad \frac{d\ell}{d\lambda}=\frac{r}{\lambda}-T,\qquad \frac{d^2\ell}{d\lambda^2}=-\frac{r}{\lambda^2}.

When \(r>0\) and \(T>0\), setting the score to zero gives the maximum likelihood estimate. The negative second derivative confirms that the stationary point is a maximum.

\widehat{\lambda}=\frac{r}{T} =\frac{\text{number of exact events}}{\text{total observed time or amount}}.

Why the formula holds: Each exact observation supplies one factor of \(\lambda\), while each exact or censored record supplies an exponential factor involving its full observed duration. Thus the product has \(r\) factors of \(\lambda\) and exponential term \(e^{-\lambda T}\). Differentiating its log gives \(r/\lambda-T\); the zero-score solution is \(r/T\), and the second derivative is negative there. Censored observations therefore add to \(T\), but not to \(r\).

The estimate has units reciprocal to the units of \(t_i\). If time is recorded in months, \(\widehat{\lambda}\) is a rate per month; if losses are measured in thousands of dollars, it is a rate per thousand dollars. The fitted exponential mean is \(1/\widehat{\lambda}\) in the corresponding original units. Changing the units of every observation changes the numerical rate, but not the fitted distribution when interpreted consistently.

Worked Applications

Worked Example: Exact Event Times With Administrative Censoring

A hypothetical study follows five devices, with time measured in months until failure. Two devices fail at months 3 and 5. The other devices are still operating when observation ends at months 2, 4, and 6, so those three times are right-censoring times. Assume independent observations and noninformative administrative censoring. Estimate the exponential failure rate and mean lifetime.

Method and event count. The exact failures are the events, so \(r=2\). The three operating devices are censored, not failures.

Total observed time. Include every record's time, whether its outcome is exact or censored: \(T=3+5+2+4+6=20\) months.

Likelihood and estimate. The likelihood is \(L(\lambda)=\lambda^2e^{-20\lambda}\), and the log-likelihood score is \(2/\lambda-20\). Equating the score to zero gives \(\widehat{\lambda}=2/20=0.1000\) per month. The fitted mean lifetime is \(1/0.1000=10.00\) months.

Check and trap avoided. At the estimate, the score is \(2/0.1000-20=0\). Treating all five records as failures would incorrectly give \(5/20=0.2500\) per month. The censored devices contribute observed time, not event factors.

Worked Example: Losses Censored at Different Policy Limits

A hypothetical portfolio has four independent losses from a common exponential severity distribution. Amounts are recorded in thousands of dollars. Two exact losses are 12 and 18. The other two losses exceed their respective policy limits of 20 and 40, so they are right-censored at those amounts. Estimate the rate per thousand dollars and the mean loss.

Method and event count. The two exact losses are the only observed events, giving \(r=2\). Each policy-limit record establishes that the loss exceeded the limit; neither limit is an exact loss.

Total observed amount. Add both exact amounts and both censoring thresholds: \(T=12+18+20+40=90\) thousand dollars.

Likelihood and estimate. The likelihood is \(L(\lambda)=\lambda^2e^{-90\lambda}\). Its score is \(2/\lambda-90\), so \(\widehat{\lambda}=2/90=1/45=0.02222\) per thousand dollars. The fitted mean is \(1/\widehat{\lambda}=45.00\) thousand dollars.

Check and trap avoided. Substitution gives \(2/(1/45)-90=90-90=0\), as required. Replacing each censored loss by its limit as though it were exact would add two incorrect density factors; the limits contribute survival probabilities instead. The units also matter: the estimated rate is per thousand dollars, not per dollar.

Worked Example: Combining Records Across Follow-Up Groups

Two groups in a hypothetical reliability study use the same exponential model, and times are measured in days. Group A has exact event times 2, 4, and 6 days, plus censored observations at 5 and 8 days. Group B has exact event times 3 and 9 days, plus a censored observation at 7 days. Assume the common rate and noninformative censoring. Estimate the rate and mean waiting time.

Method and event count. There are three exact events in Group A and two in Group B, so the combined event count is \(r=3+2=5\). Because the model rate is common, all records contribute to the same likelihood.

Total observed time. Group A contributes \(2+4+6+5+8=25\) days. Group B contributes \(3+9+7=19\) days. Thus \(T=25+19=44\) days.

Likelihood and estimate. The combined likelihood is \(L(\lambda)=\lambda^5e^{-44\lambda}\), with score \(5/\lambda-44\). Therefore \(\widehat{\lambda}=5/44=0.1136\) per day. The fitted mean waiting time is \(1/\widehat{\lambda}=44/5=8.80\) days.

Check and trap avoided. At the estimate, \(5/(5/44)-44=44-44=0\), and the second derivative is \(-5/\lambda^2<0\). Pooling is justified here by the stated common-rate assumption; if the groups had different rates, a single pooled rate would not represent both models. The calculation also avoids counting censoring times as events.

Boundary Case and Exam Traps

The formula requires at least one observed event. If \(r=0\) and \(T>0\), then \(L(\lambda)=e^{-\lambda T}\), which decreases as \(\lambda\) increases. Over the parameter space \(\lambda>0\), the likelihood approaches its supremum as \(\lambda\) approaches zero, but zero is outside that space. There is no interior maximum likelihood estimate; a report should identify this boundary behavior rather than substitute \(r/T=0\) as a positive-rate estimate.

Common errors can often be caught by checking which part of each record represents an event and which part represents exposure:

Counting censored records as events. Only exact events contribute a factor of \(\lambda\). A censored record contributes a survival probability.
Omitting censored time from \(T\). A censored observation confirms survival through its recorded time, so that duration belongs in the exponential term.
Using a censoring threshold as an exact value. For a threshold \(c\), the observation is \(X>c\), with contribution \(S_X(c)\), not \(f_X(c)\).
Mixing units. All \(t_i\) must use the same unit before they are added. The estimated rate has the reciprocal unit.
Assuming a common rate without support. The pooled formula applies to a common exponential rate. If rates differ across groups, the common-rate model must not be imposed without justification.

The likelihood form is aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus. In an exam calculation, state the event count, total observed time, and units explicitly; those three items make both the estimate and its interpretation auditable.

Check Your Understanding

For each problem, assume independent observations, a common exponential rate within the stated sample, and noninformative right censoring. Use the indicated time or loss units.

A device study records exact failures at 4 and 6 days and right-censoring at 2 and 8 days. Find the maximum likelihood rate per day. (Answer: \(0.1000\) per day)
A hypothetical loss sample, in thousands of dollars, has one exact loss of 15 and three losses censored at 10, 20, and 25. Find the estimated rate per thousand dollars and fitted mean loss. (Answer: \(1/70=0.01429\) per thousand dollars; mean \(70.00\) thousand dollars)
A reliability sample has exact event times of 2, 5, and 9 hours and censoring times of 4 and 10 hours. Find the rate and mean waiting time. (Answer: \(0.1000\) per hour; mean \(10.00\) hours)
A portfolio measured in thousands of dollars has exact losses of 8 and 12 and right-censored losses at 6, 14, and 18. Find the estimated rate and mean loss. (Answer: \(1/29=0.03448\) per thousand dollars; mean \(29.00\) thousand dollars)
A study has exact event times of 3, 5, 7, and 9 days, with three additional records censored at 2, 4, and 6 days. Find the rate and mean. Then state whether treating all seven records as events would be correct. (Answer: \(1/9=0.1111\) per day; mean \(9.00\) days; no, only four exact events count in the numerator)

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