Exponential Severity: Tail Probabilities and Moments

Introduction

The exponential distribution is a useful severity model when claim amounts are nonnegative and decline smoothly in probability as the amount increases. Its formulas are compact, but exam calculations can still go wrong if the rate and scale are confused, or if a percentile is mistaken for a tail probability. This tutorial develops the model’s tail probabilities, quantiles, mean, and variance, with units kept explicit throughout.

The focus is on calculations from the exponential model itself, not on deciding when a real portfolio is well represented by that model. The results and problem-solving methods here are aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus.

Definition and Parameter Conventions

Let $X$ denote a hypothetical claim severity, measured in a specified unit such as dollars. An exponential severity has a rate parameter $\lambda>0$ and support $x\geq 0$. The rate has units of inverse severity: if $X$ is measured in dollars, $\lambda$ is measured per dollar.

Definition: If $X$ has an exponential distribution with rate $\lambda$, its density, cumulative distribution function, and survival function are $$ f_X(x)=\lambda e^{-\lambda x},\qquad F_X(x)=1-e^{-\lambda x},\qquad S_X(x)=P(X>x)=e^{-\lambda x},\qquad x\geq 0. $$ For $x<0$, $F_X(x)=0$ and $S_X(x)=1$. An equivalent scale parameterization uses $\theta=1/\lambda$, so that $f_X(x)=\frac{1}{\theta}e^{-x/\theta}$ for $x\geq 0$. The rate is in inverse severity units; the scale is in severity units.

The density integrates to one because $\int_0^\infty \lambda e^{-\lambda x}\,dx=1$. The CDF and survival function are complements: $F_X(x)+S_X(x)=1$ for $x\geq0$. In particular, $P(X>x)$ is a strict upper-tail probability, while $P(X\leq x)=F_X(x)$. Since the exponential model is continuous, $P(X=x)=0$; consequently, changing a strict inequality to a non-strict one at a single endpoint does not change the probability.

The two parameterizations describe the same model, not different distributions. Under the scale convention, $\theta$ is not a rate. A useful first check is therefore dimensional: multiplying $\lambda$ by an amount gives a dimensionless exponent, while dividing an amount by $\theta$ does the same.

Tail Probabilities and Quantiles

For a nonnegative threshold $x$, the survival function gives the probability that a severity exceeds that threshold. An interval probability follows by subtracting CDF values, or equivalently by subtracting survival probabilities in the reverse order. These formulas are often more direct than integrating the density separately.

Result: For $0\leq a<b$, and for a cumulative percentile $p$ with $0<p<1$, $$ P(X>x)=e^{-\lambda x},\qquad P(a<X\leq b)=e^{-\lambda a}-e^{-\lambda b},\qquad x_p=F_X^{-1}(p)=\frac{-\ln(1-p)}{\lambda}=-\theta\ln(1-p). $$ Here $x_p$ is the value with $P(X\leq x_p)=p$. Its upper-tail probability is $1-p$, not $p$.

The quantile formula follows by solving the CDF equation $F_X(x_p)=p$: $1-e^{-\lambda x_p}=p$, so $e^{-\lambda x_p}=1-p$. Taking logarithms gives $-\lambda x_p=\ln(1-p)$, hence $x_p=-\ln(1-p)/\lambda$. The logarithm has a positive argument between zero and one, so the resulting quantile is positive.

Worked Example: Combining a Tail, an Interval, and a Percentile

A hypothetical travel insurer models a covered repair severity $X$, in dollars, by an exponential distribution with rate $\lambda=0.0004$ per dollar. Calculate (1) the probability a repair exceeds $4,000, (2) the probability it is greater than $1,000 and at most $4,000, and (3) the 90th percentile. State the model and retain full precision until the final rounding.

Method and model. The assumed density is $f_X(x)=0.0004e^{-0.0004x}$ for $x\geq0$. Use the survival function for the tail, the difference of survival values for the interval, and the CDF quantile formula for the percentile. The rate is per dollar, so each exponent uses a dollar amount multiplied by $0.0004$ per dollar.

(1) Exceeding $4,000. $$ P(X>4{,}000)=e^{-(0.0004)(4{,}000)} =e^{-1.6} =0.2018965\ldots $$ Rounded to four decimal places, the probability is $0.2019$.

(2) Between $1,000 and $4,000. $$ \begin{aligned} P(1{,}000<X\leq4{,}000) &=S_X(1{,}000)-S_X(4{,}000)\\ &=e^{-(0.0004)(1{,}000)}-e^{-(0.0004)(4{,}000)}\\ &=e^{-0.4}-e^{-1.6}\\ &=0.6703200\ldots-0.2018965\ldots\\ &=0.4684235\ldots \end{aligned} $$ Rounded to four decimal places, the interval probability is $0.4684$.

(3) 90th percentile. For $p=0.90$, the upper-tail probability is $1-p=0.10$. $$ x_{0.90}=\frac{-\ln(1-0.90)}{0.0004} =\frac{-\ln(0.10)}{0.0004} =5{,}756.4627\ldots $$ The 90th percentile is $5,756.46, rounded to the nearest cent. As a check, its survival probability is $e^{-0.0004(5{,}756.4627\ldots)}=0.10$, so its CDF value is $0.90$.

Common mistake avoided: Subtracting $S_X(4{,}000)$ from $S_X(1{,}000)$ gives the probability in the specified interval. Reversing that subtraction would produce a negative number, which cannot be a probability.

Mean and Variance

For an exponential severity with rate $\lambda>0$, the first two raw moments yield its mean and variance. The derivation below also makes clear why the scale parameter equals the mean.

Result: For $X$ with density $f_X(x)=\lambda e^{-\lambda x}$ on $x\geq0$, $$ E[X]=\frac{1}{\lambda}=\theta,\qquad E[X^2]=\frac{2}{\lambda^2},\qquad \operatorname{Var}(X)=\frac{1}{\lambda^2}=\theta^2. $$ These moment formulas apply to the ordinary, untruncated exponential severity model with positive rate.

To justify the mean, use the density definition of expectation. Integration by parts, with $u=x$ and $dv=\lambda e^{-\lambda x}dx$, gives $du=dx$ and $v=-e^{-\lambda x}$. The boundary term $[-xe^{-\lambda x}]_0^\infty$ is zero, since $xe^{-\lambda x}$ tends to zero as $x$ grows. Thus:

\begin{aligned} E[X] &=\int_0^\infty x\lambda e^{-\lambda x}\,dx\\ &=[-xe^{-\lambda x}]_0^\infty+\int_0^\infty e^{-\lambda x}\,dx\\ &=0+\left[-\frac{1}{\lambda}e^{-\lambda x}\right]_0^\infty =\frac{1}{\lambda}. \end{aligned}

For the second raw moment, integration by parts with $u=x^2$ and $dv=\lambda e^{-\lambda x}dx$ gives a zero boundary term and $E[X^2]=2\int_0^\infty xe^{-\lambda x}dx$. The mean integral just evaluated is $\int_0^\infty xe^{-\lambda x}dx=1/\lambda^2$, so $E[X^2]=2/\lambda^2$. By the definition of variance, subtract the square of the mean:

\operatorname{Var}(X)=E[X^2]-(E[X])^2 =\frac{2}{\lambda^2}-\left(\frac{1}{\lambda}\right)^2 =\frac{1}{\lambda^2}.

In the scale parameterization, substituting $\lambda=1/\theta$ gives $E[X]=\theta$ and $\operatorname{Var}(X)=\theta^2$. The variance is in squared severity units: if $X$ is in dollars, its variance is in dollars squared. The standard deviation is the positive square root of the variance, $1/\lambda=\theta$, and is therefore in dollars.

Worked Example: Reading Results From a Scale Parameter

A hypothetical equipment-protection plan models repair severity $X$, in dollars, as exponential with scale $\theta=\$6{,}000$. Calculate the rate, mean, variance, standard deviation, probability of exceeding $9,000, and 80th percentile.

Method and model. The assumed density is $f_X(x)=\frac{1}{6{,}000}e^{-x/6{,}000}$ for $x\geq0$. Convert the scale to a rate before using rate-based formulas, while keeping the equivalent scale formulas available for checking.

Parameter conversion. The rate is the reciprocal of the scale: $$ \lambda=\frac{1}{\theta}=\frac{1}{6{,}000} =0.0001666667\ldots\text{ per dollar}. $$

Moments. The mean is $E[X]=\theta=\$6{,}000.00$. The variance and standard deviation are $$ \operatorname{Var}(X)=\theta^2=(6{,}000)^2=36{,}000{,}000\text{ dollars}^2,\qquad \operatorname{SD}(X)=\sqrt{36{,}000{,}000}=\$6{,}000.00. $$

Exceeding $9,000. $$ P(X>9{,}000)=e^{-9{,}000/6{,}000}=e^{-1.5} =0.2231302\ldots\approx0.2231. $$

80th percentile. The tail probability at this percentile is $1-0.80=0.20$. $$ x_{0.80}=-6{,}000\ln(0.20) =6{,}000\ln(5) =9{,}656.6275\ldots $$ The 80th percentile is $9,656.63, rounded to the nearest cent. Checking, $e^{-9{,}656.6275\ldots/6{,}000}=0.20$, which implies a CDF value of $0.80$.

Common mistake avoided: A scale of $6,000 is not a rate of 6,000 per dollar. Its reciprocal, $1/6{,}000$ per dollar, is the rate. The moment formulas also confirm the interpretation: the mean equals the scale, not the rate.

Inferring Parameters From a Percentile

A percentile can identify the rate when the exponential model is assumed and the percentile level is known. The equation must be written using the percentile’s CDF probability. Once the rate is found, the same parameter determines all other percentiles and moments.

Worked Example: Recovering the Rate From the 25th Percentile

A hypothetical cargo insurer uses an exponential severity model for a covered loss $X$, measured in dollars. The model’s 25th percentile is $2,000. Infer the rate and mean, then calculate the 75th percentile. Round monetary results to the nearest cent and the rate to four significant figures.

Method and model. Assume $X$ has density $f_X(x)=\lambda e^{-\lambda x}$ on $x\geq0$, with an unknown positive rate per dollar. At the 25th percentile, $F_X(2{,}000)=0.25$, so the corresponding survival probability is $0.75$.

Infer the rate. $$ \begin{aligned} 0.25&=1-e^{-2{,}000\lambda},\\ e^{-2{,}000\lambda}&=0.75,\\ \lambda&=\frac{-\ln(0.75)}{2{,}000} =0.0001438410\ldots\text{ per dollar} \approx0.0001438\text{ per dollar}. \end{aligned} $$

Find the mean. Since $E[X]=1/\lambda$, $$ E[X]=\frac{1}{0.0001438410\ldots} =\frac{2{,}000}{-\ln(0.75)} =6{,}952.119\ldots. $$ The mean is $6,952.12, rounded to the nearest cent.

Find the 75th percentile. The 75th percentile has CDF probability $0.75$, so its survival probability is $0.25$: $$ \begin{aligned} x_{0.75} &=\frac{-\ln(0.25)}{\lambda}\\ &=\frac{2{,}000\ln(4)}{-\ln(0.75)}\\ &=9{,}637.683\ldots. \end{aligned} $$ The 75th percentile is $9,637.68, rounded to the nearest cent. The check is $e^{-\lambda x_{0.75}}=0.25$, giving $F_X(x_{0.75})=0.75$.

Common mistake avoided: Setting the survival probability equal to $0.75$ again would give $x=-\ln(0.75)/\lambda=\$2{,}000$, which is the 25th percentile, not the 75th. A $p$th percentile uses CDF probability $p$ and tail probability $1-p$.

Common Exam Traps and Checks

Exponential calculations are short enough that a parameter or probability convention error can pass unnoticed unless the result is checked against its meaning. Before finalizing a calculation, confirm the following:

Rate and scale have opposite units. A rate in inverse dollars multiplies a dollar amount in $e^{-\lambda x}$. A scale in dollars is divided into the amount in $e^{-x/\theta}$.
A percentile is a CDF statement. The $p$th percentile satisfies $F_X(x_p)=p$ and $S_X(x_p)=1-p$. Do not set the tail equal to $p$ unless the question explicitly asks for an upper-tail probability $p$.
Interval probabilities are differences. For $a<b$, $P(a<X\leq b)=F_X(b)-F_X(a)=S_X(a)-S_X(b)$, which must be nonnegative.
Units of moments differ. The mean and standard deviation use the units of $X$; variance uses squared units. A dollar-valued variance is dimensionally incorrect.
Check limiting behavior. At $x=0$, $F_X(0)=0$ and $S_X(0)=1$; as $x$ increases, $F_X(x)$ rises and $S_X(x)$ falls. A computed tail probability above one or below zero signals an error.

The model’s simple formulas are useful precisely because a single parameter controls the tail and both moments. That connection makes internal checks possible: a rate of $\lambda$ implies mean and standard deviation $1/\lambda$, while the scale $1/\lambda$ gives the same two quantities directly.

Check Your Understanding

For each hypothetical model, state units before substituting into a formula. Round probabilities to four decimal places and monetary answers to the nearest cent where relevant.

A severity $X$ is measured in thousands of dollars and has exponential rate $0.2$ per thousand dollars. Find its mean in dollars and $P(X>\$3{,}000)$. (Answer: mean $5,000.00; probability $0.5488$.)
A hypothetical repair severity is exponential with scale $\theta=\$4{,}000$. Find $P(\$1{,}000<X\leq\$5{,}000)$ and its 90th percentile. (Answer: probability $0.4923$; 90th percentile $9,210.34.)
An exponential severity has 40th percentile $1,200. Find its rate per dollar and its mean. (Answer: $\lambda=-\ln(0.60)/1{,}200\approx0.0004257$ per dollar; mean approximately $2,349.14.)
A hypothetical loss severity is exponential with rate $0.5$ per thousand dollars. Find its variance in dollars squared, its probability of exceeding $2,000, and its 95th percentile in dollars. (Answer: variance $4{,}000{,}000$ dollars squared; probability $0.3679$; 95th percentile $5,991.46.)
A hypothetical severity has exponential mean $9,000. Find its standard deviation and the probability that it is greater than $3,000 but no more than $12,000. (Answer: standard deviation $9,000.00; probability $e^{-1/3}-e^{-4/3}\approx0.4529$.)

What You'll Learn