Scaling Exponential Severity Models

Why Scaling Changes the Rate

An exponential severity model is often specified in one monetary unit, while a calculation or policy description uses another amount or rescales losses. Multiplying every loss by a positive constant changes its numerical size, but it does not change the model’s exponential form. The key is to transform the rate in the opposite direction from the loss multiplier.

Suppose $X$ is a nonnegative loss with an exponential distribution and rate $\lambda>0$. If the reported or adjusted loss is $Z=sX$, where $s>0$, then large values of $s$ make losses larger and therefore make the transformed rate smaller. The rate’s units also change: if $X$ is measured in dollars, its rate is per dollar; if $Z$ is measured in thousands of dollars, its rate is per thousand dollars.

Definition: If $X\sim\operatorname{Exponential}(\lambda)$ in the rate parameterization, then $F_X(x)=1-e^{-\lambda x}$ for $x\geq 0$, $S_X(x)=e^{-\lambda x}$ for $x\geq 0$, and $f_X(x)=\lambda e^{-\lambda x}$ for $x\geq 0$. For $x<0$, the distribution function is zero and the density is zero. The rate $\lambda$ is positive and has units reciprocal to the units of $X$. For a positive multiplier $s$, the scaled loss $Z=sX$ has rate $\lambda/s$.

In this rate parameterization, $E[X]=1/\lambda$, $\operatorname{Var}(X)=1/\lambda^2$, and the $p$-quantile for $0<p<1$ is $-\ln(1-p)/\lambda$. Scaling multiplies the mean, standard deviation, and every quantile by $s$, while it multiplies the variance by $s^2$. These relationships make it possible to check a transformed calculation in more than one way.

Deriving the Scaled Distribution

The model assumes $X$ is exponential with rate $\lambda$, and $Z=sX$ for a fixed constant $s>0$. Since $s$ is positive, the event $Z\leq z$ is equivalent to $X\leq z/s$. This gives the transformed distribution directly:

F_Z(z)=P(Z\leq z)= \begin{cases} 0, & z<0,\\ 1-e^{-\lambda z/s}, & z\geq 0. \end{cases}

Comparing this expression with the exponential distribution function shows that $Z$ has rate $\lambda_Z=\lambda/s$. Differentiating the distribution function gives $f_Z(z)=(\lambda/s)e^{-\lambda z/s}$ for $z\geq 0$, and its survival function is $S_Z(z)=e^{-\lambda z/s}$. Thus the exponential model is preserved, but its rate is not.

Derivation: For $z\geq 0$, the density transformation can also be obtained from $x=z/s$, so $dx/dz=1/s$. Therefore $f_Z(z)=f_X(z/s)(1/s)=(\lambda/s)e^{-\lambda z/s}$. The factor $1/s$ is essential: it ensures the transformed density integrates to one.

The hazard rate is the density divided by the survival function. For $z\geq 0$, the scaled model therefore has

h_Z(z)=\frac{f_Z(z)}{S_Z(z)} =\frac{(\lambda/s)e^{-\lambda z/s}}{e^{-\lambda z/s}} =\frac{\lambda}{s}.

The hazard remains constant in the loss amount. Scaling changes its level from $\lambda$ to $\lambda/s$, but does not make it increase or decrease with the amount already observed. The positive multiplier also preserves memorylessness: for $a,t\geq 0$, $P(Z>a+t\mid Z>a)=e^{-\lambda t/s}=P(Z>t)$. This follows from the transformed survival function and requires the exponential model and a fixed positive scale.

Consequences for Moments and Quantiles

For any positive integer $k$ for which the moment exists, $Z^k=s^kX^k$. Exponential losses have finite moments of every positive integer order, so $E[Z^k]=s^kE[X^k]$. In particular, the mean is multiplied by $s$ and the variance by $s^2$. The standard deviation is multiplied by $s$, and the coefficient of variation remains one.

E[Z]=\frac{s}{\lambda},\qquad \operatorname{Var}(Z)=\frac{s^2}{\lambda^2},\qquad \operatorname{SD}(Z)=\frac{s}{\lambda},\qquad Q_Z(p)=\frac{-s\ln(1-p)}{\lambda}.

Here $Q_Z(p)$ denotes the $p$-quantile, obtained by solving $F_Z(z)=p$. The entire quantile function scales linearly: $Q_Z(p)=sQ_X(p)$. This is also a useful check on units because a quantile of $Z$ must be expressed in the same units as $Z$.

Identify the original rate and units.
Write $\lambda$ with its unit, such as per dollar or per thousand dollars. The original mean is $1/\lambda$ in the corresponding loss unit.

Apply the loss multiplier.
For $Z=sX$ with $s>0$, use $\lambda_Z=\lambda/s$. Check that the transformed rate is reciprocal to the units of $Z$.

Use the transformed model.
Calculate probabilities with $S_Z(z)=e^{-\lambda_Zz}$, moments with the transformed rate, and quantiles with $Q_Z(p)=-\ln(1-p)/\lambda_Z$.

Check the scale relationship.
The transformed mean and standard deviation should be $s$ times their original values, and the transformed variance should be $s^2$ times its original value.

Worked Calculations

Worked Example: Doubling a Dollar-Denominated Loss

A hypothetical loss $X$, measured in dollars, has an exponential distribution with rate $0.001$ per dollar. An adjustment produces $Z=2X$. Find the rate, mean, variance, probability that $Z$ exceeds \$6,000, and the 90th percentile. Give monetary results to the nearest cent and the probability to four decimals.

Method. The multiplier is $s=2$, so the transformed rate is $\lambda_Z=0.001/2=0.0005$ per dollar. Equivalently, $X$ has mean $1/0.001=\$1{,}000$, so $Z$ must have mean $2(\$1{,}000)=\$2{,}000$.

The variance is $1/(0.0005)^2=4{,}000{,}000$ dollars squared. This agrees with scaling the original variance: $2^2/(0.001)^2=4{,}000{,}000$ dollars squared.

Using the transformed survival function, $P(Z>6{,}000)=e^{-0.0005(6{,}000)}=e^{-3}=0.049787\ldots$, which is $0.0498$ to four decimals.

The 90th percentile is $Q_Z(0.90)=-\ln(1-0.90)/0.0005 =2.302585\ldots/0.0005=\$4{,}605.170\ldots$, or \$4,605.17 to the nearest cent.

Check and common mistake avoided. The transformed mean is twice the original mean, as required. Multiplying the rate by two would instead give $0.002$ per dollar and a mean of only \$500, contradicting the fact that every loss was doubled.

Worked Example: Scaling a Loss Measured in Thousands

A hypothetical loss $X$, measured in thousands of dollars, is exponential with rate $0.004$ per thousand dollars. A reporting adjustment sets $Y=0.5X$. Find the rate and mean of $Y$, $P(Y>250)$, and its 95th percentile. Report money in thousands of dollars to three decimal places and the probability to four decimals.

Method. The multiplier is $s=0.5$, so $\lambda_Y=0.004/0.5=0.008$ per thousand dollars. The original mean is $1/0.004=250$ thousand dollars, so the scaled mean is $0.5(250)=125$ thousand dollars. These units are consistent with the reciprocal rate.

For the exceedance probability, $P(Y>250)=e^{-0.008(250)}=e^{-2}=0.135335\ldots$, or 0.1353.

The 95th percentile is \[ Q_Y(0.95)=\frac{-\ln(0.05)}{0.008} =\frac{2.995732274\ldots}{0.008} =374.466534\ldots \] thousand dollars, which rounds to 374.467 thousand dollars to three decimal places.

Check and common mistake avoided. Halving all losses halves the mean from 250 to 125 thousand dollars and doubles the rate from 0.004 to 0.008 per thousand dollars. Using $0.004(0.5)=0.002$ as the rate would reverse both effects and give an incorrect 95th percentile.

Worked Example: Checking Moments, Tail, and Hazard Together

A hypothetical loss $X$, measured in dollars, has an exponential distribution with rate $0.00025$ per dollar. A fixed adjustment gives $Z=0.75X$. Find the rate, mean, variance, 75th percentile, $P(Z>6{,}000)$, and hazard rate. Round monetary results to the nearest cent and the probability to four decimals.

Method. Since $s=0.75$, the transformed rate is $\lambda_Z=0.00025/0.75=0.000333333\ldots$ per dollar. The transformed mean is $1/\lambda_Z=\$3{,}000$, which also equals $0.75(1/0.00025)=0.75(\$4{,}000)$.

The variance is $1/(\lambda_Z)^2=(3{,}000)^2=9{,}000{,}000$ dollars squared. The 75th percentile is \[ Q_Z(0.75)=\frac{-\ln(0.25)}{0.000333333\ldots} =1.386294361\ldots(3{,}000) =\$4{,}158.88 \] to the nearest cent.

The tail probability is $P(Z>6{,}000)=e^{-0.000333333\ldots(6{,}000)}=e^{-2}=0.1353$ to four decimals. The hazard is constant at $\lambda_Z=0.000333333\ldots$ per dollar, or about $0.0003333$ per dollar to four significant figures.

Check and common mistake avoided. The original mean is \$4,000, and multiplying losses by $0.75$ gives \$3,000. A rate of $0.00025(0.75)=0.0001875$ per dollar would imply a mean above \$5,000, so it cannot describe losses reduced by 25%.

Common Scaling Traps

The most frequent error is to apply the loss multiplier directly to the rate. The distribution-function derivation shows why the correct rate is $\lambda/s$: the event $sX\leq z$ uses the original cutoff $z/s$. If $s>1$, the transformed losses are larger and the rate must decrease. If $0<s<1$, the transformed losses are smaller and the rate must increase.

A second trap is to confuse rate units with loss units. A rate of $0.004$ per thousand dollars is not $0.004$ per dollar. When converting the unit of measurement, convert the rate consistently, then apply any additional multiplier. A final dimensional check should show that rate times loss is dimensionless inside the exponential.

Scaling by a fixed positive constant preserves the exponential distribution’s standardized shape: its coefficient of variation remains one, and the hazard remains constant. It does not mean that every feature remains numerically unchanged. The mean, variance, quantiles, and hazard all change according to their scale relationships. The result also does not cover a zero or negative multiplier: zero produces a degenerate loss at zero, while a negative multiplier no longer describes a nonnegative severity model.

Key takeaway: For $Z=sX$, where $X\sim\operatorname{Exponential}(\lambda)$ and $s>0$, the transformed rate is $\lambda/s$. The mean, standard deviation, and quantiles multiply by $s$; the variance multiplies by $s^2$; and the hazard remains constant at $\lambda/s$.

Check Your Understanding

For each problem, assume the stated exponential model applies and the scale factor is fixed and positive. Keep the loss units attached to rates and answers.

A hypothetical loss $X$ is exponential with rate $0.005$ per dollar, and $Z=4X$. Find the rate, mean, and variance of $Z$. (Answer: rate $0.00125$ per dollar; mean $\$800$; variance $640{,}000$ dollars squared)
A loss $X$, measured in thousands of dollars, is exponential with rate $0.02$ per thousand dollars. Let $Z=3X$. Find $P(Z>150)$. (Answer: $0.3679$)
A loss $X$ has exponential rate $0.006$ per unit, and $Z=0.4X$. Find the 90th percentile of $Z$, to three decimal places. (Answer: $153.506$ units)
A dollar-denominated loss $X$ is exponential with rate $0.0005$ per dollar, and $Z=1.2X$. Find the mean and variance of $Z$. (Answer: mean $\$2{,}400$; variance $5{,}760{,}000$ dollars squared)
A hypothetical loss $X$ is exponential with mean $\$600$. A fixed adjustment $Z=sX$ results in a mean of $\$1{,}500$. Find $s$, the rate of $Z$, $P(Z> \$3{,}000)$, and the 80th percentile of $Z$, to the nearest cent. (Answer: $s=2.5$; rate $0.0006667$ per dollar; probability $0.1353$; 80th percentile $\$2{,}414.16$)

What You'll Learn