Method-of-Moments Component Estimates

Why Estimate the Components Separately?

In the Bühlmann model, observations vary for two distinct reasons: outcomes fluctuate within a risk, and the underlying expected outcomes differ across risks. The first source is the expected process variance (EPV); the second is the variance of hypothetical means (VHM). Estimating these components separately is useful because they describe different features of the portfolio and enter later credibility calculations in different ways.

For balanced repeated-period data, analysis of variance (ANOVA) separates the total variation into within-risk and between-risk parts. The method of moments matches the expected values of the corresponding mean squares to their observed values. The key is to use the right degrees of freedom and account for the fact that between-risk variation contains both components.

Definition: Suppose there are \(R\) risks and each risk has \(n\) observations, with \(R \ge 2\) and \(n \ge 2\). Let \(X_{ij}\) be observation \(j\) for risk \(i\), and let \(\bar X_i\) be that risk’s average. Write \(\mu(\Theta_i)=E[X_{ij}\mid\Theta_i]\) and \(\sigma^2(\Theta_i)=\operatorname{Var}(X_{ij}\mid\Theta_i)\), where \(\Theta_i\) represents the risk’s underlying characteristics. Then \(\operatorname{EPV}=E[\sigma^2(\Theta_i)]\) and \(\operatorname{VHM}=\operatorname{Var}(\mu(\Theta_i))\).

The standard balanced-data assumptions for these estimates are that risks are independent draws from the same population, and observations within a risk are conditionally independent given that risk’s characteristics. Each risk has the same number of observations, measured on a comparable scale and under the same data definition. The conditional mean and variance are common across periods for a given risk. Normality is not required for the expected-mean-square equations below, though it is often used for exact ANOVA distributional results.

Mean Squares and Their Degrees of Freedom

The within-risk sum of squares measures how far observations fall from their own risk average. The between-risk sum of squares measures how far the risk averages fall from the grand average. Define \(\bar X_{\cdot\cdot}\) as the average of all \(Rn\) observations. For a balanced design, the mean squares are:

\begin{aligned} SS_W &= \sum_{i=1}^{R}\sum_{j=1}^{n}(X_{ij}-\bar X_i)^2, & MS_W &= \frac{SS_W}{R(n-1)},\\ SS_B &= n\sum_{i=1}^{R}(\bar X_i-\bar X_{\cdot\cdot})^2, & MS_B &= \frac{SS_B}{R-1}. \end{aligned}

There are \(R(n-1)\) within degrees of freedom: each of the \(R\) risks contributes \(n\) observations, but estimating its own average uses one degree of freedom. There are \(R-1\) between degrees of freedom because the \(R\) risk averages are centered around their estimated grand average. The factor \(n\) in \(SS_B\) puts between-risk variation on the same observation-level scale as within-risk variation.

Theorem: Under the balanced-data assumptions above, \(E[MS_W]=\operatorname{EPV}\) and \(E[MS_B]=\operatorname{EPV}+n\operatorname{VHM}\). Therefore, the method-of-moments estimates are \(\widehat{\operatorname{EPV}}=MS_W\) and \(\widehat{\operatorname{VHM}}=(MS_B-MS_W)/n\).

Why the Mean-Square Equations Hold

The within-risk equation follows by conditioning on a risk’s characteristics. Given \(\Theta_i\), the \(n\) observations have a common conditional mean and variance \(\sigma^2(\Theta_i)\), and are conditionally independent. The expected sum of squared deviations from their sample average is \((n-1)\sigma^2(\Theta_i)\). Summing across risks and taking expectations gives:

E[SS_W]=R(n-1)E[\sigma^2(\Theta_i)] =R(n-1)\operatorname{EPV}, \qquad E[MS_W]=\operatorname{EPV}.

For the between-risk equation, the law of total variance gives the variance of an individual observation as \(\operatorname{VHM}+\operatorname{EPV}\). Conditional independence means the risk average has process variance \(\operatorname{EPV}/n\), while its variation in underlying risk mean remains \(\operatorname{VHM}\). Thus the variance of a risk average is \(\operatorname{VHM}+\operatorname{EPV}/n\). The expected sample variance of the \(R\) risk averages is this quantity, so multiplying by \(n\), as in \(SS_B\), yields:

E[MS_B] =n\left(\operatorname{VHM}+\frac{\operatorname{EPV}}{n}\right) =\operatorname{EPV}+n\operatorname{VHM}.

The method-of-moments step now has two equations in two unknowns. Set the observed mean squares equal to their expected values and solve:

MS_W=\operatorname{EPV}, \qquad MS_B=\operatorname{EPV}+n\operatorname{VHM}.

The first equation estimates EPV directly. Subtracting it from the second gives \(MS_B-MS_W=n\operatorname{VHM}\), and division by \(n\) gives the VHM estimate. The estimate is divided by the number of observations per risk, not by a degrees-of-freedom count.

Applying the Estimates

Worked Example: Computing Both Components From Repeated Observations

A hypothetical portfolio has four risks, each observed for three periods. The observations for the risks are \((7,8,9)\), \((9,10,11)\), \((11,12,13)\), and \((13,14,15)\). Estimate EPV and VHM by the method of moments.

Method. Calculate the within and between sums of squares, divide by their respective degrees of freedom, and solve the two expected-mean-square equations. The common mistake to avoid is using the between degrees of freedom for the within mean square.

Step 1: Risk and grand averages. The four risk averages are \(8,10,12,14\). Their grand average is \((8+10+12+14)/4=11\), which is also the average of all 12 observations.

Step 2: Within mean square. Within each risk, the deviations from its average are \((-1,0,1)\), with squared deviations summing to \(1+0+1=2\). Across four risks, \(SS_W=4(2)=8\). The within degrees of freedom are \(R(n-1)=4(3-1)=8\). Hence \(MS_W=8/8=1\), so \(\widehat{\operatorname{EPV}}=1\).

Step 3: Between mean square. The squared deviations of the risk averages from 11 are \(9,1,1,9\), summing to 20. Therefore \(SS_B=n(20)=3(20)=60\). The between degrees of freedom are \(R-1=3\), so \(MS_B=60/3=20\).

Step 4: VHM estimate and check. The estimate is \((MS_B-MS_W)/n=(20-1)/3=19/3=6.333\ldots\). To check, the fitted expected mean squares are \(1\) for within and \(1+3(19/3)=20\) for between, matching the observed values. Rounded to three decimals, the estimates are EPV \(=1.000\) and VHM \(=6.333\).

Worked Example: Correcting a Degrees-of-Freedom Trap

A hypothetical study has five risks with four observations per risk. The within-risk sum of squares is 15. The between-risk mean square is 40. Estimate the two variance components.

Method. Use the balanced within degrees of freedom \(R(n-1)\), then use the mean-square difference to estimate VHM. The trap is to divide the within sum of squares by \(R-1\).

Step 1: Correct within mean square. Here \(R=5\) and \(n=4\), so the within degrees of freedom are \(5(4-1)=15\). Thus \(MS_W=15/15=1\), and \(\widehat{\operatorname{EPV}}=1\).

Step 2: Between component. Since \(MS_B=40\), \(\widehat{\operatorname{VHM}}=(40-1)/4=39/4=9.75\).

Step 3: Check and contrast the error. The fitted between mean square is \(1+4(9.75)=40\), as observed. Dividing \(SS_W\) by \(R-1=4\) would give \(15/4=3.75\), but that is not the within mean square: \(R-1\) belongs to the between calculation. The estimates are EPV \(=1\) and VHM \(=9.75\).

Worked Example: Reading an ANOVA Summary

For a hypothetical balanced data set, an ANOVA summary reports 24 within degrees of freedom and \(SS_W=60\), along with 5 between degrees of freedom and \(SS_B=90\). Each risk has the same number \(n=5\) of observations. Estimate EPV and VHM.

Method. Convert each sum of squares into a mean square using its reported degrees of freedom, then apply the moment equations. This avoids treating a sum of squares as if it were already a variance estimate.

Step 1: Verify the design and calculate mean squares. Since \(n=5\), the within degrees of freedom are \(R(5-1)=4R\). The reported value 24 implies \(R=6\). The between degrees of freedom are then \(6-1=5\), consistent with the summary. The mean squares are \(MS_W=60/24=2.5\) and \(MS_B=90/5=18\).

Step 2: Estimate the components. The EPV estimate is \(MS_W=2.5\). The VHM estimate is \((18-2.5)/5=15.5/5=3.1\).

Step 3: Check. The implied expected within mean square is \(2.5\); the implied expected between mean square is \(2.5+5(3.1)=18\). Both match the reported mean squares. The estimates are EPV \(=2.5\) and VHM \(=3.1\), in squared units of the observations.

Conditions, Interpretation, and Practical Cautions

These formulas rely on the balanced structure. If risks have different numbers of observations, the simple between mean-square equation \(E[MS_B]=\operatorname{EPV}+n\operatorname{VHM}\) does not apply with one common \(n\); an unbalanced-data method is needed. Similarly, if observations are dependent within a risk, the usual within and between expected mean squares may not hold. In either case, first check the data design rather than mechanically applying the balanced formulas.

The two estimates also have different interpretations. EPV estimates the average variance of outcomes around each risk’s own expected value. VHM estimates how much those risk-specific expected values vary around the overall mean. The between mean square alone is not a VHM estimate: it includes process variation as well as variation in hypothetical means. Subtracting the within mean square removes the process component in expectation, and dividing by \(n\) accounts for the factor in the between-risk equation.

A sample can produce \(MS_B<MS_W\), and therefore a negative method-of-moments VHM estimate. That does not mean the population variance is negative; it is a consequence of sampling variation in the observed mean squares. The raw estimate should be identified as such rather than silently changed. How boundary estimates are handled is a separate issue.

For a quick calculation audit, verify all three points: the within denominator is \(R(n-1)\); the between denominator is \(R-1\); and the VHM estimate is the mean-square difference divided by \(n\). These checks prevent the most common errors in balanced component estimation.

Check Your Understanding

For each balanced-data problem, calculate the requested component estimates. Retain unrounded values in intermediate steps.

A study has three risks with four observations per risk. The within sum of squares is 18 and the between mean square is 14. Find the method-of-moments EPV and VHM estimates. (Answer: EPV \(=2\), VHM \(=3\).)
There are four risks, each with three observations. The within sum of squares is 24, and the between mean square is 15. Estimate both components and check the fitted between mean square. (Answer: EPV \(=3\), VHM \(=4\); fitted \(MS_B=15\).)
Five risks each have two observations. The ANOVA mean squares are \(MS_W=1.8\) and \(MS_B=7.8\). Estimate EPV and VHM. (Answer: EPV \(=1.8\), VHM \(=3\).)
Six risks each have five observations. The within sum of squares is 72 and the between sum of squares is 270. Calculate the degrees of freedom, both mean squares, and the two component estimates. (Answer: within df \(=24\), between df \(=5\), \(MS_W=3\), \(MS_B=54\), EPV \(=3\), VHM \(=10.2\).)
Four risks each have three observations. The observed mean squares are \(MS_W=2.5\) and \(MS_B=12.5\). Estimate the components, and state why using \(MS_B-MS_W\) alone as VHM is incorrect. (Answer: EPV \(=2.5\), VHM \(=10/3\approx3.333\); the difference equals \(n\) times VHM, so divide by \(n=3\).)

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