Loss Distributions as Inputs to Risk Measures

Why the Loss Variable Comes First

A risk measure is only meaningful after the random quantity being measured is clear. The ground-up claim on one policy, the insurer’s payment after coverage terms, and the total payments across a portfolio are different random variables. They can have different distributions, moments, and tail risks, even when they arise from the same underlying claims.

For example, an individual claim can have a finite mean but infinite variance. A portfolio total can have an atom at zero even when each individual claim is continuous. A calculation that skips these distinctions can produce a precise-looking answer for the wrong loss variable. This tutorial focuses on specifying the loss distribution and checking its moment conditions before calculating variance, value-at-risk (VaR), or tail value-at-risk (TVaR).

Definition: A loss random variable is a specified random amount, measured over a stated unit such as one claim, one policy, or one portfolio over one year. Its distribution describes the possible amounts and their probabilities. The ground-up loss $X$ is the loss before applying policy terms. The payment per loss $Y^L$ is the insurer’s payment for one exposure, including zero when there is no payment. The aggregate loss is the sum of payments over a stated group and period.

The unit and coverage basis should be stated along with the variable. “Annual loss” could mean the loss on one policy during one year, the sum of all claims on that policy during one year, or the total portfolio payment. Those quantities are not interchangeable. If a deductible $d$ and a payment limit $u$ apply to each claim, for instance, the payment for a ground-up claim $X$ is $Y^L=\min((X-d)_+,u)$, where $(z)_+=\max(z,0)$. The distribution of $Y^L$ is therefore not generally the same as the distribution of $X$.

Distribution, Moments, and Risk Measures

Write $F_X(x)=P(X\le x)$ for the cumulative distribution function, $S_X(x)=P(X>x)$ for the survival function, and $f_X(x)$ for a density when $X$ is continuous. For a nonnegative loss, its mean is determined by the area under its survival function:

E[X]=\int_0^\infty S_X(x)\,dx.

When the second moment exists, $E[X^2]=\int_0^\infty 2xS_X(x)\,dx$ and $\operatorname{Var}(X)=E[X^2]-(E[X])^2$. These expressions make the tail’s importance visible: a distribution can have a finite mean while its second moment, and hence its variance, is infinite.

For a confidence level $p$ and a continuous distribution with strictly increasing cumulative distribution function, use $\operatorname{VaR}_p(X)=F_X^{-1}(p)$. In this setting, $\operatorname{TVaR}_p(X)=E[X\mid X>\operatorname{VaR}_p(X)]$. These expressions are sufficient for the continuous examples below. If a distribution has point masses, including an aggregate loss with positive probability of zero, tail-risk conventions require care; a single continuous-distribution formula should not be applied without checking the convention.

Moment check: A variance calculation requires a finite second moment. A finite VaR at a confidence level below 1 can exist even when variance is infinite. TVaR, understood here as a conditional tail mean for a continuous loss, requires that the expected loss in the selected tail be finite. Check these conditions from the distribution before using a formula.

From Individual Claims to Aggregate Loss

A common aggregate model is the compound-frequency model. Let $N$ be the number of claims in a fixed period, and let $X_1,X_2,\ldots$ be identically distributed claim severities. Assume that $N$ is independent of the severities and that the severities are mutually independent. The aggregate loss is $S=\sum_{j=1}^{N}X_j$, with $S=0$ when $N=0$. The frequency distribution specifies $N$; the severity distribution specifies each $X_j$.

If $E[N]$ and $E[X_1]$ are finite, then $E[S]=E[N]E[X_1]$. If, in addition, $E[N^2]$ and $E[X_1^2]$ are finite, conditioning on $N$ gives the variance formula below.

\operatorname{Var}(S) =E[N]\operatorname{Var}(X_1)+\operatorname{Var}(N)(E[X_1])^2.

To see why, condition first on $N$. Given $N=n$, the sum has mean $nE[X_1]$ and variance $n\operatorname{Var}(X_1)$. The law of total variance then gives

\begin{aligned} \operatorname{Var}(S) &=E[\operatorname{Var}(S\mid N)] +\operatorname{Var}(E[S\mid N])\\ &=E[N]\operatorname{Var}(X_1) +\operatorname{Var}(N)(E[X_1])^2. \end{aligned}

For $N\sim\operatorname{Poisson}(\lambda)$, $E[N]=\operatorname{Var}(N)=\lambda$, so $\operatorname{Var}(S)=\lambda E[X_1^2]$. In particular, when $\lambda>0$, a finite aggregate variance requires a finite severity second moment. The aggregate distribution still includes its frequency structure: for example, $P(S=0)=P(N=0)$ when severities are strictly positive.

Worked Applications

Worked Example: Exponential Loss and Tail Risk

A hypothetical one-claim loss $X$ has an exponential distribution with mean $\theta=\$1{,}000$. There is no deductible, limit, or discounting. Calculate its variance, its 90% VaR, and its 90% TVaR. Use the continuous-distribution definitions above.

Method and assumptions. For an exponential loss with mean $\theta$, $S_X(x)=e^{-x/\theta}$ for $x\ge0$, and $E[X]=\theta$, $\operatorname{Var}(X)=\theta^2$. The distribution has a finite second moment, and its continuous, strictly increasing cumulative distribution function permits the stated quantile definition.

Variance. The variance is $\theta^2=(\$1{,}000)^2=\$1{,}000{,}000$, so the standard deviation is $\$1{,}000$.

90% VaR. Since $F_X(x)=1-e^{-x/\theta}$, solve $F_X(q)=0.90$: $1-e^{-q/1000}=0.90$, hence $e^{-q/1000}=0.10$, and $q=-1000\ln(0.10)=\$2{,}302.5851$. Thus the 90% VaR is $\$2{,}302.59$, rounded to the nearest cent.

90% TVaR. For an exponential random variable, the memoryless property gives $E[X\mid X>q]=q+\theta$. Therefore, the 90% TVaR is $\$2{,}302.5851+\$1{,}000=\$3{,}302.5851$, or $\$3{,}302.59$ to the nearest cent.

Check and trap avoided. The tail mean exceeds the VaR by exactly one exponential mean, as required by memorylessness. A common mistake is to report the VaR itself as TVaR; the calculation shows that TVaR averages the losses beyond the threshold and is not just the threshold.

Worked Example: Compound Poisson Aggregate Variance

A hypothetical portfolio has a Poisson number of claims per year with parameter $\lambda=2$. Each claim severity is independent and exponentially distributed with mean $\$500$. Calculate the annual aggregate mean and variance, and the probability of no claims. There is no discounting.

Method and assumptions. Let $N\sim\operatorname{Poisson}(2)$, let the severities be independent of $N$, and set $S=\sum_{j=1}^N X_j$. For an exponential severity with mean $\theta=\$500$, $E[X_j]=\$500$, $\operatorname{Var}(X_j)=(\$500)^2=\$250{,}000$, and $E[X_j^2]=2\theta^2=\$500{,}000$.

Aggregate mean. $E[S]=E[N]E[X_j]=2(\$500)=\$1{,}000$ per year.

Aggregate variance. Using the compound-frequency variance formula gives $\operatorname{Var}(S)=2(\$250{,}000)+2(\$500)^2=\$500{,}000+\$500{,}000=\$1{,}000{,}000$. Equivalently, the Poisson formula gives $\lambda E[X_j^2]=2(\$500{,}000)=\$1{,}000{,}000$. The annual aggregate standard deviation is $\$1{,}000$.

Probability of no claims. $P(S=0)=P(N=0)=e^{-2}=0.1353$ to four decimal places, because every exponential claim severity is positive with probability one.

Check and trap avoided. The two variance calculations agree. Using only $E[N]\operatorname{Var}(X_j)$ would give $\$500{,}000$, omitting the variation in claim count; the missing term is $\operatorname{Var}(N)(E[X_j])^2=\$500{,}000$. Also, this aggregate distribution has an atom at zero, so formulas stated only for continuous loss distributions cannot automatically be used for its tail risk.

Worked Example: Pareto Loss With Infinite Variance

A hypothetical ground-up loss $X$ follows a Pareto Type I distribution with minimum amount $m=\$100$ and shape parameter $\alpha=1.5$. Its survival function is $S_X(x)=(m/x)^\alpha$ for $x\ge m$. There are no coverage modifications or discounting. Determine whether its mean and variance are finite, then calculate its 90% VaR and TVaR.

Method and assumptions. For this Pareto distribution, the $r$th moment is finite only when $r<\alpha$. When finite, $E[X^r]=\alpha m^r/(\alpha-r)$. Since $1<1.5$, the mean exists; since $2$ is not less than $1.5$, the second moment and variance are infinite.

Mean. $E[X]=\alpha m/(\alpha-1)=1.5(\$100)/0.5=\$300$. The variance is infinite, not zero and not a finite number that can be found by subtracting moments.

90% VaR. For $q\ge m$, solve $S_X(q)=0.10$: $(100/q)^{1.5}=0.10$, so $q=100(0.10)^{-1/1.5}=100(10^{2/3})=\$464.1589$. The 90% VaR is $\$464.16$, rounded to the nearest cent.

90% TVaR. For a Pareto Type I loss with $\alpha>1$, integration of the conditional tail density gives $E[X\mid X>q]=\alpha q/(\alpha-1)$. Here this is $1.5(\$464.1589)/0.5=3(\$464.1589)=\$1{,}392.4767$, or $\$1{,}392.48$ to the nearest cent.

Check and trap avoided. The mean and this conditional tail mean are finite, while the variance is infinite. The mistake to avoid is assuming that a finite VaR or TVaR implies finite variance. The moment condition for variance is stricter here than the conditions needed for these particular tail calculations.

Exam Checks and Common Errors

Before calculating, write down the random variable in words and then in notation. State whether it is per loss, per payment, or aggregate; identify the time period and coverage basis; and record the assumptions connecting frequency and severity. Only then select a distribution formula. This sequence is especially useful when a problem supplies parameters for ground-up claims but asks about insurer payments or portfolio totals.

Mixing up severity and aggregate loss. A severity model describes one claim; an aggregate model also needs the frequency distribution.
Dropping zeros from the variable. The payment per loss includes exposures with no payment. A payment per payment is conditional on a positive payment and has a different distribution.
Assuming every variance exists. Check the second-moment condition. For Pareto Type I, the variance is finite only for $\alpha>2$, while the mean is finite for $\alpha>1$.
Using a continuous formula without checking the distribution. A compound model can have probability mass at zero even when each severity has a density.
Using a severity variance as aggregate variance. Claim-count variability contributes an additional term unless frequency is fixed.

Key takeaway: A risk measure belongs to a specified random loss, not merely to a named distribution. Define the loss, determine its distribution, check the required moment conditions, and only then calculate variance, VaR, or TVaR. These foundations are aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus.

Check Your Understanding

For each hypothetical scenario, identify the relevant distribution or moment condition and calculate the requested quantity. Round money to the nearest cent where needed.

A one-claim loss is exponential with mean $\$400$, with no coverage modification or discounting. Find its variance, 95% VaR, and 95% TVaR. (Answer: variance $=\$160{,}000$; VaR $=\$1{,}198.29$; TVaR $=\$1{,}598.29$.)
A portfolio has $N\sim\operatorname{Poisson}(3)$ claims per year, each with deterministic severity $\$200$, independently of $N$. Find the aggregate mean, variance, and probability of no claims. (Answer: mean $=\$600$; variance $=\$120{,}000$; probability of no claims $=e^{-3}=0.0498$ to four decimal places.)
A ground-up loss is Pareto Type I with minimum $\$50$ and shape $\alpha=2.5$. Determine whether its mean and variance are finite, and calculate both. (Answer: both are finite; mean $=\$83.33$; variance $=\$5{,}555.56$.)
A frequency count has $E[N]=4$ and $\operatorname{Var}(N)=4$. Independent claim severities have mean $\$300$ and variance $\$90{,}000$. Find the aggregate mean and variance, assuming the required moments exist. (Answer: mean $=\$1{,}200$; variance $=\$720{,}000$.)
A ground-up loss is uniform on $[0,\$1{,}000]$. The hypothetical insurer payment per loss is $Y^L=\min((X-\$200)_+,\$500)$. Find $P(Y^L=0)$, $P(Y^L=\$500)$, $E[Y^L]$, and $\operatorname{Var}(Y^L)$. (Answer: $P(Y^L=0)=0.2$; $P(Y^L=\$500)=0.3$; $E[Y^L]=\$275$; variance $=\$41{,}041.67$.)

What You'll Learn