Why the Conditioning Matters
A portfolio can contain risks with different underlying claim tendencies. A risk parameter, denoted by \(\Theta\), represents the risk-specific feature that determines a claim distribution in the model. Once a value of \(\Theta\) is specified, the claim outcome \(X\) has a conditional distribution. Across a population of risks, however, \(\Theta\) varies, so the distribution of \(X\) also reflects the mix of risks in that population.
This distinction matters in ratemaking and credibility work. A mean or variance calculated for a fixed risk parameter describes variation for that risk under its conditional model. An unconditional mean or variance describes outcomes across the population, including the fact that different risks may have different conditional means. The population mix must therefore be specified before an unconditional moment can be calculated.
For a discrete population, the probabilities assigned to the risk types must sum to one. They should represent the population being studied: for example, the share of policies in each type if the target is a randomly selected policy. A different target population, such as a randomly selected exposure unit, may have different weights. For a continuous \(\Theta\), the corresponding averages are integrals over its distribution.
Two Results for Moving Between Conditional and Unconditional Moments
Provided the relevant expectations exist, the unconditional mean is the expected value of the conditional mean. If \(E[X^2]\) is finite, the unconditional variance is the expected conditional variance plus the variance of the conditional mean:
The first term in the variance formula is the average variation within a fixed risk type. The second term is the variation in risk-specific means across the population. The second term is nonnegative, so the unconditional variance cannot be less than the average conditional variance.
A short derivation shows why both terms are needed. Let \(m(\Theta)=E[X\mid\Theta]\), and write \(\mu=E[X]=E[m(\Theta)]\). Add and subtract \(m(\Theta)\) inside the centered outcome:
Squaring and taking expectations gives the sum of the two squared terms and a cross-product. The cross-product has expectation zero because, conditional on \(\Theta\), \(m(\Theta)-\mu\) is fixed and \(E[X-m(\Theta)\mid\Theta]=0\):
This identity is a moment calculation, not an assumption that risks have identical parameters or that their conditional variances are equal. In each application, first identify the conditional distribution and the population weights; then calculate the two variance contributions.
Worked Conditional and Unconditional Calculations
Worked Example: A Claim Count Across Two Risk Types
A hypothetical book contains 40% low-volume risks and 60% high-volume risks. For one fixed policy period, the claim count \(N\), conditional on risk type, is Poisson with parameter \(\lambda=1\) for a low-volume risk and \(\lambda=3\) for a high-volume risk. Find the conditional means and variances, then the unconditional mean and variance for a randomly selected policy. Use the Poisson facts \(E[N\mid\Theta]=\lambda\) and \(\operatorname{Var}(N\mid\Theta)=\lambda\).
Method and assumptions. The risk type is selected with probabilities 0.40 and 0.60, and the stated Poisson model applies within each type. We calculate the conditional moments first, then average over types.
For the low-volume type, the conditional mean and variance are both 1. For the high-volume type, both are 3. Thus the unconditional mean is
The average conditional variance is
The conditional means are 1 and 3, with population mean 2.20. Their variance is
Therefore,
Check. For a Poisson variable, the conditional second moment is \(\lambda+\lambda^2\). Its population average here is \(0.40(1+1^2)+0.60(3+3^2)=0.40(2)+0.60(12)=8\). Subtracting \(E[N]^2=2.20^2=4.84\) gives \(8-4.84=3.16\), as required. The common mistake avoided is reporting 2.20, the average conditional variance, as the unconditional variance.
Worked Example: A Claim Indicator With Different Risk Probabilities
In a hypothetical portfolio, 60% of policies have a conditional probability of a claim of 0.01, and 40% have a conditional probability of 0.06. Let \(I\) equal 1 if a policy has a claim during the period and 0 otherwise. Calculate the unconditional claim probability, the average conditional variance, and the unconditional variance.
Method and assumptions. Within each risk type, \(I\) is Bernoulli with its stated claim probability \(p\), so \(E[I\mid\Theta]=p\) and \(\operatorname{Var}(I\mid\Theta)=p(1-p)\). The type weights describe the randomly selected policy population.
The unconditional claim probability, which is also \(E[I]\), is
The conditional variances are \(0.01(0.99)=0.0099\) and \(0.06(0.94)=0.0564\). Their weighted average is
The conditional means are 0.01 and 0.06, while their population mean is 0.03. The variance of these conditional means is
Thus the unconditional variance is \(0.0285+0.0006=0.0291\). As a check, unconditionally \(I\) is still a zero-one variable with success probability 0.03, so \(\operatorname{Var}(I)=0.03(0.97)=0.0291\).
Common mistake avoided. Using only the weighted conditional variances gives 0.0285, not the unconditional variance. The missing 0.0006 is the variation in claim probability across risk types.
Worked Example: Exponential Claim Severities in a Mixed Population
Suppose a hypothetical insurer models a full claim severity \(X\), in dollars, as exponential conditional on risk type. Twenty-five percent of risks have conditional mean severity $400, and 75% have conditional mean severity $800. For an exponential distribution with mean \(m\), the variance is \(m^2\). Find the unconditional mean and variance.
Method and assumptions. Each stated exponential mean is the scale parameter and the conditional variance is the square of that mean. The type proportions are the weights for a randomly selected claim from the specified population. We assume the stated mixture directly describes that claim population.
The unconditional mean is
The conditional variances are \(400^2=160{,}000\) and \(800^2=640{,}000\) dollars squared. Their weighted average is
The variance of the conditional means is
Therefore, the unconditional variance is \(520{,}000+30{,}000=550{,}000\) dollars squared. A direct check uses \(E[X^2\mid\Theta]=\operatorname{Var}(X\mid\Theta)+E[X\mid\Theta]^2=2m^2\) for an exponential distribution. Hence \(E[X^2]=0.25(2\cdot400^2)+0.75(2\cdot800^2)=1{,}040{,}000\), and \(E[X^2]-E[X]^2=1{,}040{,}000-700^2=550{,}000\).
Common mistake avoided. Averaging only the two conditional variances produces $520,000 squared, which omits the $30,000 squared contribution from differences in risk-specific mean severity.
Calculation Sequence and Interpretation
A reliable approach is to keep the two sources of variation separate until the final step. For a discrete risk mix, the following sequence makes the weighting and units explicit.
State what is being selected at random and the probability of each risk type. Do not silently substitute policy weights for claim weights, or vice versa.
For each risk type, use its stated conditional distribution to find \(E[X\mid\Theta]\) and \(\operatorname{Var}(X\mid\Theta)\).
Weight each conditional mean by the population probability to obtain \(E[X]\).
Average the conditional variances, calculate the variance of the conditional means, and add the results.
Where practical, calculate \(E[X^2]-E[X]^2\) from conditional second moments and confirm that it agrees with the total variance.
A conditional variance answers a within-risk question: if the risk parameter were held fixed, how much would outcomes vary? The unconditional variance answers a broader question: how much do outcomes vary when both the claim process and the risk mix are allowed to vary? The difference between those questions is not a technical detail; it changes the quantity used in a calculation.
Common Exam Traps
- Stopping after averaging conditional variances. This gives \(E[\operatorname{Var}(X\mid\Theta)]\), not generally \(\operatorname{Var}(X)\). Include \(\operatorname{Var}(E[X\mid\Theta])\).
- Using the wrong mixture weights. A weighted average is meaningful only when the weights describe the population named in the question. Check whether the unit is a policy, exposure, or claim.
- Taking the variance of a weighted mean instead of the weighted variance. For discrete types, calculate \(\sum_j w_j(m_j-\bar m)^2\), where \(w_j\) is the type probability and \(\bar m=\sum_jw_jm_j\). Do not omit the squared deviations.
- Mixing units. A variance has squared units. For dollar severities, the variance is in dollars squared; it is not itself a dollar amount.
- Assuming a conditional model is automatically an unconditional model. A distribution specified given \(\Theta\) describes a fixed risk type. Across types, outcomes follow a mixture distribution, which need not have the same form.
Check Your Understanding
For each problem, identify the conditional moments before combining them across the stated risk population.
- A policy has a claim indicator that is conditionally Bernoulli with probability 0.03 for 20% of policies and 0.07 for 80%. Find the unconditional mean and variance of the indicator. (Answer: mean 0.062; variance 0.058156.)
- A hypothetical claim count is conditionally Poisson with parameter 0.5 for 25% of risks and parameter 2 for 75%. Find its unconditional mean and variance. (Answer: mean 1.625; variance 2.046875.)
- Conditional claim severity is exponential with mean $300 for half the claims and $900 for half the claims. Find the unconditional mean and variance in dollars and dollars squared. (Answer: mean $600; variance 540,000 dollars squared.)
- For 60% of risks, \(X\) is conditionally normal with mean 100 and variance 400; for the remaining 40%, it is conditionally normal with mean 200 and variance 900. Find the unconditional mean and variance. (Answer: mean 140; variance 3,000.)
- A student calculates an average conditional variance of 0.0285 for a Bernoulli claim indicator with claim probabilities 0.01 for 60% of policies and 0.06 for 40%. What additional term must be included, and what is the correct unconditional variance? (Answer: add variance of conditional means 0.0006; unconditional variance 0.0291.)