Brehm Ch 3

Question 1

Describe 4 organizational details to be addressed early in IRM startup.

Answer 1

1. Organization chart: modeling team reporting line, solid line vs dotted line reporting
2. Functions represented: reserving, pricing, finance, planning, UW, risk
3. Resource commitment: mix of skill set (actuarial, UW, communication), full time vs part time
4. Critical Roles and Responsibilities: control of input parameter, control of output data, analyses and uses of output
5. Purpose: goal of the model to quantify variation around plan or provide distribution of results
6. Scope: prospective UW year only or including reserves, assets and operational risks. Low detail (whole company) or high detail (specific segment).

Question 2

Provide a recommendation for reporting relationship of IRM startup.

Answer 2

The reporting line for the IRM team is less important than ensuring they report to a leader who is fair and balanced.

Question 3

Provide a recommendation for resource commitment of IRM startup.

Answer 3

Team should have a full-time commitment to the implementation.

Question 4

Provide a recommendation for inputs and outputs of IRM.

Answer 4

Should be controlled similarly to the general ledger or reserving systems.

Question 5

Provide a recommendation for initial scope of IRM.

Answer 5

Prospective underwriting period, variation around plan.

Question 6

Describe 4 parameter development details to be addressed in IRM development.

Answer 6

1. Modelling software: capabilities, scalability, learning curve, integration with other systems, output management.
2. Developing input parameters: process is heavily data driven, requires expert opinion, many functional areas should be involved
3. Correlations: LOB representatives cannot set cross-line parameters, corporate-level ownership of these parameters
4. Validation and testing: no existing IRM with which to compare, multi-metric testing is required, iterative testing with increasing scope and detail.

Question 7

Provide a recommendation for modelling software when developing IRM.

Answer 7

Capabilities of the modelling team should determine how much is pre-built and how much the team builds.

Question 8

Provide a recommendation for IRM parameter development.

Answer 8

Include expert opinion from underwriting claims, planning and actuarial.

Question 9

Provide a recommendation for correlation in IRM development.

Answer 9

Modelling team recommends assumptions, which are owned at the corporate level (CRO/CEO/CUO)

Question 10

Provide a recommendation for validation of IRM.

Answer 10

Validate and test over an extended period.

Question 11

Describe 4 model implementation details to be addressed.

Answer 11

1. Priority setting: importance of priority, approach and style (ask vs mandate), priority and timeline must be driven from the top.
2. Interest and impact: implement communication and education plans across enterprise
3. Pilot test: assign multidisciplinary team to provide real data and real analysis on company as a whole or on one specific segment
4. Education process: run in parallel with pilot test, bring leadership to same point of understanding regarding probability and statistics.

Question 12

Provide a recommendation for priority setting in implementation of IRM.

Answer 12

Top management should set the priority for implementation.

Question 13

Provide a recommendation for communication during IRM implementation.

Answer 13

Regular communication to broad audiences.

Question 14

Provide a recommendation for pilot testing during IRM implementation.

Answer 14

Do pilot testing to prepare stakeholders for the magnitude of the change.

Question 15

Provide a recommendation for education during IRM implementation.

Answer 15

Bring leadership to a base level of understanding about the model.

Question 16

Describe 3 IRM integration and maintenance details to be addressed

Answer 16

1. Cycle: integrate model runs into major corporate calendar and ensure output support major company decisions.
2. Updating: determine frequency and magnitude of updates.
3. Controls: ensure there is centralized storage and control on inputs/outputs, ensure there is an endorsed set of analytical templates used to manipulate IRM outputs for various purposes.

Question 17

Provide a recommendation for cycle in integration and maintenance of IRM.

Answer 17

Integrate into the corporate calendar (at least for planning)

Question 18

Provide a recommendation for updating IRM during integration and maintenance.

Answer 18

Major updates to inputs no more frequently than semiannual.

Question 19

Provide a recommendation for controls during integration and maintenance of IRM.

Answer 19

Maintain centralized control of inputs, outputs and templates.

Question 20

Contrast the impact of parameter risk on small versus large companies

Answer 20

A small insurer already has significant uncertainty, so the added impact of parameter risk is not too large.

For a large company, without parameter risk, the loss ratio modelled is unrealistically stable. Parameter risk is not diversified away with more insureds, so it significantly increases uncertainty for a large insurer.

Question 21

Calculate the coefficient of variation (CV) of total losses (S)

Answer 21

CV(S) = ((V(N)/E(N) + CV^2(X))/E(N))^0.5

CV(X) = SD(X)/E(X)

Note:
E(S) = E(N)E(X)
V(S) = E(N)V(X) + V(N)*E^2(X)

Question 22

Does the CV of total losses results in more risk for small or large companies?

Answer 22

Smaller companies

Question 23

Describe a simple trend model

Answer 23

To project future levels of loss costs, a trend line is often fit to loss cost history.

Prediction intervals can be placed around this projection to provide a quantification of projection risk.

Question 24

Provide 2 disadvantages of the simple trend model.

Answer 24

1. Loss cost data is based on historical claims that have not settled, which adds uncertainty.
2. Assumes a single constant trend for historical data that will continue into the future.

Question 25

Explain why placing prediction intervals around projected losses may be too narrow.

Answer 25

In the projection period, the projection uncertainty is a combination of uncertainty in each historical point AND uncertainty in fitted trend line.

Thus, the spread in the prediction intervals increases in the projection period.

Actuary’s prediction intervals may be too narrow due to missing uncertainty associated with historical data.

Question 26

Describe 2 approaches to model claim severity trend.

Answer 26

1. Modelled from insurance data with no regard to general inflation (exceptions include prop insurance auto collision where specific inflation indices may be used)
2. Correct payment data using general inflation indices and model the residual superimposed inflation. Any subsequent projection is a projection of superimposed inflation only.
   A separate projection of general inflation is required.

Question 27

Explain why projecting superimposed inflation and general inflation separately is advantageous.

Answer 27

It reflects the dependency between claim severity trend and general inflation.

Most enterprise risk models include a macroeconomic model, which includes future inflation rates.

It is essential that claim severity trend model reflects appropriate dependencies between claims severity trend and inflation.

In doing so, inflation uncertainty is incorporated into projection risk.

Question 28

Describe the primary difference between simple trend model and time series model.

Answer 28

Simple trend model assumes there is a single underlying trend rate that has been constant throughout the historical period and will remain constant in the future.
The only potential error is misestimation of the rate.
This is flawed assumption that usually only holds true over limited time periods.

AR-1 reflects more uncertainty than a simple trend model and models trend as a time series rather than a constant.

Question 29

Describe the first order autocorrelated time series (AR-1)

Answer 29

AR-1 is a mean-reverting time series.

The true mean is unknown and estimated from data (similar to simple trend model).

However, the AR-1 model also includes an autocorrelation coefficient and an annual disturbance distribution.

Question 30

Compare prediction intervals between simple trend model and time series.

Answer 30

In the simple trend model, the prediction intervals widen with time due to uncertainty in estimated trend rate.

In the AR-1 model, the prediction intervals widen with time as well, but the effect is more pronounced and the prediction intervals are wide. This is due to additional uncertainty of AR-1 process.

Question 31

Briefly explain why a simple trend model might not be appropriate for a long-tailed LOB.

Answer 31

Because long projection period is required, thus simple trend model will likely understate the projection risk.

Question 32

Describe a consequence of parametrizing time series with limited data.

Answer 32

If the time period of the data is too limited to exhibit a range of behaviours, the resulting model will be limited as well and will understate the projection risk.

One should ensure that datasets are large and time periods are long.

Question 33

Explain how to include parameter estimation risk in ERM.

Answer 33

1. Parameter estimates are calculated using MLE.
2. Fit a joint lognormal (or normal) distribution to the covariance matrix from MLE.
3. In ERM model, each simulation draws a random sample from joint lognormal distribution to be the parameters for the loss distribution.

This approach adds variability to the parameter used in model.

Question 34

Briefly explain 3 key elements of uncertainty inherent to the loss modelling process. (3 types of parameter risk)

Answer 34

1. Model risk
   Risk that selected distribution is incorrect
2. Estimation risk
   Risk that form and parameters of frequency and severity distributions are incorrect
3. Projection risk
   Risk that projected parameter changes and trends into the future are incorrect

Question 35

Explain how to include model risk in ERM.

Answer 35

1. Assign probabilities of being “correct” to each of the better-fitting distributions.
2. For each simulation:
   a) select a distribution from set of distributions
   b) Select the parameters from the lognormal distribution of parameters (add estimation risk)
   c) Use this fixed distribution for all losses in the simulated scenario.

Question 36

When is using MLE difficult?

Answer 36

The best-fitting parameters can be difficult to determine if the likelihood surface is very flat near the maximum.

When the surface is flat near the max, a wide range of parameter sets have almost the same likelihood. The set that maximizes likelihood might not be any better than one that has slightly smaller likelihood.

When surface is sharply peaked at max, the correct parameter set is easier to determine.

Question 37

Explain why incorporating dependency in an ERM is important.

Answer 37

Each individual model may be realistic, but if dependencies between different lines and risks are unrealistic, then ERM model in aggregate will be unrealistic.

Question 38

Describe 2 problems arising when normality assumption is used for small dataset.

Answer 38

For large datasets, the parameter distributions in MLE procedure are multivariate normal.

For smaller data sets, normality assumption creates problems:
1. The standard deviation of the parameters can be high enough to produce negative parameter values with significant probability
2. The distribution of the parameters may be heavy-tailed (bivariate normal is not heavy-tailed)
3. Simulation tests on small samples have found that log-normal distributions provide a good fit to the parameter distributions.

Question 39

Provide 3 statistics that penalize models that use a large number of parameters.

Answer 39

1. AIC
2. BIC
3. HQIC

A low statistic indicates a better fit and less complexity.

Question 40

Describe 2 best practices when selecting a model.

Answer 40

1. Alternative models should always be tested for validity against available data.
2. Common sense judgment should be used to ensure that chosen model form is consistent with underlying process it represents.
3. Although parsimony is desirable, too much parsimony can produce unrealistically stable results. Some model complexity is required to ensure that model is capturing the true uncertainty of underlying process.

Question 41

Define a copula

Answer 41

A copula is a function that combines individual marginal distributions into a multivariate distribution.

F(x,y) = C(F(x), F(y)) = C(u,v)

Thus, C is a function on the unit square (square defined by (0,0), (0,1), (1,0), (1,1))

Question 42

Contrast correlation and dependency

Answer 42

Correlation uses a single value and does not differentiate between different levels of dependency in different parts of the distribution.

Dependency between different risks and lines may be higher in the tail than the rest of the distribution.

Question 43

Describe Frank Copula

Answer 43

1. Light-tailed
2. Weak correlation, thus graph is sparse.
3. g_z = exp(-az)-1
4. C(u,v) = -ln(1+g_u * g_v / g_1)/a
5. C1 can be inverted
6. R(1) = 0

Question 44

Describe Gumbel Copula

Answer 44

1. Heavier tail than Frank, thus better candidate for insurance losses.
2. Asymmetric with more weight in right tail.
3. C(u,v) = exp(-((-ln(u))^a + (-ln(v))^a)^1/a)
4. tau(a) = 1-1/a
5. C1 is not invertible, thus cannot be easily simulated
6. R(1) positive

Question 45

Describe the heavy right tail (HRT) copula

Answer 45

1. Less correlation in left tail and more correlation in right tail.
2. C1 is invertible
3. C(u,v) = u + v - 1 + ((1-u)^-1/a+(1-v)^-1/a-1)^-a
4. tau(a) = (2a+1)^-1
5. R(1) positive

Question 46

Place Frank, normal, Gumbel and HRT copula in increasing order based on their right tail correlation.

Answer 46

1. Frank (smallest)
2. Normal
3. Gombel
4. HRT (greatest)

Question 47

Describe tau(a)

Answer 47

Tau is a measure of correlation of copulas.

When comparing between copulas, we must ensure they have the same tau.

Question 48

Describe the normal copula

Answer 48

1. Heavier tail than Frank, but less than Gumbel and HRT.
2. C1 is invertible
3. tau(a) = 2arcsin(a)/pi
4. R(1) = 0

C(u,v) formula would be given (too complex)

Question 49

Provide 2 advantages of normal copula

Answer 49

1. Easy simulation method
2. Generalizes to multi-dimensions (can be used for more than 2 dimensions)

Question 50

How do you determine if good fit based on calculated functions.

Answer 50

If calculated function for a dataset closely matches a specific copula, then that copula might be a good candidate for the data.

Question 51

Define the tail concentration functions

Answer 51

Provide a quantification of tail strength.

Left tail concentration:
L(z) = C(z,z)/z
L(z) = P(U<z|V<z)
L(z) = P(U<z,V<z)/z

Right tail concentration:
R(z) = (1-2z+C(z,z))/(1-z)
R(z) = P(U>z|V>z)
R(z) = P(U>z,V>z)/(1-z)

Note that L(z) and R(z) formulas are not given in the text, thus would be given on exam.

Question 52

Define the tail dependence parameters (L, R and LR)

Answer 52

L = L(0) is the lower tail dependence parameter.
If L(0) positive then there is strong evidence for left tail correlation.

R = R(1) is the upper tail dependence parameter.
If R(1) positive, then there is strong evidence for right tail correlation.

Since L and R are only useful below 0.5 and above 0.5 respectively, we combine them into an LR function which is L below 0.5 and R above 0.5.

Question 53

Explain why tail dependence parameters can be misleading. Recommend a solution to this problem

Answer 53

Because the slopes of L and R are indirectly steep close to 0 and 1, respectively.

Thus, these functions may indicate no tail correlation when significant correlation is present just prior to 0 and 1.

A solution is to look at the function at values slightly below 1 and assess the strength of the dependence at those values.

Question 54

Describe t-Copulas.

Answer 54

t-Copulas (just like normal) can be used to combine more than 2 random variables.

They take correlation matrices as inputs which makes them highly flexible.

Question 55

Contrast Normal and t-Copulas.

Answer 55

Normal copula is uncorrelated for very large and very small losses.

Since t-copula has an additional parameter for tail heaviness, it can be strongly correlated in the tails if desired.

For large n dimensions, the t-copula approaches the normal copula.

Question 56

List 3 ways to compare goodness of fit of copulas.

Answer 56

1. Left and Right tails concentration functions
2. J function: compare fitted j to empirical j function to evaluate which copula fits best
3. Chi function: compare fitted chi function to empirical chi function

Question 57

Provide 2 benefits of using copulas to express correlation from joint loss distributions.

Answer 57

1. Copula allows recognition of varying correlations at different levels of a distribution
2. Copula can join any distributions, regardless of what family they are from

Question 58

Explain why we want right tail correlation when modelling insurance loss at portfolio level.

Answer 58

When a large loss occurs, it’s likely that multiple lines will be impacted at the same time.

This correlation needs to be reflected by incorporating more weight into the right tail of multivariate distribution describing the sum of the lines.

Question 59

Describe how parameter risk can be reduced.

Answer 59

Parameter risk can be reduced by improving the quality of the data underlying the parameter estimation.

Question 60

Which of the Normal and HRT copula will produce the highest expected value.

Answer 60

The expected values will be the same because copulas only impact correlation and variance, expectations remain unchanged.

Question 61

Which of the Normal and HRT copula will produce the highest VaR.

Answer 61

The HRT will produce the higher VaR because there is more correlation in the right tail.

Question 62

Discuss an example of macroeconomic tail dependency ERM model should incorporate.

Answer 62

High general inflation would impact both UW losses and loss reserve developments.

Question 63

Discuss an example of insurance tail dependency ERM model should incorporate.

Answer 63

Home and Auto have low correlation normally, but an extreme event (e.g. hurricane) would cause large losses for both.