Exponential MLE With Left-Truncated Losses

Why the Entry Threshold Changes the Likelihood

Suppose a hypothetical insurer records a loss only if it exceeds a known reporting threshold. The available observations are then not a complete sample from the loss distribution: each recorded loss is known to have passed the threshold. The likelihood must reflect that selection. For an exponential model, the resulting calculation is especially useful because the observed amount above the threshold has the same exponential rate as the original loss.

This tutorial considers exact loss amounts observed above one common, fixed entry threshold. It is aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus.

Definition: Let \(X\) be a loss with exponential density \(f_X(x)=\lambda e^{-\lambda x}\) for \(x\geq 0\), where \(\lambda>0\) is the rate. A left-truncated sample at threshold \(c\geq 0\) consists of exact observations sampled from the conditional distribution of \(X\) given \(X>c\). The threshold \(c\) is known and common to all observations. Define each observation's excess above entry as \(z_i=x_i-c\).

The strict inequality \(X>c\) describes eligibility for observation; with a continuous distribution, the probability of \(X=c\) is zero. The model assumes independent observations, the same exponential rate for every loss, and selection solely by the stated threshold. It does not describe right-censored amounts, rounded values, or records selected using different thresholds.

Conditional Density and Likelihood

For a loss recorded above \(c\), the density is the original density divided by the probability of passing the entry rule. Since \(S_X(c)=P(X>c)=e^{-\lambda c}\), for \(x>c\),

f_{X\mid X>c}(x) =\frac{f_X(x)}{P(X>c)} =\frac{\lambda e^{-\lambda x}}{e^{-\lambda c}} =\lambda e^{-\lambda(x-c)}.

For independent exact observations \(x_1,\ldots,x_n\), all above \(c\), the likelihood is the product of these conditional densities. Writing \(z_i=x_i-c\) and \(Z=\sum_{i=1}^n z_i\),

L(\lambda) =\prod_{i=1}^n \lambda e^{-\lambda(x_i-c)} =\lambda^n e^{-\lambda Z}, \qquad \ell(\lambda)=n\log\lambda-\lambda Z.

The threshold enters through the excesses, not as an additional unconditioned contribution from the original loss amounts. Differentiating the log-likelihood gives the score and its derivative:

U(\lambda)=\frac{d\ell}{d\lambda}=\frac{n}{\lambda}-Z, \qquad \frac{d^2\ell}{d\lambda^2}=-\frac{n}{\lambda^2}<0.

Setting the score equal to zero yields the maximum likelihood estimate. The negative second derivative confirms that the log-likelihood is strictly concave for \(\lambda>0\), so this stationary point is the unique maximum when \(Z>0\).

\widehat{\lambda}=\frac{n}{Z} =\frac{n}{\sum_{i=1}^n(x_i-c)}.

Why the Formula Works

The conditional density also explains why the estimate uses the average excess rather than the average recorded loss. Given that the loss passed the threshold, the density of its excess \(X-c\) is

f_{X-c\mid X>c}(z)=\lambda e^{-\lambda z},\qquad z>0.

Thus the excesses \(z_i\) behave like independent exponential observations with rate \(\lambda\). Their likelihood is \(\lambda^n e^{-\lambda\sum z_i}\), which gives the score equation \(n/\lambda=\sum z_i\). Equivalently, the fitted rate makes the mean excess \(1/\widehat{\lambda}\) equal to the sample mean excess \(Z/n\). This memoryless property is specific to the exponential model; it is not a general rule for other severity distributions.

If \(Z>0\), the likelihood tends to zero as \(\lambda\) approaches zero or infinity, and strict concavity gives a unique interior maximum. In an ordinary continuous sample with every \(x_i>c\), \(Z\) is positive. If all observations were exactly at \(c\), the stated exact continuous-data model would not describe those records.

Worked Example: Three Losses Above a Reporting Threshold

A hypothetical portfolio records exact losses in thousands of dollars only when they exceed \(c=8\). The recorded losses are \(11\), \(14\), and \(17\) thousand dollars. Estimate the exponential rate and the mean excess above the threshold.

Method. Use the conditional likelihood for left-truncated exact observations. First calculate each excess, in thousands of dollars: \(11-8=3\), \(14-8=6\), and \(17-8=9\).

The total excess is \(Z=3+6+9=18\) thousand dollars, and the sample size is \(n=3\). Therefore,

\widehat{\lambda}=\frac{3}{18} =\frac{1}{6} =0.1667\text{ per thousand dollars}.

The fitted mean excess is \(1/\widehat{\lambda}=6\) thousand dollars. As a likelihood check, the score at the estimate is \(3/(1/6)-18=18-18=0\), and the second derivative is negative. The estimate is therefore the likelihood maximum.

Common mistake avoided. The excesses are \(3,6,9\), whose sum is \(18\), not \(6\). Using \(3/6\) would incorrectly give a rate of \(0.5\) per thousand dollars. The correct rate is \(1/6\) per thousand dollars.

Applying the Estimate and Keeping Units Straight

The rate has units reciprocal to the units used for the losses. If losses are entered in thousands of dollars, the estimated rate is per thousand dollars, and \(1/\widehat{\lambda}\) is in thousands of dollars. Changing the loss unit changes the numerical rate but not the fitted distribution in monetary terms. Always subtract the threshold in the same units as the recorded losses.

For example, after fitting the conditional model, the probability that a recorded loss exceeds \(c+a\), given that it exceeds \(c\), is \(e^{-\widehat{\lambda}a}\) for \(a\geq0\). This follows from the exponential survival function applied to the excess \(X-c\). It is a conditional probability among losses that pass the entry threshold, not the probability that an arbitrary loss exceeds \(c+a\).

Worked Example: Estimating a Rate in Thousands of Dollars

A hypothetical equipment-loss study includes exact losses exceeding \(40\) thousand dollars. Four recorded losses, also in thousands, are \(44,49,55,\) and \(62\). Estimate the rate and use it to calculate the fitted conditional probability that a recorded loss exceeds \(55\) thousand dollars.

Method. Form the excesses over the entry threshold: \(44-40=4\), \(49-40=9\), \(55-40=15\), and \(62-40=22\). Their total is \(Z=4+9+15+22=50\) thousand dollars, with \(n=4\).

\widehat{\lambda}=\frac{4}{50}=0.08 \text{ per thousand dollars}.

Exceeding \(55\) thousand dollars means exceeding the entry threshold by a further \(a=55-40=15\) thousand dollars. Hence the fitted conditional probability is

P(X>55\mid X>40) =e^{-\widehat{\lambda}(55-40)} =e^{-0.08(15)} =e^{-1.2} \approx 0.3010.

As a check, the fitted mean excess is \(1/0.08=12.5\) thousand dollars, and the score is \(4/0.08-50=0\). The rate and excess are in reciprocal and direct thousands-of-dollars units, respectively.

Common mistake avoided. The probability is conditional on exceeding \(40\), so the additional amount is \(55-40=15\), not \(55\). Using \(e^{-0.08(55)}\) would incorrectly measure from zero rather than from the entry threshold.

A Complete Likelihood Check

A final calculation can verify both the excess total and the fitted rate. Suppose a hypothetical study records losses in thousands of dollars above \(c=12\): \(15,17,21,27,\) and \(32\). The excesses are \(3,5,9,15,\) and \(20\), with total \(Z=52\) and \(n=5\). The estimate is \(5/52=0.09615\) per thousand dollars, or \(0.0962\) per thousand dollars to four decimal places.

At this estimate, \(n/\widehat{\lambda}=5/(5/52)=52=Z\), so the score is zero. The second derivative is \(-5/(5/52)^2<0\), confirming a maximum. The fitted mean excess is \(52/5=10.4\) thousand dollars. For a recorded loss to exceed \(30\) thousand dollars, its excess must exceed \(30-12=18\) thousand dollars. The fitted conditional probability is

P(X>30\mid X>12) =e^{-(5/52)(18)} =e^{-45/26} \approx 0.1772.

The probability is rounded to four decimals. It is conditional on entry above \(12\) thousand dollars, consistent with the sample used to estimate the rate.

Exam Checks and Common Traps

Identify the selection rule.
Confirm that every exact observation was included because it exceeded the same known threshold \(c\), and that the sample is modeled conditionally on that selection.

Calculate excesses in consistent units.
For every observation, subtract \(c\) from \(x_i\), then add the excesses to obtain \(Z\).

Estimate and verify.
Calculate \(n/Z\), state the reciprocal loss unit, and check \(n/\widehat{\lambda}-Z=0\).

The central exam trap is treating a selected sample as though it were a complete sample from zero. A complete-sample exponential likelihood would involve the sum of the observed losses, whereas this conditional likelihood involves \(\sum(x_i-c)\). Using the former ignores the fact that losses at or below \(c\) could not appear in the data. Another trap is treating a known threshold as an estimated model parameter: in this setting, \(c\) is fixed and known, while \(\lambda\) is estimated.

Key takeaway: For independent exact exponential losses observed only above a common known threshold \(c\), maximize the conditional likelihood using the excesses \(x_i-c\). The maximum likelihood estimate is the number of observations divided by their total excess, with rate units reciprocal to the loss units.

Check Your Understanding

For each problem, use the conditional exponential likelihood and keep the loss units consistent.

A hypothetical loss sample is recorded only above \(4\) thousand dollars. The losses are \(7,10,\) and \(13\) thousand dollars. Find the rate estimate in reciprocal thousands of dollars. (Answer: \(1/6\), or approximately \(0.1667\), per thousand dollars.)
Exact losses above \(25\) thousand dollars are \(31,37,42,\) and \(50\) thousand dollars. Find the rate estimate and fitted mean excess. (Answer: \(1/15\approx0.0667\) per thousand dollars; mean excess \(15\) thousand dollars.)
A hypothetical sample is selected above \(100\) dollars and has exact losses of \(108,115,127,\) and \(150\) dollars. Find the rate estimate per dollar. (Answer: \(4/100=0.04\) per dollar.)
Losses in thousands of dollars are recorded above \(10\); the five values are \(15,18,22,31,\) and \(34\). Find the rate estimate and fitted mean excess. (Answer: \(1/14\approx0.0714\) per thousand dollars; mean excess \(14\) thousand dollars.)
A sample of losses in thousands of dollars is recorded above \(5\), with values \(7,9,12,\) and \(16\). Estimate the rate and the fitted probability that a recorded loss exceeds \(11\) thousand dollars. (Answer: \(\widehat{\lambda}=1/6\) per thousand dollars; probability \(e^{-1}\approx0.3679\).)

What You'll Learn