Categorical Severity Maximum Likelihood

Why Categorized Severity Data Need Their Own Likelihood

A severity data set does not always record each loss as an exact dollar amount. It may instead record whether a loss falls below a threshold, between two thresholds, or above a limit. When the category is all that is observed, the likelihood must use the probability of the observed category—not a density evaluated at an invented representative loss such as the category midpoint.

This tutorial treats complete observations: every loss is observed and assigned to exactly one of a specified set of categories. The central task is to estimate the probabilities of those categories from their counts. These estimates describe the categorized data; by themselves, they do not determine how losses are distributed within a category.

Definition: Let $C_1,\ldots,C_k$ be mutually exclusive categories whose union includes every possible outcome in the model. Write $p_j=P(X\in C_j)$, where $X$ is the loss. For $n$ independent losses, let $n_j$ be the number assigned to category $C_j$, so that $\sum_{j=1}^k n_j=n$. The categorical severity likelihood is the probability of the observed category records, viewed as a function of $p_1,\ldots,p_k$.

The categories may be labels, such as “minor,” “moderate,” and “major,” or intervals of loss amounts. For example, if $X$ has a continuous distribution and categories include losses in $[a,b)$, the probability of that category is $F_X(b)-F_X(a)$. If the categories do not cover all possible outcomes, include an additional category for the omitted outcomes. Otherwise, the category probabilities do not sum to one and the categorical model is incomplete.

Likelihood and the Unrestricted Maximum

For an ordered record of $n$ category assignments, independence gives the likelihood

L(p_1,\ldots,p_k)=\prod_{j=1}^k p_j^{n_j}, \qquad p_j\geq 0,\quad \sum_{j=1}^k p_j=1.

If only the counts are recorded, the probability of those counts includes the multinomial coefficient $n!/(n_1!\cdots n_k!)$. That factor does not depend on the probabilities, so it does not change the maximum likelihood estimates. The log likelihood for the ordered records is $\ell=\sum_j n_j\log p_j$, for positive probabilities. Categories with zero counts are handled at the boundary, as discussed below.

Theorem: If the category probabilities are unrestricted apart from $p_j\geq 0$ and $\sum_j p_j=1$, their maximum likelihood estimates are $\widehat p_j=n_j/n$, including $\widehat p_j=0$ when $n_j=0$.

The result follows from maximizing the log likelihood subject to the probabilities summing to one. When every count is positive, introduce a multiplier $\lambda$ for the constraint. The first-order conditions are

\frac{\partial}{\partial p_j} \left(\sum_{j=1}^k n_j\log p_j+\lambda\left(1-\sum_{j=1}^k p_j\right)\right) =\frac{n_j}{p_j}-\lambda=0.

Thus $p_j=n_j/\lambda$. Summing these expressions over categories gives $1=(\sum_j n_j)/\lambda=n/\lambda$, so $\lambda=n$ and $p_j=n_j/n$. This is a maximum: the log likelihood is concave in the probabilities, and when all counts are positive it is strictly concave on the feasible interior. Consequently, the stationary point is the unique maximum.

If one or more counts are zero, the log-likelihood derivative for that category cannot be used at $p_j=0$. Instead, assign probability zero to each zero-count category and apply the same argument to the categories with positive counts. This boundary choice maximizes the likelihood: allocating positive probability to a category with no observations takes probability away from categories that did occur, reducing their likelihood contribution.

Worked Examples

Worked Example: Estimating Three Loss-Band Probabilities

A hypothetical insurer records 40 complete claims in three exhaustive bands: less than $500, $500 to less than $1,500, and $1,500 or more. The respective counts are 18, 14, and 8. Estimate the category probabilities and calculate the maximized log likelihood for the ordered category records.

Method. Treat the category probabilities as unrestricted, subject only to being nonnegative and summing to one. The sample size is $n=18+14+8=40$.

Estimates. Divide each count by 40:

\widehat p_1=\frac{18}{40}=0.45,\qquad \widehat p_2=\frac{14}{40}=0.35,\qquad \widehat p_3=\frac{8}{40}=0.20.

The check is $0.45+0.35+0.20=1.00$. The fitted expected counts in a sample of 40 are $40(0.45)=18$, $40(0.35)=14$, and $40(0.20)=8$, matching the observed counts.

Maximized log likelihood. For the ordered records, substitute the estimates into $\ell=\sum_j n_j\log p_j$:

\begin{aligned} \widehat\ell &=18\log(0.45)+14\log(0.35)+8\log(0.20)\\ &\approx18(-0.798507)+14(-1.049822)+8(-1.609438)\\ &\approx-14.3731-14.6975-12.8755\\ &\approx-41.9461. \end{aligned}

The result is a log likelihood, not a probability, so it may be negative. If the likelihood for the unordered counts is used instead, its log includes the additional constant $\log(40!/(18!14!8!))$; the maximizing estimates remain unchanged. This method avoids treating band midpoints as if they were the observed exact losses.

Worked Example: A Category With No Observations

In a hypothetical portfolio, 25 complete losses are classified as below $1,000, from $1,000 to less than $4,000, or $4,000 and above. There are 6, 19, and 0 losses in those categories, respectively. Find the maximum likelihood estimates and the log likelihood for the ordered category records.

Method. This is an unrestricted categorical model, but the third category has zero observations. The estimates are the category proportions, including the boundary estimate for that third category:

\widehat p_1=\frac{6}{25}=0.24,\qquad \widehat p_2=\frac{19}{25}=0.76,\qquad \widehat p_3=\frac{0}{25}=0.

The estimates sum to $0.24+0.76+0=1$. The ordered-record likelihood at these estimates is

L(\widehat p_1,\widehat p_2,\widehat p_3) =(0.24)^6(0.76)^{19}(0)^0 =(0.24)^6(0.76)^{19}.

The factor for the third category is omitted because there are no observations in that category; equivalently, its contribution to the product is 1. The corresponding log likelihood is

\begin{aligned} \widehat\ell &=6\log(0.24)+19\log(0.76)\\ &\approx6(-1.427116)+19(-0.274437)\\ &\approx-8.562696-5.214303\\ &\approx-13.7770. \end{aligned}

A zero estimate does not prove the category is impossible in the population. It is the maximum likelihood estimate in this unrestricted model for this particular sample. Replacing it with a small positive number would be a different procedure, not the unrestricted maximum likelihood estimate. The common mistake avoided here is to take the logarithm of a zero probability for a category that has no observed records; its count is zero, so it contributes no log-likelihood term.

Worked Example: Estimating Probabilities Under a Constraint

A hypothetical severity model has three categories with probabilities $p_1=\theta$, $p_2=2\theta$, and $p_3=1-3\theta$, where $0\leq\theta\leq 1/3$. From 40 complete losses, the category counts are 12, 18, and 10. Estimate $\theta$ and the three category probabilities.

Method. The probabilities are constrained by the stated model, so the unrestricted estimates $12/40$, $18/40$, and $10/40$ cannot simply be used. The likelihood, omitting the count-only multinomial coefficient, is

L(\theta)=\theta^{12}(2\theta)^{18}(1-3\theta)^{10} =2^{18}\theta^{30}(1-3\theta)^{10}.

For an interior value of $\theta$, the log likelihood is $18\log 2+30\log\theta+10\log(1-3\theta)$. Differentiate and set the score to zero:

\begin{aligned} \frac{d\ell}{d\theta} &=\frac{30}{\theta}-\frac{30}{1-3\theta}=0,\\ 30(1-3\theta)&=30\theta,\\ \widehat\theta&=\frac14=0.25. \end{aligned}

The resulting probability estimates are $\widehat p_1=0.25$, $\widehat p_2=2(0.25)=0.50$, and $\widehat p_3=1-3(0.25)=0.25$. They are nonnegative and sum to 1. The second derivative is

\frac{d^2\ell}{d\theta^2} =-\frac{30}{\theta^2}-\frac{90}{(1-3\theta)^2}<0 \qquad (0<\theta<1/3).

Thus the interior stationary point is a maximum; at either endpoint, the likelihood is zero because at least one category with a positive observed count has probability zero. This example shows why the count proportions need not be the estimates when category probabilities are linked by a model constraint.

What the Estimates Do and Do Not Say

For unrestricted category probabilities, the estimate $\widehat p_j=n_j/n$ is also the observed proportion in category $C_j$. If a question asks for the probability of a union of categories, add their estimated probabilities. For example, if categories 2 and 3 represent losses at or above a specified threshold, the estimated probability of exceeding that threshold is $(n_2+n_3)/n$, provided the category definitions match the threshold event.

Categorizing losses discards information about their values within each band. The fitted probabilities alone do not provide a unique estimate of $E[X]$, $\operatorname{Var}(X)$, or a within-category severity distribution. Such quantities require additional assumptions or information. In particular, replacing each band by its midpoint and fitting a continuous severity model treats constructed values as observed data and is not the categorical likelihood method.

A further check is to confirm that the categories form a partition and that all records are complete. If a record is missing its category, or the observation only tells us that a loss lies in one of several possible categories, its likelihood contribution is not necessarily the same as a fully observed category assignment. Do not silently include such a record as though its category were known.

Common Mistake: Forgetting the Probability Constraint

Suppose the observed counts are 18, 14, and 8. Reporting those counts as the probability estimates is not valid: $18+14+8=40$, while probabilities must sum to one. The correction is to divide every count by the same sample size, giving $18/40=0.45$, $14/40=0.35$, and $8/40=0.20$. A related error is to estimate each probability without imposing the sum-to-one constraint. The category estimates are linked: increasing one probability leaves less total probability available for the others.

Key takeaway: With complete, independent category records and unrestricted category probabilities, each maximum likelihood estimate is the category count divided by the total count. Include every possible outcome in the partition, allow zero estimates for empty categories, and re-solve the likelihood when the model imposes relationships among probabilities.

Check Your Understanding

For each problem, treat the listed records as complete and use the stated category model.

A hypothetical portfolio has 20 losses in three exhaustive bands, with counts 9, 6, and 5. Estimate the three unrestricted probabilities and the expected number in the third band in a future sample of 32 losses. (Answer: $0.45,\ 0.30,\ 0.25$; expected count $32(0.25)=8$.)
Among 20 complete losses in three categories, the counts are 0, 14, and 6. Find the unrestricted maximum likelihood estimates and identify the boundary estimate. (Answer: $(0,\ 0.70,\ 0.30)$; the first probability estimate is on the boundary at zero.)
For four exhaustive loss bands, the counts are 5, 15, and 10 across the first three bands, with a total sample size of 30. Estimate the four unrestricted probabilities if the fourth band has 0 observations, and estimate the probability of being in band 2 or band 3. (Answer: $(1/6,\ 1/2,\ 1/3,\ 0)$; pooled estimate $5/6$.)
A constrained model has probabilities $(\theta,2\theta,1-3\theta)$, with $0\leq\theta\leq1/3$. Counts from 40 complete observations are 12, 18, and 10. Find the maximum likelihood estimate of $\theta$ and the fitted probabilities. (Answer: $\widehat\theta=0.25$; fitted probabilities $(0.25,\ 0.50,\ 0.25)$.)
Four exhaustive loss bands are $[0,100)$, $[100,300)$, $[300,600)$, and $[600,\infty)$. Counts from 20 complete losses are 7, 8, 3, and 2. Estimate the probability of a loss of at least 300 using the unrestricted categorical model. (Answer: $(3+2)/20=0.25$.)

What You'll Learn