Exponential Severity And Constant Hazard

Introduction

The exponential distribution is a useful model for a nonnegative claim severity when the probability of surviving an additional amount of loss does not depend on how much loss has already occurred. This tutorial focuses on the model’s defining rate and the consequences of that definition: its density, survival function, moments, quantiles, and hazard.

An exponential model is determined by one positive parameter. Once its rate is known, the entire distribution follows. Conversely, a reported mean, percentile, or tail probability can be used to identify that rate, provided the exponential assumption is appropriate. The calculations below use hypothetical claim severities and state the units of the rate explicitly.

Definition: A nonnegative loss $X$ has an exponential distribution with rate $\lambda>0$, written $X\sim\operatorname{Exponential}(\lambda)$, if its cumulative distribution function is $$ F_X(x)= \begin{cases} 0, & x<0,\\ 1-e^{-\lambda x}, & x\ge 0. \end{cases} $$ The rate $\lambda$ is measured in inverse units of severity. For example, if $X$ is measured in dollars, $\lambda$ is measured per dollar.

Density and Survival Function

For $x>0$, the density is the derivative of the cumulative distribution function. At $x=0$, the distribution has no point mass; the displayed density formula also gives the right-hand density there.

f_X(x)=\frac{d}{dx}F_X(x)=\lambda e^{-\lambda x},\qquad x\ge 0.

The survival function gives the probability that the loss exceeds a specified amount. Since $S_X(x)=1-F_X(x)$, for $x\ge 0$,

S_X(x)=P(X>x)=e^{-\lambda x}.

These formulas satisfy the basic probability checks. The density is nonnegative, and its integral over the nonnegative support is $\int_0^\infty \lambda e^{-\lambda x}\,dx=1$. Also, $F_X(0)=0$, $F_X(x)$ increases toward 1, and $S_X(x)$ decreases from 1 toward 0.

Worked Example: Recovering the Distribution From the Mean

A hypothetical portfolio models claim severity $X$, in dollars, as exponential with mean $1,200. Find the rate, variance, median, and probability that a claim exceeds $2,400. Name the method and identify a common error to avoid.

Method: parameterize from the mean, then apply the survival and quantile formulas. For an exponential distribution, $E[X]=1/\lambda$. Thus $\lambda=1/1{,}200=0.000833333\ldots$ per dollar.

The variance is $1/\lambda^2$, so $\operatorname{Var}(X)=1/(0.000833333\ldots)^2=1{,}440{,}000$ dollars squared.

For the median $m$, the probability of exceeding $m$ is $0.5$. Hence $e^{-\lambda m}=0.5$, giving $m=\ln(2)/\lambda=1{,}200\ln(2)=831.7766\ldots$ dollars.

The exceedance probability at $2,400 is $S_X(2{,}400)=e^{-(0.000833333\ldots)(2{,}400)}=e^{-2}=0.135335\ldots$. Rounded at the final step, the rate is 0.0008333 per dollar, the variance is 1,440,000 dollars squared, the median is $831.78, and the exceedance probability is 0.1353.

Check: The median is below the mean, as is expected for an exponential distribution, and the tail probability is positive and less than 1. A common error is to use the mean as the rate; the mean is $1/\lambda$, not $\lambda$.

Mean and Variance

The mean can be obtained directly from the density:

E[X]=\int_0^\infty x\lambda e^{-\lambda x}\,dx=\frac{1}{\lambda}.

For the second moment, integration by parts gives $E[X^2]=2/\lambda^2$. Therefore, $\operatorname{Var}(X)=E[X^2]-(E[X])^2=2/\lambda^2-1/\lambda^2=1/\lambda^2$. The standard deviation is also $1/\lambda$, so the mean and standard deviation are equal. The units provide a useful check: the mean has severity units, while the variance has squared-severity units.

Quantiles and Recovering the Rate

For $0<p<1$, the $p$-quantile $x_p$ solves $F_X(x_p)=p$. Substituting the exponential CDF and rearranging gives

p=1-e^{-\lambda x_p} \quad\Longrightarrow\quad x_p=\frac{-\ln(1-p)}{\lambda}.

The same relation can be rearranged to identify $\lambda$ from a quantile: $\lambda=-\ln(1-p)/x_p$. From a survival probability $S_X(a)=s$, where $a\ge 0$ and $0<s<1$, the corresponding rate is $\lambda=-\ln(s)/a$, provided $a>0$. Always check that the logarithm’s argument and the resulting rate are valid.

Worked Example: Using a Tail Probability to Find the Rate

A hypothetical severity model is exponential, and the probability that a loss exceeds $400 is 0.6. Find the rate, mean, 90th percentile, and hazard. State why the tail information determines the rate.

Method: solve the survival equation for the rate. The given statement is $S_X(400)=0.6$, so $e^{-400\lambda}=0.6$. Taking natural logarithms gives $\lambda=-\ln(0.6)/400=0.001277064\ldots$ per dollar.

The mean is the reciprocal of the rate: $E[X]=1/\lambda=400/[-\ln(0.6)]=783.0461\ldots$ dollars, or $783.05.

For the 90th percentile $x_{0.9}$, the survival probability is $1-0.9=0.1$. Therefore, $x_{0.9}=-\ln(0.1)/\lambda=400\ln(10)/[-\ln(0.6)]=1{,}803.0303\ldots$ dollars, or $1,803.03.

The hazard, derived below, is constant and equal to $\lambda$, or 0.001277 per dollar to four significant figures. The tail information determines the rate because the exponential survival function at 400 is $e^{-400\lambda}$, which is one-to-one in $\lambda$.

Check: The 90th percentile exceeds $400, consistent with only 60% of losses exceeding $400. Do not substitute $0.6$ as the probability below $400: the given value is the survival probability.

Constant Hazard

For a continuous nonnegative severity with density $f_X(x)$ and positive survival function $S_X(x)$, the hazard at $x$ is the instantaneous rate of crossing severity level $x$, conditional on having exceeded that level. It is defined by

h_X(x)=\frac{f_X(x)}{S_X(x)}.

For the exponential model, substitution gives

h_X(x)=\frac{\lambda e^{-\lambda x}}{e^{-\lambda x}}=\lambda,\qquad x\ge 0.

Thus the hazard is constant over the severity scale. Its units are inverse severity units, not probability units. A hazard of 0.001 per dollar is not a probability of 0.001 of exceeding any particular dollar amount. Probabilities over a finite interval must be calculated from the distribution.

Derivation: For a continuous distribution with differentiable positive survival function, $S_X(x)=1-F_X(x)$, so $S_X'(x)=-f_X(x)$. Consequently, $$ h_X(x)=\frac{f_X(x)}{S_X(x)} =-\frac{d}{dx}\ln S_X(x). $$ If the hazard is the constant $\lambda$, then $\frac{d}{dx}\ln S_X(x)=-\lambda$. Integrating from 0 to $x$, and using $S_X(0)=1$, gives $\ln S_X(x)=-\lambda x$ and hence $S_X(x)=e^{-\lambda x}$. This shows why a constant hazard leads to the exponential survival function.

Worked Example: From the Median to Other Distribution Quantities

A hypothetical equipment-repair severity is exponential, and its median is $1,000. Find the rate, mean, 75th percentile, probability of exceeding $3,000, and hazard. Explain the check on the resulting tail probability.

Method: use the median as a quantile to recover the rate. The median condition is $F_X(1{,}000)=0.5$, equivalently $S_X(1{,}000)=0.5$. Thus $\lambda=\ln(2)/1{,}000=0.000693147\ldots$ per dollar.

The mean is $1/\lambda=1{,}000/\ln(2)=1{,}442.6950\ldots$ dollars, or $1,442.70.

For the 75th percentile, the survival probability is 0.25. Therefore, $x_{0.75}=-\ln(0.25)/\lambda=\ln(4)/\lambda=2{,}000$ dollars, since $\ln(4)=2\ln(2)$.

The probability of exceeding $3,000 is $e^{-3{,}000\lambda}=e^{-3\ln(2)}=1/8=0.1250$. The hazard is $\lambda=0.0006931$ per dollar to four significant figures.

Check: The 75th percentile is twice the median, and the probability of exceeding three times the median is one eighth. Both follow from the displayed formulas. The hazard is a rate per dollar, whereas 0.1250 is a probability.

When the Model Is Useful—and When to Be Careful

The exponential model is mathematically simple: one parameter determines the whole distribution, and its constant hazard gives a compact description of how the tail declines. That simplicity also imposes a strong shape restriction. For example, its mean and standard deviation must be equal, and its quantiles have fixed ratios. Data that do not support those features may require another severity model.

Before using an exponential formula, confirm that the variable is a nonnegative continuous severity, the rate is positive, and all amounts use consistent units. If severity is measured in thousands of dollars, a rate stated per dollar must be converted before combining it with that severity scale. Also distinguish a density value from a probability, and a hazard rate from a finite-interval probability.

Rate versus mean: $\lambda$ is the reciprocal of the mean, not the mean itself.
Tail versus cumulative probability: $S_X(x)=P(X>x)$, while $F_X(x)=P(X\le x)$.
Quantile versus tail level: for a $p$-quantile, use $1-p$ inside the survival equation.
Units: if $X$ is in dollars, $\lambda$ is per dollar and $\operatorname{Var}(X)$ is in dollars squared.
Hazard versus probability: $h_X(x)=\lambda$ has inverse-severity units; calculate actual probabilities with $F_X$ or $S_X$.

Key takeaway: For $X\sim\operatorname{Exponential}(\lambda)$, $f_X(x)=\lambda e^{-\lambda x}$, $S_X(x)=e^{-\lambda x}$, $E[X]=1/\lambda$, $\operatorname{Var}(X)=1/\lambda^2$, $x_p=-\ln(1-p)/\lambda$, and $h_X(x)=\lambda$. These results are aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus.

Check Your Understanding

For each hypothetical severity, identify the rate carefully and keep the stated dollar units throughout.

A claim severity is exponential with mean $2,500. Find its rate and the probability of exceeding $5,000. (Answer: $\lambda=0.0004$ per dollar; $P(X>5{,}000)=e^{-2}=0.1353$.)
An exponential severity has $S_X(600)=0.75$. Find its rate and mean. (Answer: $\lambda=-\ln(0.75)/600=0.0004795$ per dollar; $E[X]=600/[-\ln(0.75)]=\$2{,}085.64$.)
The 90th percentile of an exponential loss is $2,400. Find the rate and mean. (Answer: $\lambda=\ln(10)/2{,}400=0.0009594$ per dollar; $E[X]=2{,}400/\ln(10)=\$1{,}042.31$.)
An exponential repair cost has median $700. Find the probability that it exceeds $2,100. (Answer: $\lambda=\ln(2)/700$ per dollar and $P(X>2{,}100)=e^{-3\ln(2)}=0.1250$.)
A severity model has constant hazard 0.002 per dollar. Find its mean and the probability that severity exceeds $1,000. (Answer: $E[X]=1/0.002=\$500.00$; $P(X>1{,}000)=e^{-2}=0.1353$.)

What You'll Learn