Testing Frequency-Severity Independence

Why Test Frequency-Severity Independence?

Operational loss models often describe annual loss using a count of events and a severity for each event. A common simplifying assumption is that the count is independent of the severity distribution: in a year with many events, each individual loss is assumed to have the same distribution as in a year with few events. This assumption can make a model easier to fit, but it is not guaranteed by the data or by the definitions of frequency and severity.

Dependence can arise for plausible reasons. A control failure may cause several events and make each event more costly; alternatively, an increase in small, routine incidents may raise the count while leaving severe losses unchanged. A common driver, such as transaction volume or a system outage, can affect both. Testing for dependence therefore means looking for changes in the severity distribution across frequency conditions, not merely calculating one overall correlation.

Definition: Let $N$ be the number of operational losses in a fixed exposure period, and let $X_1,\ldots,X_N$ be their severities. Frequency-severity independence means that, conditional on $N=n$, the severity marks have the same distribution as they do for other values of $n$, and that the event severities are independent of $N$. A zero correlation between $N$ and a severity summary does not, by itself, establish this independence.

The observational unit must match the question. A useful unit may be a fixed year, business unit-year, or another exposure block observed consistently. A severity summary for a block with no events is undefined, so an analysis of mean severity by count generally concerns positive-count blocks; the probability of zero events remains part of the frequency model. When the data are grouped into categories, the test can detect differences between those categories but may miss patterns within them.

Testing Association in Grouped Data

A contingency-table test is useful when each observation can be classified both by frequency and by severity. Under the null hypothesis of independence, the expected number in a cell is the product of its row and column totals divided by the overall total. The Pearson chi-square statistic compares observed and expected counts. Its usual large-sample reference distribution requires independent observational units and sufficiently large expected cell counts; for a two-way table with $r$ rows and $c$ columns, the degrees of freedom are $(r-1)(c-1)$.

E_{ij}=\frac{(\text{row total}_i)(\text{column total}_j)}{n}, \qquad \chi^2=\sum_{i=1}^{r}\sum_{j=1}^{c}\frac{(O_{ij}-E_{ij})^2}{E_{ij}}.

Worked Example: Testing Frequency and Severity Categories

A hypothetical operational-risk team has 200 business-unit-years with at least one event. For each unit-year, it records whether the count is low or high and whether the mean severity is below or above a pre-set threshold. The observed table is:

	Below-threshold mean severity	Above-threshold mean severity	Total
Low count	60	40	100
High count	30	70	100
Total	90	110	200

Method and null hypothesis. Use the Pearson chi-square test. The null hypothesis is that count category and mean-severity category are independent across these unit-years. The alternative is that they are associated.

Expected counts. For the low-count, below-threshold cell, the expected count is $100(90)/200=45$. The expected count in the low-count, above-threshold cell is $100(110)/200=55$. Since both row totals are 100, the expected counts for the high-count row are also 45 and 55. All expected counts are at least 45, so the usual large-sample approximation is reasonable, assuming the unit-years are independent.

Test statistic. The four contributions are $(60-45)^2/45=5$, $(40-55)^2/55=4.0909$, $(30-45)^2/45=5$, and $(70-55)^2/55=4.0909$. Thus $\chi^2=18.1818$. The degrees of freedom are $(2-1)(2-1)=1$. At the 5% level, the critical value is approximately 3.841, so the statistic is in the rejection region; the p-value is approximately 0.00002.

Conclusion and limitation. These hypothetical data provide strong evidence of an association between the two categories. High-count unit-years are more often in the above-threshold severity category than independence would predict. This result does not show that high frequency causes high severity. It also applies to the chosen categories and to positive-count unit-years, not automatically to zero-event periods or to every severity feature.

Rank Association and Small Samples

When frequency and a continuous severity summary are recorded for each exposure block, rank methods can identify a monotone association without requiring normally distributed data. Spearman’s rank correlation is the ordinary correlation of the two sets of ranks. With no ties and $n$ paired observations, it can also be calculated from rank differences $d_i$ using $1-6\sum d_i^2/[n(n^2-1)]$. This statistic measures monotone association; it is not a general test of independence.

For a small sample, an exact permutation test can compare the observed statistic with the values obtained by permuting one set of ranks while holding the other fixed. Under the null, this requires exchangeability of the paired observations: the pairing must carry no information, and the observations must not have an unaccounted-for time or exposure structure. A small sample may have little power even when the observed association looks substantial.

Worked Example: Exact Rank Test for Four Unit-Years

In a hypothetical pilot study, four comparable unit-years have event counts 1, 2, 3, and 4. Their respective mean severities are $70, $130, $110, and $220. Test for a monotone association using Spearman’s rank correlation and a two-sided exact permutation test at the 5% level.

Rank the observations. The count ranks are 1, 2, 3, 4. The severity ranks in the same order are 1, 3, 2, 4. The rank differences are 0, -1, 1, 0, so $\sum d_i^2=2$.

Calculate the statistic. With $n=4$, Spearman’s statistic is $1-6(2)/[4(4^2-1)]=1-12/60=0.8$. The observed rank association is positive.

Calculate the exact p-value. There are $4!=24$ equally likely permutations of the severity ranks under the null. The observed statistic is at least as extreme in absolute value for 4 permutations with correlation at least 0.8 and 4 with correlation at most -0.8. Therefore the two-sided exact p-value is $8/24=0.3333$.

Conclusion and common mistake avoided. At the 5% level, do not reject independence based on this test: the observed statistic is large, but the exact p-value is not small. Treating a high sample correlation as proof of dependence would overstate what four observations establish. Conversely, failure to reject is not proof of independence; the sample has limited power.

How Dependence Changes Aggregate Loss Moments

Let annual aggregate loss be $L=\sum_{i=1}^{N}X_i$, with $L=0$ when $N=0$. Suppose that, conditional on $N=n$, the severities are independent and identically distributed with conditional mean $m_n$ and variance $v_n$. These assumptions allow the severity distribution to change with the count, while retaining conditional independence among event severities.

Result: Under the conditional model just stated, $$ E[L]=E[Nm_N], \qquad \operatorname{Var}(L)=E[Nv_N]+\operatorname{Var}(Nm_N). $$ If frequency and severity are independent, so $m_n=\mu$ and $v_n=\sigma^2$ for every positive $n$, these reduce to $E[L]=E[N]\mu$ and $\operatorname{Var}(L)=E[N]\sigma^2+\operatorname{Var}(N)\mu^2$.

Why the result holds. Conditional on $N=n$, the sum has mean $nm_n$. Conditional independence of the $n$ severities makes its variance the sum of their variances, $nv_n$. The law of total expectation gives $E[L]=E[E(L\mid N)]=E[Nm_N]$. The law of total variance then gives $\operatorname{Var}(L)=E[\operatorname{Var}(L\mid N)]+\operatorname{Var}(E[L\mid N])=E[Nv_N]+\operatorname{Var}(Nm_N)$. Under independence, $Nm_N=N\mu$, so $\operatorname{Var}(N\mu)=\mu^2\operatorname{Var}(N)$, yielding the familiar formula.

Worked Example: Comparing Dependent and Independent Models

Consider a hypothetical annual model. The count $N$ is 1 or 2, each with probability 0.5. Given $N=1$, the single severity is $90 or $110 with equal probability. Given $N=2$, each of the two conditionally independent severities is $180 or $220 with equal probability. Calculate the aggregate mean and variance under this dependence, then compare the variance with an independent model that uses the event-weighted pooled severity distribution.

Conditional severity moments. For $N=1$, the severity mean is $(90+110)/2=100$, and its variance is $[(90-100)^2+(110-100)^2]/2=100$. For $N=2$, the severity mean per event is $(180+220)/2=200$, and the variance per event is $[(180-200)^2+(220-200)^2]/2=400$.

Dependent aggregate mean. Given $N=1$, $E[L\mid N=1]=100$; given $N=2$, $E[L\mid N=2]=2(200)=400$. Thus $E[L]=0.5(100)+0.5(400)=250$.

Dependent aggregate variance. The conditional second moment is $E[L^2\mid N=n]=\operatorname{Var}(L\mid N=n)+E[L\mid N=n]^2$. For $N=1$, it is $100+100^2=10{,}100$. For $N=2$, conditional independence gives variance $2(400)=800$, so the second moment is $800+400^2=160{,}800$. Therefore $E[L^2]=0.5(10{,}100)+0.5(160{,}800)=85{,}450$, and $\operatorname{Var}(L)=85{,}450-250^2=22{,}950$.

Independent-model comparison. To pool severities by event, not by year, the $N=1$ severity distribution receives weight $0.5(1)/E[N]=1/3$, while the $N=2$ severity distribution receives weight $0.5(2)/E[N]=2/3$. The pooled severity mean is $(1/3)(100)+(2/3)(200)=500/3$. Its second moment is $(1/3)(10{,}100)+(2/3)(40{,}400)=30{,}300$, so its variance is $30{,}300-(500/3)^2=22{,}700/9$. Also $E[N]=1.5$ and $\operatorname{Var}(N)=0.25$. The independent-model aggregate variance is therefore $1.5(22{,}700/9)+0.25(500/3)^2=10{,}727.7778$. Its mean is $1.5(500/3)=250$.

Check and interpretation. Both models have the same expected aggregate loss, $250, because the pooled severity distribution was weighted per event. But the dependent model’s variance is $22,950, compared with $10,727.7778 under independence. The difference is $22{,}950-10{,}727.7778=12{,}222.2222$. The independent assumption understates variability in this constructed example because the higher-count years also have higher-severity event distributions. The common mistake is to pool severity distributions equally by year: that would give the wrong event-level distribution when years have different counts.

Interpreting a Test in a Risk Model

A test result is evidence about a specified model and data construction, not a final verdict on all frequency-severity dependence. Before drawing a conclusion, check the exposure basis, event definition, and treatment of thresholds. The earlier material on cleaning, truncation, and censoring matters here: inconsistent reporting thresholds can create apparent changes in severity across count groups. Changes in business volume or control environment can also create association without a direct causal relationship between count and severity.

Check the unit of analysis. Repeated observations from one unit may not be independent. Treating them as independent can make a test appear more conclusive than it is.
Check both conditional patterns and the overall test. A significant test says that some aspect of the distributions differs; it does not identify which mechanism or quantify a causal effect.
Do not equate zero correlation with independence. A nonlinear or tail-only relationship can have little linear correlation. Conversely, common trends can produce correlation without direct frequency-to-severity influence.
Carry material dependence into model sensitivity. If conditional severity changes with count, compare aggregate predictions under plausible conditional models rather than relying only on a pooled independent fit.

A non-significant test does not establish independence, especially with sparse data or few high-count years. A defensible modeling decision considers the test’s power, data limitations, operational explanations, and the effect of reasonable dependence assumptions on aggregate loss predictions. This approach is aligned with the CAS Exam 9 (Risk Management for Actuaries) syllabus.

Check Your Understanding

For each problem, identify the relevant assumption or calculation before checking the answer.

A team tests whether annual event count and average severity category are independent using a table of positive-event years only. What does the test say about zero-event years? (Answer: It does not test zero-event years; the severity summary is undefined there, so zero-count probability must be assessed separately.)
A hypothetical 80-unit-year table has observed counts 24 and 16 in the first row, and 10 and 30 in the second row. Calculate the four expected counts and the Pearson statistic. (Answer: Expected counts are 17, 23, 17, and 23; $\chi^2=10.0256$ with 1 degree of freedom, so independence is rejected at the 5% level.)
Annual count is 1 or 3 with equal probability. Conditional on count 1, mean event severity is $50; conditional on count 3, it is $100. What is expected annual aggregate loss? (Answer: $0.5(1)(50)+0.5(3)(100)=\$175$.)
Suppose a frequency-severity test is not significant, but there are only three high-count years. Does this establish independence, and what is a key modeling response? (Answer: No; low power may conceal dependence. Assess plausible conditional relationships and test aggregate predictions under sensitivity assumptions.)
In the worked four-unit-year rank example, what conclusion follows from the exact two-sided permutation p-value at the 5% level? (Answer: The p-value is $8/24=0.3333$, so do not reject independence at 5%; the small sample does not establish that the variables are independent.)

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