Deriving Mixture Variance From Component Variation

Why Mixture Variance Has Two Parts

A severity portfolio can contain distinct groups of risks. For example, claims from different types of insured property may have different average sizes and different amounts of variation around those averages. A finite mixture model represents this heterogeneity by first selecting a component and then drawing a loss from that component’s distribution.

The variance of the resulting loss reflects both the spread of losses within each component and the differences among component means. Averaging the component variances alone misses the second source of variation. This tutorial derives the decomposition and shows how to calculate and check it.

Definition: Let $Z$ identify the component, with $P(Z=j)=w_j$, where $w_j\ge 0$ and $\sum_{j=1}^k w_j=1$. Conditional on $Z=j$, let the loss $X$ have mean $\mu_j=E[X\mid Z=j]$ and variance $\sigma_j^2=\operatorname{Var}(X\mid Z=j)$. Assume the component means and variances are finite. The mixture mean is $\mu=E[X]=\sum_{j=1}^k w_j\mu_j$, and its variance is $\operatorname{Var}(X)=\sum_{j=1}^k w_j\sigma_j^2+\sum_{j=1}^k w_j(\mu_j-\mu)^2$.

The first sum is the within-component variation: it averages each component’s variance using the probability of selecting that component. The second sum is the between-component variation: it measures how far each component mean lies from the overall mixture mean, again weighted by the component probability. Both terms are nonnegative.

The mixture is not generally obtained by simply averaging component distributions’ variances. If all component means are equal, the between-component term is zero. Otherwise, differences among those means add variance, even if each component has little internal spread.

Derivation From Conditional Expectation

The derivation uses the law of total variance. It follows directly by expanding the squared deviation of $X$ from its overall mean. Conditional on component $j$, write $X-\mu=(X-\mu_j)+(\mu_j-\mu)$. The first term has conditional mean zero, so the cross-product vanishes after taking the conditional expectation.

\begin{aligned} \operatorname{Var}(X) &=E[(X-\mu)^2]\\ &=E\!\left[E\!\left[(X-\mu)^2\mid Z\right]\right]\\ &=E\!\left[E\!\left[(X-\mu_Z+\mu_Z-\mu)^2\mid Z\right]\right]\\ &=E\!\left[\operatorname{Var}(X\mid Z)+(\mu_Z-\mu)^2\right]\\ &=E[\operatorname{Var}(X\mid Z)]+\operatorname{Var}(E[X\mid Z]). \end{aligned}

In the fourth line, the omitted cross-product has conditional expectation zero because $E[X-\mu_Z\mid Z]=0$. For a finite mixture, $E[\operatorname{Var}(X\mid Z)]=\sum_j w_j\sigma_j^2$, while $\operatorname{Var}(E[X\mid Z])=\sum_j w_j(\mu_j-\mu)^2$. Substitution gives the stated formula.

Result: For a finite mixture with finite component second moments, $\operatorname{Var}(X)=E[\operatorname{Var}(X\mid Z)]+\operatorname{Var}(E[X\mid Z])$. This is also called the law of total variance. Its first term is average conditional variance; its second term is the variance of the conditional means.

A useful equivalent calculation follows from $E[X^2]=\operatorname{Var}(X)+E[X]^2$. Since $E[X^2\mid Z=j]=\sigma_j^2+\mu_j^2$,

\operatorname{Var}(X) =\sum_{j=1}^k w_j(\sigma_j^2+\mu_j^2) -\left(\sum_{j=1}^k w_j\mu_j\right)^2.

This second-moment form is often convenient as an independent check. It also makes clear why averaging component variances is insufficient: the component means contribute to the second moment as well.

Calculating the Decomposition

Check the weights.
Confirm that the component probabilities are nonnegative and sum to one.

Calculate the mixture mean.
Use $\mu=\sum_j w_j\mu_j$, keeping the units consistent.

Calculate within- and between-component terms.
Find $\sum_jw_j\sigma_j^2$ and $\sum_jw_j(\mu_j-\mu)^2$ separately.

Add and check.
Add the two terms, then verify the result using the second-moment formula when practical.

Worked Example: Two Severity Groups With Different Means

A hypothetical insurer models a claim severity $X$ in dollars as a two-component mixture. Component A is selected with probability $0.7$; conditional on A, $X$ equals \$800 or \$1,200 with equal probability. Component B is selected with probability $0.3$; conditional on B, $X$ equals \$2,600 or \$3,400 with equal probability. There is no interest or mortality assumption in this severity calculation. Find the mixture mean and variance.

Method. Calculate each component’s conditional mean and variance, then apply the within-plus-between decomposition.

For A, the conditional mean is $(800+1{,}200)/2=\$1{,}000$. Its conditional variance is \[ \frac{1}{2}(800-1{,}000)^2+\frac{1}{2}(1{,}200-1{,}000)^2 =40{,}000\text{ dollars}^2. \] For B, the conditional mean is $(2{,}600+3{,}400)/2=\$3{,}000$. Its conditional variance is \[ \frac{1}{2}(2{,}600-3{,}000)^2+\frac{1}{2}(3{,}400-3{,}000)^2 =160{,}000\text{ dollars}^2. \]

The mixture mean is $0.7(1{,}000)+0.3(3{,}000)=\$1{,}600$. The within-component term is $0.7(40{,}000)+0.3(160{,}000)=28{,}000+48{,}000=76{,}000\text{ dollars}^2$.

The between-component term is \[ 0.7(1{,}000-1{,}600)^2+0.3(3{,}000-1{,}600)^2 =0.7(360{,}000)+0.3(1{,}960{,}000) =252{,}000+588{,}000=840{,}000\text{ dollars}^2. \] Thus the mixture variance is $76{,}000+840{,}000=916{,}000\text{ dollars}^2$.

Check. The component second moments are $40{,}000+1{,}000^2=1{,}040{,}000$ and $160{,}000+3{,}000^2=9{,}160{,}000$. Their weighted average is $0.7(1{,}040{,}000)+0.3(9{,}160{,}000)=3{,}476{,}000$. Subtracting the squared mixture mean, $1{,}600^2=2{,}560{,}000$, gives $916{,}000\text{ dollars}^2$, as required. The common mistake avoided here is reporting only $76{,}000$, which omits variation between the group means.

Worked Example: Three Components in Thousands of Dollars

A hypothetical portfolio has three equally structured discrete component models, with component probabilities $0.25$, $0.50$, and $0.25$. All amounts below are in thousands of dollars. Conditional on component 1, the loss is 1 or 3 with equal probability; conditional on component 2, it is 2 or 6 with equal probability; conditional on component 3, it is 5 or 11 with equal probability. Find the mixture mean and variance. No interest or mortality model is involved.

The conditional means are $\mu_1=2$, $\mu_2=4$, and $\mu_3=8$, in thousands of dollars. For each two-point distribution, the outcomes are equally spaced around the mean. The conditional variances are therefore $\sigma_1^2=1$, $\sigma_2^2=4$, and $\sigma_3^2=9$, in thousands of dollars squared. For example, component 3 has variance $\frac12(5-8)^2+\frac12(11-8)^2=9$.

The mixture mean is $0.25(2)+0.50(4)+0.25(8)=0.5+2+2=4.5$ thousand dollars. The within-component term is $0.25(1)+0.50(4)+0.25(9)=0.25+2+2.25=4.5$ thousand dollars squared.

The between-component term is \[ 0.25(2-4.5)^2+0.50(4-4.5)^2+0.25(8-4.5)^2 =0.25(6.25)+0.50(0.25)+0.25(12.25) =1.5625+0.125+3.0625=4.75. \] The variance is $4.5+4.75=9.25$ thousand dollars squared, or $\$9{,}250{,}000$ when expressed in dollars squared.

Check. The weighted second moment is $0.25(1+2^2)+0.50(4+4^2)+0.25(9+8^2)=1.25+10+18.25=29.5$ thousand dollars squared. Subtracting $4.5^2=20.25$ gives $9.25$, agreeing with the decomposition. The common mistake avoided is mixing dollars with thousands of dollars: variance units are squared, so converting the unit multiplies the numerical variance by $1{,}000^2$.

Worked Example: Equal Component Variances Do Not Mean Equal Mixture Variance

A hypothetical severity model selects component A or B with probability $0.5$ each. Conditional on A, the loss in dollars is \$200 or \$800 with equal probability. Conditional on B, it is \$1,200 or \$1,800 with equal probability. There is no interest or mortality assumption. Calculate the mixture variance and identify the contribution from component separation.

For A, the conditional mean is $(200+800)/2=\$500$, and its variance is $\frac12(200-500)^2+\frac12(800-500)^2=90{,}000\text{ dollars}^2$. For B, the conditional mean is $(1{,}200+1{,}800)/2=\$1{,}500$, and its variance is $\frac12(1{,}200-1{,}500)^2+\frac12(1{,}800-1{,}500)^2=90{,}000\text{ dollars}^2$.

The mixture mean is $0.5(500)+0.5(1{,}500)=\$1{,}000$. The within-component term is $0.5(90{,}000)+0.5(90{,}000)=90{,}000\text{ dollars}^2$. The between-component term is $0.5(500-1{,}000)^2+0.5(1{,}500-1{,}000)^2=125{,}000+125{,}000=250{,}000\text{ dollars}^2$. Thus, the mixture variance is $340{,}000\text{ dollars}^2$.

Check. The weighted second moment is $0.5(90{,}000+500^2)+0.5(90{,}000+1{,}500^2)=170{,}000+1{,}170{,}000=1{,}340{,}000$. Subtracting $1{,}000^2=1{,}000{,}000$ gives $340{,}000\text{ dollars}^2$. Although the component variances are identical, the mixture variance is much larger because the component means are separated. Averaging the conditional variances would incorrectly give $90{,}000$ as the total.

Conditions, Interpretation, and Exam Traps

The decomposition requires a genuine partition of the modeled population: exactly one component is selected, the weights are probabilities that sum to one, and each component has a finite second moment. It applies whether the conditional severities are discrete or continuous. The component label need not be directly observed; it is enough that the model defines the conditional distributions and selection weights.

A negative between-component contribution is impossible. If a calculation gives one, check that the weights are valid, that the overall mean was calculated correctly, and that each squared deviation is formed relative to that overall mean. Keep units consistent: if component means are in dollars and variances in thousands of dollars squared, the terms cannot be added until converted to a common unit.

Leaving out between-component variation. The weighted average of conditional variances is only the within-component term, not the full mixture variance.
Using unweighted component means. The mixture mean is $\sum_j w_j\mu_j$, not the simple average of the component means unless all weights are equal.
Squaring too late or too early. The between term is the weighted average of squared deviations, $\sum_jw_j(\mu_j-\mu)^2$, not the square of a weighted average of deviations.
Confusing variance with standard deviation. The formula adds variances, not standard deviations. Take a square root only if a standard deviation is requested.

For a severity model, this distinction matters whenever a portfolio combines risks with different underlying means. A portfolio can have substantial overall dispersion even when each subgroup is relatively homogeneous. The separation into within- and between-component variation makes that source of risk explicit, and the second-moment identity provides a practical arithmetic check.

Check Your Understanding

For each mixture, calculate the overall mean and variance. Treat each stated component variance as a conditional variance.

A two-component model has weights $0.6$ and $0.4$, component means 10 and 20, and component variances 4 and 9. Calculate the within and between terms. (Answer: mean $14$; within variance $6$; between variance $24$; total variance $30$.)
A hypothetical severity model, measured in dollars, has weights $0.2$ and $0.8$, means \$100 and \$150, and variances $100$ and $400$ dollars squared. Calculate its total variance. (Answer: mean $\$140$; within variance $340$, between variance $400$, and total variance $740$ dollars squared.)
Two equally likely components have means 2 and 6 and variances 1 and 9. Find the mixture variance and explain numerically why averaging the component variances is insufficient. (Answer: mean $4$; within variance $5$; between variance $4$; total variance $9$, rather than $5$.)
Two components have weights $0.75$ and $0.25$, means 40 and 100, and variances 64 and 144. Find the mixture mean and variance. (Answer: mean $55$; within variance $84$; between variance $675$; total variance $759$.)
A three-component model has weights $0.2,0.3,0.5$, means 1, 3, 5, and variances 1, 4, 9. Calculate the mean, both variance terms, and total variance. (Answer: mean $3.6$; within variance $5.9$; between variance $2.44$; total variance $8.34$.)

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