Normal Approximation From Aggregate Moments

Why Approximate an Aggregate Loss?

An aggregate loss may combine a random number of claims with random amounts per claim. A recursion can produce probabilities for a discrete aggregate distribution, but a normal approximation can provide a quicker estimate when the aggregate is sufficiently spread out. The essential inputs are the aggregate mean and variance, not just the mean and variance of an individual claim.

This tutorial develops the normal approximation for cumulative probabilities and for point probabilities on a lattice. The point-probability calculation uses a local density approximation; it is distinct from adjusting a cumulative-probability threshold. The approximation is not exact, and its usefulness depends on the shape and spread of the aggregate distribution.

Definition: Let $N$ be a nonnegative integer-valued claim count, and let $X_1,X_2,\ldots$ be independent, identically distributed claim severities, independent of $N$. The aggregate loss is $S=\sum_{j=1}^{N}X_j$, with $S=0$ when $N=0$. Write $m=E[X]$, $\sigma_X^2=\operatorname{Var}(X)$, and $q=E[X^2]=\sigma_X^2+m^2$.

Aggregate Moments and the Normal Model

Conditional on $N=n$, the aggregate is the sum of $n$ independent severities. Therefore $E[S\mid N=n]=nm$ and $\operatorname{Var}(S\mid N=n)=n\sigma_X^2$. Applying the laws of total expectation and total variance gives the aggregate moments.

\mu_S=E[S]=E[N]m, \qquad \sigma_S^2=\operatorname{Var}(S) =E[N]\sigma_X^2+\operatorname{Var}(N)m^2.

An equivalent expression, using $q=E[X^2]$, is

\operatorname{Var}(S) =E[N]q+\bigl(\operatorname{Var}(N)-E[N]\bigr)m^2.

The first variance formula separates two sources of variation: random claim amounts, represented by $E[N]\sigma_X^2$, and random claim count, represented by $\operatorname{Var}(N)m^2$. In particular, using only the expected number of claims to scale the severity variance omits the second source.

Normal approximation: If $\sigma_S^2>0$ and the aggregate distribution is sufficiently spread out for a normal model to be reasonable, approximate $S$ by $N(\mu_S,\sigma_S^2)$. For a real threshold $x$, this gives $P(S\leq x)\approx\Phi\!\left((x-\mu_S)/\sigma_S\right)$, where $\Phi$ is the standard normal cumulative distribution function and $\sigma_S=\sqrt{\sigma_S^2}$. This is an approximation, not an identity.

The approximation is most credible when the aggregate combines enough independent contributions to have a reasonably smooth, approximately symmetric distribution. A large variance alone does not guarantee accuracy: a low claim frequency, a large probability of no claims, or highly uneven severities can leave the aggregate strongly nonnormal. Check that the mean, variance, claim-count convention, and units all match the model before standardizing a threshold.

Why the Aggregate Variance Formula Holds

The variance result follows directly from conditioning on the claim count. Given $N=n$, the $n$ severities are independent, so their means add and their variances add. The law of total variance then accounts for both variation within a fixed claim count and variation between different claim counts.

Condition on the count.
For $N=n$, $S$ is a sum of $n$ independent severities, so $E[S\mid N=n]=nm$ and $\operatorname{Var}(S\mid N=n)=n\sigma_X^2$.

Average the conditional mean.
$E[S]=E[E[S\mid N]]=E[Nm]=E[N]m$.

Combine the two variance components.
$\operatorname{Var}(S)=E[\operatorname{Var}(S\mid N)]+\operatorname{Var}(E[S\mid N])=E[N]\sigma_X^2+\operatorname{Var}(N)m^2$.

For example, if $N$ is Poisson with parameter $\lambda$, then $E[N]=\operatorname{Var}(N)=\lambda$. Substituting into the aggregate variance formula yields $\operatorname{Var}(S)=\lambda(\sigma_X^2+m^2)=\lambda E[X^2]$. This simplification depends on the Poisson count having equal mean and variance; it should not be used automatically for other frequency models.

Cumulative Probabilities and Lattice Point Probabilities

For a cumulative probability, standardize the threshold using the aggregate mean and standard deviation, then evaluate the standard normal CDF. If the requested event is an interval, subtract two cumulative approximations. For example, $P(a<S\leq b)$ is approximated by $\Phi((b-\mu_S)/\sigma_S)-\Phi((a-\mu_S)/\sigma_S)$.

A point probability requires different treatment. If $S$ is supported on a lattice with maximal span $h$, its possible values lie on a grid separated by $h$. A local normal approximation estimates the probability at a grid point $s$ by the normal density there, multiplied by the grid width:

P(S=s)\ \approx\ h\,\frac{1}{\sigma_S} \phi\!\left(\frac{s-\mu_S}{\sigma_S}\right),

where $\phi(z)=\exp(-z^2/2)/\sqrt{2\pi}$ is the standard normal density. The factor $h$ matters because a density is probability per unit of loss, whereas a point probability is dimensionless. For a continuous aggregate distribution, an individual point has probability zero, so this lattice formula does not apply. The local approximation is most useful when the distribution has enough spread and the point is in a region where the normal density represents its shape reasonably well.

Worked Example: Poisson Frequency With Two Possible Claim Sizes

A hypothetical portfolio has $N\sim\operatorname{Poisson}(100)$ claims. Independently, each claim is \$1 or \$3 with probability $0.5$ each. Approximate (a) $P(S\leq \$220)$ and (b) $P(S=\$210)$.

Method and moments. For one claim, $m=0.5(1)+0.5(3)=2$. Also, $E[X^2]=0.5(1^2)+0.5(3^2)=5$, so $\sigma_X^2=5-2^2=1$. The Poisson count has mean and variance 100. Thus

\mu_S=100(2)=200, \qquad \sigma_S^2=100(1)+100(2^2)=500, \qquad \sigma_S=\sqrt{500}\approx22.3607.

(a) Cumulative probability. The standardized threshold is $(220-200)/\sqrt{500}=20/22.3607\approx0.8944$. Therefore $P(S\leq \$220)\approx\Phi(0.8944)\approx\mathbf{0.8145}$, to four decimal places.

(b) Point probability. The severity values are integer dollars, so the aggregate is on a lattice with span $h=\$1$. At $s=210$, $z=(210-200)/\sqrt{500}\approx0.4472$. The local normal estimate is

P(S=210)\approx \frac{1}{\sqrt{500}}\phi(0.4472) \approx \frac{e^{-0.1}}{\sqrt{2\pi(500)}} = \frac{0.9048}{56.0499} \approx\mathbf{0.0161}.

Both estimates use the same aggregate moments, but the point calculation also uses the lattice span. The method avoids treating a density value by itself as a point probability.

Worked Example: Fixed Severity With Binomial Frequency

A hypothetical contract portfolio has $N\sim\operatorname{Binomial}(80,0.25)$ claims, each with a fixed payment of \$4. Approximate (a) $P(S\leq \$100)$ and (b) $P(S=\$80)$.

Method and moments. The count has $E[N]=80(0.25)=20$ and $\operatorname{Var}(N)=80(0.25)(0.75)=15$. Since $X=\$4$ with certainty, $m=4$ and $\sigma_X^2=0$. Hence

\mu_S=20(4)=80, \qquad \sigma_S^2=20(0)+15(4^2)=240, \qquad \sigma_S=\sqrt{240}\approx15.4919.

(a) Cumulative probability. The standardized threshold is $(100-80)/\sqrt{240}=20/15.4919\approx1.2910$. Thus $P(S\leq \$100)\approx\Phi(1.2910)\approx\mathbf{0.9016}$, to four decimal places.

(b) Point probability. The aggregate is $4N$, so its lattice span is $h=\$4$. The requested point is the mean, giving $z=(80-80)/\sqrt{240}=0$. Therefore

P(S=80)\approx 4\frac{1}{\sqrt{240}}\phi(0) = \frac{4}{\sqrt{2\pi(240)}} \approx\mathbf{0.1030}.

This point estimate uses a \$4 span, not a \$1 span: only multiples of \$4 are possible. The count variance contributes the entire aggregate variance here because the individual severity is fixed.

Worked Example: Negative Binomial Convention and Mixed Severity

A hypothetical claim count $N$ is the number of failures before the third success in independent trials, with success probability $p=0.4$. Each claim pays \$2 or \$6 with probability $0.5$ each, independently of the count and other severities. Approximate (a) $P(S\leq \$32)$ and (b) $P(S=\$18)$.

Method and count convention. For the number of failures before $r$ successes, $E[N]=r(1-p)/p$ and $\operatorname{Var}(N)=r(1-p)/p^2$. Here $r=3$, so

E[N]=\frac{3(0.6)}{0.4}=4.5, \qquad \operatorname{Var}(N)=\frac{3(0.6)}{0.4^2} =\frac{1.8}{0.16}=11.25.

For severity, $m=0.5(2)+0.5(6)=4$, $E[X^2]=0.5(4)+0.5(36)=20$, and $\sigma_X^2=20-4^2=4$. The aggregate moments are

\mu_S=4.5(4)=18, \qquad \sigma_S^2=4.5(4)+11.25(4^2)=18+180=198, \qquad \sigma_S=\sqrt{198}\approx14.0712.

(a) Cumulative probability. The standardized threshold is $(32-18)/\sqrt{198}=14/14.0712\approx0.9949$. Thus $P(S\leq \$32)\approx\Phi(0.9949)\approx\mathbf{0.8400}$, to four decimal places.

(b) Point probability. Both severity amounts are multiples of \$2, and all possible aggregate amounts lie on a \$2 lattice. At the mean, $z=0$, so

P(S=18)\approx 2\frac{1}{\sqrt{198}}\phi(0) = \frac{2}{\sqrt{2\pi(198)}} \approx\mathbf{0.0567}.

This calculation uses the failures-before-successes convention explicitly. Reversing the convention changes the claim-count moments and therefore changes both aggregate estimates.

Common Mistakes and Checks

Moment-based calculations are short, but a mistaken model assumption can carry through every subsequent step. Before evaluating a normal probability, check whether the claim-count parameterization and severity units match the problem, and whether a requested probability is cumulative or at a single lattice point.

Using the wrong negative binomial convention. If $N$ counts failures before $r$ successes, use $E[N]=r(1-p)/p$. For the third example, the incorrect failures-before-successes mean calculation $3(0.4)/(0.6)=2$ would give aggregate mean $2(4)=8$, rather than the correct mean $18$. The count variances would also differ.
Dropping count variability. The term $\operatorname{Var}(N)m^2$ remains even when claim severities are fixed. Omitting it would incorrectly assign zero variance to the binomial example.
Using the wrong lattice span. A point estimate at a multiple of \$4 in the fixed-severity example must use $h=\$4$, not $h=\$1$.
Confusing density and probability. The normal density at a point is not itself an approximation to a discrete point mass. For a lattice local approximation, multiply the density by the lattice span.
Reporting an approximation as exact. A normal model may be inaccurate for a highly skewed aggregate or an aggregate with substantial probability at zero. Label the result as an approximation and check whether the model is plausible.

Key takeaway: First calculate $\mu_S=E[N]E[X]$ and $\sigma_S^2=E[N]\operatorname{Var}(X)+\operatorname{Var}(N)(E[X])^2$. Use these to standardize cumulative thresholds. For a discrete point probability, also identify the aggregate lattice span and multiply the normal density by that span.

Check Your Understanding

For each question, use the stated frequency and severity model, and report probabilities to four decimal places.

A hypothetical portfolio has $N\sim\operatorname{Poisson}(50)$, and each claim is exactly \$2. Use the normal approximation to estimate $P(S\leq \$110)$. (Answer: $0.7602$)
A hypothetical count is $N\sim\operatorname{Binomial}(40,0.5)$, and every claim pays \$3. Use a lattice local normal approximation for $P(S=\$60)$. (Answer: $0.1261$)
A hypothetical count is the number of failures before two successes, with success probability $p=0.4$. Each claim pays exactly \$5. Use the normal approximation to estimate $P(S\leq \$25)$. (Answer: $0.7673$)
A hypothetical portfolio has $N\sim\operatorname{Poisson}(25)$. Each claim is \$1 or \$5 with probability $0.5$ each. Use the lattice local normal approximation for $P(S=\$75)$. (Answer: $0.0221$)

What You'll Learn