A Worked Anderson-Darling Test

Why Work Through the Whole Test?

The Anderson-Darling test measures how far a sample departs from a proposed continuous severity model, with particular sensitivity to departures in the tails. A complete test requires more than substituting observations into a formula: the observations must be ordered, the null model must be used to transform them, and the reference values must match the way the model was specified.

The examples below use hypothetical, independent, fully observed severity samples. In each test, the null distribution is continuous and its parameters are specified in advance. This matters: reference values for a fully specified model are not automatically appropriate when parameters have been estimated from the same sample.

Definition: Let $X_1,\ldots,X_n$ be an independent sample from a continuous null cumulative distribution function $F_0$, and write the ordered observations as $X_{(1)}\le\cdots\le X_{(n)}$. The Anderson-Darling statistic is $$ A^2=-n-\frac{1}{n}\sum_{i=1}^{n}(2i-1)\left[\ln F_0(X_{(i)})+\ln\left(1-F_0(X_{(n+1-i)})\right)\right]. $$ Large values of $A^2$ indicate greater disagreement with the null model. The logarithms require probabilities strictly between zero and one.

Transform, Order, and Pair Both Tails

For each ordered loss, calculate $U_{(i)}=F_0(X_{(i)})$. If the null model is correct and continuous, the probability integral transform makes each un-ordered $U$ uniform on $(0,1)$. The transformed values remain ordered because a cumulative distribution function is nondecreasing. The test therefore measures whether the transformed sample looks consistent with a uniform sample.

Each summand uses two observations: the lower-tail probability $F_0(X_{(i)})$ and the upper-tail probability $1-F_0(X_{(n+1-i)})$. The weights $2i-1$ increase from 1 to $2n-1$; this structure is why the statistic is sensitive to tail departures. Pairing the observations in reverse order is essential.

Reference values for a fully specified continuous model: Standard asymptotic critical values commonly used for $A^2$ are approximately 1.933 at the 10% level, 2.492 at the 5% level, 3.070 at the 2.5% level, and 3.857 at the 1% level. Reject the null hypothesis at a chosen significance level when the statistic exceeds its critical value. These are not the appropriate unadjusted values when parameters are estimated from the tested sample; that setting requires suitable adjusted or simulated calibration.

The standard decision is to reject or not reject the null hypothesis. A result below the critical value means there is insufficient evidence to reject the specified model at that significance level; it does not prove that the model is true. Use the calibration stated in a problem, since finite-sample or parameter-estimation adjustments can change the reference values.

Why the Statistic Uses These Terms

The probability integral transform explains why the test can be calculated from $F_0(X_{(i)})$ rather than from loss amounts in their original units. Under the continuous null, the transformed data have the uniform distribution, regardless of the particular severity scale.

A useful way to understand the tail weighting is the weighted squared discrepancy between the empirical distribution and the null distribution. For transformed observations $U_{(i)}$, the Anderson-Darling statistic can be written as

A^2=n\int_0^1 \frac{\left(F_n(u)-u\right)^2}{u(1-u)}\,du,

where $F_n(u)$ is the empirical cumulative distribution function of the transformed sample. The factor $1/[u(1-u)]$ is larger near 0 and 1 than in the middle, so discrepancies in either tail receive more weight. Evaluating this integral over the intervals separated by the ordered transformed observations gives the logarithmic order-statistic formula above. This also explains why the formula contains both $\ln U_{(i)}$ and $\ln(1-U_{(n+1-i)})$: the two terms account for the lower and upper tails.

Specify the null model.
Identify $F_0$, its parameters, and whether those parameters were fixed independently of the tested sample.

Order and transform the data.
Sort the losses, then calculate $F_0(X_{(i)})$ for every ordered observation.

Evaluate the paired logarithms.
For each index $i$, use the lower-tail value at $i$ and the upper-tail value at $n+1-i$, with weight $2i-1$.

Compare with a matching reference value.
Use the significance level and calibration appropriate to the null model, then state the decision in context.

Worked Tests

Worked Example: Testing a Specified Exponential Severity Model

A hypothetical portfolio has four ordered losses of $100\ln(10/9)$, $100\ln(5/4)$, $100\ln 2$, and $100\ln 5$ dollars. Test the fully specified exponential severity model with mean $\theta=100$ dollars at the 5% level. Use the standard asymptotic critical value 2.492.

Method and assumptions. The losses are independent, continuous, and fully observed. Under the null, $F_0(x)=1-e^{-x/100}$ for $x\ge0$; the mean is known, not estimated from these four observations.

Transform the ordered losses. Applying $F_0$ gives $1/10,1/5,1/2,4/5$, or $0.10,0.20,0.50,0.80$. For example, for the first loss, $e^{-x/100}=9/10$, so $F_0(x)=1-9/10=0.10$.

Calculate the weighted logarithm sum. With $n=4$, the weights are 1, 3, 5, and 7. Pair each lower-tail transformed value with the upper tail of the observation in reverse order:

\begin{aligned} S &=\ln(0.10)+\ln(1-0.80)\\ &\quad+3\left[\ln(0.20)+\ln(1-0.50)\right]\\ &\quad+5\left[\ln(0.50)+\ln(1-0.20)\right]\\ &\quad+7\left[\ln(0.80)+\ln(1-0.10)\right]\\ &=\ln(0.02)+3\ln(0.10)+5\ln(0.40)+7\ln(0.72)\\ &=-3.912023-6.907755-4.581454-2.299528\\ &=-17.700760. \end{aligned}

Therefore,

A^2=-4-\frac{-17.700760}{4}=0.425190.

Decision. Rounded to four decimal places, $A^2=0.4252$. Since $0.4252<2.492$, do not reject the exponential model at the 5% level. The sample does not provide sufficient evidence against this specified model at that level.

Common mistake avoided. The calculation includes the upper-tail term for every order statistic. Using only $\ln F_0(X_{(i)})$ omits half of the statistic and can even produce an impossible negative value.

Worked Example: A Uniform Severity Model

A hypothetical severity model is uniform on $(0,200)$ dollars. Four ordered losses are $20,70,130,180$ dollars. Calculate $A^2$ and decide at the 5% level using 2.492.

Method and assumptions. The losses are independent and fully observed, and the uniform distribution and both endpoints are specified in advance. For this model, $F_0(x)=x/200$ when $0<x<200$.

The ordered transformed observations are $0.10,0.35,0.65,0.90$. For instance, $F_0(70)=70/200=0.35$. Using the weights 1, 3, 5, and 7 and pairing the tails in reverse order gives

\begin{aligned} S &=\ln(0.10)+\ln(1-0.90) +3\left[\ln(0.35)+\ln(1-0.65)\right]\\ &\quad+5\left[\ln(0.65)+\ln(1-0.35)\right] +7\left[\ln(0.90)+\ln(1-0.10)\right]\\ &=\ln(0.01)+3\ln(0.1225)+5\ln(0.4225)+7\ln(0.81)\\ &=-4.605170-6.298933-4.307829-1.475047\\ &=-16.686979. \end{aligned}

Thus $A^2=-4-(-16.686979)/4=0.171745$, which rounds to $0.1717$ to four decimal places. Since $0.1717<2.492$, do not reject the specified uniform model at the 5% level.

Common mistake avoided. The transformed observations are $0.10,0.35,0.65,0.90$; they must be calculated from the stated losses and endpoints, not replaced by a different, more evenly spaced uniform sample.

Worked Example: A Near-Boundary Tail Departure

For a hypothetical severity sample of size four, the ordered observations transform under a fully specified continuous null model to $0.000005,0.40,0.60,0.80$. Calculate the statistic and decide at 5% using 2.492.

The transforms are already ordered. Pairing each value with the complementary tail of its reverse-order partner gives the products $0.000005(1-0.80)=0.000001$, $0.40(1-0.60)=0.16$, $0.60(1-0.40)=0.36$, and $0.80(1-0.000005)=0.799996$. Hence

\begin{aligned} S&=\ln(0.000001)+3\ln(0.16)+5\ln(0.36)+7\ln(0.799996)\\ &=-13.815511-5.497744-5.108256-1.562040\\ &=-25.983551, \end{aligned}

and

A^2=-4-\frac{-25.983551}{4}=2.495888.

Rounded to four decimal places, $A^2=2.4959$. The unrounded statistic exceeds 2.492, so reject the null model at the 5% level. This is a close decision: keeping adequate precision matters, but the comparison here is still above the stated critical value.

Common mistake avoided. The first transformed value is extremely small, and its logarithm has a large negative contribution. Replacing it with zero is not valid because $\ln(0)$ is undefined; use the actual probability supplied.

Common Exam Traps

The arithmetic is short, but several interpretation and setup errors can invalidate an otherwise careful calculation.

Omitting a tail. The formula includes both $\ln F_0(X_{(i)})$ and $\ln[1-F_0(X_{(n+1-i)})]$. Using only one set of logarithms is not the Anderson-Darling statistic.
Pairing in the wrong direction. The upper-tail term uses the reverse-order observation $X_{(n+1-i)}$, not $X_{(i)}$ again.
Using the wrong reference values. Critical values depend on whether the distribution is fully specified or its parameters were estimated. Do not use fully specified-model values without checking this condition.
Reversing the decision rule. Large statistics indicate greater disagreement, so reject when $A^2$ exceeds the critical value.
Calling non-rejection proof of fit. A test may fail to detect a departure, especially with a small sample. State only that the evidence is insufficient to reject at the chosen level.
Rounding too early. Retain precision through the logarithm sum and statistic, then round the reported statistic. For a close comparison, use the unrounded value to decide.

Key takeaway: A complete Anderson-Darling test for continuous severity data requires ordered observations, null-model probability transforms, both weighted tails, and a reference value appropriate to the model specification. The statistic gives a test decision, not proof that a severity model is correct.

Check Your Understanding

For the first four questions, assume independent, fully observed samples and fully specified continuous null models. Use 2.492 as the 5% critical value where a decision is requested. Round statistics to four decimal places.

A sample of size two has ordered transformed observations $0.25,0.75$. Calculate $A^2$ and state the 5% decision. (Answer: $A^2=0.2493$; do not reject.)
A sample of size three has ordered transformed observations $0.20,0.50,0.80$. Calculate $A^2$ and state the 5% decision. (Answer: $A^2=0.2031$; do not reject.)
A hypothetical exponential severity model has known mean $\theta=250$ dollars. The ordered losses are $250\ln(4/3)$, $250\ln 2$, and $250\ln 4$ dollars. Find their transformed values and calculate $A^2$. (Answer: $0.25,0.50,0.75$; $A^2=0.2694$.)
A fully specified continuous null model gives ordered transforms $0.000005,0.40,0.60,0.80$ for a sample of size four. Calculate the statistic and make the 5% decision. (Answer: $A^2=2.4959$; reject.)
A severity distribution has parameters estimated using the same observations that are tested. Is the standard fully specified-model critical value automatically appropriate? (Answer: No; use an appropriate adjusted or simulated calibration for the fitted-parameter setting.)

What You'll Learn