Common Errors in Loss Distribution Calculations

Why Loss Calculations Need a Diagnostic Check

A loss calculation can look plausible and still be wrong because a probability table is not normalized, a rate is applied to the wrong exposure period, or a claim count is confused with a claim amount. These errors can flow into estimates of expected loss, variability, and risk measures. A reliable check begins by asking what each number represents, what its units are, and which assumptions support the calculation.

This tutorial focuses on diagnosing those errors rather than developing a new loss model. In a compound frequency-severity model, $N$ is the number of losses and $X_1,X_2,\ldots$ are individual loss amounts. Aggregate loss is $S=\sum_{i=1}^{N}X_i$, with $S=0$ when $N=0$. Unless stated otherwise, the severity amounts are independent and identically distributed, independent of $N$, and have finite means and variances.

Definition: A discrete probability distribution is normalized when its probabilities are nonnegative and sum to one. In a frequency-severity model, frequency describes the random number of claims; severity describes the loss amount for an individual claim. Their units and roles are different and should not be interchanged.

Check Probability Normalization Before Calculating

For a discrete loss $X$ taking values $x_j$ with probabilities $p_j$, the probabilities must satisfy $\sum_j p_j=1$. The expected loss is then $E[X]=\sum_j x_jp_j$. If reported weights do not sum to one but are intended to be proportional probabilities, divide each weight by their total before calculating probabilities or expectations. Do not silently treat unnormalized weights as probabilities.

Worked Example: Repairing an Unnormalized Loss Table

A hypothetical insurer models a single policy loss as $0, $1,000, or $5,000. A draft table assigns respective weights 0.20, 0.50, and 0.28. Treat the weights as proportional probabilities. Find the normalized probabilities, the probability of a loss of at least $5,000, and the expected loss.

Method and normalization check. First add the weights: $0.20+0.50+0.28=0.98$. Since 0.98 is not one, these weights cannot be used directly as probabilities. Divide each weight by 0.98:

p_0=\frac{0.20}{0.98}=0.2040816,\qquad p_{1000}=\frac{0.50}{0.98}=0.5102041,\qquad p_{5000}=\frac{0.28}{0.98}=0.2857143.

The normalized probabilities sum to $0.2040816+0.5102041+0.2857143=1.0000001$, with the small excess due to displayed rounding. Using the exact fractions, their sum is $(0.20+0.50+0.28)/0.98=1$.

Probability of at least $5,000. Only the $5,000 outcome qualifies, so the probability is $0.28/0.98=0.2857143$, or 0.2857 to four decimals.

Expected loss. The zero-loss outcome contributes zero: $E[X]=0(0.20/0.98)+1{,}000(0.50/0.98)+5{,}000(0.28/0.98)$. The numerator is $0+500+1{,}400=1{,}900$, so $E[X]=1{,}900/0.98=\$1{,}938.7755$, or $1,938.78 to the nearest cent.

Common mistake avoided. Using the unnormalized weights gives an apparent expected loss of $1,900, but the weights total only 0.98. The corrected expectation is $1,900 divided by 0.98, not $1,900.

Track Units and Keep Frequency Separate from Severity

A frequency rate is meaningful only together with its exposure unit. A rate of 0.04 claims per policy-year is not a rate of 0.04 claims per month. Multiplying a rate by exposure produces an expected claim count; multiplying that expected count by mean severity produces expected aggregate loss. Severity is measured in currency per claim, while aggregate loss is measured in currency for the full modeled portfolio and period.

Units are especially useful when calculating variance. If severity is measured in dollars, its variance is measured in dollars squared. Aggregate variance is also in dollars squared. It is not an expected dollar amount and should not be added to a mean.

Worked Example: Applying a Claim Rate to the Correct Exposure

A hypothetical portfolio has 250 vehicle-years of exposure over a stated one-year period. Claim frequency is Poisson with rate 0.04 claims per vehicle-year. Individual claim severity has mean $12,000 and variance $64,000,000. Assume claim severities are independent and identically distributed, independent of frequency, with finite variance. Calculate the expected number of claims, expected aggregate loss, and aggregate variance.

Step 1: Convert exposure into the modeled claim count. The rate is per vehicle-year, and the exposure is 250 vehicle-years. Thus the Poisson parameter for the portfolio period is $0.04\times250=10$. Therefore $E[N]=\operatorname{Var}(N)=10$.

Step 2: Calculate expected aggregate loss. The compound mean is $E[S]=E[N]E[X]$. Substitution gives $10\times\$12{,}000=\$120{,}000$.

Step 3: Calculate aggregate variance. First find the severity second moment: $E[X^2]=\operatorname{Var}(X)+(E[X])^2=\$64{,}000{,}000+(\$12{,}000)^2$. Since $(12{,}000)^2=144{,}000{,}000$, $E[X^2]=\$208{,}000{,}000$ in squared-dollar units. For Poisson frequency, $\operatorname{Var}(S)=E[N]E[X^2]$, so the aggregate variance is $10\times\$208{,}000{,}000=\$2{,}080{,}000{,}000$, in squared dollars.

Check and common mistake avoided. The expected claim count is 10, not 0.04: the rate must be multiplied by 250 vehicle-years. The resulting mean is in dollars, while the variance is in squared dollars. Multiplying 10 by the severity variance alone would omit the variability attributable to random claim counts.

Read Discrete Quantiles From the Cumulative Distribution

For a loss distribution with cumulative distribution function $F_X(x)=P(X\le x)$, a commonly used population quantile convention is the generalized inverse $q_p=\inf\{x:F_X(x)\ge p\}$, for $0<p<1$. In a discrete distribution, the cumulative probability can jump over $p$; the quantile is the loss value at the first point where the cumulative probability reaches or exceeds $p$. Sample quantiles can also depend on the stated empirical convention, so a calculation should identify which convention it uses.

At a probability atom, strict and non-strict exceedance probabilities differ. In particular, $P(X>x)$ excludes losses equal to $x$, while $P(X\ge x)$ includes them. Check the inequality in the question before reading a tail probability from a table.

Worked Example: Finding a Discrete Loss Quantile Without Reversing the Tail

A hypothetical annual loss $L$ has the following distribution: $0 with probability 0.70, $100 with probability 0.20, $400 with probability 0.08, and $1,000 with probability 0.02. Using the generalized-inverse convention, find the 95th percentile. Then find $P(L\ge \$400)$ and $P(L>\$400)$.

Method. Form the cumulative probabilities in increasing loss order: $F_L(0)=0.70$, $F_L(100)=0.70+0.20=0.90$, $F_L(400)=0.90+0.08=0.98$, and $F_L(1{,}000)=1.00$.

The cumulative probability at $100 is 0.90, below 0.95; at $400 it is 0.98, at least 0.95. Thus the 95th percentile is $400.

For $P(L\ge \$400)$, include both the $400 and $1,000 outcomes: $0.08+0.02=0.10$. For $P(L>\$400)$, include only the $1,000 outcome: $0.02$. The check is that the difference, 0.08, equals the probability mass at exactly $400.

Common mistake avoided. The 95th percentile is not found by looking for a loss with 95% probability above it. The specified convention accumulates probability from the low-loss end until it reaches 0.95. Also, replacing $L\ge \$400$ with $L>\$400$ would incorrectly omit the 0.08 probability at $400.

Verify Compound Aggregate Formulas

For $S=\sum_{i=1}^{N}X_i$, let $E[N]=m$, $\operatorname{Var}(N)=v_N$, $E[X_i]=\mu$, and $\operatorname{Var}(X_i)=\sigma^2$. Under the independence and identical-distribution assumptions stated above, the aggregate mean and variance are:

E[S]=m\mu,\qquad \operatorname{Var}(S)=m\sigma^2+v_N\mu^2.

The formula requires finite moments and independence of severities from frequency. If frequency and severity are dependent, or severities are dependent across claims, these formulas need not apply as written. The variance formula has two components: variability in claim amounts for a given count and variability in the count itself.

Condition on the claim count.
Given $N=n$, the sum has conditional mean $n\mu$ and conditional variance $n\sigma^2$, because it is a sum of $n$ independent severities.

Average the conditional variance.
$E[\operatorname{Var}(S\mid N)]=E[N\sigma^2]=m\sigma^2$.

Measure variation in the conditional mean.
$\operatorname{Var}(E[S\mid N])=\operatorname{Var}(N\mu)=v_N\mu^2$.

Combine the components.
The law of total variance gives $\operatorname{Var}(S)=E[\operatorname{Var}(S\mid N)]+\operatorname{Var}(E[S\mid N])=m\sigma^2+v_N\mu^2$.

Worked Example: Checking the Compound Variance Arithmetic

Suppose $E[N]=5$, $\operatorname{Var}(N)=7$, $E[X_i]=\$300$, and $\operatorname{Var}(X_i)=\$900$. Assume independent and identically distributed severities, independent of $N$, with finite variance. Find the aggregate mean and variance.

Method. Apply the compound mean and variance formulas. The mean is $E[S]=5\times\$300=\$1{,}500$.

For variance, calculate each term separately. The claim-amount term is $5\times\$900=\$4{,}500$. The claim-count term is $7\times(\$300)^2=7\times\$90{,}000=\$630{,}000$. Adding gives $\operatorname{Var}(S)=\$4{,}500+\$630{,}000=\$634{,}500$, in squared dollars.

Check and common mistake avoided. The aggregate variance is $634,500 squared, not $3,195,000 squared: direct substitution gives $4,500 plus $630,000, which equals $634,500. The mean, $1,500, is in dollars and must not be added to the variance. Also, replacing $\operatorname{Var}(N)=7$ with $E[N]=5$ in the second term would give the wrong model calculation.

A Practical Error-Checking Routine

When reviewing a loss calculation, use a consistent sequence rather than relying on whether the result seems reasonable. This is particularly useful when a problem combines exposure, claim counts, individual losses, or a discrete table.

Check the inputs. Confirm that probabilities sum to one and that stated weights are actually probabilities rather than unnormalized weights.
Write the units beside each quantity. A rate per exposure unit must be multiplied by exposure in that same unit; a variance has squared units.
Identify the random variable. Determine whether a number describes claim frequency, individual severity, or aggregate loss.
State the convention. For a discrete quantile, specify the cumulative-distribution rule; for exceedance, distinguish $>$ from $\ge$.
Substitute before simplifying. Show the separate terms in an aggregate formula and check their arithmetic independently.

These checks support later risk-measure calculations without replacing the definitions of those measures. In particular, the discrete quantile calculation above is a distribution check: it does not remove the need to state a quantile convention when one is requested.

Key takeaway: A defensible loss calculation makes its probability total, exposure basis, frequency-severity roles, quantile convention, model assumptions, and units visible. Those details often reveal an error before it affects a final risk measure.

Check Your Understanding

For each problem, identify the relevant check before calculating. Answers use the stated conventions and assumptions.

A hypothetical loss takes values $0, $200, and $1,000 with proportional weights 0.30, 0.40, and 0.25. Normalize the weights and find the expected loss and probability of a loss of at least $1,000. (Answer: normalized probabilities 0.3158, 0.4211, and 0.2632; expected loss $347.37; probability 0.2632.)
A Poisson claim rate is 0.03 claims per vehicle-year for a portfolio with 400 vehicle-years during the modeled period. Mean claim severity is $7,500. Under the compound model assumptions, find the expected claim count and expected aggregate loss. (Answer: expected claim count 12; expected aggregate loss $90,000.)
A loss is $0 with probability 0.60, $50 with probability 0.25, and $300 with probability 0.15. Using $q_p=\inf\{x:F_X(x)\ge p\}$, find the 90th percentile and both $P(X\ge \$300)$ and $P(X>\$300)$. (Answer: 90th percentile $300; $P(X\ge \$300)=0.15$; $P(X>\$300)=0$.)
For a compound model, $E[N]=4$, $\operatorname{Var}(N)=6$, $E[X_i]=\$500$, and $\operatorname{Var}(X_i)=\$2,500$. Assuming independent and identically distributed severities independent of $N$, find aggregate mean and variance. (Answer: mean $2,000; variance $1,510,000 squared dollars.)
Claim frequency is Poisson with mean 3 over the modeled period. Independent claim severities have mean $200 and variance $1,600. Find expected aggregate loss and aggregate variance, checking the severity second moment. (Answer: expected aggregate loss $600; $E[X^2]=\$41,600$ squared dollars; aggregate variance $124,800 squared dollars.)

What You'll Learn