Worked Example: Complete Loss Distribution And Survival

Introduction

A complete sample of exact losses gives a direct way to describe how the observed losses are distributed. Each observation contributes the same amount of probability mass, so repeated loss amounts create jumps in the empirical distribution. Those jumps matter when a question asks for a quantile or an exceedance rate: the answer can change depending on whether the threshold is included.

This tutorial uses only exact, observed loss amounts. The calculations are descriptive summaries of that sample. Interpreting them as estimates of a target population distribution requires the sample to be representative of that population; a sample containing censored or missing losses cannot simply be treated as complete.

Definition: Let $X_1,\ldots,X_n$ be $n$ exact observed losses, with each observation included once and given equal weight. The empirical distribution function at $x$ is $$ \widehat F_X(x)=\frac{1}{n}\sum_{i=1}^{n} I(X_i\le x), $$ where $I(X_i\le x)$ is 1 when $X_i\le x$, and 0 otherwise. Under the convention $S_X(x)=P(X>x)$, the empirical survival function is $$ \widehat S_X(x)=\frac{1}{n}\sum_{i=1}^{n} I(X_i>x)=1-\widehat F_X(x). $$

Turning Exact Losses into a Distribution

The empirical distribution can be recorded as a table of distinct observed loss amounts and their relative frequencies. If an amount $x_j$ occurs $m_j$ times, its empirical probability mass is $m_j/n$. The masses are nonnegative and sum to 1, because every one of the $n$ observations is counted exactly once. The empirical distribution function accumulates those masses from the smallest loss upward.

For any threshold $x$, $\widehat F_X(x)$ counts observations at or below the threshold. By contrast, $\widehat S_X(x)$ counts observations strictly above it. Thus, an observation equal to $x$ belongs to the distribution function count, not to the strict exceedance count. If a question instead asks for the fraction at least $x$, that is a different count: it includes observations equal to $x$.

Definition: For $0<p\le 1$, the empirical $p$-quantile is the smallest $x$ at which the empirical distribution reaches or exceeds $p$: $$ \widehat Q_X(p)=\inf\{x:\widehat F_X(x)\ge p\}. $$ If the losses are ordered as $X_{(1)}\le \cdots\le X_{(n)}$, then $\widehat Q_X(p)=X_{(\lceil np\rceil)}$, where $\lceil np\rceil$ is the smallest integer at least $np$. This convention selects an observed loss; other sample-quantile conventions may interpolate between observations.

Why the Empirical Formulas Work

Each observed loss represents one of the $n$ equally weighted sample outcomes. To estimate the fraction of outcomes at or below a threshold, count how many satisfy $X_i\le x$, then divide by the total $n$. This is exactly the indicator sum in $\widehat F_X(x)$. The same reasoning counts strict exceedances using $X_i>x$.

For every observation, exactly one of $X_i\le x$ and $X_i>x$ is true. Therefore their indicator values sum to 1. Summing this identity across the sample and dividing by $n$ gives the complement relationship:

\widehat F_X(x)+\widehat S_X(x) =\frac{1}{n}\sum_{i=1}^{n}\left[I(X_i\le x)+I(X_i>x)\right] =\frac{1}{n}\sum_{i=1}^{n}1 =1.

The quantile formula follows from the same cumulative count. At the $k$th ordered observation, at least $k$ observations are at or below $X_{(k)}$, so $\widehat F_X(X_{(k)})\ge k/n$. Taking $k=\lceil np\rceil$ ensures $k/n\ge p$. At any smaller observed value there are at most $k-1$ observations at or below it, so the empirical distribution is less than $p$. Thus $X_{(\lceil np\rceil)}$ is the smallest value reaching probability $p$, including when values are tied.

Worked Example: Constructing the Distribution from Ten Exact Losses

A hypothetical insurer reviews ten exact claim amounts, in dollars: $80, 120, 80, 200, 120, 150, 300, 80, 250, 120$. Find the empirical probability mass at each distinct amount, the empirical distribution values, the median under the stated quantile convention, the 75th percentile, and the strict exceedance rate above $120.

Method and frequency table. There are $n=10$ observations. Sorting them gives $80,80,80,120,120,120,150,200,250,300$. The frequencies and cumulative counts are:

Loss amount	Frequency	Empirical mass	Cumulative count	$\widehat F_X(x)$
$80	3	3/10	3	0.3
$120	3	3/10	6	0.6
$150	1	1/10	7	0.7
$200	1	1/10	8	0.8
$250	1	1/10	9	0.9
$300	1	1/10	10	1.0

The empirical masses sum to $3/10+3/10+1/10+1/10+1/10+1/10=10/10=1$. The empirical distribution is 0 below $80, takes the table's cumulative values at the listed amounts, and remains 1 at and above $300.

Median. For $p=0.5$, $\lceil np\rceil=\lceil 10(0.5)\rceil=5$. The fifth ordered loss is $X_{(5)}=\$120$, so $\widehat Q_X(0.5)=\$120$.

75th percentile. For $p=0.75$, $\lceil np\rceil=\lceil 7.5\rceil=8$. The eighth ordered loss is $X_{(8)}=\$200$, so $\widehat Q_X(0.75)=\$200$.

Strict exceedance above $120. The losses strictly greater than $120 are $150, $200, $250, and $300. Therefore $\widehat S_X(120)=4/10=0.4$. As a check, $\widehat F_X(120)=6/10=0.6$, and $0.6+0.4=1$. The common mistake avoided here is counting the three losses equal to $120 as exceedances.

Thresholds, Ties, and Quantiles

A threshold can divide the sample in two different ways depending on the wording. Under the conventions used here, the empirical distribution counts $X_i\le x$, strict exceedance counts $X_i>x$, and inclusive exceedance counts $X_i\ge x$. Their relationship is:

\frac{\#\{i:X_i\ge x\}}{n} =1-\frac{\#\{i:X_i<x\}}{n}, \qquad \frac{\#\{i:X_i>x\}}{n}=1-\widehat F_X(x).

These distinctions have practical consequences. If a contract or report defines a threshold as “more than $x,” an observation exactly equal to $x$ does not qualify. If the wording is “at least $x,” it does. Always translate the words into an inequality before counting. When there are no observations exactly at the threshold, strict and inclusive counts happen to agree; that coincidence does not make the conventions interchangeable.

Quantiles also respond to ties and jumps. The empirical distribution increases by the relative frequency at each distinct loss amount. A requested probability can fall inside such a jump; the generalized-inverse convention selects the loss value at the jump, not an interpolated amount. This is why the definition of quantile must be stated when reporting results. In particular, the median under this convention is the smallest amount at which the empirical distribution reaches 0.5; it need not be the average of the two middle observations for an even-sized sample.

Worked Example: Strict and Inclusive Exceedance Rates

A hypothetical portfolio has eight exact losses, in thousands of dollars: $40, 60, 60, 90, 110, 110, 160, 240$. Calculate the empirical distribution at $110 thousand, the strict exceedance rate above that threshold, the inclusive rate at or above it, and the 75th percentile.

Method and ordered data. The listed data are already ordered, and $n=8$. At or below 110 there are six observations: 40, 60, 60, 90, 110, and 110. Therefore $\widehat F_X(110)=6/8=0.75$.

Strict exceedance. Only 160 and 240 are strictly greater than 110. The rate is $2/8=0.25$, which agrees with $1-\widehat F_X(110)=1-0.75=0.25$.

Inclusive exceedance. The observations at least 110 are 110, 110, 160, and 240. The rate is $4/8=0.50$. The difference between the inclusive and strict rates is $2/8=0.25$, exactly the empirical mass at 110.

75th percentile. Here $\lceil np\rceil=\lceil 8(0.75)\rceil=6$. The sixth ordered loss is $110 thousand, so $\widehat Q_X(0.75)=\$110{,}000$. The check is that six of eight losses are at or below this amount, giving 0.75. The mistake avoided is treating “at least 110” as if it meant “greater than 110.”

Worked Example: A Quantile at a Jump in the Empirical Distribution

A hypothetical sample contains twelve exact losses, in dollars: $0,0,50,75,75,100,125,125,125,200,300,500$. Find the empirical distribution and survival values at $125, the 75th percentile, and the 80th percentile. Explain how the tied losses affect the answer.

Method and counts at $125. There are nine losses at or below $125: the two zeros, $50, two $75 losses, $100, and three $125 losses. Thus $\widehat F_X(125)=9/12=0.75$. There are three losses strictly above $125: $200, $300, and $500. Thus $\widehat S_X(125)=3/12=0.25$. The values add to 1, as required.

75th percentile. The index is $\lceil 12(0.75)\rceil=\lceil9\rceil=9$. The ninth ordered loss is $125, so $\widehat Q_X(0.75)=\$125$.

80th percentile. The index is $\lceil 12(0.80)\rceil=\lceil9.6\rceil=10$. The tenth ordered loss is $200, so $\widehat Q_X(0.80)=\$200$. The empirical distribution jumps from $9/12=0.75$ at $125 to $10/12$ at $200; it has no observed loss amount between those points. The quantile convention therefore selects $200 rather than interpolating. The common mistake avoided is assuming that a percentile must be an observed fraction exactly equal to its requested probability.

Common Mistakes and Interpretation

Reversing the threshold inequality. $\widehat F_X(x)$ counts losses at or below $x$; $\widehat S_X(x)$ counts losses strictly above $x$. Check the inequality before tallying.
Dropping ties from the count. Every observation equal to a threshold contributes to $X_i\le x$ and $X_i\ge x$, but not to $X_i>x$.
Using an unstated quantile rule. The generalized-inverse convention selects $X_{(\lceil np\rceil)}$. A convention that averages middle observations or interpolates can give a different answer.
Treating incomplete records as exact losses. These formulas use actual observed loss amounts. Replacing a censored observation with its censoring threshold changes the data and does not recover the empirical distribution of exact losses.
Overstating the population conclusion. The empirical distribution describes the observed sample exactly. Its use as an estimate for a broader population depends on the sampling process and whether the records represent that population.

Key takeaway: For a complete sample of exact losses, count observations and divide by the sample size. The empirical distribution uses $X_i\le x$, strict survival uses $X_i>x$, and the empirical $p$-quantile under the generalized-inverse convention is $X_{(\lceil np\rceil)}$. State the threshold and quantile conventions whenever ties can affect the result.

Check Your Understanding

Use the empirical distribution and generalized-inverse quantile convention defined above. Treat each listed loss as an exact observation.

A sample of five losses is $25,40,40,70,110$. Find $\widehat F_X(40)$, the strict exceedance rate above 40, and the inclusive rate at or above 40. (Answer: $3/5=0.6,\ 2/5=0.4,\ 3/5=0.6$, respectively.)
For ordered losses $30,50,50,80,120,150$, find the empirical 50th and 90th percentiles. (Answer: $50$ and $150$.)
A sample is $20,30,30,60,90,90,140,200$. Find $\widehat F_X(90)$, the strict exceedance rate above 90, and the inclusive rate at or above 90. (Answer: $6/8=0.75,\ 2/8=0.25,\ 4/8=0.50$.)
For the ordered sample $10,20,40,40,70$, find the 60th percentile and the empirical distribution value at that loss amount. (Answer: the 60th percentile is $40$, and $\widehat F_X(40)=4/5=0.8$.)
A sample of ten losses is $25,50,75,75,90,100,125,125,160,200$. Find the strict exceedance rate above 90 and the inclusive rate at or above 90. (Answer: $5/10=0.5$ and $6/10=0.6$, respectively.)

What You'll Learn