Empirical Distributions From Grouped Loss Data

Why Grouped Loss Data Need Careful Interpretation

Loss records are sometimes reported as counts in ranges rather than as exact amounts. A summary might say how many losses were at most $100, above $100 and at most $250, or above $250 and at most $500. These grouped counts can support useful probability summaries, but they do not reveal the individual amounts inside each range.

The central distinction is between what the counts determine and what they leave unknown. Counts determine the relative frequency in each bin, and therefore the cumulative relative frequency at bin boundaries when the boundary convention is specified. They do not, by themselves, determine the distribution within a bin. Any more detailed result, such as a precise average or a probability at a threshold inside a bin, must be treated as an estimate or a bound.

Definition: Suppose $n$ observed losses are assigned to disjoint bins, and bin $j$ contains $n_j$ losses. Its relative frequency is $n_j/n$. If the bins cover all observed losses, their relative frequencies sum to 1. A grouped empirical distribution records these bin counts but not the exact loss amounts within the bins.

For clarity, this tutorial uses bins of the form $[0,b_1]$, $(b_1,b_2]$, and so on. Thus, a loss exactly equal to a boundary is assigned to the bin ending at that boundary. This convention matters for empirical data, since observations can tie at a boundary. The bin width is the difference between its upper and lower endpoints; including or excluding an endpoint does not change that width.

Relative Frequencies, Cumulative Frequencies, and Histogram Heights

The relative frequency in a bin is the fraction of observed losses in that interval. Adding relative frequencies from the first bin through a given bin gives the fraction at or below that bin’s upper boundary. In the empirical distribution, that cumulative fraction is the probability assigned to the boundary and all smaller losses.

A histogram’s height is not generally the relative frequency itself. For unequal-width bins, use relative frequency divided by bin width. The height then has units of inverse dollars, and the area of each bar equals the bin’s relative frequency. This allows bins of different widths to be compared without suggesting that a wider bin has a higher concentration of observations merely because it covers more dollars.

\text{Relative frequency in bin }j=\frac{n_j}{n}, \qquad \text{relative-frequency density in bin }j =\frac{n_j/n}{w_j},

Here $w_j$ is the bin width in dollars. The bar’s area is $w_j(n_j/n)/w_j=n_j/n$. Summing the bar areas over bins gives $\sum_j n_j/n=1$, provided the bins include every observation. This proves why density heights, not raw relative frequencies, are required for a relative-frequency histogram when bin widths differ.

Worked Example: Building a Relative-Frequency Histogram

A hypothetical portfolio has 25 observed losses grouped as follows: 8 losses in $[0,100]$, 12 in $(100,250]$, and 5 in $(250,500]$, with amounts in dollars. Find each relative frequency, each histogram height, and the total histogram area.

Method. Divide each count by $n=25$ to obtain the bin’s relative frequency. Then divide that result by the bin width to obtain the height.

The bin widths are $100-0=100$, $250-100=150$, and $500-250=250$ dollars. The relative frequencies are $8/25=0.32$, $12/25=0.48$, and $5/25=0.20$. They sum to $0.32+0.48+0.20=1.00$.

The first histogram height is $0.32/100=0.0032$ per dollar. The second is $0.48/150=0.0032$ per dollar. The third is $0.20/250=0.0008$ per dollar.

The respective bar areas are $100(0.0032)=0.32$, $150(0.0032)=0.48$, and $250(0.0008)=0.20$. Their total is $0.32+0.48+0.20=1.00$, as required.

Check and common mistake. The areas match the bin relative frequencies and add to 1. A common error is to use the counts divided by the total as the bar heights. Those values are probabilities, not densities; with unequal bin widths, using them as heights would give incorrect areas.

What Cumulative Probabilities Do and Do Not Tell You

With bins defined by their endpoints, the grouped counts determine the empirical cumulative probability at each bin boundary. For example, if the first two bins contain 8 and 12 losses out of 25, then the observed fraction at or below the second upper boundary is $(8+12)/25$. At a threshold inside a bin, however, the counts do not say how many losses fall below that threshold.

The uncertainty can be stated as bounds. Suppose $x$ lies inside bin $j$. All losses in earlier bins are at or below $x$, and no losses in later bins are at or below $x$. The losses in bin $j$ could be on either side of $x$. Therefore, the grouped counts alone bound the empirical cumulative probability at $x$ between the cumulative relative frequency before bin $j$ and the cumulative relative frequency through bin $j$.

\frac{\sum_{k<j} n_k}{n} \ \leq\ F_{\mathrm{emp}}(x) \ \leq\ \frac{\sum_{k\leq j} n_k}{n}, \qquad x\text{ inside bin }j.

These bounds can be wide, especially when a bin contains many observations. A midpoint allocation or a uniform-within-bin assumption may produce a single estimate, but neither is implied by the grouped counts. The assumption must be stated, and the result should not be presented as an exactly observed probability.

Worked Example: Locating a Grouped Empirical Median

A hypothetical set of 20 claim severities has 4 losses in $[0,50]$, 10 in $(50,150]$, and 6 in $(150,300]$. Find what the grouped data establish about the empirical median, and calculate a median estimate under a uniform-within-bin assumption.

Method. The empirical median is the $10$th ordered observation for this even-sized sample under the usual lower-middle order-statistic convention; equivalently, it is the smallest loss at which the empirical cumulative proportion reaches at least $10/20=0.50$.

At $50, the cumulative proportion is $4/20=0.20$. At $150, it is $(4+10)/20=14/20=0.70$. Thus the $10$th ordered loss is in $(50,150]$. The grouped counts do not identify its exact value.

For an estimate, assume losses are distributed uniformly within the bin $(50,150]$. The target cumulative proportion $0.50$ is $0.50-0.20=0.30$ above the cumulative proportion at $50. The bin contains $10/20=0.50$ of the observations, so the target is $0.30/0.50=0.60$ of the way through that bin. The estimated median is therefore $50+0.60(150-50)=50+60=\$110$.

Check and common mistake. The exact grouped-data conclusion is that the median lies in $(50,150]$; $110 is only the estimate under the stated uniform assumption. Treating $110 as an observed median would claim more precision than the data provide.

Approximating Moments with Bin Midpoints

A common summary of grouped losses is the midpoint estimate of the empirical mean. Assign every loss in a bin the midpoint of that bin, multiply each midpoint by its bin count, and divide the total by $n$. This is a convenient approximation, not an exact recovery of the sample mean. It is most credible when bins are relatively narrow and losses are not concentrated at one end of a bin.

The limits of this method can be made explicit. If all losses in a bin lie between its lower and upper endpoints, then the sample mean lies between the weighted average of the lower endpoints and the weighted average of the upper endpoints. Endpoint inclusion determines whether a bound can be attained exactly; conservative non-strict bounds remain valid either way.

Worked Example: Midpoint Mean and Its Range

A hypothetical dataset contains 10 losses: 2 in $[0,100]$, 5 in $(100,200]$, and 3 in $(200,400]$. Calculate the midpoint estimate of the mean and give conservative bounds for the actual empirical mean.

Method. Use bin midpoints $50$, $150$, and $300$ dollars for the estimate. For bounds, use the lower and upper endpoints in each bin, weighted by the counts.

The midpoint estimate is $(2(50)+5(150)+3(300))/10=(100+750+900)/10=1750/10=\$175$.

Using lower endpoints gives $(2(0)+5(100)+3(200))/10=(0+500+600)/10=\$110$. Using upper endpoints gives $(2(100)+5(200)+3(400))/10=(200+1000+1200)/10=\$240$. Hence the actual empirical mean is between $110 and $240, inclusive as conservative bounds.

Check and common mistake. The midpoint estimate, $175, is inside the bounds. A frequent error is to report it as the exact mean. The counts do not reveal whether losses cluster near the lower endpoints, the upper endpoints, or somewhere between them.

Which Distributional Details Are Lost?

Grouping preserves the number of losses in each interval, but discards their exact positions within those intervals. Consequently, the same grouped table can be consistent with very different sets of exact losses. Those sets can have different means, variances, quantiles, and probabilities at thresholds inside a bin, even though their bin counts are identical.

The following distinction is useful when interpreting a grouped summary:

Quantity	What grouped counts establish	What may require an assumption
Relative frequency by bin	Exact count divided by total count	Nothing further, if all counts and the total are known
Cumulative frequency at a bin boundary	Exact cumulative count under the stated endpoint convention	Nothing further, if bins are complete and disjoint
Probability at a threshold inside a bin	A lower and upper bound based on the bin counts	A point estimate requires information or an assumption about within-bin losses
Mean or quantile within a bin	Generally an interval or bound	A single estimate may use midpoints or interpolation
Exact values, ties, and within-bin shape	Not recoverable from counts alone	Requires exact records or additional information

A histogram is also an approximation to the distributional shape. Its bars summarize concentrations over intervals; they do not show the exact empirical steps or reveal where observations lie within a bar. In particular, equal histogram heights in two bins do not mean that the same exact losses occurred in each bin. They mean the relative frequencies per dollar are equal under the chosen binning.

Common Exam Traps

Ignoring bin width. Relative frequency is bar area, not necessarily bar height. Divide by the width to get a density height.
Confusing an exact boundary result with an interior-threshold result. With a defined endpoint convention, cumulative counts at bin boundaries are known. A threshold inside a bin generally has only bounds.
Calling a midpoint result exact. A midpoint mean, quantile interpolation, or within-bin uniform estimate adds an assumption not contained in the counts.
Forgetting to check total mass. Relative frequencies should sum to 1, and a complete relative-frequency histogram should have total area 1.
Using dollars inconsistently. A height computed per dollar must be multiplied by a width measured in dollars to obtain a dimensionless area.

Key takeaway: Grouped loss data determine bin frequencies and cumulative proportions at specified bin boundaries. Within-bin probabilities and exact loss summaries are not identified without additional data or assumptions. State clearly whether a result is exact, bounded, or estimated.

Check Your Understanding

For each question, distinguish an exact grouped-data result from a bound or an assumption-based estimate.

A hypothetical sample of 40 losses has counts 6, 14, and 20 in $[0,100]$, $(100,300]$, and $(300,800]$, respectively. Find the three relative frequencies, the histogram heights per dollar, and the area of each bar. (Answer: Relative frequencies $0.15, 0.35, 0.50$; heights $0.0015, 0.00175, 0.0010$ per dollar; areas $0.15, 0.35, 0.50$, totaling $1.00$.)
A sample of 50 losses has counts 15, 25, and 10 in $[0,200]$, $(200,500]$, and $(500,1000]$. Give the cumulative proportions at $200 and $500. Bound the proportion of losses in $(250,400]$ using only the grouped counts. (Answer: $F_{\mathrm{emp}}(200)=0.30$, $F_{\mathrm{emp}}(500)=0.80$; the proportion in $(250,400]$ is between $0$ and $25/50=0.50$, inclusive as conservative bounds.)
A sample of 32 losses has counts 5, 11, and 16 in $[0,40]$, $(40,120]$, and $(120,320]$. In which interval must the 16th ordered loss lie? (Answer: $(40,120]$; the grouped counts do not determine its exact value.)
A sample of 10 losses has counts 3, 4, and 3 in $[0,50]$, $(50,150]$, and $(150,350]$. Calculate the midpoint mean estimate and conservative bounds for the actual mean. (Answer: Midpoint estimate $=\$122.50$; actual mean is between $\$65$ and $\$180$, inclusive as conservative bounds.)
Two bins each contain 10 losses, but the first bin is $[0,100]$ and the second is $(100,500]$. Are the relative-frequency histogram heights equal? (Answer: No. The heights are $0.10/100=0.001$ and $0.10/400=0.00025$ per dollar, respectively; equal bin frequencies do not imply equal density heights when widths differ.)

What You'll Learn