Frequency Changes From Changing Risk Mix

Why Risk Mix Changes Frequency

A portfolio’s claim frequency depends not only on the frequency of each kind of risk, but also on how many risks of each kind are insured. If a book shifts toward classes with higher expected claim counts, its overall expected count can rise even when the frequency within every class stays unchanged. Separating these effects is useful when comparing portfolios or assessing a proposed change in underwriting mix.

The calculation also depends on what is known about the composition. If the numbers of risks in each class are fixed, the portfolio count has one variance. If each risk’s class is itself randomly assigned, uncertainty in the composition contributes additional variation. These are different models, even when they have the same expected class proportions.

Definition: Suppose a portfolio contains risk classes \(j=1,\ldots,k\). Let \(w_j\) be the proportion of risks in class \(j\), where \(w_j\geq 0\) and \(\sum_{j=1}^k w_j=1\). Let \(\lambda_j\) be the expected claim count per unit exposure in class \(j\). The weighted average frequency is \(\bar{\lambda}=\sum_{j=1}^k w_j\lambda_j\). For \(M\) equal-exposure risks, the expected total count is \(M\bar{\lambda}\), provided the class proportions describe those risks.

Here, frequency is an expected number of claims per unit exposure, not a probability unless the model and exposure make it one. For example, a Poisson count can exceed one. The weighted average is a statement about the mean; by itself, it does not always determine the variance of the portfolio count.

Expected Counts When the Composition Is Fixed

First suppose that the portfolio has a known, fixed number \(n_j\) of risks in each class. Assume that, conditional on class, each risk has an independent Poisson claim count with mean \(\lambda_j\) over the exposure period. The total count \(N\) is the sum of these class-specific counts. Its expectation is the sum of the class expectations. Its variance is also the sum of the class variances, because the counts are independent.

E[N]=\sum_{j=1}^k n_j\lambda_j, \qquad \operatorname{Var}(N)=\sum_{j=1}^k n_j\lambda_j.

Thus, with a fixed class composition, the total is Poisson with mean \(\sum_j n_j\lambda_j\), under the stated assumptions. Equivalently, if \(M=\sum_j n_j\) and \(w_j=n_j/M\), its mean and variance are both \(M\bar{\lambda}\). The mix affects the total by changing the weighted average frequency.

Worked Example: Comparing Two Fixed Portfolio Mixes

A hypothetical insurer compares two portfolios of 800 one-year policies. Lower-frequency risks have a Poisson claim count with mean 0.10 per policy-year, and higher-frequency risks have a Poisson claim count with mean 0.30. Counts are independent conditional on class. The first portfolio has 75% lower-frequency risks; the second has 60%. Find the expected count and variance for each portfolio.

Method. The numbers of risks in each class are fixed, so calculate the mean by summing class count times class frequency. Under independent Poisson counts, the total variance equals the total mean.

First portfolio. There are \(800(0.75)=600\) lower-frequency risks and \(800(0.25)=200\) higher-frequency risks. The expected count is

$$ 600(0.10)+200(0.30)=60+60=120. $$

Its variance is also 120 claims squared, since the total is Poisson under the model.

Second portfolio. There are \(800(0.60)=480\) lower-frequency risks and \(800(0.40)=320\) higher-frequency risks. The expected count is

$$ 480(0.10)+320(0.30)=48+96=144. $$

Its variance is also 144 claims squared. The expected count rises by \(144-120=24\), or \(24/120=0.20\), a 20% increase. The class-specific frequencies did not change; the shift toward the higher-frequency class explains the increase. Check: the weighted frequencies are \(0.75(0.10)+0.25(0.30)=0.15\) and \(0.60(0.10)+0.40(0.30)=0.18\); multiplying each by 800 gives 120 and 144.

Common mistake avoided: applying one class’s frequency to all 800 policies. The correct calculation weights each class frequency by the number of risks in that class.

When Random Risk Mix Adds Variation

Now consider a different setup: each risk is independently assigned to a class, with probability \(w_j\) of belonging to class \(j\). The class is not fixed in advance for the risk being modelled. Conditional on its class, its count is still Poisson with mean \(\lambda_j\). The count for one randomly selected risk is therefore a mixture of Poisson distributions.

Its mean is the weighted average \(\bar{\lambda}\). Its variance is larger than that mean when the class-specific frequencies differ. The reason is that the count varies both within a class and because the risk might belong to a class with a different frequency.

Theorem: For one risk with random class \(J\), where \(P(J=j)=w_j\), and with \(N\mid J=j\sim\operatorname{Poisson}(\lambda_j)\), the mean and variance are

E[N]=\bar{\lambda}=\sum_j w_j\lambda_j, \qquad \operatorname{Var}(N)=\bar{\lambda}+\sum_jw_j(\lambda_j-\bar{\lambda})^2.

The second term is the variance of the class-specific means under the class weights. It is zero if all classes have the same frequency.

To justify the variance formula, condition on the class. Within class \(j\), the Poisson variance equals the Poisson mean, so \(E[\operatorname{Var}(N\mid J)]=\sum_jw_j\lambda_j=\bar{\lambda}\). The variance of the conditional mean is \(\operatorname{Var}(E[N\mid J])=\sum_jw_j(\lambda_j-\bar{\lambda})^2\). The law of total variance states that the unconditional variance is the sum of these two terms. This gives the result.

\operatorname{Var}(N) =E[\operatorname{Var}(N\mid J)] +\operatorname{Var}(E[N\mid J]).

For \(M\) independent risks, each independently assigned a class according to the same weights, the total count has mean \(M\bar{\lambda}\) and variance \(M\) times the one-risk variance. This extra variance is appropriate for random class assignment. It should not be added when the portfolio’s class counts are known and fixed: in that case, sum the class-specific Poisson variances instead.

Worked Example: Fixed Counts Versus Random Class Assignment

A hypothetical group has 100 risks. Each risk is either lower-frequency, with a Poisson mean of 0.10 claims, or higher-frequency, with a Poisson mean of 0.50 claims for the period. The higher-frequency proportion is 30%. Find the mean and variance of the total count under (a) a fixed composition of 70 lower-frequency and 30 higher-frequency risks, and (b) independent random class assignment with those same probabilities.

Method. Use the class-specific Poisson sums for the fixed composition. For random assignment, first find the mean and between-class variance for one risk, then multiply both by 100.

(a) Fixed composition. The expected count is \(70(0.10)+30(0.50)=7+15=22\). The counts are independent Poisson counts, so

\operatorname{Var}(N)=70(0.10)+30(0.50)=22.

(b) Random assignment. The weighted mean per risk is \(0.70(0.10)+0.30(0.50)=0.07+0.15=0.22\). The between-class variance per risk is

$$ 0.70(0.10-0.22)^2+0.30(0.50-0.22)^2 =0.70(0.0144)+0.30(0.0784) =0.01008+0.02352=0.0336. $$

The one-risk variance is \(0.22+0.0336=0.2536\). For 100 independent risks, the total mean is \(100(0.22)=22\) and the total variance is \(100(0.2536)=25.36\).

The two models have the same mean but different variances: 22 for fixed class counts and 25.36 for independent random assignment. Check: the additional 3.36 in the random-assignment variance is \(100(0.0336)\), the variation due solely to class assignment. Common mistake avoided: adding this term to the fixed-composition model, where the class counts do not vary.

Measuring the Effect of a Mix Shift

For two classes, let \(p\) be the proportion in the higher-frequency class, with frequency \(\lambda_H\), and let the remaining proportion \(1-p\) have frequency \(\lambda_L\). The weighted average is

\bar{\lambda}=(1-p)\lambda_L+p\lambda_H =\lambda_L+p(\lambda_H-\lambda_L).

If \(\lambda_H>\lambda_L\), increasing \(p\) increases the expected count. For a change from \(p_1\) to \(p_2\) among \(M\) equal-exposure risks, the expected-count change is \(M(p_2-p_1)(\lambda_H-\lambda_L)\). This identifies the effect of the mix change while holding the class-specific frequencies and total exposure constant.

With random class assignment, the between-class variance for one risk in the two-class case is \(p(1-p)(\lambda_H-\lambda_L)^2\). This follows because the weighted variance of two values is \((1-p)(\lambda_L-\bar{\lambda})^2+p(\lambda_H-\bar{\lambda})^2\), which simplifies to the stated expression. The term depends on both the frequency gap and the proportions. It is zero when the mix contains only one class, and it is largest at \(p=0.5\) for a fixed frequency gap.

Worked Example: A Mix Shift with Random Assignment

A hypothetical portfolio has 1,000 equal-exposure risks. Each risk is independently assigned to a lower-frequency class with Poisson mean 0.04 or a higher-frequency class with Poisson mean 0.16. Compare the mean and variance when the higher-frequency proportion is 25% and when it is 40%.

Method. For each mix, calculate the weighted mean and the between-class variance per risk. Under independent random assignment, multiply their sum by 1,000 to obtain the total variance.

At 25% higher-frequency risks. The mean per risk is \(0.75(0.04)+0.25(0.16)=0.03+0.04=0.07\), so the total expected count is \(1{,}000(0.07)=70\). The frequency gap is \(0.16-0.04=0.12\), and the between-class variance per risk is \(0.25(0.75)(0.12)^2=0.1875(0.0144)=0.0027\). Thus, the total variance is \(1{,}000(0.07+0.0027)=72.7\).

At 40% higher-frequency risks. The mean per risk is \(0.60(0.04)+0.40(0.16)=0.024+0.064=0.088\), so the total expected count is 88. The between-class variance per risk is \(0.40(0.60)(0.12)^2=0.24(0.0144)=0.003456\). The total variance is \(1{,}000(0.088+0.003456)=91.456\).

The change in expected count is \(88-70=18\). The change in variance is \(91.456-72.7=18.756\). Check: using the two-class mean formula, the change in mean per risk is \((0.40-0.25)(0.16-0.04)=0.018\), or 18 claims across 1,000 risks. Common mistake avoided: reporting 70 and 88 as the variances as well as the means. Those would be the variances for fixed class counts; with random assignments, the between-class terms also contribute.

Common Exam Traps and Practical Interpretation

Confusing a mix with a frequency. A percentage such as 30% describes the proportion in a class. It is not a claim frequency. Multiply class frequencies by their exposure weights to get the portfolio average.
Changing the mix and the class rates at once. To isolate the effect of mix, hold the class-specific frequencies and total exposure fixed. If class rates also change, calculate and describe both effects.
Using the wrong variance model. Fixed class counts with independent Poisson counts give variance equal to the total mean. Independently random class assignments add the weighted variance of class means.
Applying a proportion to the wrong total. Confirm that proportions sum to one and that the number of risks in each class is consistent with the total portfolio size.
Calling every mixed count Poisson. A mixture of Poisson counts with different means generally is not Poisson. Its variance exceeds its mean when the class means vary and classes are randomly assigned.

Key takeaway: The weighted average frequency determines the expected count. To determine variance, first state whether class counts are fixed or whether class membership is random. The latter creates additional variation whenever class-specific frequencies differ.

Check Your Understanding

For each problem, identify whether the composition is fixed or random before calculating. Claim counts described as Poisson are independent conditional on class.

A fixed portfolio has 400 lower-frequency risks at 0.06 expected claims each and 100 higher-frequency risks at 0.26 each. Find the total expected count and variance. (Answer: mean 50; variance 50.)
Each of 250 risks is independently assigned to a class: 60% have Poisson mean 0.20 and 40% have Poisson mean 0.50. Find the total expected count and variance. (Answer: mean 80; variance 85.4.)
A fixed portfolio contains 300 risks at frequency 0.08 and 200 risks at frequency 0.18. Find its expected count and variance. (Answer: mean 60; variance 60.)
For 1,000 risks independently assigned to one of two classes, the lower-frequency rate is 0.05 and the higher-frequency rate is 0.25. Find the total variance when the higher-frequency proportion is 20%, then when it is 50%. (Answer: 96.4 at 20%; 160 at 50%.)
A fixed portfolio of 600 risks shifts from 80% lower-frequency risks to 70%. The lower-frequency rate is 0.03 and the higher-frequency rate is 0.15. Find the expected count before and after the shift and the increase. (Answer: 32.4 before; 39.6 after; increase 7.2 claims.)

What You'll Learn