Likelihood for Grouped Loss Data¶

Definition¶

Suppose independent losses are observed only as counts $n_j$ in disjoint intervals $(c_{j-1},c_j]$, $j=1,\ldots,k$$1,\dots ,k$, covering the observation space. Under cumulative distribution function (CDF) $F_\theta$, interval probability is $p_j(\theta)=F_\theta(c_j)-F_\theta(c_{j-1})$$F_{\theta }(c_{j-1})$. The likelihood is proportional to $\prod_j p_j(\theta)^{n_j}$.

Interpretation¶

The observed event is membership in a bin. Replacing each loss by a bin midpoint and using a density discards the actual observation mechanism.

![Exponential density split at c_1=100 into the bins (0,100] and (100,∞), shaded with bin probabilities 0.6 and 0.4 and labelled with the observed counts 6 and 4.](../../assets/svg/grouped-data-likelihood.svg)

The figure draws the Examples case. Only the bin of each loss is recorded, so each observation contributes the area of its bin, $p_1=1-\mathrm{e}^{-100\lambda}$ or $p_2=\mathrm{e}^{-100\lambda}$${\mathrm{e}}^{-100\lambda }$, not a density height, and the likelihood is proportional to $p_1^{6}p_2^{4}$. Maximizing $6\log(1-q)+4\log q$ in $q=\mathrm{e}^{-100\lambda}$ gives $q=0.4$$0.4$, that is $\hat\lambda=\log(2.5)/100\approx0.0092$$0.0092$. The drawn density uses this value, so the shaded areas are 0.6 and 0.4, the observed proportions $6/10$ and $4/10$.

Maximum likelihood estimation must use the probability of the recorded bin. Multiplying those bin probabilities, with each raised to its observed count, gives the objective to maximize; point-density likelihoods describe different data.

Key Results¶

• $\displaystyle \ell(\theta)=\sum_jn_j\log p_j(\theta)+C$$C$, where $C$ is independent of $\theta$.
• For multinomial counts, the constant factor is $n!/\prod_jn_j!$, with $\displaystyle n=\sum_jn_j$$\sum _j{n}_j$.
• If the bins omit part of the population, include an outside bin or condition on the sampling region.
• Grouping can reduce information compared with exact observations.

Examples¶

If 6 exponential losses are in $(0,100]$ and 4 exceed 100, the likelihood is proportional to $(1-\mathrm{e}^{-100\lambda})^6(\mathrm{e}^{-100\lambda})^4$${\mathrm{e}}^{-100\lambda }{)}^6({\mathrm{e}}^{-100\lambda }{)}^4$.

Prerequisites¶

• Maximum Likelihood Estimation
• Cumulative Distribution Function

Related Concepts¶

• Likelihood for Censored Loss Data
• Likelihood for Truncated Loss Data

Sources¶

• Risk Insurance · Lecture 5

Aliases: interval-count likelihood, binned insurance data, grouped data likelihood, likelihood for grouped data Keywords: multinomial likelihood, bin probabilities, p_j = F(c_j) - F(c_{j-1}), histogram data, interval data Symbols: infinity, lambda, product, sum, theta

Last updated: 17 October 2026 21:53