Exponential Severity Models

Introduction

The exponential distribution is a basic continuous model for a nonnegative loss. Its simple formulas make it useful for practicing the main calculations with a severity distribution: probabilities, moments, and quantiles. The parameter may be reported as a rate or as a scale, so a correct calculation begins by identifying which convention is being used and what units go with it.

This tutorial considers a single hypothetical loss $X$, measured in a stated unit such as dollars or thousands of dollars. It does not assume that every real loss pattern follows an exponential distribution; it shows how to calculate accurately when that model is specified.

Definition and Parameterizations

Definition: A loss $X$ has an exponential distribution with rate $\lambda>0$ if it is continuous, takes values $x\ge 0$, and has density $f_X(x)=\lambda e^{-\lambda x}$ for $x\ge 0$, with $f_X(x)=0$ for $x<0$. The rate $\lambda$ is measured in inverse loss units. The equivalent scale parameter is $\theta=1/\lambda$, measured in loss units; in that parameterization, $f_X(x)=\frac{1}{\theta}e^{-x/\theta}$ for $x\ge 0$.

The two parameterizations describe the same family of distributions, not two different models. For example, a rate of $0.002$ per dollar corresponds to a scale of $500$ dollars. A rate is larger when the typical loss is smaller; a scale is larger when the typical loss is larger. Always carry the units through the calculation: if $X$ is measured in thousands of dollars, then $\lambda$ is per thousand dollars and $\theta$ is in thousands of dollars.

\lambda=\frac{1}{\theta} \qquad\qquad f_X(x)=\lambda e^{-\lambda x} =\frac{1}{\theta}e^{-x/\theta},\quad x\ge 0

There is no probability mass at $X=0$; the model is continuous. The density value $f_X(0)=\lambda$ is not the probability of a zero loss. The density and survival function determine probabilities over intervals and beyond thresholds, respectively.

Distribution and Survival Functions

For $x<0$, the cumulative distribution function is zero, because the loss cannot be negative. For $x\ge 0$, integrate the density from zero to $x$:

F_X(x)=P(X\le x)=\int_0^x \lambda e^{-\lambda t}\,dt =\left[-e^{-\lambda t}\right]_0^x =1-e^{-\lambda x}.

The survival function is $S_X(x)=P(X>x)=1-F_X(x)$. Thus, for $x\ge 0$, it has a particularly convenient exponential form. Because the distribution is continuous, using $X\ge x$ instead of $X>x$ gives the same probability.

F_X(x)= \begin{cases} 0, & x<0,\\ 1-e^{-\lambda x}, & x\ge 0, \end{cases} \qquad S_X(x)=e^{-\lambda x},\quad x\ge 0.

For $0\le a<b$, subtract cumulative probabilities to obtain an interval probability. Equivalently, subtract the survival probability at the upper endpoint from that at the lower endpoint. The two expressions agree:

P(a<X\le b)=F_X(b)-F_X(a) =e^{-\lambda a}-e^{-\lambda b}.

Moments and Quantiles

For an exponential loss with rate $\lambda$, its raw moment of positive integer order $k$ is $E[X^k]=k!/\lambda^k$. In scale notation this is $k!\theta^k$. In particular, the mean is $\theta=1/\lambda$, and the variance is $\theta^2=1/\lambda^2$. The second raw moment is $E[X^2]=2\theta^2$, which is not the same as the variance.

E[X^k]=\frac{k!}{\lambda^k}=k!\theta^k,\qquad E[X]=\frac{1}{\lambda}=\theta,\qquad \operatorname{Var}(X)=\frac{1}{\lambda^2}=\theta^2.

A $p$-quantile $q_p$, for $0<p<1$, satisfies $F_X(q_p)=p$. Solve this equation for $q_p$. The logarithm is the natural logarithm. Quantiles are in the same loss units as $X$; $q_0=0$, while the distribution has no finite $q_1$.

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

Calculation guide: Identify the parameterization and loss unit first. Use $F_X(x)=1-e^{-\lambda x}$ for a probability at or below a threshold, $S_X(x)=e^{-\lambda x}$ for an upper tail, and $q_p=-\ln(1-p)/\lambda$ for a quantile. Convert a scale to a rate, or vice versa, before substituting if necessary.

Worked Calculations

Worked Example: Interval Probability and Moments in Dollars

A hypothetical insurer models the amount of an individual repair claim by $X\sim\operatorname{Exp}(\lambda)$, with rate $\lambda=0.002$ per dollar. Calculate the probability that a claim is between $200 and $700, then find the mean and variance. State the variance in squared dollars.

Method and parameter check. The loss is measured in dollars, so the rate is $0.002$ per dollar. The corresponding scale is $\theta=1/0.002=500$ dollars. Use the exponential interval formula and moment formulas.

Interval probability. The exponents are $\lambda(200)=0.002(200)=0.4$ and $\lambda(700)=0.002(700)=1.4$. Therefore,

P(200<X\le 700)=e^{-0.4}-e^{-1.4} =0.6703200-0.2465970 =0.4237230.

The probability to four decimal places is $0.4237$.

Mean and variance. The mean is $1/\lambda=1/0.002=500$ dollars. The variance is $1/\lambda^2=1/(0.002)^2=250{,}000$ squared dollars. As a check, the variance also equals $\theta^2=500^2=250{,}000$ squared dollars.

Common mistake avoided: The interval probability is not $F_X(700)$; that would include every loss from zero through $700. Subtracting $F_X(200)$, or equivalently using the difference of survival probabilities, isolates the requested interval. The calculated probability is positive and less than one, as required.

Worked Example: Scale Parameter and a High Quantile

A hypothetical equipment-loss model specifies $X\sim\operatorname{Exp}(\theta)$, with scale $\theta=2.5$ thousand dollars. Find the probability that a loss exceeds $4,000, the mean loss, and the 90th percentile. Give monetary answers in thousands of dollars.

Method and unit check. Write losses in thousands of dollars: the threshold is $4$, and $\theta=2.5$ thousand dollars. The equivalent rate is $\lambda=1/2.5=0.4$ per thousand dollars.

Upper-tail probability. Apply the survival function at $x=4$:

P(X>4)=e^{-0.4(4)}=e^{-1.6}=0.2018965\approx 0.2019.

Mean. The mean equals the scale, so $E[X]=2.5$ thousand dollars.

90th percentile. Set $p=0.90$ in the quantile formula:

q_{0.90}=-2.5\ln(1-0.90) =-2.5\ln(0.10) =5.7564627\text{ thousand dollars}.

To the nearest cent, this is $5,756.46. As a check, $\lambda q_{0.90}=0.4(5.7564627)=2.3025851=-\ln(0.10)$, so the model gives $P(X\le q_{0.90})=0.90$.

Common mistake avoided: The scale is in thousands, so inserting a dollar threshold of $4{,}000$ directly into $e^{-x/\theta}$ would mix units. Either use $x=4$ and $\theta=2.5$ in thousands, or convert the scale to $2,500 and use $x=4{,}000$ in dollars.

Worked Example: Recovering the Rate from a Tail Probability

A hypothetical loss model is exponential with an unknown rate per dollar. The probability of a loss exceeding $600 is $0.25$. Determine the rate, mean, probability the loss is at most $300, and 95th percentile.

Method. Use the specified survival probability to solve for the rate, then use that rate consistently in the other formulas.

Rate. Since $S_X(600)=e^{-600\lambda}=0.25$,

\lambda=-\frac{\ln(0.25)}{600} =\frac{\ln 4}{600} =0.00231049\text{ per dollar}.

Mean. The mean is the reciprocal of the rate: $E[X]=600/\ln 4=432.8085$ dollars, or $432.81 to the nearest cent.

Probability at most $300. Here $300\lambda=(\ln4)/2=\ln2$, so

P(X\le 300)=1-e^{-300\lambda} =1-e^{-\ln2} =1-\frac12 =0.5000.

95th percentile. Use $1-p=0.05$:

q_{0.95} =-\frac{\ln(0.05)}{\lambda} =\frac{600\ln20}{\ln4} =1{,}296.58\text{ dollars}.

Check and common mistake avoided: Substitution gives $S_X(600)=e^{-\ln4}=1/4$, matching the given probability. The equation uses the survival probability $0.25$, not the cumulative probability at $600, which is $0.75$.

Worked Example: Distinguishing the Second Moment from Variance

A hypothetical component-loss amount has distribution $X\sim\operatorname{Exp}(\lambda)$, where $\lambda=0.004$ per dollar. Find $E[X^2]$, $\operatorname{Var}(X)$, and $P(X>250)$. Use the raw-moment formula and verify the variance from the first two raw moments.

Second raw moment. For $k=2$, $E[X^2]=2/\lambda^2$. Thus,

E[X^2]=\frac{2}{(0.004)^2} =\frac{2}{0.000016} =125{,}000\text{ dollars}^2.

Variance check. The mean is $1/0.004=250$ dollars, so its square is $62{,}500$ squared dollars. Therefore,

\operatorname{Var}(X)=E[X^2]-(E[X])^2 =125{,}000-62{,}500 =62{,}500\text{ dollars}^2.

This agrees with $1/\lambda^2=62{,}500$ squared dollars. Finally, the upper-tail probability is $P(X>250)=e^{-0.004(250)}=e^{-1}=0.3679$ to four decimal places.

Common mistake avoided: $E[X^2]$ is not the variance. Subtracting the squared mean is necessary; the check confirms both calculations agree.

Common Exam Traps and Checks

Exponential calculations are short, but a small parameter or interpretation error can change the result substantially. Before finalizing an answer, check the following:

Rate versus scale: If the question gives a scale $\theta$, use $e^{-x/\theta}$ or first convert to $\lambda=1/\theta$. Do not use the scale as though it were a rate.
Units: The product $\lambda x$ must be dimensionless. A rate per dollar must be paired with a dollar amount; a rate per thousand dollars must be paired with an amount in thousands.
Tail versus cumulative probability: $P(X>x)=e^{-\lambda x}$, while $P(X\le x)=1-e^{-\lambda x}$. They are complements, not interchangeable expressions.
Raw second moment versus variance: Use $\operatorname{Var}(X)=E[X^2]-(E[X])^2$. For an exponential loss, $E[X^2]=2/\lambda^2$, whereas the variance is $1/\lambda^2$.
Quantile complement: For a $p$-quantile, use $-\ln(1-p)/\lambda$, not $-\ln(p)/\lambda$. The latter corresponds to a different tail probability.

Key takeaway: Once the exponential parameterization and units are clear, the density, survival function, moments, and quantiles follow from a small set of formulas. A useful final check is to substitute a calculated quantile back into $F_X$, or a calculated threshold back into $S_X$, and verify that it gives the requested probability.

Check Your Understanding

For each hypothetical loss model, identify the parameterization and units before calculating. Probabilities are rounded to four decimal places; monetary answers are rounded to the nearest cent unless stated otherwise.

A claim amount has exponential rate $0.005$ per dollar. Find $P(100<X\le 400)$ and $E[X]$. (Answer: $0.4712$; $200.00)
A loss has exponential scale $3$ thousand dollars. Find its variance in squared thousands of dollars and $P(X>6)$. (Answer: $9$ thousand-dollars squared; $0.1353$)
An exponential loss in dollars satisfies $P(X>500)=0.20$. Find its rate and mean. (Answer: $\lambda=\ln(5)/500=0.003219$ per dollar; mean $=\$310.67$)
A loss has exponential rate $0.0015$ per dollar. Find its 75th percentile and $P(X\le 200)$. (Answer: $924.20; $0.2592$)
A loss measured in thousands of dollars has exponential rate $0.25$ per thousand dollars. Find $E[X^2]$, $\operatorname{Var}(X)$, and $P(2<X\le 8)$. (Answer: $32$ thousand-dollars squared; $16$ thousand-dollars squared; $0.4712$)

What You'll Learn