How Inflation Scales Severity Moments and Quantiles

Why Scaling Severity Matters

When claim amounts are moved to a higher price level, a deterministic inflation factor multiplies each loss. This changes not just the average claim but also the spread and every quantile of the severity distribution. Keeping those effects separate is useful: variance scales differently from standard deviation, while quantiles scale in the same way as individual losses.

The results below concern a ground-up loss $X$ multiplied by a positive, fixed factor. They do not require an interest or mortality model. This severity-scaling topic is aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

Definition: Let $X$ be a nonnegative random loss and let $c>0$ be a deterministic trend factor. The loss at the trended price level is $Z=cX$. For example, an increase of 8% corresponds to $c=1.08$, not $c=0.08$. The factor is dimensionless; $X$ and $Z$ are expressed in the same monetary unit.

How Moments and Spread Change

For any positive integer $r$ for which the $r$th raw moment exists, the $r$th raw moment of the trended loss is obtained by multiplying the original moment by $c^r$. The mean is the first raw moment, so it scales by $c$. Variance is different: it scales by $c^2$, and standard deviation therefore scales by $c$.

E[Z^r]=E[(cX)^r]=c^rE[X^r], \qquad E[Z]=cE[X].

The variance result follows directly from the definition. Centering the scaled loss at its own mean gives $Z-E[Z]=cX-cE[X]=c(X-E[X])$. Squaring this difference brings in the factor $c^2$.

\begin{aligned} \operatorname{Var}(Z) &=E\!\left[(Z-E[Z])^2\right]\\ &=E\!\left[(cX-cE[X])^2\right]\\ &=E\!\left[c^2(X-E[X])^2\right]\\ &=c^2\operatorname{Var}(X). \end{aligned}

Since $c>0$, taking the nonnegative square root gives $\operatorname{SD}(Z)=c\operatorname{SD}(X)$. These formulas assume the relevant moments exist. If $X$ has an infinite mean or variance, multiplying by a finite positive $c$ does not make that moment finite.

Theorem: For $Z=cX$, where $c>0$ is deterministic, $E[Z]=cE[X]$, $\operatorname{Var}(Z)=c^2\operatorname{Var}(X)$, and $\operatorname{SD}(Z)=c\operatorname{SD}(X)$, whenever the stated moments exist. More generally, $E[Z^r]=c^rE[X^r]$ whenever the $r$th raw moment exists.

A quick scale check helps catch errors. If $c=1.10$, the mean and standard deviation increase by 10%, but the variance is multiplied by $1.10^2=1.21$, so it increases by 21%. Variance is measured in squared monetary units, which is consistent with its squared scaling factor.

How Quantiles Change

For a distribution function $F_X$, define its $p$-quantile, for $0<p<1$, by the generalized inverse $Q_X(p)=\inf\{x:F_X(x)\ge p\}$. This definition works for both continuous and discrete severity distributions. Because $c$ is positive, multiplying by $c$ preserves the ordering of all loss values.

F_Z(z)=P(cX\le z)=F_X\!\left(\frac{z}{c}\right), \qquad Q_Z(p)=cQ_X(p), \quad 0<p<1.

To see why the quantile formula holds, consider the set of losses at or below the $p$-quantile. The transformation $x\mapsto cx$ sends each such loss to $cx$ and preserves which losses are below or above it. Equivalently, $F_Z(cQ_X(p))=F_X(Q_X(p))\ge p$, while any smaller candidate $z<cQ_X(p)$ has $z/c<Q_X(p)$ and cannot reach cumulative probability $p$. Thus the smallest value reaching probability $p$ is $cQ_X(p)$.

For continuous distributions, this is often seen by solving $F_Z(z)=p$. The generalized-inverse definition is important when there are probability atoms: a quantile can stay at one loss amount across a range of probability levels. It still scales by $c$, but one must not treat the quantile as if every probability level corresponded to a unique loss.

Worked Applications

Worked Example: Uniform Severity With a High Quantile

A hypothetical portfolio has ground-up loss $X\sim\operatorname{Uniform}(0,12000)$, in dollars. Trend losses by a factor $c=1.10$. Find the trended mean, variance, standard deviation, and 90th percentile. This is a severity-only model; no interest or mortality assumption is involved.

Method. Calculate the original uniform moments and quantile, then apply the scaling formulas. The common mistake to avoid is multiplying variance by 1.10 instead of by $1.10^2$.

For a uniform loss on $[0,a]$, $E[X]=a/2$, $\operatorname{Var}(X)=a^2/12$, and $Q_X(p)=pa$. With $a=12000$:

E[X]=\frac{12000}{2}=6000,\qquad \operatorname{Var}(X)=\frac{12000^2}{12} =\frac{144000000}{12}=12000000.

The original standard deviation is $\sqrt{12000000}=3464.1016$ dollars. The original 90th percentile is $Q_X(0.90)=0.90(12000)=10800$ dollars. Now apply $c=1.10$:

\begin{aligned} E[Z]&=1.10(6000)=6600,\\ \operatorname{Var}(Z)&=1.10^2(12000000) =1.21(12000000)=14520000,\\ \operatorname{SD}(Z)&=1.10(3464.1016)=3810.5118,\\ Q_Z(0.90)&=1.10(10800)=11880. \end{aligned}

To check the result directly, $Z$ is uniform on $[0,13200]$. Its mean is $13200/2=6600$, its variance is $13200^2/12=14520000$, and its 90th percentile is $0.90(13200)=11880$. Rounded to the nearest cent, the standard deviation is $3,810.51; the other requested amounts are $6,600.00, $14,520,000.00 in squared dollars, and $11,880.00. The direct calculations agree with the scaling method.

Worked Example: Exponential Severity and an Upper Quantile

Suppose a hypothetical ground-up loss $X$ has an exponential distribution with mean $5,000. Trend the severity by $c=1.20$. Find the trended mean, variance, standard deviation, and 80th percentile. There is no interest or mortality component in this severity model.

Method. Use the exponential mean as its scale parameter, derive the original quantile from its survival function, and then scale. The common mistake to avoid is using $\ln(0.80)$ rather than the tail probability $1-0.80$ when finding the 80th percentile.

For an exponential loss with mean $\theta=5000$, $E[X]=\theta$ and $\operatorname{Var}(X)=\theta^2$. Therefore:

E[X]=5000,\qquad \operatorname{Var}(X)=5000^2=25000000,\qquad \operatorname{SD}(X)=5000.

Its survival function is $P(X>x)=e^{-x/5000}$ for $x\ge0$. At the 80th percentile, the probability of exceeding the quantile is $0.20$, so $e^{-Q_X(0.80)/5000}=0.20$. Solving gives:

Q_X(0.80)=-5000\ln(0.20)=8047.1896.

Now apply the trend factor to the mean, variance, standard deviation, and quantile:

\begin{aligned} E[Z]&=1.20(5000)=6000,\\ \operatorname{Var}(Z)&=1.20^2(25000000) =1.44(25000000)=36000000,\\ \operatorname{SD}(Z)&=1.20(5000)=6000,\\ Q_Z(0.80)&=1.20(8047.1896)=9656.6275. \end{aligned}

Check the quantile using the trended exponential mean $1.20(5000)=6000$: $-6000\ln(0.20)=9656.6275$, as obtained by scaling the original quantile. Rounded to the nearest cent, the trended 80th percentile is $9,656.63; the mean is $6,000.00, variance is 36,000,000.00 squared dollars, and standard deviation is $6,000.00.

Worked Example: Discrete Severity and a Probability Atom

A hypothetical claim has ground-up severity $X=\$1,000$ with probability 0.75 and $X=\$5,000$ with probability 0.25. Trend every loss by $c=1.10$. Find the mean, variance, standard deviation, the 70th percentile, and the 90th percentile. No interest or mortality assumption is used.

Method. Calculate the original moments from the probability-weighted values, then transform the outcomes and quantiles. The common mistake to avoid is assigning the 70th percentile to the higher loss: cumulative probability has already reached 0.75 at the lower loss.

The original mean and second raw moment are:

\begin{aligned} E[X]&=0.75(1000)+0.25(5000)=750+1250=2000,\\ E[X^2]&=0.75(1000^2)+0.25(5000^2)\\ &=0.75(1000000)+0.25(25000000)\\ &=750000+6250000=7000000. \end{aligned}

Thus $\operatorname{Var}(X)=E[X^2]-E[X]^2=7000000-2000^2=7000000-4000000=3000000$, and $\operatorname{SD}(X)=\sqrt{3000000}=1732.0508$ dollars. After trending, the two possible losses are $1.10(1000)=1100$ and $1.10(5000)=5500$, with the same probabilities. The scaled moments are:

\begin{aligned} E[Z]&=1.10(2000)=2200,\\ \operatorname{Var}(Z)&=1.10^2(3000000)=1.21(3000000)=3630000,\\ \operatorname{SD}(Z)&=1.10(1732.0508)=1905.2559. \end{aligned}

The cumulative probability at $1,100 is 0.75. Therefore the smallest loss whose cumulative probability is at least 0.70 is $1,100, so $Q_Z(0.70)=1100$. The cumulative probability at $1,100 is still below 0.90, while at $5,500 it is 1.00, so $Q_Z(0.90)=5500$.

As a final check, the trended distribution has mean $0.75(1100)+0.25(5500)=825+1375=2200$. Its second raw moment is $0.75(1100^2)+0.25(5500^2)=907500+7562500=8470000$, giving variance $8470000-2200^2=8470000-4840000=3630000$. The results agree. Rounded to the nearest cent, the standard deviation is $1,905.26; the mean is $2,200.00, variance is 3,630,000.00 squared dollars, and the quantiles are $1,100.00 and $5,500.00.

Interpretation and Common Traps

A deterministic trend changes the scale of the loss distribution without changing its probability structure. In particular, the same probability remains attached to each corresponding scaled loss, and the order of losses is preserved. The examples show why the quantile result works for both smooth and discrete distributions.

The formulas apply to a simple multiplication of the loss by a positive constant. If a loss is also modified by a fixed deductible or limit, the resulting payment is a different transformation; it should not automatically be treated as the original severity multiplied by $c$. In that setting, whether trend is applied before or after a coverage term matters. The present scaling results alone do not resolve that coverage question.

Scaling variance linearly. The correct formula is $\operatorname{Var}(cX)=c^2\operatorname{Var}(X)$, not $c\operatorname{Var}(X)$.
Confusing a rate and a factor. A 10% increase means multiplying by 1.10. Multiplying by 0.10 would retain only one-tenth of the original loss.
Using the wrong exponential tail probability. For a $p$-quantile, the exponential survival probability is $1-p$, not $p$.
Ignoring jumps in a discrete distribution. Use the smallest loss at which the cumulative probability reaches or exceeds $p$; do not interpolate between possible loss values.

Key takeaway: Under deterministic positive scaling $Z=cX$, the mean, standard deviation, and every generalized-inverse quantile scale by $c$; the variance scales by $c^2$. State the factor clearly, and use the quantile definition carefully when the severity distribution has atoms.

Check Your Understanding

For each hypothetical severity model, apply the deterministic scaling rules and check any quantile against the model’s distribution.

A loss has mean $800 and variance 160,000. Its 75th percentile is $1,200. After multiplying every loss by 1.05, find the mean, variance, standard deviation, and 75th percentile. (Answer: mean $840; variance 176,400; standard deviation $420; 75th percentile $1,260.)
Let $X\sim\operatorname{Uniform}(0,4000)$ dollars, and set $Z=1.15X$. Find $E[Z]$, $\operatorname{Var}(Z)$, $\operatorname{SD}(Z)$, and $Q_Z(0.90)$. (Answer: mean $2,300; variance 1,763,333.33 squared dollars; standard deviation $1,327.91; 90th percentile $4,140.)
A loss is $2,000 with probability 0.60 and $8,000 with probability 0.40. Multiply it by 0.90. Find the trended mean, variance, standard deviation, and 50th percentile. (Answer: mean $3,960; variance 6,998,400 squared dollars; standard deviation $2,645.45; 50th percentile $1,800.)
An exponential loss has mean $2,500. Multiply it by 1.20. Find the trended mean, variance, standard deviation, and 95th percentile. Use $Q_X(p)=-\theta\ln(1-p)$. (Answer: mean $3,000; variance 9,000,000 squared dollars; standard deviation $3,000; 95th percentile $8,987.20.)

What You'll Learn