Why Grouping Changes the Likelihood
A recorded loss is not always an exact observation. An adjuster may report a loss to the nearest thousand dollars, or a data extract may give only the number of claims in each loss band. In either case, the observation says that a loss fell in an interval; it does not say that the loss equaled the interval midpoint. Fitting a continuous severity model therefore requires probabilities of intervals.
This distinction matters because a likelihood is built from the probability of what was observed. For an exact observation from a continuous distribution, the likelihood contribution is a density value. For an interval observation, it is the probability that the loss lies in that interval. Replacing that probability by a density at the midpoint is an approximation, not the same likelihood.
If each claim's interval is recorded separately, the likelihood is the product of the corresponding interval probabilities, assuming the claims are independent and share the stated severity model. If only the counts in categories are available, the likelihood is multinomial: it is proportional to the product of each category probability raised to its observed count. The multinomial coefficient may be included for the likelihood of the counts, but it does not depend on the model parameters and so does not change the maximum-likelihood estimate.
Building the Grouped-Data Likelihood
Let the intervals be nonoverlapping categories that cover the possible observations, and let \(n_j\) be the number of observations in category \(j\). Under an independent, identically distributed continuous model with parameter \(\theta\), write \(p_j(\theta)\) for the probability of category \(j\). Then the grouped-count likelihood is \(L(\theta)=\frac{n!}{\prod_j n_j!}\prod_j p_j(\theta)^{n_j}\), where \(n=\sum_j n_j\). The log-likelihood is often easier to maximize: \(\ell(\theta)=\text{constant}+\sum_j n_j\log p_j(\theta)\). The constant is the log of the multinomial coefficient.
For a category \([a_j,\infty)\), use \(p_j(\theta)=1-F_X(a_j;\theta)\). A category that begins at zero uses \(F_X(b_j;\theta)\) when the model is supported on nonnegative losses. Unequal interval widths require no special correction: each probability already accounts for the width through the CDF difference.
Worked Example: Estimating an Exponential Rate from Grouped Counts
A hypothetical portfolio has 10 independent losses, measured in thousands of dollars. The reported categories and counts are \([0,1)\): 4 losses; \([1,2)\): 4 losses; and \([2,\infty)\): 2 losses. Assume an exponential severity distribution with rate \(\lambda>0\) per thousand dollars, so \(F_X(x)=1-e^{-\lambda x}\) for \(x\ge0\). Estimate \(\lambda\) by maximum likelihood and calculate the likelihood of the observed counts.
Method. Form each category probability from the exponential CDF, then maximize the grouped-count likelihood. This uses interval events, rather than pretending each loss equals a category midpoint.
The probabilities are \(p_1=1-e^{-\lambda}\), \(p_2=e^{-\lambda}-e^{-2\lambda}\), and \(p_3=e^{-2\lambda}\). Set \(q=e^{-\lambda}\), so \(0<q<1\). Then \(p_1=1-q\), \(p_2=q(1-q)\), and \(p_3=q^2\). Omitting the parameter-free multinomial coefficient, the likelihood is \(L(q)\propto(1-q)^4[q(1-q)]^4(q^2)^2=q^8(1-q)^8\).
The log-likelihood apart from a constant is \(8\log q+8\log(1-q)\). Its derivative is \(8/q-8/(1-q)\), which is zero at \(q=1/2\); the second derivative, \(-8/q^2-8/(1-q)^2\), is negative throughout \(0<q<1\). Thus the maximum occurs at \(q=1/2\), giving \(\hat\lambda=-\log(1/2)=\log 2=0.6931\) per thousand dollars. The fitted mean severity is \(1/\hat\lambda=1.4427\) thousand dollars.
At this estimate the category probabilities are \(0.5,\ 0.25,\ 0.25\). The likelihood with the multinomial coefficient omitted is \(0.5^4(0.25)^4(0.25)^2=0.0000152588\) (approximately). The multinomial coefficient is \(10!/(4!4!2!)=3150\), so the likelihood of the observed counts is approximately \(3150(0.0000152587891)=0.0480652\). The fitted expected counts are \(10(0.5)=5,\ 10(0.25)=2.5,\) and \(10(0.25)=2.5\), which sum to 10 as required.
Check and mistake avoided. The category probabilities sum to \(0.5+0.25+0.25=1\). A common error is to use \(f_X(0.5)\), \(f_X(1.5)\), and a guessed value for the last category as if the recorded data were exact losses. Those density values are not the probabilities of the reported ranges.
Why Midpoint Substitution Is Only an Approximation
For a finite interval \([a,b)\) of width \(h=b-a\), let \(m=(a+b)/2\). The exact contribution is the integral of the density across the whole interval. If the density is smooth and changes little across a narrow interval, a midpoint approximation is \(\int_a^b f_X(x;\theta)\,dx\approx h f_X(m;\theta)\). This is the midpoint rule for approximating an integral. It is not an identity, and its accuracy depends on how much the density curves and changes over the interval.
The approximation can distort an estimate because maximum likelihood depends on how each contribution changes with the parameter. The exact likelihood uses the derivative of an interval probability; the midpoint approximation uses the derivative of a density at one point. Those derivatives need not give the same maximizing parameter. The discrepancy can be material for wide intervals, steeply changing densities, or intervals in a tail. An open-ended interval has no finite midpoint at all, so it must be handled with a survival probability rather than a midpoint density.
Worked Example: Rounded Losses and a Midpoint-Based Estimate
Three hypothetical losses are reported to the nearest thousand dollars as \(0.5,\ 1.5,\) and \(2.5\) thousand dollars. Assume independent losses from an exponential distribution with rate \(\lambda\) per thousand dollars. Find the maximum-likelihood estimate using the actual rounding intervals, then compare it with the estimate obtained by treating the reported values as exact.
Method. A value rounded to the nearest thousand represents a one-thousand-dollar-wide interval. The three observations therefore correspond to \([0,1), [1,2), [2,3)\), respectively. The rounding intervals, not their centers, determine the exact likelihood.
Set \(q=e^{-\lambda}\). The three interval probabilities are \(1-q,\ q(1-q),\) and \(q^2(1-q)\). Since there is one observation in each interval, the grouped likelihood is \(L(q)=(1-q)\,q(1-q)\,q^2(1-q)=q^3(1-q)^3\). Its log-likelihood derivative is \(3/q-3/(1-q)\), which vanishes at \(q=1/2\); the second derivative is negative, so this is the maximum. Hence the grouped-data estimate is \(\hat\lambda=\log 2=0.6931\) per thousand dollars. The likelihood at that estimate is \(0.5(0.25)(0.125)=0.015625\).
If the three reported centers are incorrectly treated as exact observations, the exponential density likelihood is proportional to \(\lambda^3 e^{-\lambda(0.5+1.5+2.5)}=\lambda^3e^{-4.5\lambda}\). Its log-likelihood derivative is \(3/\lambda-4.5\), giving \(\hat\lambda_{\text{mid}}=3/4.5=2/3=0.6667\). This corresponds to an estimated mean of \(1.5\) thousand dollars, compared with \(1/\log 2=1.4427\) thousand dollars from the interval likelihood. The midpoint calculation is not the exact likelihood estimate.
As a check on the distinction, at \(\lambda=2/3\), \(q=e^{-2/3}\approx0.5134\). The exact interval likelihood is \((1-q)\,[q(1-q)]\,[q^2(1-q)]\approx0.01559\), below its maximum \(0.015625\) at \(\lambda=\log 2\). The midpoint estimate does not maximize the exact grouped likelihood. The two estimates happen to be fairly close here; that is not a guarantee for wider intervals or more strongly curved densities.
Unequal Intervals, Open Ends, and Rounded Values
The interval-likelihood method applies without requiring equal-width bands. For example, a record in \([0,0.5)\) contributes \(F_X(0.5)\), while a record in \([0.5,1.5)\) contributes \(F_X(1.5)-F_X(0.5)\). A record in \([1.5,\infty)\) contributes \(1-F_X(1.5)\). All three contributions are probabilities, even though the first two widths differ and the last interval is unbounded.
The interval boundaries must reflect the recording rule. If a loss is rounded to the nearest unit \(h\), an interior reported value \(r\) typically represents \([r-h/2,r+h/2)\). For nonnegative losses, the lowest interval must be adjusted to respect the support: a reported zero cannot represent negative losses. If the recording system instead truncates or rounds by a different convention, use its actual intervals. Do not infer the intervals from the printed values alone.
Worked Example: Evaluating a Likelihood with Unequal Categories
A hypothetical data extract reports six independent losses, in thousands of dollars, as counts in \([0,0.5)\): 2; \([0.5,1.5)\): 3; and \([1.5,\infty)\): 1. For this calculation, take the exponential rate to be the specified value \(\lambda=1\) per thousand dollars. Find the category probabilities and the likelihood of the observed counts.
Method. Calculate each category probability from the CDF at its actual boundaries, including the survival probability for the unbounded category. The rate is specified here, so this is a likelihood evaluation rather than a parameter estimate.
The first probability is \(p_1=1-e^{-0.5}=0.3934693\). The second is \(p_2=e^{-0.5}-e^{-1.5}=0.6065307-0.2231302=0.3834005\). The final probability is \(p_3=e^{-1.5}=0.2231302\). Their sum is \(0.3934693+0.3834005+0.2231302=1.0000000\), up to rounding.
Omitting the multinomial coefficient, the likelihood is \(p_1^2p_2^3p_3\), or approximately \((0.3934693)^2(0.3834005)^3(0.2231302)=0.0019469\). The coefficient for counts \(2,3,1\) is \(6!/(2!3!1!)=60\), so the likelihood of the counts is approximately \(60(0.0019469)=0.1168\). The factor of 60 accounts for the different orderings of the six observations among the categories.
Check and mistake avoided. The three probabilities sum to one, and the counts sum to six. Treating the final category as a finite-width band would omit losses above an arbitrary endpoint; its probability is the full survival probability at 1.5.
Practical Checks and Common Traps
Before maximizing a grouped-data likelihood, verify that the categories are mutually exclusive and cover the relevant support, that the counts sum to the sample size, and that the model probabilities sum to one. These checks often reveal a missing tail category or a boundary entered in the wrong units. Keep the severity unit consistent: a rate per thousand dollars must be paired with boundaries measured in thousands of dollars.
- Using a midpoint as an exact observation. A rounded report identifies a range. Use the CDF difference for that range unless a midpoint approximation is explicitly intended.
- Forgetting interval width. A narrow and a wide category do not generally have the same probability. Use the probability across each stated interval, not just its center.
- Handling the last category as an ordinary finite bin. For \([a,\infty)\), use \(1-F_X(a)\). There is no finite midpoint.
- Confusing grouped counts with individually labeled intervals. For parameter estimation, the multinomial coefficient does not affect the estimate, but it is part of the likelihood of the observed counts.
- Using inconsistent units. If losses are recorded in thousands, convert any dollar boundaries before evaluating \(F_X\) or \(S_X\).
Grouped and rounded severity data still contain useful information, but less detail than exact observations. The appropriate likelihood preserves exactly what is known: each loss belongs to its reported interval. Midpoint substitution can be a convenient approximation for narrow, well-behaved bins, but the interval-probability likelihood is the principled method for fitting the specified continuous model.
Check Your Understanding
For each problem, identify the relevant interval probabilities before calculating. Answers are given at the end of each item.
- An exponential severity model has mean 2 thousand dollars. Find the probability that a loss lies in the reported band \([1,3)\) thousand dollars. (Answer: \(e^{-1/2}-e^{-3/2}\approx0.3834\).)
- Under an exponential model with rate \(\lambda\) per unit, grouped counts in \([0,1),[1,2),[2,\infty)\) are \(3,5,2\). Find the maximum-likelihood estimate of \(\lambda\). (Answer: \(\hat\lambda=-\log(9/17)\approx0.6360\) per unit.)
- One loss is recorded in each rounded category \([0,1),[1,2),[2,3)\). Under an exponential model, state the likelihood apart from a parameter-free factor in terms of \(q=e^{-\lambda}\), and give the maximum-likelihood estimate of \(\lambda\). (Answer: \(L(q)\propto q^3(1-q)^3\), so \(\hat\lambda=\log 2\approx0.6931\) per unit.)
- For an exponential model with rate \(1\) per unit, find the probability of the band \([0.5,1.5)\). (Answer: \(e^{-0.5}-e^{-1.5}\approx0.3834\).)
- Grouped exponential observations have counts \(2,1,1\) in \([0,2),[2,4),[4,6)\), respectively. Find the exact grouped-data estimate and the estimate obtained by treating the category centers \(1,3,5\) as exact values. (Answer: Exact grouped estimate \(\hat\lambda=-\frac12\log(3/7)\approx0.4236\) per unit; midpoint estimate \(\hat\lambda=1/2.5=0.4000\) per unit.)