Plots for Diagnosing Loss Distribution Shape

Why Plot Loss Distributions?

A numerical summary compresses a loss sample into a few figures; a plot can show where observations concentrate, how quickly the upper tail thins, and whether a small number of losses sit far from the rest. These are useful diagnostics when describing observed losses or assessing whether a candidate distribution is plausible. A plot does not, by itself, establish which model is correct.

This tutorial uses hypothetical loss data. Unless otherwise stated, losses are nonnegative amounts measured in thousands of dollars. Histograms describe the overall sample shape. Empirical survival plots focus on exceedances: how much of the sample lies above a given loss threshold. Changing the vertical or horizontal scale can make tail patterns easier to compare.

Definition: For observed losses $x_1,\ldots,x_n$, the empirical survival function at threshold $x$ is $$ \widehat{S}(x)=\frac{1}{n}\sum_{j=1}^{n} I(x_j>x), $$ where $I(x_j>x)$ is 1 when observation $x_j$ exceeds $x$, and 0 otherwise. Thus, $\widehat{S}(x)$ is the observed fraction of losses strictly greater than $x$. A histogram groups observations into intervals and displays the count, relative frequency, or density in each interval.

Reading Histograms

A histogram partitions the horizontal loss scale into adjacent bins and shows how many observations fall in each. Its broad features may suggest symmetry, skewness, a concentration of small losses, multiple clusters, or an unusually distant observation. These are visual clues, not conclusions about the process that generated the data.

The choice of bins matters. Wider bins smooth over local variation; narrower bins can make random fluctuations look like meaningful structure. The horizontal scale and vertical scale matter too. A count histogram uses the number of observations in each bin, while a relative-frequency histogram reports the fraction in each bin. If bin widths differ, density is the appropriate vertical measure for comparing areas: count divided by the total sample size and bin width. With equal-width bins, counts and relative frequencies have the same general shape.

Worked Example: Diagnosing a Long Right Tail in a Histogram

A hypothetical insurer records 12 claim severities, in thousands of dollars: 2, 3, 4, 4, 5, 6, 7, 8, 9, 10, 12, and 30. Using bins $[0,10)$, $[10,20)$, $[20,30)$, and $[30,40)$, calculate the bin counts and relative frequencies, then describe what the histogram suggests. The method is to classify each observation once, then divide each count by the sample size. This avoids treating a bin boundary as belonging to both neighboring bins.

First bin, $[0,10)$. Its observations are 2, 3, 4, 4, 5, 6, 7, 8, and 9, giving a count of 9 and relative frequency $9/12=0.75$.

Second bin, $[10,20)$. Its observations are 10 and 12, giving a count of 2 and relative frequency $2/12=0.1667$, rounded to four decimals.

Third bin, $[20,30)$. There are no observations, giving a count of 0 and relative frequency 0.

Fourth bin, $[30,40)$. The observation 30 belongs here, giving a count of 1 and relative frequency $1/12=0.0833$, rounded to four decimals. The counts sum to $9+2+0+1=12$, and the relative frequencies sum to 1.0000.

The histogram would show most observations below 10, with one isolated loss at 30 and a gap immediately below it. This is consistent with a long right tail or a potential outlier, but the sample is too small to settle which interpretation is appropriate. The bin conventions matter: because the intervals include their left endpoint and exclude their right endpoint, the loss of 10 belongs in $[10,20)$, not $[0,10)$.

Empirical Survival Plots

An empirical survival plot places the threshold $x$ on the horizontal axis and $\widehat{S}(x)$ on the vertical axis. As the threshold rises, the fraction of observations exceeding it cannot increase. The graph is therefore a nonincreasing step function. Ties produce steps at the tied loss values; they do not represent multiple distinct thresholds.

This plot makes the upper tail visible without hiding it among the many small observations. For example, the value at a threshold of 100,000 dollars is the fraction of the sample with losses greater than 100,000 dollars. It is important to read the threshold and the inequality exactly: with the strict-exceedance definition above, losses equal to the threshold are not counted.

Worked Example: Calculating an Empirical Survival Plot With Ties

A hypothetical portfolio has 10 recorded claim amounts, in thousands of dollars: 1, 1, 2, 2, 2, 4, 5, 8, 10, and 20. Calculate $\widehat{S}(x)$ at thresholds $x=1,2,5,10,$ and 20, and identify the last positive plotted value. The method is to count losses strictly greater than each threshold and divide by 10. This avoids the common error of including losses tied with the threshold.

At $x=1$, the losses exceeding 1 are 2, 2, 2, 4, 5, 8, 10, and 20. Therefore, $\widehat{S}(1)=8/10=0.8000$.

At $x=2$, the losses exceeding 2 are 4, 5, 8, 10, and 20. Therefore, $\widehat{S}(2)=5/10=0.5000$.

At $x=5$, the losses exceeding 5 are 8, 10, and 20. Therefore, $\widehat{S}(5)=3/10=0.3000$.

At $x=10$, only 20 exceeds 10, so $\widehat{S}(10)=1/10=0.1000$. At $x=20$, no observation exceeds 20, so $\widehat{S}(20)=0/10=0.0000$. The last positive value at these thresholds is 0.1000 at $x=10$. The counts fall from 8 to 5 to 3 to 1 to 0, as required for a nonincreasing survival plot.

Using Log Scales to Inspect Tail Decay

A log scale can make differences in tail behavior easier to see when survival probabilities span a wide range. Two common views are a semilog plot, with a logarithmic vertical axis and an ordinary horizontal loss axis, and a log-log plot, with logarithmic scales on both axes. In either view, zero survival cannot be plotted on a logarithmic vertical axis because $\log(0)$ is undefined. Empirical curves may reach zero once the threshold is at least as large as the sample maximum.

The shape of the plotted curve can be compared with standard candidate distributions. For an exponential loss with rate $\lambda>0$, the survival function is $S_X(x)=e^{-\lambda x}$ for $x\geq 0$. Taking natural logarithms gives $\log S_X(x)=-\lambda x$, a straight line on a semilog plot. For a Pareto loss with minimum value $x_m>0$ and shape parameter $\alpha>0$, the survival function is $S_X(x)=(x_m/x)^\alpha$ for $x\geq x_m$. Taking logarithms gives $\log S_X(x)=\alpha\log x_m-\alpha\log x$, a straight line on a log-log plot.

Why these plots help: The straight-line patterns follow directly from the survival formulas. Exponential decay is linear in $x$ after taking the log of survival; Pareto decay is linear in $\log x$ after taking the log of survival. A noticeably curved pattern can signal that the candidate model does not describe the plotted range well, although sampling variability can also create curvature.

Worked Example: Comparing Exponential and Pareto Tail Patterns

Consider two hypothetical candidate severity models, with amounts measured in thousands of dollars. Model E is exponential with rate $\lambda=0.1$ per thousand dollars. Model P is Pareto with $x_m=5$ and $\alpha=2$. Compare their survival probabilities at 10, 20, and 40, then identify which transformed plot should be linear for each model. Assume the stated models apply over the evaluated ranges and use natural logarithms. This comparison avoids inferring a model from the visual steepness of an untransformed plot alone.

For Model E, $S_X(x)=e^{-0.1x}$. At $x=10$, the exponent is $-0.1(10)=-1$, so $S_X(10)=e^{-1}=0.3679$. At $x=20$, the exponent is $-2$, so $S_X(20)=e^{-2}=0.1353$. At $x=40$, the exponent is $-4$, so $S_X(40)=e^{-4}=0.0183$. Probabilities are rounded to four decimals.

For Model P, $S_X(x)=(5/x)^2$. At $x=10$, $S_X(10)=(5/10)^2=0.5^2=0.2500$. At $x=20$, $S_X(20)=(5/20)^2=0.25^2=0.0625$. At $x=40$, $S_X(40)=(5/40)^2=0.125^2=0.0156$, rounded to four decimals.

Model E has $\log S_X(x)=-0.1x$, so its semilog survival plot is a straight line with slope $-0.1$. Model P has $\log S_X(x)=2\log 5-2\log x$, so its log-log survival plot is a straight line with slope $-2$. The probabilities at 40 happen to be close, but that single comparison does not make the models equivalent: the way survival changes across thresholds and the transformed shape are also relevant.

A Practical Diagnostic Sequence

A disciplined reading keeps the plot, the data, and the proposed model distinct. Use the following sequence to organize the diagnosis.

Check what is plotted.
Confirm the loss units, sample size, bin boundaries or threshold convention, and whether the vertical axis shows counts, relative frequencies, density, or survival.

Describe the broad shape.
Look for concentration, skewness, gaps, clusters, and unusually large observations. Treat these as observations about this sample, not automatic claims about the underlying loss process.

Inspect the upper tail.
Use survival values or a survival plot to see how many losses remain above increasing thresholds. A log vertical scale can display small positive survival values more clearly.

Compare the appropriate transformed pattern.
Check semilog survival for exponential-like decay and log-log survival for Pareto-like decay, while allowing for sampling noise and the limited observed range.

Check data limitations before concluding.
Ask whether the observed losses are censored, truncated, grouped, or affected by a reporting threshold. Those features can change the plotted pattern without reflecting the full underlying severity distribution.

Common Traps and What a Plot Cannot Establish

A frequent calculation error is to count observations equal to a threshold in the empirical survival value. In the sample above, a mistaken calculation at $x=2$ might include the three losses of 2 and report $8/10=0.8000$. That is incorrect for the strict-exceedance definition: the correct count is the five losses 4, 5, 8, 10, and 20, giving $\widehat{S}(2)=5/10=0.5000$. If a different convention is desired, it must be stated explicitly and used consistently.

Over-interpreting bins. A small change in bin width can hide or create apparent clusters. Check whether the broad conclusion persists under reasonable bin choices.
Calling one large loss an error. An extreme point may be a genuine loss, a recording problem, or evidence of a distinct mechanism. Investigate it rather than deleting it automatically.
Reading a zero on a log axis. Zero survival cannot appear on a log vertical scale. Omitting it from the transformed plot does not mean the tail probability is literally negative or that the underlying model has no tail.
Treating a visual match as proof. A short, noisy empirical tail can look nearly straight under more than one transformation. A plot is a diagnostic, not a formal goodness-of-fit conclusion.

Plot choice also cannot repair incomplete data. If large claims are censored at a policy limit, or small claims are absent because of a reporting threshold, a histogram or empirical survival plot of the recorded observations may not represent the full loss distribution. The observation process must be understood before using the shape to support a candidate model.

Key takeaway: Histograms summarize where losses fall; empirical survival plots show the proportion exceeding each threshold. Semilog and log-log views highlight different tail patterns. Interpret all of them with attention to definitions, scales, sampling variability, and how the data were observed. These diagnostic skills are aligned with the CAS Exam 9 (Risk Management for Actuaries) syllabus.

Check Your Understanding

For each problem, state the calculation or diagnostic reasoning before checking the final answer.

A hypothetical sample of eight losses, in thousands of dollars, is 1, 2, 2, 4, 7, 9, 10, and 13. Using bins $[0,5)$, $[5,10)$, and $[10,15)$, find the count and relative frequency in each bin. (Answer: Counts 4, 2, and 2; relative frequencies 0.5000, 0.2500, and 0.2500.)
For losses 3, 3, 6, 9, 12, 12, 18, and 30, calculate the empirical survival at thresholds 9 and 12 using strict exceedance. (Answer: $\widehat{S}(9)=4/8=0.5000$ and $\widehat{S}(12)=2/8=0.2500$.)
A candidate exponential severity model has mean 8 thousand dollars. Under an exponential model, calculate survival at 16 thousand dollars. (Answer: $\lambda=1/8$, so $S_X(16)=e^{-16/8}=e^{-2}=0.1353$.)
A candidate Pareto model has $x_m=4$ and $\alpha=2$. Calculate survival at 16 thousand dollars and state which transformed survival plot is linear under the model. (Answer: $S_X(16)=(4/16)^2=0.0625$; the log-log survival plot is linear.)
A log-scale empirical survival plot reaches zero beyond the largest observed claim. Can those zero values be plotted on a logarithmic vertical axis, and does their absence establish that the underlying model has no tail? (Answer: No; $\log(0)$ is undefined, and their absence does not establish that the underlying model has no tail.)

What You'll Learn