Empirical Claim Quantiles and Order Statistics

Why Quantiles Depend on the Convention

A claim quantile describes a location in the ordered distribution of claim amounts. For example, a 75th percentile is a value at or above which roughly the upper quarter of observations lie, with the precise meaning depending on the quantile convention. In a finite sample, there may be no observation at exactly the requested percentile rank. A calculation must therefore specify how it chooses a value from the ordered data.

This tutorial uses the inverse empirical distribution convention as its default: the empirical $p$-quantile is the smallest amount at which the empirical distribution function reaches or exceeds $p$. Under this convention, a percentile rank between observations is rounded up to the next order-statistic position; the two surrounding observations are not averaged. Some procedures instead interpolate. An answer should follow the convention stated in the problem, and should name the convention when it is not specified.

Definition: Given $n$ observed claim amounts $X_1,\ldots,X_n$, the order statistics are those observations arranged in nondecreasing order: $$ X_{(1)} \le X_{(2)} \le \cdots \le X_{(n)}. $$ Here $X_{(k)}$ is the observation in position $k$ after sorting, including repeated observations when claims are tied. The empirical distribution function is $$ \widehat{F}_n(x)=\frac{1}{n}\sum_{j=1}^{n} \mathbf{1}(X_j\le x), $$ where $\mathbf{1}(X_j\le x)$ equals 1 when $X_j\le x$, and 0 otherwise. For $0<p\le 1$, the inverse empirical distribution quantile is $$ \widehat{Q}(p)=\inf\{x:\widehat{F}_n(x)\ge p\}=X_{(\lceil np\rceil)}. $$

This definition assumes that the observations being summarized are fully observed claim amounts measured on a comparable basis. The empirical distribution itself can be calculated for any finite set of values. Interpreting it as information about an underlying claim severity distribution typically also assumes that the observations are a representative sample from that distribution. The formula above does not require a parametric model for claim severity.

Finding the Empirical Quantile

To calculate the inverse empirical distribution quantile, first sort every observation, retaining duplicates. Then multiply the sample size $n$ by the requested percentile $p$, expressed as a decimal. The required position is $\lceil np\rceil$, where the ceiling function rounds up to the next integer (or leaves an integer unchanged). Finally, read the value at that position in the sorted sample.

Sort the observations.
List the claim amounts from smallest to largest. Do not remove repeated amounts.

Calculate the rank.
Write the requested percentile as $p$ between 0 and 1, and calculate $np$.

Apply the convention.
For the inverse empirical distribution convention, take position $k=\lceil np\rceil$.

Read and check the value.
The quantile is $X_{(k)}$. Confirm that the empirical distribution function reaches $p$ there and is still below $p$ immediately before the quantile.

The calculation is a step-function calculation, not a search for an exact percentile among the observations. If $np$ is not an integer, the empirical distribution function can jump from below $p$ to above $p$ at an observed claim amount. The inverse convention selects the first value at which that jump has reached $p$.

Worked Example: A 75th Percentile With Eight Claims

A hypothetical insurer records eight fully observed claim amounts, in dollars: $410, 125, 680, 305, 520, 275, 410, 760$. Find the empirical 75th percentile using the inverse empirical distribution convention. State the order statistic and verify the result.

Method. Sort the observations, calculate the ceiling rank, and check the empirical distribution immediately before and at the selected amount.

The sorted claims are $$ 125,\ 275,\ 305,\ 410,\ 410,\ 520,\ 680,\ 760. $$ There are $n=8$ observations and $p=0.75$, so $$ np=8(0.75)=6,\qquad k=\lceil 6\rceil=6. $$ The sixth order statistic is $X_{(6)}=\$520$. Thus the empirical 75th percentile is $\$520$.

Check. Five of the eight claims are strictly below $\$520$, so $\widehat{F}_8(x)=5/8=0.625$ for values immediately below $\$520$. At $\$520$, six claims are at or below that amount, so $\widehat{F}_8(520)=6/8=0.75$. The first point at which the empirical distribution reaches 0.75 is therefore $\$520$.

The repeated $\$410$ amounts occupy positions four and five. They are both retained; removing one would change the sample size and could change the percentile rank. A common mistake is to treat the two tied claims as one observation.

Why the Ceiling Rank Works

The ceiling formula follows directly from the definition of the empirical quantile. For $k=\lceil np\rceil$, the definition of the ceiling gives $k-1<np\le k$. At $X_{(k)}$, at least $k$ observations are at or below the value, so the empirical distribution is at least $k/n$, which is at least $p$. At a value strictly below $X_{(k)}$, at most $k-1$ observations can be at or below it, so the empirical distribution is at most $(k-1)/n$, which is less than $p$.

\widehat{F}_n(X_{(k)})\ge \frac{k}{n}\ge p, \qquad x<X_{(k)}\ \Longrightarrow\ \widehat{F}_n(x)\le \frac{k-1}{n}<p.

This argument also explains how ties behave. If several claims have the same value, the empirical distribution can jump by more than $1/n$ at that value. The quantile is still the smallest claim amount at which the cumulative proportion reaches $p$; it is not necessary for the empirical distribution to equal $p$ exactly.

Worked Example: A Percentile Rank Between Two Observations

A hypothetical equipment insurer observes nine claim amounts, in thousands of dollars: $90, 140, 115, 260, 175, 140, 310, 205, 155$. Find the empirical 60th percentile using the inverse empirical distribution convention, and show why the noninteger rank does not call for averaging.

Method. Use the ceiling rank, then check the cumulative proportions on either side of the selected order statistic.

In sorted order, the claims are $$ 90,\ 115,\ 140,\ 140,\ 155,\ 175,\ 205,\ 260,\ 310. $$ Here $n=9$ and $p=0.60$, so $$ np=9(0.60)=5.4,\qquad k=\lceil 5.4\rceil=6. $$ The sixth order statistic is $X_{(6)}=175$, so the empirical 60th percentile is $\$175{,}000$.

Five of the nine observations are at or below $\$155{,}000$, giving an empirical proportion of $5/9=0.5556$ to four decimal places. At $\$175{,}000$, six observations are at or below the amount, giving $6/9=0.6667$. The empirical distribution has crossed 0.60 at the sixth observation.

The value $np=5.4$ is a rank calculation, not a claim amount and not an instruction to average $X_{(5)}$ and $X_{(6)}$. Averaging those observations would give $(155+175)/2=165$ thousand dollars, but that is not the inverse empirical distribution quantile. The common mistake avoided here is confusing a fractional rank with a request to interpolate.

When a Problem Specifies Interpolation

Some sample-quantile procedures define a value between adjacent order statistics. One possible linear interpolation convention, used here only when explicitly requested, sets $$ h=1+(n-1)p. $$ If $h=j+g$, where $j$ is an integer and $0\le g<1$, the interpolated quantile is $$ (1-g)X_{(j)}+gX_{(j+1)}. $$ If $h$ is an integer, this rule selects $X_{(h)}$. This is a different convention from the inverse empirical distribution quantile. Neither result should be presented as though it followed from the other convention.

The practical rule is to identify the convention before calculating. If the problem defines an empirical quantile as the smallest $x$ with $\widehat{F}_n(x)\ge p$, use $\lceil np\rceil$. If it specifies an interpolation rule, use its rank and interpolation formula. If no interpolation rule is stated, state the convention used in the answer.

Worked Example: Applying an Explicit Linear Interpolation Rule

A hypothetical claims review has six observed amounts, in thousands of dollars: $18, 24, 31, 39, 52, 80$. The review instructions explicitly define the 70th percentile using $h=1+(n-1)p$ and linear interpolation between adjacent order statistics. Calculate that percentile, then compare it with the inverse empirical distribution quantile.

Method. Apply the stipulated interpolation rule first. Then calculate the inverse empirical distribution quantile separately to make the convention difference clear.

The observations are already sorted, so $n=6$ and $p=0.70$. Under the stated interpolation rule, $$ h=1+(6-1)(0.70)=1+3.5=4.5. $$ Thus $j=4$ and $g=0.5$. The adjacent observations are $X_{(4)}=39$ and $X_{(5)}=52$, in thousands of dollars. The interpolated value is $$ (1-0.5)(39)+0.5(52)=19.5+26=45.5. $$ The requested interpolated 70th percentile is therefore $\$45{,}500$.

Under the inverse empirical distribution convention, the rank would instead be $$ \lceil np\rceil=\lceil 6(0.70)\rceil=\lceil 4.2\rceil=5, $$ so that quantile would be $X_{(5)}=\$52{,}000$. The two results differ because the conventions define the sample quantile differently. The mistake to avoid is using the interpolation result when the problem asks for the inverse empirical distribution quantile, or silently substituting the inverse convention when interpolation is explicitly required.

Check. Under the specified interpolation rule, $45.5$ lies between $39$ and $52$, halfway between them, as required by $g=0.5$. Under the inverse convention, the fifth observation is the first point where the empirical distribution reaches at least 0.70: four of six observations give $4/6<0.70$, while five give $5/6\ge0.70$.

Common Mistakes and Practical Checks

Empirical quantile errors often come from a correct formula applied to the wrong rank, or from using a different convention without saying so. Before reporting a result, check the following:

Convert the percentile to a proportion. Use $p=0.80$ for the 80th percentile, not $p=80$.
Sort all observations and keep ties. The order statistics refer to positions in the full sample.
Do not round a fractional rank down under the inverse convention. If $np=5.4$, the required position is $\lceil 5.4\rceil=6$, not 5.
Do not interpolate unless the convention calls for it. Interpolation can produce a value that was not observed, while the inverse empirical distribution quantile is an observed order statistic.
Use units consistently. If claims are recorded in thousands of dollars, convert the final amount to dollars only if the requested reporting unit requires it.

A useful final check is to count observations at or below the selected value. For the inverse empirical distribution convention, that fraction must be at least $p$; immediately below the selected value, the fraction must be less than $p$. This check catches many rank and rounding errors without requiring any additional model assumptions.

Key takeaway: For the inverse empirical distribution convention, sort the sample and select $X_{(\lceil np\rceil)}$. If a problem specifies interpolation, apply that stated rule instead and identify it clearly; the conventions need not produce the same claim amount.

Check Your Understanding

Use the inverse empirical distribution convention unless a question explicitly specifies another rule. Amounts are in dollars unless stated otherwise.

A sample of seven claims is $12, 7, 19, 7, 25, 14, 10$. Find the empirical median, defined here as the $p=0.50$ quantile. (Answer: $\$12$)
Ten claim amounts, in thousands of dollars, are $4, 10, 7, 15, 9, 11, 18, 6, 14, 20$. Find the empirical 90th percentile. (Answer: $\$18{,}000$)
Five claim amounts are $3, 8, 8, 12, 21$. Find the empirical 40th percentile and account for the tied observations. (Answer: $\$8$)
Eight amounts, in thousands of dollars, are $2, 5, 9, 11, 14, 18, 23, 30$. A report explicitly uses $h=1+(n-1)p$ with linear interpolation to calculate the 25th percentile. Find the interpolated value. (Answer: $\$8{,}000$)
Six sorted claims, in thousands of dollars, are $4, 8, 12, 16, 20, 28$. Using the inverse empirical distribution convention, calculate the 50th percentile and explain why it differs from the value produced by averaging the two middle observations. (Answer: $\$12{,}000$; the inverse convention selects $X_{(3)}$, whereas averaging the middle observations gives $\$14{,}000$)

What You'll Learn