Tutorials › CAS Exam 9 - Risk Management for Actuaries › Backtesting and Outcome Analysis for Capital Models

CAS Exam 9 · Capital Model Design · Tutorial 345 of 995

Backtesting and Outcome Analysis for Capital Models

Use exceedance and calibration diagnostics to assess whether emerging losses are consistent with capital-model forecasts, and interpret limited backtesting evidence cautiously.

Advanced 12 min read

What You'll Learn

  • Define forecast thresholds and exceedance indicators consistently across periods
  • Calculate expected exceedances under explicit calibration and independence assumptions
  • Apply a binomial tail diagnostic and interpret its limitations
  • Use probability integral transforms and interval coverage as complementary checks
  • Handle quantile ties and discrete predictive distributions correctly
  • Explain why a short record cannot establish that a capital model is valid

Why Compare Forecasts With Outcomes?

A capital model describes a distribution of possible losses, not just a single capital number. Backtesting compares forecasts made before outcomes were known with the losses that subsequently emerged. The comparison can reveal patterns worth investigating: too many threshold breaches, losses repeatedly near the forecast tail, or outcomes that do not look consistent with the predicted distribution. It cannot, by itself, prove that a model is valid or invalid. In particular, a short run of annual outcomes gives limited evidence about rare losses.

A useful comparison first makes the forecast and the outcome commensurable. The forecast and observed loss should cover the same time horizon, business scope, valuation basis, and treatment of expenses, recoveries, and risk mitigation. If a forecast is for annual net losses, comparing it with quarterly gross losses does not test the forecast that was made. Backtesting is most informative when the predictions are recorded before the outcomes occur and the comparison rules are set in advance.

Definition: For period \(t\), let \(X_t\) be the realized loss on the same basis and over the same horizon as the model forecast. Let \(F_t(x)\) be the forecast cumulative distribution function (CDF) for that loss, formed using information available when the forecast was made. For a probability level \(p\), define the forecast quantile \(q_t=F_t^{-1}(p)\), using the convention \(F_t^{-1}(p)=\inf\{x:F_t(x)\ge p\}\). A strict exceedance indicator is \(I_t=\mathbf{1}\{X_t>q_t\}\), which equals 1 when the realized loss is strictly greater than the forecast threshold and 0 otherwise.

Exceedance Frequency and Its Assumptions

For a continuous forecast distribution that is correctly calibrated, \(P(X_t>q_t)=1-p\). Thus a 99th-percentile threshold has a 1% strict-exceedance probability, and a 95th-percentile threshold has a 5% probability. This probability statement depends on both calibration and the distributional assumptions. It is not a consequence of the quantile label alone.

If the forecast has point masses, the strict-exceedance probability need not equal \(1-p\). With the quantile convention above, \(F_t(q_t)\ge p\), so \(P(X_t>q_t)=1-F_t(q_t)\le 1-p\). A tie at the threshold is not a strict exceedance. Consequently, before comparing an observed count with a nominal exceedance rate, specify how ties and discrete forecasts are handled.

Theorem: Suppose each forecast has strict-exceedance probability \(r=1-p\), and the exceedance indicators across \(n\) periods are independent. Then the total number of exceedances \(N=\sum_{t=1}^{n} I_t\) has a binomial distribution with parameters \(n\) and \(r\). Its expected value is \(nr\), and its variance is \(nr(1-r)\). The result applies, for example, to independent outcomes under correctly calibrated continuous predictive distributions. It does not apply just because each threshold is labelled a \(p\)-quantile: calibration, the strict-exceedance probability, and independence must also be justified.

The expectation follows by adding indicator expectations: \(E[N]=\sum_t E[I_t]=nr\). Independence gives \(\operatorname{Var}(N)=\sum_t\operatorname{Var}(I_t)=nr(1-r)\), because each indicator is Bernoulli with probability \(r\) and all covariance terms are zero. This explains why a count near its expected value is not a requirement in a short sample. The count is random and may differ materially from its mean even when the assumptions hold.

If the periods are independent but have different strict-exceedance probabilities \(r_t\), the count is generally Poisson-binomial rather than binomial. Its expectation is \(\sum_t r_t\), and its variance is \(\sum_t r_t(1-r_t)\). If outcomes are dependent over time, even these variance calculations need covariance terms: \(\operatorname{Var}(N)=\sum_t\operatorname{Var}(I_t)+2\sum_{s<t}\operatorname{Cov}(I_s,I_t)\). Clustering of exceedances can therefore make a simple binomial comparison misleading.

Worked Example: A Small-Sample Exceedance Count

A hypothetical insurer records ten independent annual net losses. Before each year begins, its model forecasts a continuous, correctly calibrated 90th-percentile annual-loss threshold. Two of the ten realized losses exceed their respective thresholds. Calculate the expected count and the probability, under the model assumptions, of observing at least two exceedances. State what this diagnostic does and does not show.

Method. Because each forecast is continuous and calibrated, the strict-exceedance probability is \(r=1-0.90=0.10\). Independence and the common probability give \(N\sim\operatorname{Binomial}(10,0.10)\).

The expected count is \(E[N]=nr=10(0.10)=1.00\) exceedance. For the observed count of two, calculate the upper-tail probability:

$$ P(N\ge 2)=1-P(N=0)-P(N=1) $$
$$ P(N\ge 2)=1-(0.90)^{10}-\binom{10}{1}(0.10)(0.90)^9 =1-0.3486784401-0.387420489 =0.2639010709. $$

To four decimal places, the probability is 0.2639. The count of two is not unusual under this model and these assumptions. This is not proof of calibration: the test has only ten observations and low ability to detect many departures. Common mistake avoided: treating two breaches as automatically unacceptable because they exceed the expected count of one. The expected count is an average over repeated samples, not a maximum permitted count.

Complementary Calibration Diagnostics

A threshold count compresses each period's outcome to a zero or one. It discards how far an exceedance lies beyond the threshold and what happened in the rest of the distribution. Other diagnostics can therefore add useful information, provided their assumptions and limitations are also made clear.

For a continuous predictive CDF, the probability integral transform is \(U_t=F_t(X_t)\). If the full forecast distribution is correctly calibrated for the period's outcome, \(U_t\) is uniformly distributed on \([0,1]\). This follows because, for \(0\le u\le1\), \(P(F_t(X_t)\le u)=u\) when the predictive CDF is continuous and correctly describes the outcome. A collection of PIT values can be examined for concentration near 0 or 1, which may indicate tail errors, or for other departures from uniformity. If the forecasts are conditional on past information, the interpretation also depends on using the appropriate conditional CDFs; serial dependence can undermine tests that assume independent PIT values.

In discrete distributions, the ordinary PIT is not generally uniform because outcomes can produce repeated CDF values. A randomized PIT can address this: for observed \(X_t\), set \(U_t=F_t(X_t^-)+V_t\{F_t(X_t)-F_t(X_t^-)\}\), where \(F_t(X_t^-)\) is the left limit of the CDF and \(V_t\) is an independent uniform random variable on \([0,1]\). This distributes the probability mass at each outcome across its CDF jump. The diagnostic must be interpreted with the randomization and the discrete model in mind.

Coverage of forecast intervals is another straightforward check. For example, if a model produces a central 90% predictive interval each year, the observed losses should fall inside such intervals about 90% of the time over repeated comparable periods, under correct calibration. Record both overall coverage and which tail contains any misses: equal overall coverage can hide too many low-tail misses and too many high-tail misses. Interval coverage is still a sample statistic, not a verdict from a single observation.

Worked Example: Checking Thresholds With Changing Forecasts

A hypothetical portfolio has five independent annual loss forecasts, each with a continuous and correctly calibrated 95th-percentile threshold. The forecast thresholds, in millions of dollars, are 10, 12, 12, 10, and 14. The realized losses are 8, 11, 13, 9, and 15 million dollars. Count the exceedances, compare the count with its expectation, and calculate the probability of at least two exceedances under the stated assumptions.

Method. Compare each outcome with its own period's threshold, using strict greater-than. The first two comparisons are \(8<10\) and \(11<12\), so neither is an exceedance. The third is \(13>12\), the fourth is \(9<10\), and the fifth is \(15>14\). Therefore, \(N=2\).

Each threshold has strict-exceedance probability \(r=1-0.95=0.05\), so \(E[N]=5(0.05)=0.25\). Under independence, \(N\sim\operatorname{Binomial}(5,0.05)\), and:

$$ P(N\ge2)=1-(0.95)^5-\binom{5}{1}(0.05)(0.95)^4 =1-0.7737809375-0.2036265625 =0.0225925. $$

To four decimal places, this upper-tail probability is 0.0226. Under the specified assumptions, two or more exceedances are relatively unusual, so the pattern merits investigation; it does not establish that the model is invalid. Common mistake avoided: comparing every loss with one fixed threshold. Each forecast was different, so the outcome must be compared with the threshold issued for that year.

Worked Example: A Quantile Tie in a Discrete Forecast

A hypothetical annual loss forecast assigns probability 0.10 to a loss of $0 and probability 0.90 to a loss of $1 million. Using \(F^{-1}(p)=\inf\{x:F(x)\ge p\}\), find the 95th-percentile threshold and its strict-exceedance probability. Explain why a 5% binomial backtest is not justified using that strict threshold.

Method. The CDF is \(F(0)=0.10\) and \(F(1\text{ million})=1.00\). Since \(F(0)<0.95\) but \(F(1\text{ million})\ge0.95\), the 95th-percentile threshold is \(q=1\) million dollars.

A strict exceedance requires \(X>1\) million dollars. Neither possible outcome is greater than that amount, so \(P(X>q)=0\). This differs from \(1-p=0.05\). A binomial calculation with a 5% strict-exceedance probability would therefore use the wrong probability. A test could instead be designed around the model's actual probability of the specified event, or use an explicitly defined tie-handling rule if a nominal 5% rate is needed. Common mistake avoided: assuming that every \(p\)-quantile automatically has strict-exceedance probability \(1-p\). That equality requires, for example, a continuous calibrated distribution.

Interpreting Evidence From Limited Observations

A backtest asks whether an observed pattern is consistent with a forecast under stated assumptions. It does not show that the assumptions are true, and a non-unusual result does not confirm the model. For rare thresholds, the expected number of breaches may be far below one over the available record. For example, even a correctly calibrated 99th-percentile forecast produces only one expected strict exceedance in 100 independent years. The absence of a breach in a much shorter record is weak evidence about the far tail.

A surprising count is a signal to examine, not a standalone conclusion. Consider whether periods are independent, forecasts were formed using only information available at the time, exposure and loss definitions remained consistent, and any model or portfolio changes affected comparability. Also inspect the sizes and direction of misses and diagnostics beyond counts. The earlier material on sparse data, parameter uncertainty, scenarios, stress tests, and sensitivity analysis provides relevant context: a model's apparent fit to a short record may reflect chance, while uncertainty in assumptions can remain substantial.

1
Fix the comparison basis.
Match each realized loss to its forecast horizon, portfolio scope, and gross or net definition.
2
State the event and assumptions.
Specify the quantile convention, whether exceedance is strict, the relevant event probability, and whether independence is assumed.
3
Calculate the diagnostic.
Compare the count or other statistic with its distribution under the stated forecast assumptions, not with an informal expectation of what “should” have happened.
4
Interpret proportionately.
Describe whether the evidence is unusual under the assumptions, then consider sample size, dependence, ties, and complementary diagnostics before drawing conclusions.

Exam trap: writing \(E[N]=n(1-p)\) without specifying what the indicator counts and why its probability is \(1-p\). Under a strict rule \(I_t=\mathbf{1}\{X_t>q_t\}\), this formula is justified when \(P(X_t>q_t)=1-p\) for every period, as with continuous calibrated forecasts at that quantile. It is not generally justified for discrete forecasts with mass at the threshold. Backtesting is strongest when the event definition, assumptions, calculation, and limits of inference are all stated explicitly.

Key takeaway: Backtesting compares forecasts with outcomes through defined diagnostics. An exceedance count is interpretable only after specifying its probability and dependence assumptions; complementary checks add context, while a limited record cannot establish model validity.

Check Your Understanding

For each problem, state any assumption needed to interpret the result and distinguish a diagnostic signal from proof of model validity.

  1. Twenty independent forecasts are continuous and correctly calibrated at the 99th percentile. What is the expected number of strict exceedances? (Answer: 0.20 exceedances.)
  2. Ten independent, continuous, calibrated 90th-percentile forecasts produce three strict exceedances. Calculate the probability of at least three under the binomial model. (Answer: \(1-(0.9)^{10}-10(0.1)(0.9)^9-\binom{10}{2}(0.1)^2(0.9)^8=0.0702\), to four decimals.)
  3. A discrete forecast assigns probabilities 0.20, 0.50, and 0.30 to losses of $0, $4 million, and $10 million, respectively. Using the stated inverse-CDF convention, find its 70th-percentile threshold and strict-exceedance probability. (Answer: \(q=4\) million dollars and \(P(X>q)=0.30\), equal to \(1-0.70\) in this case.)
  4. Five independent forecasts each have a continuous, calibrated 95th-percentile threshold. The observed count is two. What is \(P(N\ge2)\), to four decimals, under the binomial model? (Answer: \(1-(0.95)^5-5(0.05)(0.95)^4=0.0226\).)
  5. A model issues a 90% predictive interval of $5 million to $95 million in each of three years. The outcomes are $4 million, $48 million, and $99 million. How many fall inside, and what proportion is that? (Answer: 1 of 3, or 0.3333 to four decimals; this small sample alone does not establish miscalibration.)
Clinical Biostats