Conditional Severity Above A Threshold

Why Condition on a Threshold?

In severity analysis, it is often useful to describe losses only after they have crossed a specified threshold. For example, an analyst might ask how large a loss tends to be among losses exceeding a reporting level. This is a conditional probability question: the population of possible losses has been restricted to the event \(X>d\).

The inequality matters. Conditioning on \(X>d\) excludes any probability mass exactly at \(d\), while conditioning on \(X\ge d\) includes it. For a continuous severity model, a single point has probability zero, so the distinction does not change the calculations. For a discrete model, it can change them substantially.

Definition: Let \(X\) be a loss amount and let \(d\) be a threshold such that \(P(X>d)>0\). The conditional severity above \(d\) is the distribution of \(X\) given the event \(X>d\). Its conditional distribution function is \(P(X\le x\mid X>d)\). All probabilities, means, and quantiles in this tutorial refer to this conditional distribution unless stated otherwise.

The condition \(P(X>d)>0\) is essential: if no loss can exceed \(d\), the conditioning event has probability zero and the conditional distribution is not defined. The threshold is a selection rule, not a change to the loss amount. In particular, the conditional variable remains \(X\), rather than \(X-d\).

General Conditional Distribution

Write \(F_X(x)=P(X\le x)\). For \(x>d\), the event \(\{X\le x,\ X>d\}\) is the same as \(\{d<X\le x\}\). Its probability is \(F_X(x)-F_X(d)\). Dividing by the probability of the conditioning event gives the general conditional distribution:

P(X\le x\mid X>d)= \begin{cases} 0, & x\le d,\\[4pt] \dfrac{F_X(x)-F_X(d)}{1-F_X(d)}, & x>d, \end{cases} \qquad \text{provided }1-F_X(d)>0.

Here \(1-F_X(d)=P(X>d)\), so the formula uses the strict exceedance probability. If \(X\) is continuous at \(d\), then \(P(X=d)=0\), and \(F_X(d)\) can also be replaced by the left limit \(F_X(d-)\). For a discrete severity with an atom at \(d\), that replacement would alter the conditioning event and is generally incorrect.

A related expression is convenient for exceedance probabilities. For \(x\ge d\), let \(S_X(x)=P(X>x)=1-F_X(x)\). Then

P(X>x\mid X>d)=\frac{S_X(x)}{S_X(d)}, \qquad x\ge d,\quad S_X(d)>0.

This ratio follows because \(X>x\) implies \(X>d\) whenever \(x\ge d\). Thus the intersection of the two events is simply \(X>x\). For a conditional quantile at probability level \(p\), use the inverse conditional distribution: \(q_p=\inf\{x:P(X\le x\mid X>d)\ge p\}\). In a discrete model, the conditional distribution can jump, so the quantile need not have conditional cumulative probability exactly \(p\).

Worked Example: A Discrete Severity Above a Threshold

A hypothetical equipment-loss model has \(P(X=2)=0.20\), \(P(X=5)=0.50\), and \(P(X=9)=0.30\), with amounts measured in thousands of dollars. Given that \(X>4\), find (a) the probability of the conditioning event, (b) the conditional distribution, (c) \(P(X>7\mid X>4)\), and (d) the conditional mean and 0.60 quantile.

Method and event probability. Apply the strict condition directly to each support value. The value 2 is excluded, while 5 and 9 are retained because both exceed 4. Therefore,

$$ P(X>4)=P(X=5)+P(X=9)=0.50+0.30=0.80. $$

Conditional distribution. Divide each retained probability by 0.80:

P(X=5\mid X>4)=\frac{0.50}{0.80}=0.625,\qquad P(X=9\mid X>4)=\frac{0.30}{0.80}=0.375.

The conditional probabilities sum to \(0.625+0.375=1.000\), as required. The conditional distribution places no probability at 2.

Conditional exceedance probability. Among the retained values, only 9 exceeds 7. Hence,

P(X>7\mid X>4)=\frac{0.30}{0.80}=0.375.

Conditional mean. Weight the retained amounts by their conditional probabilities:

E[X\mid X>4]=5(0.625)+9(0.375)=3.125+3.375=6.500.

Thus the conditional mean is 6.500 thousand dollars. For the 0.60 quantile, the conditional cumulative probability at 5 is 0.625, which is already at least 0.60. Therefore \(q_{0.60}=5\) thousand dollars.

Check and common mistake avoided. The conditional probabilities total 1, and the mean lies between the retained amounts 5 and 9. A common error is excluding the mass \(P(X=5)=0.50\) from the denominator. Since \(5>4\), that mass is included; the correct denominator is \(0.50+0.30=0.80\), not 0.30.

Conditional Severity in Continuous Models

For a continuous severity with density \(f_X\), the conditional density above \(d\) is obtained by restricting the original density to values greater than \(d\) and dividing by the probability of that region:

f_{X\mid X>d}(x)= \begin{cases} \dfrac{f_X(x)}{S_X(d)}, & x>d,\\[4pt] 0, & x\le d. \end{cases}

The denominator normalizes the density: integrating the numerator over \(x>d\) gives \(P(X>d)=S_X(d)\), so the conditional density integrates to 1. Its mean can be calculated as

E[X\mid X>d]=\frac{\int_d^\infty x f_X(x)\,dx}{S_X(d)},

when the numerator is finite. This formula is simply the continuous expectation restricted to the conditioning event, divided by that event's probability.

Worked Example: Exponential Severity Above a Reporting Level

Suppose a hypothetical loss \(X\), measured in thousands of dollars, has an exponential distribution with rate \(\lambda=0.02\) per thousand dollars. Find (a) \(P(X>100\mid X>25)\), (b) \(E[X\mid X>25]\), and (c) the 0.90 conditional quantile. Round monetary amounts to two decimals in thousands of dollars and probabilities to four decimals.

Model and method. Under this exponential model, \(S_X(x)=e^{-\lambda x}\) for \(x\ge0\). Use the conditional survival ratio for the probability. For the mean and quantile, first obtain the conditional distribution.

P(X>x\mid X>25)=\frac{e^{-0.02x}}{e^{-0.02(25)}} =e^{-0.02(x-25)},\qquad x\ge25.

This is the survival function of \(25+Z\), where \(Z\) is exponential with rate 0.02. Therefore the conditional distribution of \(X\) is a shift by 25 of an exponential random variable with mean \(1/0.02=50\) thousand dollars.

(a) Conditional probability. The difference between 100 and 25 is 75 thousand dollars:

P(X>100\mid X>25)=e^{-0.02(100-25)} =e^{-1.5}=0.2231.

(b) Conditional mean. The shifted exponential has mean \(25+50=75\) thousand dollars:

E[X\mid X>25]=25+\frac{1}{0.02}=25+50=75.00.

(c) Conditional 0.90 quantile. Set the conditional cumulative probability equal to 0.90. Equivalently, set the conditional survival probability equal to \(1-0.90=0.10\):

e^{-0.02(q_{0.90}-25)}=0.10, \qquad q_{0.90}=25-\frac{\ln(0.10)}{0.02} =140.13.

The answers are 0.2231, 75.00 thousand dollars, and 140.13 thousand dollars. As a check, the quantile exceeds the threshold, and substituting it into the conditional survival function gives 0.10. A common mistake is to calculate \(e^{-0.02(100)}\) for part (a); that is the unconditional probability \(P(X>100)\), not the probability conditional on already exceeding 25.

Pareto Type I Above a Threshold

A Pareto Type I severity has scale parameter \(\theta>0\), shape parameter \(\alpha>0\), and support \(x\ge\theta\). Its survival function is \(S_X(x)=(\theta/x)^\alpha\) for \(x\ge\theta\). For a threshold \(d\ge\theta\), the conditional survival function follows directly from the ratio formula:

P(X>x\mid X>d) =\frac{(\theta/x)^\alpha}{(\theta/d)^\alpha} =\left(\frac{d}{x}\right)^\alpha,\qquad x\ge d.

Thus, conditional on exceeding \(d\), the severity has a Pareto Type I form with scale \(d\) and the same shape \(\alpha\). To obtain its mean, use the conditional density \(\alpha d^\alpha x^{-\alpha-1}\) for \(x>d\):

E[X\mid X>d] =\int_d^\infty x\alpha d^\alpha x^{-\alpha-1}\,dx =\alpha d^\alpha\int_d^\infty x^{-\alpha}\,dx =\frac{\alpha d}{\alpha-1},\qquad \alpha>1.

The condition \(\alpha>1\) is necessary for a finite conditional mean. If \(0<\alpha\le1\), the integral diverges, so the conditional mean is infinite. The conditional \(p\)-quantile, for \(0<p<1\), solves \(1-(d/x)^\alpha=p\), giving \(q_p=d/(1-p)^{1/\alpha}\).

Worked Example: Pareto Conditional Probability, Mean, and Quantile

A hypothetical portfolio severity in thousands of dollars follows a Pareto Type I model with scale \(\theta=10\) and shape \(\alpha=3\). Among losses exceeding \(d=20\), calculate (a) the probability of exceeding 40, (b) the conditional mean, and (c) the 0.90 conditional quantile. Round monetary amounts to two decimals in thousands of dollars and the probability to four decimals.

Model check. Since \(20\ge10\), the threshold lies within the Pareto support. Also, \(\alpha=3>1\), so the conditional mean is finite.

(a) Conditional probability. Apply the ratio of survival probabilities:

P(X>40\mid X>20) =\frac{(10/40)^3}{(10/20)^3} =\left(\frac{20}{40}\right)^3 =\frac18=0.1250.

(b) Conditional mean. The conditional Pareto scale is \(d=20\), with shape 3:

E[X\mid X>20]=\frac{3(20)}{3-1} =\frac{60}{2}=30.00.

(c) Conditional quantile. Set the conditional survival probability equal to 0.10:

\left(\frac{20}{q_{0.90}}\right)^3=0.10, \qquad q_{0.90}=\frac{20}{(0.10)^{1/3}} =20(10)^{1/3}=43.09.

The conditional probability is 0.1250, the mean is 30.00 thousand dollars, and the quantile is 43.09 thousand dollars. The probability is plausible because 40 is twice the conditional scale 20, producing a survival probability of \((1/2)^3\). A common mistake is to use the original scale 10 in the conditional quantile formula; once conditioning on \(X>20\), the conditional survival ratio has scale 20.

What Conditioning Changes—and What It Does Not

Conditioning changes the probability distribution by restricting attention to losses above the threshold and renormalizing their probabilities. It does not subtract the threshold from each loss. For example, \(E[X\mid X>d]\) is the mean loss amount among selected losses, whereas a mean calculated for \(X-d\) would describe a different quantity.

This distinction is useful when interpreting severity data that are observed only after crossing a reporting level. It also prevents confusion with contract payment variables: a conditional severity describes the original loss given an event, while a payment variable describes an amount determined by a coverage rule. Those are different random variables and require different calculations.

Identify the event.
Write the condition exactly, including whether it is \(X>d\) or \(X\ge d\).

Find its probability.
Use \(P(X>d)=1-F_X(d)\), checking for a mass at the threshold when the model is discrete.

Condition and normalize.
Divide probabilities or the restricted density by the probability of the conditioning event.

Calculate and check.
Use the conditional distribution for the requested probability, mean, or quantile; verify that probabilities are valid and quantiles exceed the threshold.

The most reliable exam habit is to write the conditioning event before substituting into a formula. For discrete severities, list the retained support values explicitly. For continuous severities, use the survival ratio or normalized density and check that the threshold is in the model's support. These steps make strict inequalities and denominators much harder to misread.

Key takeaway: For a strict threshold, normalize by \(P(X>d)\), exclude any atom at \(d\), and calculate with the conditional distribution of \(X\) itself. The continuous survival ratio \(S_X(x)/S_X(d)\) is especially efficient for conditional exceedance probabilities.

Check Your Understanding

For each problem, apply the strict condition shown and keep monetary units consistent.

A discrete severity takes values 1, 4, and 8 with probabilities 0.10, 0.60, and 0.30, respectively. Given \(X>3\), find the conditional mean and the conditional probability that \(X=8\). (Answer: mean \(16/3=5.3333\); probability \(1/3=0.3333\).)
A loss in thousands of dollars is exponential with rate \(\lambda=0.05\) per thousand dollars. Given \(X>10\), find \(P(X>30\mid X>10)\) and the conditional mean. (Answer: probability \(e^{-1}=0.3679\); mean \(30.00\) thousand dollars.)
A Pareto Type I loss has scale \(\theta=5\) and shape \(\alpha=2\). Given \(X>10\), find \(P(X>20\mid X>10)\) and the conditional mean. (Answer: probability \(0.2500\); mean \(20.00\).)
A loss is uniformly distributed on \([0,100]\), with amounts in thousands of dollars. Given \(X>40\), find the conditional mean and \(P(X>80\mid X>40)\). (Answer: mean \(70.00\) thousand dollars; probability \(1/3=0.3333\).)
A discrete severity has \(P(X=3)=0.25\), \(P(X=6)=0.35\), and \(P(X=10)=0.40\). Given \(X>6\), find the conditioning-event probability, the conditional mean, and the 0.80 conditional quantile. (Answer: event probability \(0.40\); mean \(10\); 0.80 quantile \(10\).)

What You'll Learn