Why a Gaussian Model Can Be Awkward for Positive Outcomes
Costs, claim amounts, and durations are often positive and right-skewed: many observations are modest, while a few are much larger. A Gaussian model can still be useful in some settings, but its assumptions have consequences that may conflict with this data shape. In particular, a Gaussian response has support over the entire real line, its variance is constant in the usual Gaussian GLM, and its errors are symmetric about their means.
These are not merely formal details. A fitted Gaussian model for cost can assign probability to negative costs, and a constant-variance specification can miss a spread that grows with the expected cost. A positive-response model can address some of these concerns, but the choice needs to reflect the response distribution as well as the mean. A link that makes a fitted mean positive does not, by itself, make the response positive.
The earlier material on matching variance growth to a family motivates comparing the observed spread with the mean. Here the emphasis is on what a Gaussian model implies for positive, skewed outcomes, and on diagnostic evidence that may make those implications uncomfortable. No single sign—negative fitted values, increasing spread, or an asymmetric residual sample—settles family choice on its own.
Negative Values: A Consequence of Gaussian Support
For a Gaussian response with mean \(\mu>0\) and standard deviation \(\sigma>0\), the probability of a negative outcome is determined by the distance from the mean to zero in standard-deviation units. Writing \(\Phi\) for the standard normal cumulative distribution function gives the following result.
To verify the directions of change, differentiate with respect to \(\sigma\) and \(\mu\), respectively. If \(\phi_N\) is the standard normal density, then \[ \frac{\partial}{\partial \sigma}\Phi\left(-\frac{\mu}{\sigma}\right) =\phi_N\left(\frac{\mu}{\sigma}\right)\frac{\mu}{\sigma^2}>0, \qquad \frac{\partial}{\partial \mu}\Phi\left(-\frac{\mu}{\sigma}\right) =-\frac{1}{\sigma}\phi_N\left(\frac{\mu}{\sigma}\right)<0. \] Thus the negative probability is especially concerning when the standard deviation is large relative to the mean. It is not enough to observe that the fitted mean is positive.
Worked Example: Negative Probabilities for Cost and Duration
Suppose a Gaussian model gives a predicted medical cost of \(\$100\) with standard deviation \(\$80\). The standardized distance from the mean to zero is \(100/80=1.25\), so \[ P(Y<0)=\Phi(-1.25)\approx 0.1056. \] The model assigns about 10.6% probability to a negative cost. Whether that is tolerable depends on the use of the model, but it is not a negligible probability if predictions are meant to describe individual costs.
Now suppose a Gaussian duration model predicts a mean of 10 days and a standard deviation of 2 days. Here \(10/2=5\), so \[ P(Y<0)=\Phi(-5)\approx 0.000000287. \] The probability of a negative duration is tiny under this prediction. The Gaussian support is still mathematically inconsistent with strictly positive durations, but the practical impact of that inconsistency is much smaller at this mean-to-spread ratio.
A positive-response model can assign zero probability to negative outcomes if its response distribution has positive support. A positive link, such as \(\mu_i=\exp(\eta_i)\), ensures only that the model’s mean is positive. If the response distribution itself has support over negative values, the link alone does not remove negative outcomes.
Mean-Dependent Spread and the Gaussian Variance Assumption
In the usual Gaussian GLM, \(\operatorname{Var}(Y_i\mid x_i)=\phi\) for every observation. This is a constant-variance statement: changing the predictors may change the mean, but it does not change the conditional variance under this model. Positive outcomes often show a different pattern, with larger means accompanied by greater spread. That pattern is evidence to examine, not proof that a particular alternative family is correct.
Indeed, at predictors \(x_1,x_2\) with distinct positive means \(\mu_1\ne\mu_2\), the data variance values are \(c\mu_1^2\) and \(c\mu_2^2\). These differ, since both means are positive. A Gaussian GLM with common dispersion would require both values to equal \(\phi\), which is impossible. This is a mismatch between the assumed variance pattern and the stated data pattern; it does not establish that the variance relation is known exactly in a real dataset.
Worked Example: Comparing Cost Spread Across Mean Levels
Suppose three groups of policyholders have the following observed average costs and standard deviations.
| Group | Average cost | Standard deviation | Variance |
|---|---|---|---|
| A | \(\$100\) | \(\$20\) | \(400\) |
| B | \(\$200\) | \(\$40\) | \(1{,}600\) |
| C | \(\$400\) | \(\$80\) | \(6{,}400\) |
The variances are \(20^2=400\), \(40^2=1{,}600\), and \(80^2=6{,}400\). Dividing each by the squared group mean gives \[ \frac{400}{100^2}=0.04,\qquad \frac{1{,}600}{200^2}=0.04,\qquad \frac{6{,}400}{400^2}=0.04. \] In these summaries, variance grows in proportion to the squared mean. A Gaussian GLM with common dispersion does not reproduce that pattern when the group means differ. In practice, group sizes and sampling variability matter: three sample variances alone do not establish a population variance law. They do, however, motivate checking whether residual spread changes systematically with fitted mean.
Residual Asymmetry: What It Shows and What It Does Not
A Gaussian model implies conditionally symmetric errors: the distribution of \(Y_i-\mu_i\), given the predictors, is symmetric around zero. In particular, its third central moment is zero. This follows because a symmetric density assigns equal contributions to positive and negative values when integrating the odd function \(e^3\). A right tail of unusually large positive residuals is therefore in tension with the Gaussian description.
In a fitted model, residuals are not simply independent draws from the error distribution: fitting the coefficients imposes constraints on them, and their variability can differ across observations. Residual plots and summaries are diagnostic tools, not direct observations of independent errors. One unusually large residual may be influential, reflect an omitted feature of the data, or arise by chance. Look for a pattern and consider it alongside support and spread.
Worked Example: A Large Cost and an Asymmetric Residual Sample
Consider ten positive costs, in dollars: \(80, 90, 100, 110, 120, 130, 140, 150, 160, 500\). For an intercept-only Gaussian model, the fitted mean is the sample average. The sum is \(1{,}580\), so the fitted mean is \(1{,}580/10=158\).
Subtracting 158 from each observation gives residuals \(-78,-68,-58,-48,-38,-28,-18,-8,2,342\). They sum to zero, as they must for this intercept-only least-squares fit. Eight residuals are negative and two are positive; one positive residual is especially large.
The sum of the cubes of the negative residuals is \(-1{,}177{,}856\). The positive cubes sum to \(2^3+342^3=40{,}001{,}696\). Thus the sum of all residual cubes is \[ 40{,}001{,}696-1{,}177{,}856=38{,}823{,}840>0. \] The positive third-moment direction reflects the long right tail in this sample. It is a warning that a symmetric Gaussian description may be inadequate, not a formal test that proves a different family must be used. With only ten observations, the tail pattern is also uncertain.
Putting the Diagnostics Together
For a positive, right-skewed outcome, assess three connected implications rather than relying on a single rule. First, calculate whether the Gaussian model gives material probability to impossible negative values at relevant fitted means. Second, compare residual spread across the range of fitted means; a systematic increase conflicts with the constant-variance Gaussian specification. Third, inspect residual shape for persistent asymmetry, while remembering that individual residuals are affected by fitting and may be noisy.
A positive-response GLM may better reflect the support and mean–variance pattern, but it is not automatically adequate. Its assumptions still need to be checked against the scientific problem and observed data. Conversely, a Gaussian model is not ruled out merely because the outcome is positive or somewhat skewed. If negative probability is negligible in the prediction range, spread is reasonably stable, and residuals do not show a consequential pattern, the Gaussian model may provide a useful approximation for its intended purpose.
The practical question is not whether one model is universally acceptable. It is whether the model’s support, spread, and residual shape are credible for the intended predictions or inferences. For positive costs and durations, checking these features makes the comparison concrete: the negative tail can be quantified, spread can be compared across mean levels, and asymmetry can be examined without treating any one diagnostic as conclusive.
Check Your Understanding
Use the model implications in this tutorial to answer the following questions.
- A Gaussian cost model has fitted mean \(\$60\) and standard deviation \(\$30\). Write the expression for the probability it assigns to a negative cost, and evaluate it approximately.
- Why does a positive fitted mean from a log link not, by itself, guarantee that every response generated by the model is positive?
- A Gaussian GLM has one common dispersion, while observed variance appears to increase across groups with distinct means. What assumption is in tension with this pattern?
- Why is a large positive residual informative about right-tail asymmetry but not, on its own, proof that the Gaussian model is wrong?
- For a positive duration outcome, name the three model features considered here that should be checked before deciding whether a Gaussian description is adequate.