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Testing Reserve Models

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Presentation on theme: "Testing Reserve Models"— Presentation transcript:

1 Testing Reserve Models
Gary G Venter Guy Carpenter Instrat

2 Testing Methodology Tests based on goodness of fit of models
Test by SSE adjusted for number of parameters Can also test significance of parameters and validity of model assumptions

3 Issues to Be Tested Is development multiplicative?
Is development contingent on previous losses? Are there diagonal effects in the triangle?

4 Is Development Multiplicative?
Convenient assumption Allows for differences in levels among accident years Sometimes too restrictive for actual data Additive models particularly useful when triangle is built from on-level loss ratios On leveling puts in known adjustments so parameters of model do not have to fit them Parameters can then fit development patterns only Can test significance of additive and multiplicative parameters, as well as goodness of fit

5 Development Contingent on Previous Losses?
Age-to-age factors typically applied to previous losses But in B-F models they are applied to expected Can test which is better Contingent vs. non-contingent a fundamental categorization of models

6 Diagonal Effects in Triangle?
Calendar year inflation can affect all accident years Changes in claims department practices can affect selected years Can test significance of diagonal parameters

7 Notation c(w,d): cumulative loss from accident year w as of age d
q(w,d): incremental loss for accident year w at age d q(w,d) = c(w,d) – c(w,d-1) f(d): factor applied to c(w,d) to estimate q(w,d+1)

8 Model Formulas Chain ladder Parameterized B-F Cape Cod Additive
q(w,d+1) = f(d)c(w,d) + e [f(d) means different factor for each lag] Parameterized B-F q(w,d+1) = f(d)h(w) + e [h(w) means different ultimate for each year] Cape Cod q(w,d+1) = f(d)h + e [h with no w means all years have the same h] Additive q(w,d+1) = g(d) + e [project constant loss for each year varying by lag] (Additive = Cape Cod) Separation q(w,d+1) = f(d)h(w)g(w+d+1) + e OR: q(w,d+1) = f(d)c(w,d)g(w+d+1) + e OR: q(w,d+1) = f(d)h(w)+g(w+d+1) + e OR: q(w,d+1) = f(d)c(w,d)+g(w+d+1) + e

9 Flavors of the Chain Ladder q(w,d+1) = f(d)c(w,d) + e
E(e) = 0, but what about variance of e? Jth flavor has standard deviation(e) = c(w,d)j Divide both sides by c(w,d)j to get sd = , which is regression assumption Can do similarly for other models Murphy PCAS 1994 shows that this adjusted regression based on various j’s reproduces common estimation methods for development factors But it may be hard to tell what j is

10 Hard to say j  0 for this sample data

11 RAA Fac GL 16

12 ISO Zone Rate Trucks $1M Limit or Less
Incremental Paid Loss and LAE

13 RAA: Statistical Significance of Factors
Regressions of incremental on previous cumulative Constants almost all significant, no factors are Constants even more significant without factors

14 RAA: 0 to 1 Is Constant + Random

15 ISO: Statistical Significance of Factors
No constants are significant Large factors are More factors are significant without constants

16 ISO: Factor Times Lag 0 a Good Predictor of Lag 1

17 Testing Fit of Models Multiply SSE by a factor that increases with more parameters, like: AIC: e2p/n 4% BIC or SIC: sqrt(n)2p/n 8% HQIC: ln(n)2p/n 6% OS: (n-p) %-8% for p=1…24 (percent shown is needed improvement in SSE for an extra parameter for a sample of 50 observations) IC formulas are for nExp(IC/n) with normal residuals (A=Akaike, B=Bayesian, HQ=Hannan-Quinn, OS=Old Standby)

18 Fitting BF Parameters – RAA Case
19

19 Reducing # of Parameters
CC makes all accident year factors the same Useful for stable data or on-level loss ratios Some factors may be the same If difference between factors is not significant, can make them the same Can fit lines or curves to some of the parameters: f(d) = (1+i)-d

20 Goodness of Fit Comparisons - RAA OS Adjusted SSE’s
SSE Model Params Generation Formula 157, CL 9 q(w,d+1) = f(d)c(w,d) + e 81,167 BF 18 q(w,d+1) = f(d)h(w) + e 75, CC 9 q(w,d+1) = f(d)h + e 57, CC q(w,d+1) = .78d e 52,360 BF-CC 9 q(w,d+1) = f(d)h(w) + e h5 - h9 same, h3=(h2+h4)/2, f1=f2, f6=f7, f8=f9 44, Sep q(w,d+1) = f(d)h(w)g(w+d+1)+e 4 unique age factors, 2 diagonal, 1 acc. yr. Diagonal factors fit like B-F, with 3 step iteration

21 Goodness of Fit Comparisons - ISO OS Adjusted SSE’s
SSE Model Params Generation Formula 59, CL q(w,d+1) = f(d)c(w,d) + e 69,302 BF q(w,d+1) = f(d)h(w) + e 302, CC q(w,d+1) = g(d) + e 54, Sep q(w,d+1) = f(d)c(w,d) diagonal additive dummies g(w+d+1)+e 42,558 Reduced 13 same as separation 7-8, 8-9, 9-10, factors all equal, plus 4 diagonal dummies

22 Regression Array with Diagonal Dummies
3 8 9 7 10 7

23 Which Diagonals? Too many would over determine regression
4,5,10,11 most likely suspects Reversals in alternative years No apparent trend

24 Regression Coefficients for Dummies Full Factor Model
Similar but slightly more significant in combined factor model

25 Issues to Be Tested Is development multiplicative?
RAA: no, ISO: yes Is development contingent on previous losses? Are there diagonal effects in the triangle? Some high and low diagonals in both, but no trends

26 finis


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