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Exam SRM · Regression Diagnostics · Tutorial 95 of 926

Tests of Residual Normality and Their Limitations

Learn how to interpret two residual-normality tests, calculate the classic Jarque-Bera statistic, and put statistical significance in context.

Intermediate 10 min read

What You'll Learn

  • State the null and alternative hypotheses for Shapiro-Wilk and skewness-kurtosis tests.
  • Interpret the Shapiro-Wilk statistic and its p-value without relying on a universal cutoff for the statistic.
  • Calculate the classic Jarque-Bera statistic from sample skewness and kurtosis.
  • Use the large-sample chi-square reference distribution and assess its limitations for regression residuals.
  • Explain why rejecting normality does not by itself identify a cause or settle whether a model is useful.

Why Test Residual Normality?

A regression model can produce residuals that look roughly symmetric in a histogram yet still depart from a normal shape in their tails. Formal tests provide a way to assess such departures, but their results need context: a small p-value is evidence against a specified null model, not a diagnosis of what went wrong or a complete verdict on the regression.

For the classical linear regression model, write the response for observation \(i\) as \(Y_i\), its fitted value as \(\hat{Y}_i\), and its residual as \(e_i=Y_i-\hat{Y}_i\). The normality condition concerns the errors conditional on the predictors: under the classical model, \(\varepsilon_i\mid X\sim N(0,\sigma^2)\). This condition is distinct from linearity and constant variance. It supports exact small-sample t and F inference when the other model assumptions also hold; it is not required merely for the OLS coefficients to be calculated.

Definition: A residual-normality test evaluates a null hypothesis that the observations being tested are consistent with a normal distribution. In regression, those observations are usually fitted-model residuals. Because residuals are calculated after estimating the regression, they are not generally an independent sample from a normal population, even when the model errors are normal.

That last distinction matters. If the design matrix is \(X\), the OLS residual vector is \((I-H)\boldsymbol{\varepsilon}\), where \(H\) is the projection matrix. Under normal errors its covariance matrix is \(\sigma^2(I-H)\), not \(\sigma^2 I\). Thus residuals can be correlated and can have different variances. Tests designed for an independent sample from a normal distribution are useful diagnostics, but their ordinary reference distributions may not be exact for fitted residuals. A model-specific calibration, such as simulation under the fitted model, can be useful when precise calibration is important.

Shapiro-Wilk Test

The Shapiro-Wilk test is designed to detect departures from normality, including departures in shape that may not be obvious in a small histogram. For an independent sample \(x_1,\ldots,x_n\), sort the values to obtain \(x_{(1)}\leq\cdots\leq x_{(n)}\). Its statistic has the form below. The coefficients \(a_i\) depend on the sample size and on expected order statistics from a standard normal distribution; they are supplied by statistical software or tables.

$$ W=\frac{\left(\sum_{i=1}^{n}a_i x_{(i)}\right)^2} {\sum_{i=1}^{n}(x_i-\bar{x})^2} $$

The numerator measures how closely the ordered observations align with the pattern expected under normality, while the denominator measures their total squared spread. Values of \(W\) closer to 1 are generally more consistent with normality. However, there is no universal rule such as “accept normality whenever \(W>0.95\)”: interpretation uses the p-value for the relevant sample size and test calibration.

Hypotheses: For the Shapiro-Wilk test, \(H_0\) is that the sampled values come from a normal distribution; \(H_1\) is that they do not. At significance level \(\alpha\), reject \(H_0\) when the reported p-value is at most \(\alpha\). Failure to reject means the test has not provided sufficient evidence against normality; it does not prove that the distribution is normal.

Worked Example: Interpreting a Shapiro-Wilk Output

A hypothetical analyst fits a regression to 24 policy records and runs a Shapiro-Wilk test on its residuals. Software reports \(W=0.901\) and \(p=0.032\). At \(\alpha=0.05\), state the decision and its meaning.

Method. Compare the p-value with the stated significance level. The test’s null hypothesis is that the tested values are consistent with a normal distribution.

Decision. Since \(0.032<0.05\), reject the null hypothesis at the 5% level. The residuals provide evidence of departure from normality under the test’s reference calibration.

Interpretation. This result does not show which aspect of the distribution is responsible, nor does it establish that a particular regression assumption caused the result. The analyst should inspect the residual distribution and consider the model and the intended inference. Because these are fitted residuals, the ordinary independent-sample calibration should also be understood as an approximation unless the procedure accounts for regression fitting.

Common mistake avoided. Do not interpret \(W=0.901\) using an invented universal cutoff, and do not say the model is “invalid” solely because the p-value is below 0.05. The decision is evidence against the test’s null, not a complete diagnostic judgment.

Omnibus Skewness-Kurtosis Tests

An omnibus skewness-kurtosis test combines evidence from two features of distribution shape. Skewness measures asymmetry; kurtosis measures tail weight and peakedness relative to a normal distribution. The classic Jarque-Bera test uses sample skewness \(S\) and Pearson sample kurtosis \(K\), for which a normal distribution has population values 0 and 3, respectively. Equivalently, excess kurtosis is \(g_2=K-3\).

Definition: The classic Jarque-Bera statistic is
$$ JB=\frac{n}{6}\left(S^2+\frac{(K-3)^2}{4}\right) =\frac{n}{6}\left(S^2+\frac{g_2^2}{4}\right). $$
Under the null hypothesis of normality, its large-sample reference distribution is chi-square with 2 degrees of freedom. This is an asymptotic approximation, not an exact finite-sample distribution. Other omnibus tests and software may use different definitions or corrections; use the statistic and reference distribution specified in the problem.

The classic formula uses skewness and kurtosis calculated with the stated convention. For hand calculations here, use moment estimates with divisor \(n\). If software provides a statistic directly, do not recompute it using a different finite-sample convention.

Why the Jarque-Bera Reference Distribution Has Two Degrees of Freedom

The large-sample justification follows from the behavior of sample skewness and excess kurtosis under normality. For an independent sample from a normal distribution, their approximate large-sample variances are \(6/n\) and \(24/n\), and they are asymptotically independent. Standardizing each gives two approximately standard normal quantities:

$$ Z_1=\sqrt{\frac{n}{6}}S, \qquad Z_2=\sqrt{\frac{n}{24}}g_2. $$

The sum of squares of two independent standard normal quantities has a chi-square distribution with 2 degrees of freedom. Substituting the expressions above gives

$$ Z_1^2+Z_2^2 =\frac{n}{6}S^2+\frac{n}{24}g_2^2 =\frac{n}{6}\left(S^2+\frac{g_2^2}{4}\right) =JB. $$

This explains both the formula and the reference distribution. It also shows why the result is asymptotic: the standardizations are only approximately normal in finite samples. For a chi-square distribution with 2 degrees of freedom, the upper-tail probability is particularly simple:

$$ P(\chi^2_2\geq JB)=\exp(-JB/2). $$

Worked Example: Calculating a Jarque-Bera Test

In a hypothetical regression analysis, \(n=50\) residuals have sample skewness \(S=-0.60\) and Pearson kurtosis \(K=4.20\). Use the classic large-sample Jarque-Bera test at \(\alpha=0.05\).

Method. Calculate excess kurtosis, substitute the summaries in the classic statistic, then use the chi-square reference distribution with 2 degrees of freedom.

Step 1: Excess kurtosis. The normal reference value for Pearson kurtosis is 3, so

$$ g_2=K-3=4.20-3=1.20. $$

Step 2: Square and combine the components.

$$ S^2=(-0.60)^2=0.36, \qquad \frac{g_2^2}{4}=\frac{(1.20)^2}{4}=\frac{1.44}{4}=0.36, $$

so the sum in parentheses is \(0.36+0.36=0.72\).

Step 3: Calculate the statistic.

$$ JB=\frac{50}{6}(0.72)=8.3333\ldots(0.72)=6.0000. $$

Step 4: Calculate the approximate p-value and decide.

$$ p\approx \exp(-6.0000/2)=\exp(-3)=0.0498. $$

To four decimal places, \(p=0.0498\). Since \(0.0498<0.05\), reject the null at the 5% level using the stated large-sample reference. The result is just below the significance threshold, so the evidence should not be described as overwhelming.

Common mistake avoided. Do not subtract 3 twice: if the supplied value is Pearson kurtosis \(K=4.20\), first convert it to excess kurtosis \(g_2=1.20\), then use \(g_2^2/4\). If the supplied value is already excess kurtosis, do not subtract 3 from it.

Worked Example: Reading a Borderline Omnibus Result

A hypothetical software output reports \(JB=6.0250\) for a residual-normality test. Using the classic chi-square reference with 2 degrees of freedom, determine the approximate p-value and the decisions at 5% and 1%.

Method. For 2 degrees of freedom, calculate the upper-tail probability as \(\exp(-JB/2)\).

Step 1: Substitute the statistic.

$$ p\approx \exp(-6.0250/2)=\exp(-3.0125)=0.0491684\ldots $$

Rounded to four decimal places, \(p=0.0492\). At 5%, \(0.0492<0.05\), so reject the null using this reference distribution. At 1%, \(0.0492>0.01\), so do not reject.

Interpretation and check. This is a borderline result at 5%, not a universal finding that the regression is unusable. The calculation is checked against the chi-square-2 upper-tail formula; rounding the unrounded probability to four decimal places gives 0.0492.

Common mistake avoided. Do not round the statistic or intermediate exponential too early. In particular, \(\exp(-3.0125)\) rounds to 0.0492, not 0.0491.

Why Significance Is Not a Complete Diagnostic Judgment

A p-value answers a limited question: if the null model and the test’s calibration were appropriate, how unusual would a statistic at least this extreme be? It does not measure the size or practical importance of the departure, establish the cause, or tell whether the regression’s predictions are useful. Sample size also matters. With a large sample, a modest departure can be detected; with a small sample, a test may lack power to reveal a meaningful departure.

For example, suppose a hypothetical set of \(n=400\) residuals has \(S=0.25\) and \(K=3.40\). The excess kurtosis is \(g_2=3.40-3=0.40\), so

$$ JB=\frac{400}{6}\left((0.25)^2+\frac{(0.40)^2}{4}\right) =\frac{400}{6}(0.0625+0.0400) =6.8333\ldots $$

The large-sample p-value is \(\exp(-6.8333\ldots/2)=\exp(-3.4167\ldots)\approx0.0328\). This rejects at 5%, but not at 1%. The test indicates evidence of a departure in skewness and/or kurtosis; it does not say that either departure is large enough to make a particular inference materially unreliable.

Conversely, a non-significant result is not proof of normality. In a small sample, the test may have limited ability to detect departures. A test statistic can also be sensitive to a few unusual observations, while a single omnibus p-value does not reveal whether the issue is asymmetry, tail behavior, or something else.

Interpret a test alongside the regression context and graphical checks such as the residual histogram or a normal probability plot. Also consider the other model conditions: a normality test does not test linearity, constant variance, or independence, and rejection does not identify which condition is responsible. The residual plots and the earlier material on regression diagnostics help organize that broader assessment. This careful separation of test evidence from diagnostic conclusions is aligned with the SOA Exam SRM (Statistics for Risk Modeling) syllabus.

Practical Interpretation Checklist

1
Identify what was tested.
Confirm whether the values are raw observations, residuals, or another transformation. For regression residuals, remember that fitting affects their distribution.
2
State the null and reference.
For Shapiro-Wilk, the null is normality. For classic Jarque-Bera, the null is normality with a large-sample chi-square reference having 2 degrees of freedom.
3
Compare the p-value with the stated level.
Reject when \(p\leq\alpha\); otherwise, fail to reject. Do not treat failure to reject as proof of normality.
4
Assess the evidence in context.
Consider sample size, plots, the size and likely source of any departure, and the inferential purpose. A test result alone does not provide a full model assessment.

Check Your Understanding

For each problem, use the stated test and reference distribution; distinguish the formal decision from what can be concluded about the regression.

  1. A hypothetical residual sample has \(n=30\), \(S=0.20\), and Pearson kurtosis \(K=3.40\). Calculate the classic Jarque-Bera statistic and its approximate p-value, then state the 5% decision. (Answer: \(JB=0.4000\), \(p\approx0.8187\); fail to reject at 5%.)
  2. A second hypothetical sample has \(n=80\), \(S=0.50\), and Pearson kurtosis \(K=4.00\). Calculate \(JB\) and the approximate p-value, then state the 5% decision. (Answer: \(JB=6.6667\), \(p\approx0.0357\); reject at 5% using the large-sample reference.)
  3. For 18 fitted residuals, Shapiro-Wilk software reports \(W=0.972\) and \(p=0.180\). At \(\alpha=0.05\), what is the decision, and what should not be claimed? (Answer: Fail to reject normality; do not claim that normality has been proved.)
  4. A classic Jarque-Bera output is \(JB=4.6052\). Use the 2-degree-of-freedom upper-tail formula to find the approximate p-value and make the 5% decision. (Answer: \(p\approx\exp(-2.3026)=0.1000\); fail to reject at 5%.)
  5. A large hypothetical regression sample gives a Jarque-Bera p-value of 0.004. State the formal 5% decision and one reason this result alone does not establish that the model is unusable. (Answer: Reject the normality null at 5%; the p-value does not measure the practical size or cause of the departure, and model usefulness requires broader diagnostic context.)
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