Severity Quantiles and Percentiles

Why Severity Quantiles Matter

A mean summarizes the average severity, but it does not describe where a chosen fraction of losses fall. A severity quantile answers a threshold question: what loss amount is at or below a specified proportion of the severity distribution? For example, an insurer might use a high percentile to describe a large-loss threshold in a hypothetical portfolio. The percentile is an amount of loss; the probability associated with it is a proportion of outcomes.

Quantiles are straightforward to calculate when the cumulative distribution function (CDF) is continuous and strictly increasing. They require a careful convention when the CDF has jumps, as it does in a discrete model. This tutorial uses the generalized inverse CDF, which gives one consistent definition for both cases.

Definition: Let $X$ be a loss severity with CDF $F_X(x)=P(X\le x)$. For $0<p<1$, the $p$-quantile is $$ x_p=F_X^{-1}(p)=\inf\{x:F_X(x)\ge p\}. $$ Here $F_X^{-1}$ denotes the generalized inverse, not necessarily an ordinary inverse function. The $100p$th percentile is $x_p$. In this convention, the median is $x_{0.5}$, the 50th percentile.

The infimum in the definition selects the smallest threshold at which the CDF reaches or exceeds $p$. Thus a quantile need not have exactly $p$ of the probability at or below it. In particular, a discrete CDF can jump from below $p$ to above $p$. The definition still selects the first loss amount at that jump.

Reading a Quantile as a Threshold

For a quantile $x_p$, the defining threshold interpretation is that the cumulative probability has reached $p$ at $x_p$, while it has not reached $p$ at any smaller threshold. For a usual right-continuous CDF, this can be expressed using the CDF and its left limit:

F_X(x_p)\ge p \qquad\text{and}\qquad F_X(x_p-)\le p.

The first inequality says $P(X\le x_p)\ge p$. The second says the probability strictly below the quantile does not exceed $p$. For a continuous, strictly increasing CDF, there is no jump at $x_p$, so both sides meet at $p$: $F_X(x_p)=p$. For a discrete CDF, the inequalities may be strict on one or both sides.

Set the percentile level.
Convert a stated percentile to a probability $p$. For example, the 80th percentile uses $p=0.80$.

Use the CDF convention.
Find the smallest $x$ for which $F_X(x)\ge p$. If the CDF is continuous and strictly increasing, solve $F_X(x)=p$.

Check the threshold.
Verify that the CDF at the selected amount is at least $p$, and that a smaller amount has not already reached $p$.

A tail probability can also be used, provided its direction is translated correctly. Since $S_X(x)=1-F_X(x)$, a $p$-quantile of a continuous distribution satisfies $S_X(x_p)=1-p$. Thus the 95th percentile corresponds to a 5% upper-tail probability, not a 95% upper-tail probability. For a discrete model, the generalized inverse definition remains the safest check because a tail probability may jump across the target level.

Quantiles for Continuous Severity Models

When a CDF is continuous and strictly increasing on the relevant range, finding a quantile means setting the CDF equal to the desired percentile probability and solving for the loss amount. The resulting amount must still be in the model’s support. If the CDF is not strictly increasing, an ordinary algebraic inverse may be ambiguous or unavailable; use the generalized inverse definition instead.

Worked Example: Percentiles for a Bounded Continuous Loss

A hypothetical claim severity $X$, measured in dollars, is uniformly distributed from $2,000 to $12,000, inclusive of its support endpoints. Find the median and the 80th percentile, and check what fraction of losses exceed the 80th-percentile amount.

Method and model. Assume $X$ has a continuous uniform distribution on $[2{,}000,12{,}000]$. Its CDF within the support is

F_X(x)=\frac{x-2{,}000}{12{,}000-2{,}000} =\frac{x-2{,}000}{10{,}000}, \qquad 2{,}000\le x\le12{,}000.

For the median, set $F_X(x)=0.50$:

\frac{x-2{,}000}{10{,}000}=0.50, \qquad x-2{,}000=5{,}000, \qquad x=7{,}000.

Therefore, the median is $7,000. For the 80th percentile, set $F_X(x)=0.80$:

\frac{x-2{,}000}{10{,}000}=0.80, \qquad x-2{,}000=8{,}000, \qquad x=10{,}000.

The 80th-percentile severity is $10,000. The CDF check gives $F_X(10{,}000)=8{,}000/10{,}000=0.80$, so the probability of a loss at or below $10,000 is 0.80. Because this model is continuous, the probability of a loss strictly above $10,000 is $1-0.80=0.20$.

Common mistake avoided: A percentile is not the percentage multiplied by the upper endpoint. The model starts at $2,000, so the calculation uses the CDF over the $10,000-wide support, then adds the lower endpoint.

Quantiles for Discrete Severity Models

For a discrete severity model, list the possible loss amounts in increasing order and accumulate their probabilities. The $p$-quantile is the first listed amount whose cumulative probability is at least $p$. It is not generally the loss amount at which the cumulative probability is exactly $p$, because the cumulative probability may jump over $p$.

Worked Example: Finding a Discrete Median and Percentile

A hypothetical equipment policy has severity distribution $X$ with the following possible claim amounts and probabilities. Find the median and the 80th percentile.

Loss amount $x$	$P(X=x)$	$F_X(x)=P(X\le x)$
$0	0.12	0.12
$1,000	0.28	0.40
$3,000	0.40	0.80
$8,000	0.20	1.00

Method. The probabilities sum to $0.12+0.28+0.40+0.20=1.00$, so they define the full distribution. The cumulative values are found by successive addition: at $0 the total is $0.12$; at $1,000 it is $0.12+0.28=0.40$; at $3,000 it is $0.40+0.40=0.80$; and at $8,000 it is $0.80+0.20=1.00$.

For the median, $p=0.50$. The cumulative probability at $1,000 is 0.40, which is below 0.50. At $3,000 it is 0.80, which is at least 0.50. Therefore, the smallest loss amount whose cumulative probability reaches 0.50 is $3,000:

x_{0.5}=3{,}000.

For the 80th percentile, $p=0.80$. The CDF at $1,000 is 0.40, below 0.80, and the CDF at $3,000 is exactly 0.80. Therefore,

x_{0.8}=3{,}000.

The boundary check matters: $F_X(3{,}000)=0.80$, while $F_X(1{,}000)=0.40<0.80$. A loss of $3,000 is the first amount that meets the threshold. The probability strictly below $3,000 is 0.40, so the distribution does not require the probability at or below the quantile to equal 0.50 or 0.80 in every case.

Common mistake avoided: Do not interpolate between $1,000 and $3,000 to manufacture an amount whose CDF is 0.50. This model only assigns probability to its listed amounts, and the generalized inverse selects the first amount that reaches the target.

Deriving the Quantile Check

The generalized inverse definition explains why the discrete procedure and the continuous equation both work. Let $q=\inf\{x:F_X(x)\ge p\}$. By the definition of an infimum, no value strictly below $q$ can already have CDF at least $p$; otherwise that smaller value would be in the set and $q$ would not be its infimum. Hence the left limit satisfies $F_X(q-)\le p$. Right-continuity of a CDF ensures that the CDF at the threshold reaches the target, so $F_X(q)\ge p$. This gives the two quantile checks stated above.

If $F_X$ is continuous at $q$, then $F_X(q-)=F_X(q)$. Combining the inequalities gives $F_X(q)=p$. This justifies solving an equality for a continuous, strictly increasing CDF. If instead the CDF jumps at $q$, its value can exceed $p$; the first point of that jump is still the generalized-inverse quantile.

Worked Example: Converting an Exceedance Probability to a Percentile

A hypothetical severity $X$ follows an exponential distribution with mean $5,000. Find the loss amount exceeded with probability 0.05, and identify its percentile. Round the final dollar amount to the nearest cent.

Method and model. For an exponential severity with mean $\theta=5{,}000$, assume $F_X(x)=1-e^{-x/\theta}$ for $x\ge0$, and $S_X(x)=e^{-x/\theta}$. An exceedance probability of 0.05 means $S_X(x)=0.05$; equivalently, $F_X(x)=1-0.05=0.95$. The required amount is therefore the 95th percentile.

Solve the tail equation:

e^{-x/5{,}000}=0.05, \qquad -\frac{x}{5{,}000}=\ln(0.05), \qquad x=-5{,}000\ln(0.05).

Since $\ln(0.05)=-2.9957322736\ldots$, the unrounded amount is $5{,}000(2.9957322736\ldots)=14{,}978.661368\ldots$. Rounding only the final result to two decimal places gives $14,978.66.

Check. At the unrounded amount, $S_X(x)=e^{-x/5{,}000}=0.05$, so $F_X(x)=1-0.05=0.95$. Thus 95% of losses are at or below the threshold and 5% exceed it.

Common mistake avoided: Setting the survival probability equal to 0.95 would find the amount exceeded with probability 0.95, which is a low percentile, not the requested high threshold. Translate “exceeded with probability 0.05” into $F_X(x)=0.95$ before solving.

Common Exam Traps and a Practical Check

A reliable quantile calculation keeps three items separate: the percentile probability $p$, the loss threshold $x_p$, and the probability of exceeding that threshold. The CDF concerns $P(X\le x)$, whereas the survival function concerns $P(X>x)$. For continuous distributions the equality $S_X(x_p)=1-p$ is convenient. For discrete distributions, the generalized inverse and its cumulative-probability check are more robust.

Reversing the tail. The 90th percentile has 0.90 probability at or below the amount, not 0.90 probability above it.
Assuming an exact CDF match. A discrete quantile can have $F_X(x_p)>p$; select the first amount with $F_X(x)\ge p$.
Using the wrong endpoint condition. The definition uses $P(X\le x)$. At a jump, replacing “at or below” with “strictly below” can change the selected amount.
Ignoring the support. A solution to an algebraic CDF equation must be a possible value in the model’s range.
Confusing an amount with a probability. A percentile is a loss amount; “the 80th percentile” is not itself a probability of 0.80 dollars or a payment percentage.

As a final check, substitute the proposed amount back into the CDF. Confirm $F_X(x_p)\ge p$, and for a discrete model confirm that the next smaller possible loss has cumulative probability below $p$. For a continuous model with a strictly increasing CDF, confirm that the CDF equals $p$ and that the answer lies in the support. These checks catch most percentile and tail-direction errors.

Check Your Understanding

Use the generalized inverse convention $x_p=\inf\{x:F_X(x)\ge p\}$. For money amounts, give answers to the nearest cent when needed.

A hypothetical severity is uniform on $[0,5{,}000]$. Find its median. (Answer: $2,500)
A discrete severity takes values $0, $400, $1,500, and $4,000 with respective probabilities 0.20, 0.25, 0.35, and 0.20. Find the 70th percentile. (Answer: $1,500)
A discrete severity takes values $250, $750, $1,750, and $4,000 with respective probabilities 0.15, 0.35, 0.30, and 0.20. Find the 50th and 80th percentiles. (Answer: 50th percentile = $750; 80th percentile = $1,750)
A hypothetical exponential severity has mean $3,000. Find the 90th percentile using its CDF. (Answer: $6,907.76)
A discrete severity takes values $0, $2,000, $5,000, and $10,000 with respective probabilities 0.05, 0.50, 0.35, and 0.10. Find the 95th percentile and the probability of a loss strictly greater than $5,000. (Answer: 95th percentile = $10,000; probability strictly greater than $5,000 = 0.10)

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