Empirical Quantiles From Ordered Losses

Why Empirical Quantiles Need a Convention

An empirical quantile summarizes the location of a specified fraction of observed losses. It can help describe a hypothetical portfolio’s loss distribution—for example, by identifying a loss level at or below which a given proportion of the recorded losses falls. The calculation begins with exact observations arranged from smallest to largest.

For a finite sample, however, a percentile does not always identify one uniquely agreed value. One convention selects an observed loss directly; another interpolates between nearby observations. Both can be useful, but they can give different answers. A sound calculation therefore states the convention before finding the quantile.

Definition: Let $x_1,\ldots,x_n$ be $n$ exact loss observations, and write their ordered values as $x_{(1)}\leq x_{(2)}\leq\cdots\leq x_{(n)}$. The empirical distribution function is $F_n(x)=\frac{1}{n}\sum_{j=1}^{n} I(x_j\leq x)$, where $I(\cdot)$ is 1 when its condition is true and 0 otherwise. An empirical $p$-quantile, for $0<p\leq1$, is a sample-based loss level associated with cumulative proportion $p$. Its precise value depends on the stated finite-sample quantile convention.

The order notation $x_{(j)}$ means the observation in position $j$ after sorting, not the $j$th observation as originally recorded. If losses are tied, each recorded observation still occupies a position in this ordered list. A tie can make the empirical distribution jump by more than $1/n$ at a loss value.

Two Common Finite-Sample Conventions

The inverse empirical distribution convention chooses the smallest loss value at which the empirical distribution reaches or exceeds $p$. For $0<p\leq1$, it is

Q_n^{\mathrm{inv}}(p)=\inf\{x:F_n(x)\geq p\} =x_{(\lceil np\rceil)}.

Here $\lceil a\rceil$ is the smallest integer greater than or equal to $a$. This rule always returns an observed loss. It is also called the nearest-rank rule in common percentile practice.

A second convention assigns a fractional position $h$ among the ordered values and linearly interpolates when that position is not an integer:

h=1+(n-1)p,\qquad Q_n^{\mathrm{int}}(p)=x_{(j)}+\gamma\bigl(x_{(j+1)}-x_{(j)}\bigr),

where $j=\lfloor h\rfloor$ and $\gamma=h-j$, provided $h$ is not an integer. If $h$ is an integer, the interpolated quantile is $x_{(h)}$. The floor $\lfloor h\rfloor$ is the greatest integer no larger than $h$. This interpolation convention can return a value between observed losses, even when that value was not recorded.

Key distinction: The inverse empirical distribution rule uses rank $\lceil np\rceil$. The interpolation rule uses position $1+(n-1)p$. These are different formulas and should not be combined. A quantile answer is incomplete unless its convention is clear.

Why the Inverse Rank Is the Ceiling of $np$

The empirical distribution puts mass $1/n$ at each recorded observation. When there are no ties, its cumulative proportion at the $r$th ordered observation is $r/n$. The first rank that reaches $p$ is therefore the smallest integer $r$ satisfying $r/n\geq p$. Multiplying by positive $n$ gives $r\geq np$, so the smallest eligible integer is $\lceil np\rceil$.

\frac{r}{n}\geq p \quad\Longleftrightarrow\quad r\geq np \quad\Longrightarrow\quad r=\lceil np\rceil.

With ties, the same rank formula remains valid when the ordered observations include every recorded value. The empirical distribution may reach $p$ at a tied value shared by several ranks, but the value at rank $\lceil np\rceil$ is still the smallest observed value whose cumulative empirical proportion is at least $p$.

The interpolation formula is a separate convention, not a consequence of this rank argument. Its position $h=1+(n-1)p$ places $p=0$ at the first observation and $p=1$ at the last, with linear movement across the $n-1$ gaps between them.

Worked Example: An Inverse Quantile From Five Exact Losses

A hypothetical insurer records five exact losses, in dollars: $145$, $35$, $230$, $92$, and $58$. Find the 60th percentile using the inverse empirical distribution convention, then verify that the selected value meets the definition.

Method: sort, find the ceiling rank, and check the cumulative proportion. Sorting the observations gives $x_{(1)}=35$, $x_{(2)}=58$, $x_{(3)}=92$, $x_{(4)}=145$, and $x_{(5)}=230$. Thus $n=5$ and $p=0.60$.

np=5(0.60)=3,\qquad \lceil np\rceil=3.

The inverse empirical 60th percentile is $x_{(3)}=\$92$. To check it, two of the five observations are strictly below \$92, and three are at or below \$92. Hence $F_n(92)=3/5=0.60$, while immediately below \$92 the empirical cumulative proportion is $2/5=0.40$. The first value at which the cumulative proportion reaches 0.60 is therefore \$92.

Common mistake avoided: Rounding $np$ up when it is already an integer would incorrectly produce rank 4. The ceiling of 3 is 3, not 4.

Worked Example: Interpolation With Six Losses

A hypothetical book of policies has six exact losses, recorded in thousands of dollars: $120$, $20$, $310$, $75$, $190$, and $45$. Find the 70th percentile using $h=1+(n-1)p$. Then compare it with the inverse empirical quantile at the same percentile.

Method: order the losses and interpolate at the prescribed position. In increasing order, the losses are $20$, $45$, $75$, $120$, $190$, and $310$. Thus $n=6$ and $p=0.70$. The interpolation position is

$$ h=1+(6-1)(0.70)=1+3.50=4.50. $$

The position is halfway between $x_{(4)}=120$ and $x_{(5)}=190$, in thousands of dollars. The fractional part is $\gamma=4.50-4=0.50$, so

Q_n^{\mathrm{int}}(0.70) =120+0.50(190-120) =120+35 =155

The interpolated 70th percentile is $155$ thousand dollars. For comparison, the inverse empirical rank is $\lceil6(0.70)\rceil=\lceil4.2\rceil=5$, so the inverse empirical quantile is $x_{(5)}=190$ thousand dollars. These answers differ because the conventions differ; neither calculation is an arithmetic correction of the other.

Common mistake avoided: The interpolation weight is 0.50, the fractional part of the position, not the percentile $p=0.70$. Using 0.70 would give the wrong interpolated value.

Worked Example: Sorting Eight Losses Before Using Either Rule

A hypothetical claims analyst receives eight exact loss amounts in dollars: $72$, $14$, $48$, $110$, $32$, $95$, $60$, and $20$. Find the 62.5th percentile under both the inverse empirical and interpolation conventions. State why sorting is essential.

Method: reorder the observations, then apply each convention separately. The increasing order is $14$, $20$, $32$, $48$, $60$, $72$, $95$, $110$. For the inverse empirical convention, $np=8(0.625)=5$, so the rank is $\lceil5\rceil=5$. The quantile is $x_{(5)}=\$60$.

For interpolation, the position is $h=1+(8-1)(0.625)=1+4.375=5.375$. Thus $j=5$ and $\gamma=5.375-5=0.375$. Interpolating between $x_{(5)}=\$60$ and $x_{(6)}=\$72$ gives

Q_n^{\mathrm{int}}(0.625) =60+0.375(72-60) =60+0.375(12) =60+4.50 =\$64.50.

The two answers are therefore \$60 under the inverse empirical rule and \$64.50 under the interpolation rule. The original third recorded amount was \$48, but it is not the third ordered loss; calculating positions before sorting would select the wrong values.

Common mistake avoided: Do not use the original record order as the quantile order. Quantile positions refer only to the sorted sample.

Practical Checks and Exam Traps

A reliable quantile calculation follows a short sequence. First confirm that the observations are exact losses and identify $n$ and $p$. Next sort the losses, retaining repeated observations. Then apply only the requested convention, show the rank or position, and report the units. A final check should confirm that the answer is consistent with the selected neighboring observations and the chosen rule.

State the convention.
Decide whether the task asks for the inverse empirical distribution or a particular interpolation rule.

Order the exact losses.
Write $x_{(1)}\leq\cdots\leq x_{(n)}$, keeping ties as repeated observations.

Calculate the required position.
Use $\lceil np\rceil$ for the inverse rule, or $1+(n-1)p$ for the interpolation rule.

Check and report.
Confirm the rank or interpolation endpoints, then state the quantile with its units and convention.

A frequent exam trap is to assume that an “$p$th percentile” automatically means interpolation. It does not: finite-sample definitions vary, so follow the convention given in the problem. Another trap is to use $np$ as a direct index even when it is not an integer. For example, if $np=4.2$, the inverse empirical rank is 5, not 4.2 and not 4.

For an even sample size, some familiar summaries use the midpoint of two central observations, but that fact does not determine every percentile convention. Apply the stated rule rather than replacing it with a remembered rule for the sample median. Similarly, an interpolated quantile need not be an observed claim amount, while the inverse empirical quantile always is.

These distinctions matter when reporting empirical thresholds, comparing summaries across analyses, or checking results produced by different software. A numerical answer without its convention may be ambiguous. The methods here concern complete, exact loss records; censored observations cannot simply be inserted as if their loss amounts were known exactly.

Key takeaway: For ordered exact losses, the inverse empirical $p$-quantile is $x_{(\lceil np\rceil)}$. Under the stated linear interpolation convention, use position $1+(n-1)p$ and interpolate between adjacent order statistics when the position is fractional. State which convention produced the answer.

Check Your Understanding

For each problem, apply the specified convention and keep the order of the observations explicit.

A hypothetical sample has ordered exact losses of \$12, \$24, \$39, \$81, and \$150. Find the 40th percentile using the inverse empirical distribution convention. (Answer: \$24)
A hypothetical sample has six ordered exact losses, in thousands of dollars, of $10$, $20$, $30$, $50$, $90$, and $130$. Find the median using $h=1+(n-1)p$. (Answer: 40 thousand dollars)
Seven exact losses are recorded as $95$, $8$, $140$, $41$, $23$, $70$, and $16$. Find the 80th percentile using the inverse empirical distribution convention. (Answer: \$95)
A hypothetical sample has four ordered exact losses of \$25, \$40, \$90, and \$160. Find the 25th percentile using $h=1+(n-1)p$. (Answer: \$36.25)
A hypothetical portfolio has eight ordered exact losses, in thousands of dollars, of $5$, $5$, $10$, $20$, $20$, $20$, $40$, and $80$. Find the 75th percentile under both the inverse empirical and $h=1+(n-1)p$ interpolation conventions. (Answer: 20 thousand dollars by the inverse rule; 25 thousand dollars by interpolation)

What You'll Learn

Why Empirical Quantiles Need a Convention

Two Common Finite-Sample Conventions

Why the Inverse Rank Is the Ceiling of \(np\)

Worked Example: An Inverse Quantile From Five Exact Losses

Worked Example: Interpolation With Six Losses

Worked Example: Sorting Eight Losses Before Using Either Rule

Practical Checks and Exam Traps

Check Your Understanding

Empirical Quantiles From Ordered Losses

What You'll Learn

Why Empirical Quantiles Need a Convention

Two Common Finite-Sample Conventions

Why the Inverse Rank Is the Ceiling of \(np\)

Worked Example: An Inverse Quantile From Five Exact Losses

Worked Example: Interpolation With Six Losses

Worked Example: Sorting Eight Losses Before Using Either Rule

Practical Checks and Exam Traps

Check Your Understanding

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