Brehm CH 3 Flashcards
What are some key aspects of IRM parameter development?
Modeling Software
* Capabilities
* Scalability
* Learning curve
* Integration with other systems
* Output management
Developing Input Parameters
* Process is heavily data driven
* Requires fair amount of expert opinion
* Involves many functional areas in the process
* Requires clear assignment of final ownership
Correlations
* Line of business representatives cannot set cross-line parameters in isolation
* Corporate-level ownership of these parameters required ·
Validation and Testing
* No existing IRM with which to compare
* Multi dimensional, multi metric testing is required
* Iterative testing with increasing scope and detail
What are some recommendations for IRM parameter dev?
Modeling software - Be sure to assess how much is pre-built from the many capable model offerings versus user built and make sure that aligns with the capabilities of the !RM team.
o Parameter development - Include product expertise from underwriting, claims, planning and actuarial. Develop a systematic way to capture expert risk opinion.
Correlation - Have the IRM team recommend correlation assumptions, which are ultimately owned by the CRO/CEO/CUO.
o Validation - Validate and test over an extended period. Align with educational offerings to bring all interested parties to the same basic understanding of probability and statistics
How should a company integrate and maintain an IRM?
Align input and output stages with key points on the corporate calendar
Cycle
* Integrate into major corporate calendar: planning,
reinsurance purchasing, portfolio reviews or capacity allocation
* Mandatory inclusion of IRM output as part of decision support
Updating
* Frequency and magnitude of updates
* Simple scale adjustments for minor changes
Controls
* Centralized storage and control of input sets and output sets
(date stamped)
* Endorsed set of analytical templates - avoids misuse and
abuse of info.rmation
What are three types of risk when it comes to modeling?
Estimation => due to having finite data
Projection => possible error of projecting past trends into the future
Model => choosing the wrong model(s)
What can further impact projection risk?
Having to develop claims and claim count to ultimate before assessing the trend.
More advanced regression procedures correct the prediction intervals for the effect of uncertainty in the historical points.
Severity trends and inflation
severity trends often higher than general inflation (excess known as social inflation)
for some LOBs (prop, APD) real inflation data is used instead
can correct payment data with real inflation and model excess inflation (social inflation) to add on top
claim sev trend model SHOULD reflect dependency between inflation types
AR(1) models can add in more uncertainty
Estimation risk
MLE most common methon (F/S)
large samples, MLE smallest error amongst unbiased estimators
uncertainty about parameters depends on shape of likelihood (if peak area is flat, many possible other parameters)
thats measured by the second derivative (information matrix)
issues with normal for small data: high sd means plausible negative vals also heavy tail dists can have heavy tailed parameter dists
lognormal often the best approximation for small datasets
Model risk
Hannan-Quinn Information Criterion => adds parameter penalty to negative of loglikelihood set to the log of the log of the number of obs
selected dist could be wrong. Assign probs of being right to the better fitting dists (possibly bayesian posterior dist)
simulation model can select a distribution from the set chosen, then select parameters, then model losses, then repeat to find uncertainty ranges
What is a copula?
C(u,v) is a function that expresses a joint distribution F(x,y) in terms of F(x) and F(y)
C is a distribution on the unit square [0,1]
any multivariate dist is simply a copula applied to the marginal dists
Frank Copula
g_z = exp(-az) - 1
C(u,v) = ln(1 + g_u * g_v / g1] / a
C1(u,v) = [g_u * g_v + g_v] / [g_u*g_v + g_1]
tau(a) = 1 - 4/a + 4/a^2 integral(0 to a) t / (e^t - 1) dt
Invert C1 to simulate correlated pairs (u,v)
simulate u and p as Uniform(0,1) p is conditional draw from V|u
from distribution function C1 v = c1 inverse (p|u)
once simulate u and v, invert the marginals to get X and Y
Gumbel Copula
more probability in tails than frank, asymmetric (right tail)
C(u,v) = exp(-[(-ln(u) ^ a + (-ln(v)^2] ^ (1/a)) a >= 1
tau(a) = 1 - 1/a
C1 not invertible
simulate uniform u,v : solve numerically for 1>s>0 and ln(s)s = a(s-u)
Heavy Right Tail (HRT) and Joint Burr Copula
less correlation in left tail but high correlation in right tail
C(u,v) = u + v - 1 + [(1-u)^(-1/a) + (1-v)^(-1/a) - 1]^-a | a > 0
tau(a) = 1 / (2a+1)
Joint burr is when the a parameter of both burrs is the same as HRT
Normal copula
lighter in right tail than gumbel or HRT but heavier than the frank
left tail is similar to the gumbel
C(u,v) = B(p(u), p(v) ;a)
B = bivariate standard normal p(u) = percentile function a = correlation
tau(a) = 2*arcsin(a) / pi
simulate using conditional dist C1
simulate p(u) from standard normal and then p(v) from C1
then apply standard normal to these percentiles to get u and v
How can we quantify tail strength of a copula?
R(Z) = Pr(U > z | V > z) right tail strength
L(Z) = Pr(U < z | V < z) left tail strength
limits as z goes to 1 and 0 respectively
many copulas these vals are 0, misleading because slopes can be very steep near the limit
combine to make “LR” function and plot
can also compare tail behavior by looking at contour plots
Fitting copulas to data
generally via MLE, or SSE between empirical and fitted copulas
or product of bivariate MLEs
