Scaling Claim Severities

Why Claim-Severity Scaling Matters

Aggregate losses may be reported in different currency units, adjusted by a common multiplicative factor, or compared under a scenario in which each claim amount changes proportionally. In each case, the claim count can stay exactly the same while every claim amount is rescaled. The key is to transform both the losses and any thresholds used to describe them.

This tutorial considers a random aggregate loss formed from a claim count and individual claim severities. It develops the distribution and moment transformations for multiplying every severity by a positive constant. The same results apply whether the factor represents a change in scale or a conversion between units, provided the underlying claim amounts are otherwise unchanged.

Definition: Let $N$ be a nonnegative integer-valued claim count. Let $X_1,X_2,\ldots$ be independent and identically distributed nonnegative claim severities, independent of $N$, and define the aggregate loss by $S=\sum_{j=1}^{N}X_j$, with $S=0$ when $N=0$. For a constant $c>0$, the scaled claim severities are $cX_j$, and their aggregate is $T=\sum_{j=1}^{N}cX_j=cS$. The factor changes the amount of each claim, not the number of claims.

Distribution Transformation

Write $F_S(x)=P(S\leq x)$ for the aggregate distribution function. Since $c$ is positive, multiplying both sides of an inequality by $c$ preserves its direction. Thus, for any real threshold $t$,

F_T(t)=P(T\leq t)=P(cS\leq t)=P(S\leq t/c)=F_S(t/c).

The same argument gives the strict upper-tail relation $P(T>t)=P(S>t/c)$. For an interval, both endpoints must be converted: $P(a<T\leq b)=P(a/c<S\leq b/c)$. This is often the most useful way to calculate an event involving a scaled aggregate: return to the unscaled model and use the equivalent threshold.

Scaling also moves individual probability masses. For any value $s$, $P(T=cs)=P(S=s)$. In particular, $P(T=0)=P(S=0)$, because $cS=0$ exactly when $S=0$. If $S$ has a density on a continuous portion of its distribution, the transformed density there is $f_T(t)=f_S(t/c)/c$. The factor $1/c$ makes the transformed density integrate to the same probability as before; this density formula does not replace the need to account separately for any atoms.

For a quantile defined by $Q_S(p)=\inf\{s:F_S(s)\geq p\}$, the positive scaling also gives $Q_T(p)=cQ_S(p)$. This follows because the entire distribution, including its atoms, is moved to the right by the same factor when $c>1$, or compressed toward zero when $0<c<1$.

How Aggregate Moments Change

Assume the claim severities have finite mean $\mu=E[X_1]$ and variance $\sigma^2=\operatorname{Var}(X_1)$, and the claim count has finite mean and variance. Conditioning on the claim count gives the standard random-sum formulas. The derivation also shows why the variance uses the square of the scaling factor.

Condition on the claim count.
Given $N=n$, the aggregate $T$ is a sum of $n$ independent severities $cX_j$. Therefore $E[T\mid N=n]=nc\mu$ and $\operatorname{Var}(T\mid N=n)=nc^2\sigma^2$.

Take the expectation.
By the law of total expectation, $E[T]=E[E(T\mid N)]=c\mu E[N]$.

Use total variance.
The expected conditional variance is $c^2\sigma^2E[N]$, and the variance of the conditional mean $c\mu N$ is $c^2\mu^2\operatorname{Var}(N)$. Adding them gives the aggregate variance.

E[T]=cE[S]=cE[N]\mu, \qquad \operatorname{Var}(T)=c^2\operatorname{Var}(S) =c^2\left(E[N]\sigma^2+\operatorname{Var}(N)\mu^2\right).

The first variance term reflects variation in individual severities among the claims that occur. The second reflects variation in the number of claims, each of which has mean amount $c\mu$. Both terms acquire $c^2$. These formulas require the stated independence and finite-moment assumptions; the distribution identity $T=cS$ itself does not require finite moments.

For example, if $N\sim\operatorname{Poisson}(\lambda)$, then $E[N]=\operatorname{Var}(N)=\lambda$, so $E[T]=c\lambda\mu$ and $\operatorname{Var}(T)=c^2\lambda(\sigma^2+\mu^2)=c^2\lambda E[X_1^2]$. The scaling rule works with other claim-count models as well; the relevant values of $E[N]$ and $\operatorname{Var}(N)$ come from the specified frequency model.

Worked Example: Inflating Every Claim in a Poisson Model

A hypothetical portfolio has $N\sim\operatorname{Poisson}(2)$ claims. Each original claim is exactly $500. A scenario multiplies every claim by $1.2$. Find the mean and variance of the scaled aggregate, in dollars and dollars squared, and calculate the probability that it exceeds $1,200.

Method. Keep the Poisson claim count fixed and transform the aggregate by $T=1.2S$. This avoids the mistake of treating the severity change as a change to the claim frequency.

The scaled amount per claim is $1.2(500)=\$600$, so $T=600N$. Since a Poisson random variable with parameter 2 has mean 2 and variance 2,

E[T]=600(2)=\$1{,}200, \qquad \operatorname{Var}(T)=600^2(2)=\$720{,}000.

For the strict event $T>\$1{,}200$, substitute $T=600N$: $600N>1{,}200$ is equivalent to $N>2$, or $N\geq3$. Therefore

P(T>1{,}200) =1-P(N=0)-P(N=1)-P(N=2) =1-e^{-2}\left(1+2+\frac{2^2}{2!}\right) =1-5e^{-2} \approx 0.3233.

The result is checked against the event interpretation: exactly two claims produce an aggregate of $1,200 and do not satisfy the strict inequality, while three or more claims do. The probability is reported to four decimal places.

Changing Currency Units

A currency conversion is a scaling operation, not a change in the loss model. If $S$ is measured in dollars and $T$ is measured in thousands of dollars, then $T=0.001S$. A threshold of 5 in the latter units means $5,000 in the former units. The scale factor applies to the claim amounts, aggregate amounts, and thresholds; it is squared for variances.

It is useful to write the units beside each quantity before calculating. A mean changes from dollars to thousands of dollars by multiplying by $0.001$, while a variance changes from dollars squared to thousands-of-dollars squared by multiplying by $0.001^2$. Probabilities do not have units and are unchanged when the event is translated consistently.

Worked Example: Converting a Binomial Portfolio to Thousands

A hypothetical portfolio has $N\sim\operatorname{Binomial}(3,0.4)$ claims, each exactly $2,500. Report aggregate losses in thousands of dollars. Find the mean, variance in thousands-of-dollars squared, and the probability that the reported aggregate is at least 5.

Method. Convert each claim amount first, then use the unchanged binomial frequency. The scaled severity is $0.001(2{,}500)=2.5$ thousand dollars. The claim count has mean $3(0.4)=1.2$ and variance $3(0.4)(0.6)=0.72$.

Because every claim has the same amount, the individual severity variance is zero. The aggregate mean and variance in the stated units are

E[T]=2.5(1.2)=3, \qquad \operatorname{Var}(T)=2.5^2(0.72)=6.25(0.72)=4.5.

Here the mean is 3 thousand dollars and the variance is 4.5 thousand-dollars squared. For the probability, $T=2.5N$, so $T\geq5$ is equivalent to $N\geq2$. Thus

P(T\geq5)=P(N=2)+P(N=3) =\binom{3}{2}(0.4)^2(0.6)+(0.4)^3 =0.288+0.064 =0.3520.

The answer is checked by direct unit conversion: the event means at least two $2,500 claims, or a dollar aggregate of at least $5,000. The probability is reported to four decimal places.

Scaling a Severity Distribution With More Than One Value

When individual claim amounts vary, a threshold event may depend on both the claim count and the combination of severities. The scaling identity still applies exactly: transform the threshold first, then evaluate the event under the original model. In a discrete model, it is important to handle strict and non-strict inequalities carefully, especially when a possible aggregate equals the threshold.

Worked Example: A Strict Tail With Two Possible Severities

Suppose $N\sim\operatorname{Binomial}(2,0.5)$. Independently, each claim is $100 or $300, each with probability $0.5$. Every claim is then multiplied by 2. Find the probability that the scaled aggregate is strictly greater than $400, and find its mean and variance.

Method. First use $T=2S$ to identify which original claim combinations satisfy the event. Then use the random-sum moment formula. This prevents counting an outcome equal to $400 as part of a strict upper tail.

For one claim, $E[X]=(100+300)/2=200$, and $E[X^2]=(100^2+300^2)/2=50{,}000$. Therefore $\operatorname{Var}(X)=50{,}000-200^2=10{,}000$. The binomial count has $E[N]=1$ and $\operatorname{Var}(N)=0.5$. In the original dollar units,

E[S]=1(200)=200, \qquad \operatorname{Var}(S)=1(10{,}000)+0.5(200^2)=30{,}000.

Scaling by 2 gives $E[T]=2(200)=\$400$ and $\operatorname{Var}(T)=2^2(30{,}000)=\$120{,}000$. For the probability, $T>400$ is equivalent to $S>200$.

If $N=0$, the aggregate is zero and the event fails. If $N=1$, the original severity must be $300; the probability of this case is $P(N=1)P(X=300)=0.5(0.5)=0.25$. If $N=2$, the original total exceeds $200 unless both claims are $100. The conditional probability of exceeding $200 is $1-(0.5)^2=0.75$, and $P(N=2)=0.25$. Hence

$$ P(T>400)=0.25+0.25(0.75)=0.4375. $$

The equality case is checked explicitly: with two $100 claims, the scaled aggregate is exactly $400, so it is excluded. The probability is reported to four decimal places.

Common Scaling Errors

The transformation is simple, but several errors can change the answer substantially. For example, if a claim amount is doubled, the aggregate variance is multiplied by $2^2=4$, not by 2. Applying the factor only once confuses the scaling of a standard deviation or mean with the scaling of a variance.

Another common error is to apply the factor to the number of claims. Under the model here, $N$ is unchanged: only the severities $X_j$ become $cX_j$. Changing a frequency parameter would describe a different model, not a severity rescaling.

Finally, a threshold must be expressed in the same units as the aggregate before evaluating a probability. If losses are converted to thousands of dollars but the threshold remains in dollars, the event will be off by a factor of 1,000. For $c>0$, the safest check is to write $T>t$ as $S>t/c$ and verify that both sides are in the original loss units.

Key takeaway: If every claim severity is multiplied by $c>0$ and the claim count is unchanged, then the aggregate becomes $T=cS$. Its distribution is $F_T(t)=F_S(t/c)$, its mean is multiplied by $c$, its variance by $c^2$, and each event threshold must be converted by dividing by $c$ when returning to the original aggregate.

Check Your Understanding

For each problem, keep the claim count unchanged when scaling severities and state the probability to four decimal places where requested.

A hypothetical portfolio has $N\sim\operatorname{Poisson}(2.5)$, with every claim exactly $80. Each claim is multiplied by 0.5. Find the scaled aggregate mean in dollars and variance in dollars squared. (Answer: mean $100; variance $4,000.)
A portfolio has $N\sim\operatorname{Binomial}(5,0.2)$, with every claim exactly $400. Report losses in units of $100 and find the mean, variance, and probability that the reported aggregate is at least 8. (Answer: mean 4 units; variance 12.8 units squared; probability 0.2627.)
Let $N\sim\operatorname{Poisson}(1)$, and let each independent claim be $100 or $200 with equal probability. Every claim is multiplied by 3. Find the probability that the scaled aggregate is strictly greater than $300. (Answer: $1-\frac{1.5}{e}\approx 0.4482$.)
An aggregate loss $S$, measured in dollars, has mean $2,400 and variance $1,440,000 dollars squared. A new reporting scale sets $T=0.5S$, and $P(S>1{,}800)=0.2700$. Find the mean and variance of $T$, and $P(T>900)$. (Answer: mean $1,200; variance $360,000 dollars squared; probability 0.2700.)

What You'll Learn