Two Distinct Features of Nonnormal Data
A bivariate cloud can depart from a Gaussian model in more than one way. One departure is asymmetry: the distribution has a longer or more prominent side in some direction. Another is heavy tails: observations far from the centre occur more often than a Gaussian model predicts. These features can occur separately or together. A useful diagnosis therefore asks two questions: do directional summaries indicate asymmetry, and do distances from the fitted centre have unusually heavy tails?
The relevant fitted models represent different possibilities. A Gaussian model is symmetric with Gaussian radial tails. A multivariate \(t\) model is also centrally symmetric, but has heavier tails. A skew-normal model allows asymmetry without the polynomial tails of a \(t\) distribution. A skew-\(t\) model allows both. No single statistic settles the comparison: directional and radial evidence should be read together with fitted-model diagnostics.
Directional Skewness and What It Can Show
For scalar observations \(y_1,\ldots,y_n\), write \(\bar y=n^{-1}\sum_i y_i\) and \(m_r=n^{-1}\sum_i(y_i-\bar y)^r\). The moment-based sample skewness is \(g_1=m_3/m_2^{3/2}\), provided \(m_2>0\). Positive values indicate a longer or more influential upper side in that projection; negative values indicate the opposite. In a bivariate analysis, one can inspect the two coordinates and scientifically meaningful linear combinations, rather than treating a single coordinate system as definitive.
A basic population fact clarifies the interpretation. A distribution is centrally symmetric about \(\mu\) if \(X-\mu\) and \(-(X-\mu)\) have the same distribution. When the relevant third moments exist, this symmetry forces every projected third central moment to vanish.
Proof. Let \(Y=a^\mathsf{T}(X-\mu)\). Central symmetry implies that \(Y\) and \(-Y\) have the same distribution. The stated integrability makes their third moments finite, so
Thus \(E[Y^3]=0\). The conclusion is a population statement, not a claim that every finite sample from a symmetric distribution has sample skewness exactly zero. \(\square\)
Worked Example: A Positive Skewness in One Direction
Consider five bivariate observations \((-2,-1),(-1,1),(0,-1),(1,1),(4,0)\). Examine the first coordinate \(x\). Its mean is \(\bar x=( -2-1+0+1+4)/5=0.4\), and the centered values are \(-2.4,-1.4,-0.4,0.6,3.6\).
The second and third central moments are
Therefore \(g_1=6.048/(4.24)^{3/2}\). Since \(\sqrt{4.24}\approx2.059\), the denominator is approximately \(8.73\), giving \(g_1\approx0.693\). This is evidence of positive sample skewness in the first-coordinate direction. It does not, by itself, establish that the population is skewed: with only five observations, sampling variation is substantial. Nor does it tell us whether the bivariate radial tails are heavy.
Radial Distances and Model-Specific References
A fitted location \(\hat\mu\) and positive-definite scale matrix \(\hat\Sigma\) turn each observation into a squared Mahalanobis distance \(q_i\). The matrix standardizes distances for both marginal scale and correlation. A large \(q_i\) means an observation is far from the fitted centre relative to the fitted scale, but whether it is unexpectedly large depends on the model.
For a bivariate Gaussian with location \(\mu\) and covariance matrix \(\Sigma\), \(q=(X-\mu)^\mathsf{T}\Sigma^{-1}(X-\mu)\) has a chi-square distribution with two degrees of freedom. For a bivariate \(t\), use its scale matrix \(\Sigma\) in this definition. Under the usual construction with degrees of freedom \(\nu\), \(q\) has distribution \(2F_{2,\nu}\). Here the \(t\) scale matrix is not its covariance matrix: the covariance is \(\nu\Sigma/(\nu-2)\) when \(\nu>2\).
Derivation. Let \(U=Z^\mathsf{T}\Sigma^{-1}Z\). Whitening the normal vector gives \(U\sim\chi^2_2\), and \(U\) is independent of \(W\). Substitution into the distance gives \(q=U/(W/\nu)\). By the definition of the \(F\) distribution, \((U/2)/(W/\nu)\sim F_{2,\nu}\); hence \(q=2F_{2,\nu}\). \(\square\)
Worked Example: Comparing Gaussian and t Radial Tails
Suppose eight observations have fitted bivariate squared distances \(0.2,0.4,0.6,0.8,1.0,1.2,1.4,8.0\), computed using the fitted location and scale. The largest distance is \(8\). Under a bivariate Gaussian reference, the survival probability at that value is
Now compare a bivariate \(t\) reference with \(\nu=4\). Since \(q/2\sim F_{2,4}\), the reference tail probability is \(P(F_{2,4}\geq4)\). For \(F_{2,4}\), the survival function at \(x\geq0\) is \((1+x/2)^{-2}\), so
Thus the same distance is much less surprising under the heavier-tailed \(t\) reference. This comparison is illustrative: with estimated parameters and a small sample, the nominal tail probabilities are not exact calibrated tests. It also says nothing about directional asymmetry. The distance \(q\) discards the direction in which the observation lies.
Fitted Models: Separate the Evidence Before Comparing
A practical analysis uses directional and radial diagnostics as complementary evidence. A skewness signal with ordinary radial behaviour is consistent with asymmetry without a strong heavy-tail signal. Large radial distances in several directions, together with little directional skewness, suggest heavy tails without clear asymmetry. Evidence in both suggests considering a skew-\(t\) model. These are diagnostic patterns, not proofs of the data-generating mechanism.
For a fitted Gaussian or \(t\) model, the corresponding radial reference distribution gives a direct benchmark. A skew-normal or skew-\(t\) model generally does not have that same simple chi-square or scaled-\(F\) radial law when distances are measured from its location parameter and scale matrix. In that case, compare the observed radial distribution with the fitted model’s own radial distribution, for example by simulating datasets from the fitted model and recomputing the distances. To account for parameter estimation, a useful simulation procedure refits the model to each simulated dataset before calculating its diagnostic.
Model fit can also be compared with AIC, defined as \(\mathrm{AIC}=-2\ell(\hat\theta)+2k\), where \(\ell(\hat\theta)\) is the maximized log-likelihood and \(k\) is the number of fitted parameters. AIC assesses a fit-complexity trade-off; it does not identify which feature caused one model to fit better. The directional and radial diagnostics help explain that distinction.
Worked Example: Reading Four Fitted Models Together
For a bivariate dataset with \(n=80\), suppose the largest absolute projected sample skewness across two prespecified directions is \(0.12\), while several fitted-scale distances are larger than expected under a Gaussian reference. Suppose maximum likelihood fitting gives the following illustrative results. Parameter counts include location, scale, shape where applicable, and degrees of freedom for the \(t\) models.
| Model | Fitted feature | \(k\) | \(\ell(\hat\theta)\) | AIC |
|---|---|---|---|---|
| Gaussian | Symmetric, Gaussian tails | 5 | -31.0 | 72.0 |
| t, \(\hat\nu=5.4\) | Symmetric, heavy tails | 6 | -27.2 | 66.4 |
| Skew-normal | Asymmetric, non-\(t\) tails | 7 | -27.8 | 69.6 |
| Skew-\(t\), \(\hat\nu=5.1\) | Asymmetric, heavy tails | 8 | -26.1 | 68.2 |
For the Gaussian fit, \(-2(-31.0)+2(5)=62+10=72.0\). For the \(t\) fit, \(-2(-27.2)+2(6)=54.4+12=66.4\). The skew-normal AIC is \(55.6+14=69.6\), and the skew-\(t\) AIC is \(52.2+16=68.2\). The smallest AIC is \(66.4\), for the \(t\) model.
Here the weak projected skewness and the better \(t\) fit relative to the Gaussian are consistent with heavy tails being more important than asymmetry. The skew-\(t\) has the largest maximized likelihood, but its additional parameters raise its AIC above the \(t\) model’s. These values are illustrative fit output, not calculations from the preceding five-point dataset. A different dataset could show stronger directional skewness and favour a skew-normal or skew-\(t\) fit. Model choice should also consider diagnostic plots, convergence, and whether the fitted model describes the data adequately.
Common Pitfalls and Practical Interpretation
- Calling one extreme point “skewness.” A large distance indicates extremeness relative to a chosen centre and scale; it does not establish a persistent imbalance across directions. Inspect directional projections as well.
- Using the Gaussian radial reference for every model. A \(t\) model has a heavier radial reference, and skew models may require model-specific simulation. A Gaussian-tail flag can simply reflect a wrong reference distribution.
- Reading sample skewness as a definitive population result. Third moments are sensitive to individual observations, especially in small samples. Assess several relevant directions and interpret the result with the sample size in mind.
- Treating AIC as a diagnosis of the departure. AIC ranks the fitted models in a relative complexity-adjusted comparison. It does not replace the directional and radial checks that help distinguish asymmetry from heavy tails.
A disciplined diagnosis keeps these signals separate until the evidence is combined: first examine directional imbalance, then compare radial behaviour with the fitted model’s reference, and finally assess whether models that allow asymmetry, heavy tails, or both provide a worthwhile improvement. When parameters are estimated from the same observations, simulation with refitting can make diagnostic comparisons more realistic.
Check Your Understanding
Use the definitions and calculations above to answer these questions.
- What feature of a bivariate distribution does a projected third central moment diagnose, and what does it not diagnose?
- Why does the squared distance for a bivariate \(t\) model use a scaled \(F\) reference rather than the Gaussian chi-square reference?
- An observation has a large Mahalanobis distance. What additional evidence would help distinguish asymmetry from heavy tails?
- Why might simulated radial diagnostics for a fitted skew-\(t\) model be preferable to a Gaussian chi-square comparison?
- In the four-model example, why does the skew-\(t\) model have a larger AIC than the \(t\) model despite having a higher maximized log-likelihood?