Normal-Normal Severity Components

Why Separate Process Variation From Risk Variation?

Severity observations from one risk can vary even when that risk’s underlying expected severity is fixed. They can also differ because risks have different underlying expected severities. The Normal-Normal model represents these two sources separately: variation within a risk is the process variation, while variation in expected severity across risks is the variation of hypothetical means.

This separation is useful because a single overall variance does not explain where the variation comes from. A portfolio could have substantial variation among observations of the same risk, substantial differences in average severity between risks, or both. The two components quantify these distinct sources under a specified model.

Definition: Let $\Theta$ denote the hypothetical mean severity for a randomly selected risk. In the Normal-Normal model, $\Theta \sim N(\mu,\tau^2)$, and, conditional on $\Theta$, each severity observation $X_j$ for that risk has distribution $X_j\mid\Theta \sim N(\Theta,\sigma^2)$. Observations for a risk are conditionally independent and identically distributed given $\Theta$; different risks’ hypothetical means are independent draws from the stated population distribution. The expected process variance is $\operatorname{EPV}=E[\operatorname{Var}(X_j\mid\Theta)]$. The variance of hypothetical means is $\operatorname{VHM}=\operatorname{Var}(E[X_j\mid\Theta])$.

Here $\mu$ is the average hypothetical mean across risks, $\tau^2$ is its population variance, and $\sigma^2$ is the conditional process variance. Because the conditional mean is $\Theta$ and the conditional variance is constant, the components are $\operatorname{VHM}=\tau^2$ and $\operatorname{EPV}=\sigma^2$. These quantities are variances, not standard deviations.

Deriving the Component and Total Variance Formulas

The process component follows directly from the conditional model: every risk has conditional variance $\sigma^2$, so averaging that variance over the distribution of $\Theta$ leaves it unchanged. The mean for a given risk is $E[X_j\mid\Theta]=\Theta$; its variance across risks is therefore $\operatorname{Var}(\Theta)=\tau^2$.

The two components add to the unconditional variance. To see why, write $X_j-\mu=(X_j-\Theta)+(\Theta-\mu)$, square both sides, and take expectations. The cross term has expectation zero, since $E[(X_j-\Theta)(\Theta-\mu)]=E[(\Theta-\mu)E[X_j-\Theta\mid\Theta]]=0$. Also, $E[(X_j-\Theta)^2]=E[\operatorname{Var}(X_j\mid\Theta)]=\sigma^2$, and $E[(\Theta-\mu)^2]=\tau^2$. Thus:

\begin{aligned} \operatorname{EPV} &= E[\operatorname{Var}(X_j\mid\Theta)] = \sigma^2,\\ \operatorname{VHM} &= \operatorname{Var}(E[X_j\mid\Theta]) = \operatorname{Var}(\Theta) = \tau^2,\\ \operatorname{Var}(X_j) &= \operatorname{EPV}+\operatorname{VHM} = \sigma^2+\tau^2. \end{aligned}

The unconditional mean is $E[X_j]=E[E[X_j\mid\Theta]]=E[\Theta]=\mu$. In this model, the unconditional distribution is also normal: equivalently, write $X_j=\Theta+\varepsilon_j$, where $\varepsilon_j\sim N(0,\sigma^2)$ is independent of $\Theta$. The sum of these independent normal variables has distribution $N(\mu,\sigma^2+\tau^2)$. The variance decomposition itself does not depend on normality; the stated normal distributions make this a Normal-Normal model.

Units: If $X_j$ is measured in dollars, then $\Theta$, $\mu$, and the standard deviations $\sigma$ and $\tau$ are in dollars. The components $\sigma^2$, $\tau^2$, and the unconditional variance are in dollars squared. Do not add or compare a variance in dollars squared directly with a standard deviation in dollars.

Calculating the Components

Start by identifying whether each parameter in the problem is a variance or a standard deviation. The second argument in $N(\mu,\sigma^2)$ is a variance. If a problem gives a standard deviation, square it to obtain the corresponding variance before calculating a component.

Worked Example: Components From Conditional and Population Standard Deviations

A hypothetical portfolio is modeled with a risk-specific mean severity $\Theta\sim N(240,30^2)$ dollars. Conditional on $\Theta$, each severity observation has distribution $X_j\mid\Theta\sim N(\Theta,18^2)$ dollars and is conditionally independent of the other observations for that risk. Calculate the expected process variance, the variance of hypothetical means, the unconditional variance and standard deviation, and the unconditional mean.

Method: square the stated standard deviations to obtain the variances in the model. The conditional standard deviation is $\sigma=\$18$, so $\operatorname{EPV}=\sigma^2=18^2=324$ dollars squared.

The standard deviation of hypothetical means is $\tau=\$30$, so $\operatorname{VHM}=\tau^2=30^2=900$ dollars squared.

The total variance is $324+900=1{,}224$ dollars squared. The unconditional standard deviation is $\sqrt{1{,}224}=\$34.9857\ldots$, which is $\$34.99$ rounded to the nearest cent. The unconditional mean is $\mu=\$240$.

Check: The unconditional variance exceeds the process variance because it includes the additional between-risk component: $1{,}224>324$. The component variances add exactly to the total variance. The common mistake avoided here is adding $18$ and $30$, which are standard deviations, instead of adding their squares. The correct total variance is in dollars squared; its square root is in dollars.

The interpretation is as important as the arithmetic. The value $324$ describes average variation in observations conditional on a risk’s mean severity. The value $900$ describes variation in those mean severities across risks. Neither component alone is the unconditional variance.

Worked Example: Reading Variance and Standard Deviation Inputs Correctly

For a hypothetical type of insured property, suppose the conditional severity model is $X_j\mid\Theta\sim N(\Theta,225)$ dollars, and the distribution of risk means is $\Theta\sim N(510,24^2)$ dollars. Assume the observations are conditionally independent for a risk. Find both components and the unconditional standard deviation.

Method: read the second parameter of each normal distribution as a variance. The conditional distribution has variance $225$, so $\operatorname{EPV}=225$ dollars squared. Its conditional standard deviation is $\sqrt{225}=\$15$.

The population distribution gives the standard deviation of hypothetical means as $\tau=\$24$. Therefore, $\operatorname{VHM}=24^2=576$ dollars squared.

The unconditional variance is $225+576=801$ dollars squared. The unconditional standard deviation is $\sqrt{801}=\$28.3019\ldots$, or $\$28.30$ to the nearest cent. The unconditional mean is $\$510$.

Check: The total variance $801$ is greater than either component, and its square root is a dollar amount. The common mistake avoided is treating the given conditional variance $225$ as a standard deviation and squaring it again; the model already specifies that $225$ is the variance.

Recovering a Missing Component

If the unconditional variance and one component are known, the other component is found by subtraction. This is valid under the same model and variance decomposition: $\operatorname{Var}(X_j)=\operatorname{EPV}+\operatorname{VHM}$. The resulting component must be nonnegative. A negative value indicates inconsistent inputs or a calculation error; a variance component cannot be negative.

Worked Example: Finding the Process Variance From the Total

In a hypothetical severity model, the risk means have distribution $\Theta\sim N(700,30^2)$ dollars. The unconditional severity variance is $2{,}500$ dollars squared. Conditional on $\Theta$, severity is normal with mean $\Theta$ and an unknown constant variance. Find the expected process variance and conditional standard deviation.

Method: subtract the known variance of hypothetical means from the total variance. First calculate $\operatorname{VHM}=30^2=900$ dollars squared.

Then $\operatorname{EPV}=2{,}500-900=1{,}600$ dollars squared. The conditional standard deviation is $\sigma=\sqrt{1{,}600}=\$40$.

Check: The components sum to the given total: $1{,}600+900=2{,}500$ dollars squared. The common mistake avoided is subtracting the standard deviation $\$30$ from the total variance. The subtraction must use $\tau^2=900$ dollars squared, not $\tau=30$ dollars.

Model Assumptions and Common Traps

The component formulas rely on a particular structure: every risk has a normally distributed hypothetical mean from the stated population, and observations conditional on that mean are normal with the same variance $\sigma^2$. The model also assumes that observations for a given risk are conditionally independent given its mean. If conditional variances differ by risk, then the expected process variance is the average of those conditional variances; it is not automatically a single common $\sigma^2$.

A normal distribution allows negative values. Since an actual claim severity cannot be negative, a normal model for dollar severities is appropriate only as an approximation when the negative tail is negligible for the intended use, or when it describes a quantity for which negative values are meaningful. If a normal distribution assigns material probability to negative claim amounts, its results may not be suitable as an exact severity model. A model that restricts severity to nonnegative values has different distributional assumptions and should not be assigned these Normal-Normal formulas without justification.

Confusing standard deviations and variances. Square a stated standard deviation before treating it as a component; do not square a value already identified as a variance.
Mixing units. Components for dollar observations are measured in dollars squared, while their square roots are measured in dollars.
Calling the total variance a component. The unconditional variance includes both within-risk process variation and between-risk variation in hypothetical means.
Ignoring model conditions. The simple formula $\operatorname{EPV}=\sigma^2$ assumes a common conditional variance for all risks.

Key takeaway: For $X_j\mid\Theta\sim N(\Theta,\sigma^2)$ and $\Theta\sim N(\mu,\tau^2)$, the expected process variance is $\sigma^2$, the variance of hypothetical means is $\tau^2$, and the unconditional variance is their sum. Keep the variance units consistent throughout the calculation.

Check Your Understanding

For each hypothetical model, identify the requested components or quantities. Use the stated units and distinguish variances from standard deviations.

Severity is measured in dollars, with $X_j\mid\Theta\sim N(\Theta,8^2)$ and $\Theta\sim N(260,12^2)$. Find the expected process variance, the variance of hypothetical means, and the unconditional variance. (Answer: $64$, $144$, and $208$ dollars squared, respectively.)
A hypothetical model has $\Theta\sim N(300,12^2)$, conditional variance $196$ dollars squared, and conditional mean $\Theta$. Find the expected process variance, variance of hypothetical means, unconditional variance, and unconditional mean. (Answer: $196$ dollars squared, $144$ dollars squared, $340$ dollars squared, and $\$300$.)
The unconditional severity variance is $2{,}025$ dollars squared, and the standard deviation of hypothetical means is $\$27$. Under the Normal-Normal model, find the variance of hypothetical means, expected process variance, and conditional standard deviation. (Answer: $729$ dollars squared, $1{,}296$ dollars squared, and $\$36$.)
For a model with conditional standard deviation $\$10$ and hypothetical-mean standard deviation $\$25$, calculate the fraction of unconditional variance attributable to process variation. (Answer: $100/(100+625)=0.1379$, or $13.79\%$.)
Suppose $X_1$ and $X_2$ are conditionally independent observations for the same risk, with $X_j\mid\Theta\sim N(\Theta,9^2)$ and $\Theta\sim N(120,16^2)$. Find the unconditional variance of one observation, the covariance between the two observations, and their correlation. (Answer: $337$ dollars squared, $256$ dollars squared, and $256/337=0.7596$.)

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