Moment Matching for Aggregate Losses

Why Match Moments?

An aggregate loss is the sum of losses across risks, $S=\sum_{j=1}^{n}X_j$. Its distribution may be difficult to calculate directly, but its mean and variance can often be obtained from component moments and covariances. Moment matching uses those two quantities to fit a simpler distribution, then uses that fitted distribution to estimate quantities such as a high quantile.

A moment-matched distribution is an approximation, not a reconstruction of the aggregate loss distribution. Many distributions share the same mean and variance while having different skewness and tail behavior. This distinction matters particularly for insurance losses, which are often nonnegative and right-skewed. The method is aligned with the CAS Exam 9 (Risk Management for Actuaries) syllabus.

Definition: For an aggregate loss $S$ with finite mean $m=E[S]$ and finite variance $v_S=\operatorname{Var}(S)$, a moment-matched approximation is a selected distribution whose mean is $m$ and whose variance is $v_S$. A normal approximation and a lognormal approximation are two common choices. Matching the first two moments does not, by itself, identify the distribution or guarantee accurate tail quantiles.

Aggregate Mean and Variance

The aggregate mean is the sum of the component means, whether the losses are dependent or independent. The aggregate variance also depends on the covariances between components:

E[S]=\sum_{j=1}^{n}E[X_j], \qquad \operatorname{Var}(S)=\sum_{j=1}^{n}\operatorname{Var}(X_j) +2\sum_{j<k}\operatorname{Cov}(X_j,X_k).

The covariance term must be included unless independence, or another justified zero-covariance assumption, makes it zero. Under independence, the variance reduces to the sum of component variances. For a normal approximation, write $S\approx N(m,v_S)$, where the second parameter is the variance. With $z_p$ denoting the $p$-quantile of the standard normal distribution, the approximate $p$-quantile is

\operatorname{VaR}_p(S)\approx m+z_p\sqrt{v_S}.

This formula is the quantile formula for the fitted normal distribution. It does not assert that the actual aggregate distribution is normal. If losses are nonnegative, a normal approximation can also assign positive probability to impossible negative values; the practical importance of that issue depends on the ratio of the standard deviation to the mean.

Worked Example: Including Dependence in a Normal Match

A hypothetical portfolio has two loss components. $X_1$ equals $100 with probability 0.20 and otherwise equals zero; $X_2$ equals $50 with probability 0.40 and otherwise equals zero. Their joint probabilities are $P(X_1=100,X_2=50)=0.12$, $P(X_1=100,X_2=0)=0.08$, $P(X_1=0,X_2=50)=0.28$, and $P(X_1=0,X_2=0)=0.52$. Find the mean and variance of $S=X_1+X_2$, then estimate its 95th percentile using a normal match. No interest or mortality assumptions apply.

Method. Calculate the marginal moments, calculate the covariance from the joint probabilities, and use the normal quantile formula. The stated joint probabilities sum to $0.12+0.08+0.28+0.52=1.00$, and give the specified marginal probabilities.

For a two-point loss equal to $a$ with probability $p$ and zero otherwise, the mean is $ap$ and the variance is $a^2p(1-p)$. Thus,

E[X_1]=100(0.20)=20,\qquad \operatorname{Var}(X_1)=100^2(0.20)(0.80)=1{,}600,

E[X_2]=50(0.40)=20,\qquad \operatorname{Var}(X_2)=50^2(0.40)(0.60)=600.

The expected product is $E[X_1X_2]=100(50)(0.12)=600$, while $E[X_1]E[X_2]=20(20)=400$. Therefore,

\operatorname{Cov}(X_1,X_2)=600-400=200.

It follows that $E[S]=20+20=40$ and \[ \operatorname{Var}(S)=1{,}600+600+2(200)=2{,}600. \] The standard deviation is $\sqrt{2{,}600}=50.9902$. Using $z_{0.9500}=1.644854$, the normal estimate is $40+1.644854(50.9902)=123.871$, or $123.87 to the nearest cent. The calculation is consistent: the positive covariance increases aggregate variance above the independent-components value of $1{,}600+600=2{,}200$.

Common mistake avoided: Setting covariance to zero without an independence assumption would give a variance of 2,200 and an estimated 95th percentile of about $117.14. The given joint probabilities show that the losses are dependent, so the correct covariance contribution is $2(200)=400$.

Fitting a Lognormal Distribution

A lognormal approximation is positive-valued and can represent right skew. In the convention used here, $S$ is lognormal if $\ln S\sim N(\mu_L,\sigma_L^2)$. If its desired mean and variance are $m>0$ and $v_S>0$, matching those moments gives

\sigma_L^2=\ln\!\left(1+\frac{v_S}{m^2}\right), \qquad \mu_L=\ln(m)-\frac{\sigma_L^2}{2}.

The associated fitted lognormal $p$-quantile is $\exp(\mu_L+z_p\sigma_L)$. These equations require a positive mean and positive variance. They fit the first two moments; they do not also match an arbitrary aggregate skewness.

The parameter formulas follow from the lognormal moment identities. If $\ln S\sim N(\mu_L,\sigma_L^2)$, then $E[S]=\exp(\mu_L+\sigma_L^2/2)$ and $\operatorname{Var}(S)=(\exp(\sigma_L^2)-1)\exp(2\mu_L+\sigma_L^2)$. Dividing the variance by the square of the mean gives $v_S/m^2=\exp(\sigma_L^2)-1$. Solving yields $\sigma_L^2=\ln(1+v_S/m^2)$; taking the logarithm of the mean identity then gives $\mu_L=\ln(m)-\sigma_L^2/2$.

Worked Example: Comparing Approximations for a Skewed Aggregate

Suppose a hypothetical aggregate $S=X_1+X_2$ consists of two independent exponential losses, each with mean $10. For an exponential distribution with mean $\theta$, the rate is $1/\theta$, the variance is $\theta^2$, and the survival function is $\exp(-x/\theta)$ for $x\geq0$. There is no interest or mortality component. Estimate the 95th percentile by normal and lognormal moment matching, then compare both estimates with the actual percentile.

Aggregate moments. Each component has mean 10 and variance $10^2=100$. Independence makes the covariance zero, so $E[S]=10+10=20$ and $\operatorname{Var}(S)=100+100=200$.

Normal match. The standard deviation is $\sqrt{200}=14.1421$. Thus the estimated 95th percentile is $20+1.644854(14.1421)=43.2617$, or $43.26.

Lognormal match. The variance-to-squared-mean ratio is $200/20^2=0.5$. Therefore $\sigma_L^2=\ln(1.5)=0.405465$, $\sigma_L=\sqrt{0.405465}=0.636761$, and $\mu_L=\ln(20)-0.405465/2=2.7929997$. The 95th percentile estimate is \[ \exp(2.7929997+1.644854(0.636761)) =\exp(3.840379) \approx 46.54. \]

Actual percentile. The sum of two independent exponential losses with the same mean 10 has a gamma distribution with shape 2 and scale 10. Its distribution function for $s\geq0$ is $F_S(s)=1-\exp(-s/10)(1+s/10)$. Setting $F_S(s)=0.95$ gives $\exp(-s/10)(1+s/10)=0.05$. Solving this scalar equation gives $s/10\approx4.743865$, so the actual percentile is approximately $47.44.

The normal estimate is about $4.18 below the actual percentile, and the lognormal estimate is about $0.90 below it. This example does not establish that one approximation is always superior; it shows that even a positive, skew-capable lognormal match can miss a tail quantile. The exact percentile here is available because the aggregate distribution is known; in other applications, the actual quantile may not be available in closed form.

Common mistake avoided: Treating the matched lognormal quantile as the actual aggregate quantile confuses a fitted approximation with the underlying distribution. The lognormal parameters reproduce the mean and variance, but those two moments alone do not force the fitted tail to coincide with the actual tail.

How to Use and Interpret the Approximation

A sound calculation separates the construction of the approximation from the interpretation of its result. First establish the aggregate mean and variance on a consistent loss basis. Then choose and identify the fitted distribution before calculating its quantile. Finally, assess whether the distributional shape is credible for the intended use, especially when the estimate is a high quantile.

Specify the aggregate.
Define $S$ as the sum of the losses on a common basis and identify any dependence information needed for covariances.

Calculate its moments.
Sum component means and include all pairwise covariance terms in the aggregate variance.

Select a matching family.
Use the normal formula when a symmetric approximation is defensible, or fit a lognormal when a positive, right-skewed approximation is more suitable.

Label and assess the result.
Report the quantile as an approximation and consider whether skewness, discreteness, or tail behavior could make it materially inaccurate.

The normal and lognormal matches answer different modeling needs, but neither is universally appropriate. A normal approximation can be useful when aggregation makes the distribution reasonably symmetric and the chance of negative fitted losses is negligible for the decision at hand. A lognormal approximation respects positivity and introduces right skew, but a single lognormal shape may still understate or overstate extreme losses. If the aggregate has atoms, the exact quantile is defined by the distribution function as $\inf\{s:F_S(s)\geq p\}$; a smooth fitted distribution may not reproduce that jump.

Common Exam Traps

Omitting dependence. The variance of a sum includes twice each pairwise covariance. Independence must be justified, not presumed from separate component descriptions.
Using standard deviation as variance. In $N(m,v_S)$, the second parameter is the variance. The quantile calculation uses $\sqrt{v_S}$.
Calling an approximation exact. Matching mean and variance does not match all distributional features, particularly skewness and tail weight.
Using a normal match without checking support. A normal distribution assigns some probability to negative values even when aggregate losses cannot be negative.
Rounding too early. Keep unrounded moments and fitted parameters through the quantile calculation; round only the final monetary result.

Key takeaway: Moment matching converts aggregate mean and variance into a tractable fitted distribution, not a guarantee of accurate tail risk. Compute dependence correctly, state which family is fitted, and interpret its quantiles as approximations—especially for skewed losses.

Check Your Understanding

For each problem, identify the assumptions and keep unrounded values until the final reported answer.

Two independent hypothetical losses have means $30 and $20 and variances 900 and 400. Find the aggregate mean, variance, and normal-matched 95th percentile using $z_{0.9500}=1.644854$. (Answer: mean $50, variance 1,300, normal estimate $109.31)
Two dependent hypothetical losses have means $25 and $15, variances 500 and 300, and covariance 100. Find the aggregate mean, variance, and normal-matched 95th percentile using $z_{0.9500}=1.644854$. (Answer: mean $40, variance 1,000, normal estimate $92.01)
Independent losses $X_1$ and $X_2$ equal $80 with probability 0.10 and $40 with probability 0.25, respectively, and otherwise are zero. Find the aggregate mean, variance, and normal-matched 95th percentile using $z_{0.9500}=1.644854$. (Answer: mean $18, variance 876, normal estimate $66.68)
An aggregate has mean $50 and variance 500. Fit a lognormal distribution by matching those moments and estimate its 95th percentile using $z_{0.9500}=1.644854$. (Answer: $\sigma_L^2\approx0.1823$, $\mu_L\approx3.8209$, 95th-percentile estimate approximately $92.13)

What You'll Learn