Variance Decomposition Across Risks

Why Separate the Sources of Variability?

Two insureds can have different observed losses for two distinct reasons. Even if they have the same underlying expected loss, random claim outcomes create variation from one observation to another. In addition, insureds can differ in their underlying risk levels, so their expected losses are not all the same. The law of total variance separates these sources rather than treating observed variation as one undifferentiated quantity.

In the Bühlmann model, this separation is central to interpreting the variance components. The expected process variance (EPV) describes average variation within a risk. The variance of hypothetical means (VHM) describes variation between risks’ expected values. Together, they give the unconditional variance of a single observation.

Definition: Let $\Theta$ denote a risk’s latent risk parameter, and let $X$ be a loss observation for that risk. Assume finite second moments. Define the conditional mean $m(\Theta)=E[X\mid\Theta]$ and conditional variance $s^2(\Theta)=\operatorname{Var}(X\mid\Theta)$. The Bühlmann components are $$ \operatorname{EPV}=E[s^2(\Theta)], \qquad \operatorname{VHM}=\operatorname{Var}(m(\Theta)). $$ The overall mean is $\mu=E[X]=E[m(\Theta)]$, and the total variance is $\operatorname{Var}(X)=\operatorname{EPV}+\operatorname{VHM}$.

The expectation in EPV averages conditional variances over the portfolio’s distribution of risks. It is not the variance for a particular risk unless that risk’s conditional variance happens to equal the portfolio average. VHM is the variance of the conditional means across risks; it is not the variance of observed losses.

Deriving the Total-Variance Decomposition

Write an observation as its deviation from its own risk mean plus that risk mean’s deviation from the portfolio mean:

X-\mu=\bigl(X-m(\Theta)\bigr)+\bigl(m(\Theta)-\mu\bigr).

Squaring and taking expectations produces two squared terms and a cross term. The cross term is zero because, conditional on $\Theta$, the expected deviation of $X$ from its conditional mean is zero:

E\!\left[(X-m(\Theta))(m(\Theta)-\mu)\right] =E\!\left[(m(\Theta)-\mu)E[X-m(\Theta)\mid\Theta]\right]=0.

The first squared term averages the within-risk conditional variances; the second is the variance of the risk means. Therefore,

\begin{aligned} \operatorname{Var}(X) &=E[(X-\mu)^2]\\ &=E[\operatorname{Var}(X\mid\Theta)] +\operatorname{Var}(E[X\mid\Theta])\\ &=\operatorname{EPV}+\operatorname{VHM}. \end{aligned}

This identity is the law of total variance. It requires finite second moments, but it does not require a particular probability distribution. The conditional distributions may vary by risk. The identity also does not say that the two components are equal, or that either component can be inferred from a single observed loss.

Worked Example: Decomposing One-Year Loss Variance

A hypothetical portfolio has two risk types. Type A occurs with probability $0.25$, and its annual loss has conditional mean $100 and conditional variance $400 dollars squared. Type B occurs with probability $0.75$, and its annual loss has conditional mean $160 and conditional variance $900 dollars squared. The full conditional distributions are not specified; the given moments are sufficient. Calculate the portfolio mean, EPV, VHM, and total variance.

Method. Average the conditional means to find the portfolio mean, average the conditional variances to find EPV, and calculate the variance of the conditional means for VHM.

Portfolio mean.

\mu=0.25(100)+0.75(160)=25+120=145\text{ dollars}.

EPV.

\operatorname{EPV}=0.25(400)+0.75(900)=100+675=775\text{ dollars}^2.

VHM. The type means differ from the portfolio mean by $-45$ and $15$ dollars, respectively:

\operatorname{VHM}=0.25(-45)^2+0.75(15)^2 =0.25(2025)+0.75(225)=506.25+168.75=675\text{ dollars}^2.

Thus the total variance is $775+675=1{,}450$ dollars squared. As a check, the conditional second moments are $100^2+400=10{,}400$ and $160^2+900=26{,}500$. Their portfolio average is $0.25(10{,}400)+0.75(26{,}500)=22{,}475$, and $22{,}475-145^2=22{,}475-21{,}025=1{,}450$ dollars squared. The two calculations agree.

Common mistake avoided. Using only the variance of the type means gives $675 squared and omits within-risk fluctuation. The correction is to add EPV, giving total variance of $1,450 squared.

Repeated Observations and the Variance of an Average

Suppose a risk has $n$ observations $X_1,\ldots,X_n$. To obtain a simple formula for their average, assume they are conditionally independent and identically distributed given $\Theta$, with conditional mean $m(\Theta)$ and conditional variance $s^2(\Theta)$. Conditional independence matters: it makes the conditional variance of the average equal to $s^2(\Theta)/n$.

Given $\Theta$, the average has conditional mean $m(\Theta)$ and conditional variance $s^2(\Theta)/n$. Applying total variance to the average gives

\operatorname{Var}(\bar X) =E\!\left[\frac{s^2(\Theta)}{n}\right] +\operatorname{Var}(m(\Theta)) =\frac{\operatorname{EPV}}{n}+\operatorname{VHM}.

Only the within-risk component is divided by $n$. More observations reduce the effect of random fluctuations around a risk’s mean, but they do not make genuinely different risk means identical. As $n$ increases, the variance of the average approaches VHM, not zero, when VHM is positive.

Worked Example: Variance of a Claim-Frequency Average

A hypothetical policyholder belongs to one of two latent risk types with equal probability. For type L, the probability of a claim in a year is $0.10$; for type H, it is $0.30$. In each of five years, the claim indicator $X_j$ is Bernoulli with the applicable probability, and the five indicators are conditionally independent given the type. Find the mean and variance of the five-year average claim frequency.

Method. For a Bernoulli variable with parameter $p$, the conditional mean is $p$ and the conditional variance is $p(1-p)$.

The portfolio mean is $0.5(0.10)+0.5(0.30)=0.20$. The EPV is

\operatorname{EPV}=0.5[0.10(0.90)]+0.5[0.30(0.70)] =0.5(0.09)+0.5(0.21)=0.15.

The VHM is the variance of the two conditional means around $0.20$:

\operatorname{VHM}=0.5(0.10-0.20)^2+0.5(0.30-0.20)^2 =0.5(0.01)+0.5(0.01)=0.01.

For five years, the average’s variance is

\operatorname{Var}(\bar X)=\frac{0.15}{5}+0.01=0.03+0.01=0.04.

The average claim frequency has mean $0.20$ and variance $0.04$. As a check, one annual indicator has variance $0.15+0.01=0.16$. Across five years, the shared latent type contributes covariance $0.01$ to each pair of distinct years, so

\operatorname{Var}(\bar X) =\frac{5(0.16)+5(4)(0.01)}{5^2} =\frac{0.80+0.20}{25}=0.04.

Common mistake avoided. Dividing the entire single-year variance $0.16$ by five would give $0.032$. That incorrectly treats years as unconditionally independent. They share the same risk type, so their outcomes are associated through the latent parameter.

Shared Risk Parameters and Dependence

Conditional independence does not generally imply unconditional independence. For distinct observations $X_j$ and $X_k$ from the same risk, conditional independence gives zero covariance around the conditional means, but both observations have the same random conditional mean $m(\Theta)$. In particular, under conditional independence and finite second moments,

\begin{aligned} \operatorname{Cov}(X_j,X_k) &=E[\operatorname{Cov}(X_j,X_k\mid\Theta)] +\operatorname{Cov}(E[X_j\mid\Theta],E[X_k\mid\Theta])\\ &=0+\operatorname{Var}(m(\Theta))\\ &=\operatorname{VHM}. \end{aligned}

This gives another way to understand the average-variance formula. In the variance of a sum, each observation contributes its own total variance, while each pair contributes their shared covariance. Dividing by $n^2$ yields $\operatorname{EPV}/n+\operatorname{VHM}$.

The result depends on the stated model. If observations are not conditionally independent, their conditional covariances must also be included. If the observations have different conditional means or variances by period, the simple identical-observation formula need not apply; use the appropriate conditional variance of the actual average before applying total variance.

Worked Example: Comparing Two Risk Mixes

Two hypothetical portfolios have three risk types with probabilities $0.20$, $0.50$, and $0.30$. In Portfolio A, the conditional loss means are $80, $120, and $200, and the conditional variances are 100, 400, and 900 dollars squared. Portfolio B has the same conditional loss moments but type probabilities $0.50$, $0.30$, and $0.20$, respectively. For each portfolio, calculate EPV, VHM, and total variance. The full conditional distributions are not specified.

Method. EPV is the probability-weighted average of the given conditional variances. For VHM, first calculate that portfolio’s own mean, then take the probability-weighted squared deviations of the type means from it.

Portfolio A. Its mean is

\mu_A=0.20(80)+0.50(120)+0.30(200)=16+60+60=136\text{ dollars}.

Its EPV is $0.20(100)+0.50(400)+0.30(900)=20+200+270=490$ dollars squared. Its VHM is

\begin{aligned} \operatorname{VHM}_A &=0.20(80-136)^2+0.50(120-136)^2+0.30(200-136)^2\\ &=0.20(3136)+0.50(256)+0.30(4096)\\ &=627.2+128+1228.8=1984\text{ dollars}^2. \end{aligned}

Therefore, $\operatorname{Var}_A(X)=490+1984=2474$ dollars squared.

Portfolio B. Its mean is $0.50(80)+0.30(120)+0.20(200)=40+36+40=116$ dollars. Its EPV is $0.50(100)+0.30(400)+0.20(900)=50+120+180=350$ dollars squared. Its VHM is

\begin{aligned} \operatorname{VHM}_B &=0.50(80-116)^2+0.30(120-116)^2+0.20(200-116)^2\\ &=0.50(1296)+0.30(16)+0.20(7056)\\ &=648+4.8+1411.2=2064\text{ dollars}^2. \end{aligned}

Thus, $\operatorname{Var}_B(X)=350+2064=2414$ dollars squared. The portfolio mix changes both components: B has lower EPV because more weight is assigned to the type with conditional variance 100, but higher VHM because of how its type probabilities weight the squared deviations from its own mean.

Check. For A, the weighted conditional second moment is $0.20(80^2+100)+0.50(120^2+400)+0.30(200^2+900)=20970$. Subtracting $136^2=18496$ gives 2474. For B, it is $0.50(80^2+100)+0.30(120^2+400)+0.20(200^2+900)=15870$. Subtracting $116^2=13456$ gives 2414. Both checks match.

Common mistake avoided. Reusing Portfolio A’s VHM for Portfolio B would ignore the changed risk mix. VHM is calculated using the current portfolio probabilities and that portfolio’s own mean.

Interpretation and Exam Traps

The decomposition is useful because the two components describe different sources of uncertainty. EPV concerns random outcomes for a fixed risk level; VHM concerns heterogeneity in expected outcomes across risk levels. In the Bühlmann setting, this distinction also explains why repeated observations can sharpen information about an individual risk without eliminating variation between different risks.

Do not confuse conditional and unconditional variance. A conditional variance describes one risk level; EPV averages those variances over the risk distribution.
Do not omit within-risk variability. The variance of risk means is VHM, not total variance. Add EPV to obtain the variance of one observation.
Do not divide VHM by the number of observations. Under conditional independence, averaging reduces EPV by $n$, while VHM remains.
Do not assume conditional independence means unconditional independence. A shared latent risk parameter can induce covariance across periods.
Recalculate after a change in risk mix. Both EPV and VHM depend on how risks are weighted in the portfolio.

Key takeaway: For a single Bühlmann-model observation, $\operatorname{Var}(X)=\operatorname{EPV}+\operatorname{VHM}$. For $n$ conditionally independent, identically distributed observations from the same risk, $\operatorname{Var}(\bar X)=\operatorname{EPV}/n+\operatorname{VHM}$. The first component is within-risk process variation; the second is between-risk variation in expected values.

Check Your Understanding

For each problem, identify the relevant conditional moments and apply the variance decomposition.

A hypothetical portfolio has two risk types, each with probability $0.5$. Their conditional means are $50 and $70, and their conditional variances are 36 and 64 dollars squared. Find the portfolio mean, EPV, VHM, and total variance. (Answer: Mean = $60; EPV = 50 dollars squared; VHM = 100 dollars squared; total variance = 150 dollars squared.)
A risk parameter takes values $p=0.20$ and $p=0.40$, each with probability $0.5$. Annual claim indicators are Bernoulli conditional on $p$. Find the EPV, VHM, and variance of one annual indicator. (Answer: EPV = 0.20; VHM = 0.01; total variance = 0.21.)
For the model in Question 2, suppose four annual indicators are conditionally independent given $p$. Find the variance of their average. (Answer: $0.20/4+0.01=0.06$.)
A hypothetical risk has conditional mean $90 and conditional variance 225 dollars squared with probability $0.6$, and conditional mean $150 and conditional variance 625 dollars squared with probability $0.4$. Find the mean, EPV, VHM, and total variance. (Answer: Mean = $114; EPV = 385 dollars squared; VHM = 864 dollars squared; total variance = 1,249 dollars squared.)
For a portfolio with $\operatorname{EPV}=120$ and $\operatorname{VHM}=30$, observations are conditionally independent and identically distributed within each risk. Find the variance of an average of six observations and the covariance between two distinct observations from the same risk. (Answer: Variance of the average = $120/6+30=50$; covariance = 30.)

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