Why Separate the Sources of Variance?
Suppose a portfolio contains risks with different underlying characteristics. An observed outcome, such as a claim count, varies partly because outcomes fluctuate for a given risk and partly because risks have different expected outcomes. The total variance combines these two sources. Separating them helps an actuary describe portfolio heterogeneity and is a foundation of the Bühlmann model.
The key distinction is between variation within a risk and variation between risks. A risk with a fixed underlying characteristic can still have a random outcome. Across a population of risks, the conditional means of those outcomes may also differ. The total variance decomposition gives an exact accounting of both contributions. This topic is aligned with the CAS Exam 8 (Advanced Ratemaking) syllabus.
Write \(m(\Theta)=E[X\mid\Theta]\) for the hypothetical mean associated with a given risk characteristic, and let \(\mu=E[X]\) be the unconditional mean. The law of iterated expectations gives \(E[m(\Theta)]=\mu\). Conditional variance measures the outcome's variation around \(m(\Theta)\); VHM measures how \(m(\Theta)\) varies around \(\mu\). Both components are variances on the scale of \(X\), so their units are the square of the outcome's units.
Deriving the Total Variance Identity
The identity follows by expressing an outcome's deviation from the overall mean as two deviations: its departure from its own hypothetical mean, and that hypothetical mean's departure from the portfolio mean.
Square both sides and take expectations. The two squared terms produce EPV and VHM. The cross-product has expectation zero: conditional on \(\Theta\), the quantity \(m(\Theta)-\mu\) is fixed, while \(E[X-m(\Theta)\mid\Theta]=0\). More explicitly,
The first squared term has expectation \(E[\operatorname{Var}(X\mid\Theta)]\), because the conditional mean of \(X-m(\Theta)\) is zero. The second is \(\operatorname{Var}(m(\Theta))\), because \(E[m(\Theta)]=\mu\). Therefore,
This is the law of total variance. It applies whether \(\Theta\) is discrete or continuous and does not require a particular distribution for \(X\mid\Theta\). It does require that the conditional moments and unconditional variance exist. The result is an equality, not an approximation: the unconditional variance is exactly the sum of the two components.
Calculating the Components
For a finite collection of risk types, first compute each type's conditional mean and variance, then weight them by the type probabilities. If type \(j\) occurs with probability \(w_j\), has conditional mean \(\mu_j\), and conditional variance \(\sigma_j^2\), where \(\sum_j w_j=1\), then
The weights are essential: the components describe the distribution of risks in the population, not an unweighted list of types. For a continuous \(\Theta\), the corresponding expectations and variance are calculated over its distribution.
Worked Example: A Two-Type Bernoulli Outcome
A hypothetical coverage portfolio has two risk types. Type A has probability \(0.30\) and its outcome \(X\) is Bernoulli with success probability \(0.20\). Type B has probability \(0.70\) and its outcome is Bernoulli with success probability \(0.60\). Find the overall mean, EPV, VHM, and unconditional variance, then check the result directly.
Method. Use the finite-type form of the total variance identity. For a Bernoulli outcome with success probability \(p\), the mean is \(p\) and the variance is \(p(1-p)\).
The conditional means are \(\mu_A=0.20\) and \(\mu_B=0.60\). Thus the overall mean is
The conditional variances are \(0.20(0.80)=0.16\) for Type A and \(0.60(0.40)=0.24\) for Type B. Therefore,
The deviations of the type means from the overall mean are \(0.20-0.48=-0.28\) and \(0.60-0.48=0.12\). Hence,
The total variance is \(0.216+0.03360=0.24960\). As a direct check, the unconditional outcome is itself Bernoulli: its probability of success is \(0.48\). Its variance is therefore \(0.48(0.52)=0.24960\), agreeing with the decomposition. The calculation avoids the mistake of treating the average conditional variance as the entire unconditional variance.
Worked Example: A Mixture of Continuous Loss Amounts
In a hypothetical portfolio, a covered loss amount \(X\), measured in thousands of dollars, comes from one of two risk types. Type A has probability \(0.25\) and \(X\mid A\) is uniform on \([0,2]\). Type B has probability \(0.75\) and \(X\mid B\) is uniform on \([4,8]\). Find the EPV, VHM, and total variance. Treat the stated distributions as the full conditional models.
Method. For a uniform distribution on \([a,b]\), its mean is \((a+b)/2\) and its variance is \((b-a)^2/12\). For Type A, these are \(1\) and \(2^2/12=1/3\). For Type B, they are \(6\) and \(4^2/12=4/3\).
The overall mean is
Weight the conditional variances to obtain the expected process variance:
For the variance of hypothetical means, the deviations from \(4.75\) are \(-3.75=-15/4\) and \(1.25=5/4\). Thus,
Adding the components gives
As an independent check, \(E[X^2\mid A]=1/3+1^2=4/3\), and \(E[X^2\mid B]=4/3+6^2=112/3\). Therefore \(E[X^2]=0.25(4/3)+0.75(112/3)=85/3\). Subtracting the squared overall mean gives \(85/3-4.75^2=85/3-361/16=277/48\). The answer is in squared thousands of dollars. The calculation avoids confusing variance of the type means with the variance of the loss amounts themselves.
Worked Example: A Poisson Claim Count Mixture
A hypothetical group of policies has two risk types. With probability \(0.25\), a policy has annual claim count \(X\mid\Theta\sim\operatorname{Poisson}(2)\); with probability \(0.75\), it has \(X\mid\Theta\sim\operatorname{Poisson}(6)\). Find the overall mean and variance, separating the two variance components, and verify the total using second moments.
Method. A Poisson random variable with parameter \(\lambda\) has conditional mean \(\lambda\) and conditional variance \(\lambda\). The overall mean is
Since each conditional variance equals its Poisson parameter,
The two hypothetical means differ from \(5\) by \(-3\) and \(1\), respectively. Thus,
The total claim-count variance is \(5+3=8\) claims squared. For a second-moment check, a Poisson variable with mean \(\lambda\) has \(E[X^2\mid\Theta]=\lambda+\lambda^2\). The type-specific second moments are \(2+2^2=6\) and \(6+6^2=42\), so \(E[X^2]=0.25(6)+0.75(42)=1.5+31.5=33\). Then \(\operatorname{Var}(X)=33-5^2=8\), as required. This calculation avoids assuming that a mixture of Poisson distributions must itself be Poisson; the variance check works directly without that assumption.
Interpreting the Decomposition and Avoiding Traps
EPV and VHM answer different questions. EPV asks how much outcomes vary, on average, when the risk characteristic is held fixed. VHM asks how much the conditional expected outcome varies across the risk population. A portfolio can have a large EPV and small VHM, a small EPV and large VHM, or substantial amounts of both. The identity does not require either component to dominate.
A frequent exam error is to calculate only \(E[\operatorname{Var}(X\mid\Theta)]\) and report it as \(\operatorname{Var}(X)\). For the Bernoulli example above, that incorrect calculation reports \(0.216\), omitting the \(0.03360\) attributable to differing type means. The correction is to add both components: \(0.216+0.03360=0.24960\). Another error is to weight the type variances but use an unweighted average of type means; the overall mean and both variance components must use the actual type probabilities.
A useful check is to compute \(\operatorname{Var}(X)=E[X^2]-(E[X])^2\) independently when the conditional second moments are available. A mismatch often signals an incorrect weight, a conditional variance error, or an incorrect center in the VHM calculation. Keep the scale in view: if \(X\) is a claim count, its variance is in squared counts; if \(X\) is measured in thousands of dollars, its variance is in squared thousands of dollars.
For each risk type or value of \(\Theta\), determine \(E[X\mid\Theta]\) and \(\operatorname{Var}(X\mid\Theta)\) from the stated model.
Average the hypothetical means over the risk distribution to obtain \(\mu=E[X]\).
Average the conditional variances for EPV, and calculate the variance of the hypothetical means for VHM.
Use \(\operatorname{Var}(X)=\operatorname{EPV}+\operatorname{VHM}\), then verify with unconditional moments if feasible.
Check Your Understanding
For each problem, identify the conditional moments, calculate both components, and check that their sum is the unconditional variance where requested.
- A hypothetical risk is Type A with probability \(0.50\) and Type B with probability \(0.50\). Conditional on Type A, \(X\) is Bernoulli with success probability \(0.10\); conditional on Type B, it is Bernoulli with success probability \(0.50\). Find the overall mean, EPV, VHM, and total variance. (Answer: mean \(=0.3000\), EPV \(=0.1700\), VHM \(=0.0400\), total variance \(=0.2100\))
- A hypothetical policy has a Poisson claim count with parameter \(1\) with probability \(0.20\), and parameter \(4\) with probability \(0.80\). Find the overall mean, EPV, VHM, and total variance. (Answer: mean \(=3.4\), EPV \(=3.4\), VHM \(=1.44\), total variance \(=4.84\))
- A loss amount \(X\) has a uniform distribution on \([1,3]\) with probability \(1/3\), and a uniform distribution on \([5,9]\) with probability \(2/3\). Find its overall mean, EPV, VHM, and total variance. (Answer: mean \(=16/3\), EPV \(=1\), VHM \(=50/9\), total variance \(=59/9\approx6.5556\))
- With probability \(0.25\), a hypothetical outcome is conditionally normal with mean \(10\) and variance \(4\); with probability \(0.75\), it is conditionally normal with mean \(16\) and variance \(9\). Find the overall mean and both variance components. (Answer: mean \(=14.5\), EPV \(=7.75\), VHM \(=6.75\), total variance \(=14.5\))
- Suppose \(\Theta\) is uniform on \([0,2]\), with \(E[X\mid\Theta]=3+2\Theta\) and \(\operatorname{Var}(X\mid\Theta)=1+\Theta\). Find \(E[X]\), EPV, VHM, and \(\operatorname{Var}(X)\). (Answer: mean \(=5\), EPV \(=2\), VHM \(=4/3\), total variance \(=10/3\))