How Valid Are Your Assumptions?
A Basic Introduction to Testing the Assumptions of
Loss Reserve Variability Models
Casualty Loss Reserve Seminar
Renaissance Waverly Hotel
Atlanta, Georgia
presented by
F. Douglas Ryan, FCAS, MAAA
Philip E. Heckman, ACAS, MAAA, PhD
September 11, 2006
Class for Regional Affiliates
Schedule
8:30 am Optional Class A – The mathematics of regression
Optional Class B - The reserving problem in casualty insurance
10:00
11:45
1:00
1:30
3:00
5:00
Full session 1
Break for lunch and Regional Affiliate business
Full session 2
Break out into work groups of four to six persons
Full session 3
End of class
Participants who do not wish to take an optional early session are expected to
travel to the meeting in the morning and return home the same day.
Class for Regional Affiliates
Who Is It For?
The class is designed to be offered by Regional
Affiliates to all people interested in the technical
side of the estimation of liabilities using data in
“loss development triangles”. Regional Affiliates
are encouraged to invite students and faculty in
actuarial programs, actuaries at all levels of
experience with the setting of loss reserves, and
senior management.
Class for Regional Affiliates
The Scope from 10:00 am to 5:00 pm
•
Introduce triangles of cumulative loss costs and note their
obvious features
•
Algorithms: The “actuarial methods” and Excel
instructions viewed as algorithms
•
BLUE: “Best” Linear Unbiased Estimates
•
Multiple Models and “The Reserving Problem”
•
Testing the BLUE Assumptions
•
Testing a BLUE Model: Validation
•
Model Design
Class for Regional Affiliates
The Participant Should Be Able To:
1.1 Participate in a discussion of how two
triangles differ
1.2 Contrast the insights gained from two,
three or four methods of viewing triangles
1.3 Use Chart Wizard to create “XY Scatter”
charts of possible functional relations
Class for Regional Affiliates
The Participant Should Be Able To:
2.1 Contrast the insights gained from
normalizing using exposure rather than
premium
2.2 Contrast the uses of price deflators with
the uses of time series
2.3 Appreciate the complications introduced
by introducing a triangle such as claim
counts
Class for Regional Affiliates
The Participant Should Be Able To:
3.1 Explain to a non-actuary what an
algorithm is, why the “actuarial methods”
are algorithms, and why Mack and Venter
have asked us to think of the chain-ladder
method as an algorithm.
Class for Regional Affiliates
The Participant Should Be Able To:
4.1 Answer questions about how Excel’s functions call on
algorithms and, if appropriate, draw an analogy with
Russian dolls or some other system of nesting
subroutines
4.2 Define in the Excel-user’s style (not as a statistician) the
meaning of each function: FORECAST, RSQD,
INTERCEPT, STEYX, TREND and LINEST
4.3 Give examples of what the Excel Function Wizard means
by “known_y’s” and “known_x’s”
4.4 Use each of these Excel functions for a specific, limited
purpose (although not be able yet to assemble them into
a more sophisticated algorithm)
Class for Regional Affiliates
The Participant Should Be Able To:
5.1 Using the acronym BLUE, recall the four aspects
of a regression algorithm.
5.2 Examine a triangle of data and note its qualities
as a set of “actual” observations.
5.3 Discuss the argument that regression is
inappropriate for extrapolating into the future.
5.4 Discuss the argument that big differences
between actual and expected imply a poor tool
for forecasting.
Class for Regional Affiliates
The Participant Should Be Able To:
6.1 Adjust a triangle of data (e.g., by taking logs or
using incremental data) and discuss how the
adjusted data is more or less appropriate as a set
of “actual” data.
6.2 Give examples of real-world problems for which
the linear assumption of regression is clearly
inappropriate. [This may be difficult, but if it is,
it simply shows the usefulness of linear models.]
6.3 Contrast BLUE models of paid data with BLUE
models of incurred data (paid+case reserves).
Class for Regional Affiliates
The Participant Should Be Able To:
7.1 Explain the differences between accounting
summaries of loss costs and the “sufficient
statistics” of a statistical approach such as GLM.
7.2 Explain how the “actuarial methods” rely on a
subset of the accounting data rather than the
entire triangles of paid and incurred.
7.3 Discuss the value of having a number of
estimators that reflect a diversity of possible
world-views and are independent of one another.
Class for Regional Affiliates
The Participant Should Be Able To:
8.1. List five ways that patterns in the residuals
might indicate that one or more assumptions is
not appropriate.
8.2. Create a chart in Excel of a set of residuals and
comment about what the chart says about the
appropriateness of the BLUE assumptions.
9.1. Give reasons why the fitted values should look
like the observations
9.2. Give examples of using subsets of the larger data
set to test the reasonableness of a regression
estimate
Minimum Limits Auto Dataset
California Auto Liability Insurance
Minimum Statutory Limits Policies (15/30)
Claims Screen: Claimant Is Not Represented by an Attorney
Data Reflects Only Claims Closed as of the Evaluation
Data Used with the Permission of The Qestrel Companies
AY
1998
1999
2000
2001
2002
1
Cumulative Losses
2
3
4
5
1,549,844
34,691
1,402,761
1,544,751
1,547,344
1,018,092
4,532,796
5,063,937
5,093,082
4,161,397
16,336,148
18,067,843
4,077,332
20,892,524
3,415,880
Data Limitations
• To fit on overhead – only 5 years
• More stable than “real” data
– Only one state
– Only closed claims
– Only minimum limits
Chain Ladder Projections
Wgtd Avg
f(d)
% Reported
AY
1998
1999
2000
2001
2002
1 to 2
4.6456
3.6456
0.1930
AGE TO AGE FACTORS
2 to 3
3 to 4
1.1080
1.0048
0.1080
0.0048
0.7037
0.0968
4 to 5
1.0016
0.0016
0.0048
CHAIN LADDER METHOD PROJECTED LOSSES
1
1 to 2
2 to 3
3 to 4
34,691
1,402,761
1,544,751
1,547,344
1,018,092
4,532,796
5,063,937
5,093,082
4,161,397
16,336,148
18,067,843
18,154,613
4,077,332
20,892,524
23,148,432
23,259,601
3,415,880
15,868,658
17,582,104
17,666,541
5 to Ult
0.0016
4 to 5
1,549,844
5,101,311
18,183,945
23,297,180
17,695,084
Data Adjustment
Incremental Losses
1 to 2
2 to 3
1,368,070
141,990
3,514,703
531,142
12,174,751
1,731,695
16,815,192
AY
1998
1999
2000
2001
2002
1
34,691
1,018,092
4,161,397
4,077,332
3,415,880
3 to 4
2,593
29,145
AY
1998
1999
2000
2001
2002
CHAIN LADDER METHOD FITTED INCREMENTALS
1
1 to 2
2 to 3
3 to 4
34,691
126,468
151,466
7,419
1,018,092
3,711,512
489,437
24,319
4,161,397
15,170,600
1,763,925
86,769
4,077,332
14,864,137
2,255,908
111,169
3,415,880
12,452,778
1,713,447
84,437
4 to 5
2,500
4 to 5
2,500
8,229
29,332
37,580
28,543
Severity
Exploratory Graphical Analysis
Year
Exploratory Graphical Analysis:
Residual
• A residual = Actual value for Yi – Fitted
value for Yi
• Used to assess fit of model
• There should be no pattern, just random
points
Chain Ladder Method Residual
AY
1998
1999
2000
2001
STD DEV
AY
1998
1999
2000
2001
CHAIN LADDER METHOD RESIDUALS (INCREMENTAL)
1 to 2
2 to 3
3 to 4
4 to 5
1,241,602
(9,475)
(4,825)
(0)
(196,809)
41,705
4,825
(2,995,848)
(32,230)
1,951,055
2,188,000
37,867
6,824
-
CHAIN LADDER METHOD STANDARDIZED RESIDUALS (INCREMENTAL)
1 to 2
2 to 3
3 to 4
4 to 5
0.5675
(0.2502)
(0.7071)
(0.0000)
(0.0899)
1.1014
0.7071
(1.3692)
(0.8511)
0.8917
Chain Ladder Method Residual -1
AGE 1 T O 2 CHAI N LADDER M ET HOD RESI DUAL VERSUS AGE 1 CUM ULAT I VE
( I N 0 0 0 ' s)
3,000
2,000
1,000
RESIDUAL
0
0
500
1,000
1,500
2,000
2,500
- 1,000
- 2,000
- 3,000
- 4,000
CUMULATIVE LOSS
3,000
3,500
4,000
4,500
Chain Ladder Method Residual -2
AGE 2 T O 3 CHAI N LADDER M ET HOD RESI DUAL VERSUS AGE 2 CUM ULAT I VE
( I N 0 0 0 ' s)
50
40
30
RESIDUAL
20
10
0
0
2,000
4,000
6,000
8,000
10,000
- 10
- 20
- 30
- 40
CUMULATIVE LOSS
12,000
14,000
16,000
18,000
Chain Ladder Method Residual -3
AGE 3 T O 4 CHAI N LADDER M ET HOD RESI DUAL VERSUS AGE 3 CUM ULAT I VE
( I N 0 0 0 ' s)
6
4
RESIDUAL
2
0
0
1,000
2,000
3,000
-2
-4
-6
CUMULATIVE LOSS
4,000
5,000
6,000
Chain Ladder Method Residual -4
CHAI N LADDER M ET HOD RESI DUALS OVER T I M E
( ACROSS ACCI DENT YEARS)
2.0
1.5
STD RESIDUALS
1.0
0.5
0.0
1998
1999
2000
2001
- 0.5
- 1.0
- 1.5
Age 1 t o 2
Age 2 t o 3
Age 3 t o 4
- 2.0
ACCIDENT YEAR
Chain Ladder Method Residual -5
CHAI N LADDER M ET HOD RESI DUALS OVER T I M E
( ACROSS DEVELOPM ENT PERI ODS)
2.0
1.5
STD RESIDUALS
1.0
0.5
0.0
1
2
3
4
5
- 0.5
- 1.0
- 1.5
AY 1998
AY 1999
AY 2000
- 2.0
AGE
Exploratory Graphical Analysis:
Scatterplots
• Plot of points of dependent variable vs
independent variable
• Should suggest the shape of the relationship
between the variables
– Does it look like a linear relationship?
– Does it look non-linear?
– Is any relationship suggested: If not – mean of
Y is its best estimate
Incremental vs. Cumulative - 1
AGE 1 TO 2 I NCREMENTAL VERSUS AGE 1 CUMULATI VE
( I N 0 0 0 's)
18,000
16,000
14,000
INCREMENTAL LOSS
12,000
10,000
8,000
6,000
4,000
2,000
0
0
500
1,000
1,500
2,000
2,500
CUMULATIVE LOSS
3,000
3,500
4,000
4,500
Incremental vs. Cumulative - 2
AGE 2 TO 3 I NCREMENTAL VERSUS AGE 2 CUMULATI VE
( I N 0 0 0 's)
2,000
1,800
1,600
INCREMENTAL LOSS
1,400
1,200
1,000
800
600
400
200
0
0
2,000
4,000
6,000
8,000
10,000
CUMULAT I VE LOSS
12,000
14,000
16,000
18,000
Incremental vs. Cumulative - 3
AGE 3 T O 4 I NCREM ENT AL VERSUS AGE 3 CUM ULAT I VE
( I N 0 0 0 ' s)
35
30
INCREMENTAL LOSS
25
20
15
10
5
0
0
1,000
2,000
3,000
CU MU LATI VE LO SS
4,000
5,000
6,000
Assumes a Linear Function
Linear Function
Y-Variable
Y-VAriable
Non-linear Function
X-Variable
X-Variable
Linear Regression
• A to B incremental against age A cumulative
• Use Excel function to model y = mx + b
• LINEST(known y’s, known x’s,const,stat)
– Result is a 5 x 2 array
– Arguments
•
•
•
•
known y’s – column of incremental
known x’s – column of cumulative
const – “false” sets b = 0
stat – “true” provides full array
Input Data for 1 to 2 Incremental
vs. 1 Cumulative
AY
X
Y
1998
1999
2000
2001
34,691
1,018,092
4,161,397
4,077,332
1,368,070
3,514,703
12,174,751
16,815,192
Create Array of Regression
Statistics
•
Select 5 x 2 matrix of cells
•
Type LINEST function in upper left cell
•
Keystroke <CTRL><SHIFT><ENTER>
m
SEm
r2
LINEST ARRAY TABLE (1 TO 2)
3.297
809,596 b
0.689
2,038,734 SEb
0.920
2,523,361 SEy
F Statistic
22.871
2.000 Degrees of Freedom
Reg SS 145,625,078,580,873 12,734,698,887,975 Res SS
Statistics Based on Residual
• Compute variance of regression as sum of squared
residuals divided by the degrees of freedom (the
mean square error, MSE)
– Its square root, s, is standard error of regression
• The R2 or percent of explained variance: 1-R2 =
divide MSE by total variance
• Standard deviation of constant
– Use to test significance of constant
• Standard deviation of coefficient
– Use to test significance of coefficient
Selecting Array Components
• INDEX Function in Excel will return array elements
• Form - INDEX(ARRAY, ROW #, COLUMN #)
–
–
–
–
–
INDEX(LINEST, 1, 1) = M
INDEX(LINEST, 1, 2) = B
INDEX(LINEST, 2, 1) =SEm
INDEX(LINEST, 2, 2) =SEb
INDEX(LINEST, 4, 2) =Degrees of Freedom
Test Significance Of Intercept
• Calculate Student t statistic, B / SEb
• Select Significance Level –Judgment
(selected 0.05 for this example)
• Excel Function for probability value from Student t
distribution
- TDIST(ABS(B/ SEb), Degrees Freedom, Tails)
- Set Tails = 2 For 2-tail Distribution
• Compare result to selected significance level
Results Significance of Intercept
Intercept:
Standard Error
T Statistic
Degrees of Freedom
Student t probability
Conclusion
1 to 2
809,596.433
2,038,734
0.397
2.000
0.730
Not Significant
From Zero
2 to 3
20,993.037
38,195
0.550
1.000
0.680
Not Significant
From Zero
3 to 4
-9,061.620
0
N/A
0.000
N/A
N/A
The “One Factor Model”
• Age-toAged Factor=
incrementald loss/cumulatived-1
• Incremental lossd = f(d) * cumulatived-1
• This is equivalent to weighted regression
where
–
–
–
–
–
Y is incremental at d
X is cumulative at d-1
B, the coefficient, is the (LDF – 1.0)
The regression has no constant term
Weights are xi/S xj instead of xi2/S xj2
Linear Regression Through the
Origin-Statistic Array
LINEST ARRAY TABLE (1 TO 2)
3.512073632
0.361831528
0.91324316
31.57940603
144,620,983,575,556
0.000
#N/A
2,139,999.524
3
13,738,793,893,292
Results Linear Regression
Through the Origin
Slope:
Standard Error
T Statistic
Degrees of Freedom
Student t probability
Conclusion
R Squared:
SEy
Selected f(d):
1 to 2
3.512
0.362
9.706
3.000
0.002
Significant
From Zero
0.913
2,139,999.524
2 to 3
0.107
0.002
51.969
2.000
0.000
Significant
From Zero
0.998
34,947.730
3 to 4
0.005
0.001
4.754
1.000
0.132
Not Significant
From Zero
0.897
6,023.359
1 to 2
3.512
2 to 3
0.107
3 to 4
0.005
Linear Regression Projections
CUMULATIVE LOSSES + MODELED INCREMENTAL LOSSES
1
1 to 2
2 to 3
3 to 4
34,691
1,402,761
1,544,751
1,547,344
1,018,092
4,532,796
5,063,937
5,093,082
4,161,397
16,336,148
18,067,843
18,165,559
4,077,332
20,892,524
23,123,108
23,248,164
3,415,880
15,412,702
17,058,235
17,150,490
AY
1998
1999
2000
2001
2002
AY
1998
1999
2000
2001
2002
MODELED INCREMENTAL LOSSES
1
1 to 2
2 to 3
34,691
121,837
149,765
1,018,092
3,575,616
483,943
4,161,397
14,615,132
1,744,124
4,077,332
14,319,891
2,230,584
3,415,880
11,996,822
1,645,533
4 to 5
1,549,844
5,093,082
18,165,559
23,248,164
17,150,490
3 to 4
8,354
27,387
97,716
125,056
92,255
Residual Results
AY
1998
1999
2000
2001
RESIDUALS
1 to 2
1,246,232.8090
(60,912.4134)
(2,440,380.5837)
2,495,301.5777
2 to 3
(7,775.2208)
47,199.1589
(12,428.7173)
3 to 4
(5,761.2629)
1,757.4699
STD DEV
2,109,837.6096
33,164.5420
5,316.5470
AY
1998
1999
2000
2001
STANDARDIZED RESIDUALS
1 to 2
2 to 3
0.5907
(0.2344)
(0.0289)
1.4232
(1.1567)
(0.3748)
1.1827
3 to 4
(1.0836)
0.3306
Graphs Based on Residual
•
•
•
•
Plot residual versus Y
Plot residual versus predicted Y
Plot residual versus X
Plot residual versus variables not in
regression (i.e., age, calendar year)
Chain Ladder Model Residual -1
AGE 1 T O 2 CHAI N LADDER M ODEL RESI DUAL VERSUS AGE 1 CUM ULAT I VE
( I N 0 0 0 ' s)
3,000
2,000
RESIDUAL
1,000
0
0
500
1,000
1,500
2,000
2,500
- 1,000
- 2,000
- 3,000
CUMULATIVE LOSS
3,000
3,500
4,000
4,500
Chain Ladder Model Residual -2
AGE 2 T O 3 CHAI N LADDER M ODEL RESI DUAL VERSUS AGE 2 CUM ULAT I VE
( I N 0 0 0 ' s)
60
50
40
RESIDUAL
30
20
10
0
0
2,000
4,000
6,000
8,000
10,000
- 10
- 20
CUMULATIVE LOSS
12,000
14,000
16,000
18,000
Chain Ladder Model Residual -3
AGE 3 T O 4 CHAI N LADDER M ODEL RESI DUAL VERSUS AGE 3 CUM ULAT I VE
( I N 0 0 0 ' s)
3
2
1
0
0
1,000
2,000
3,000
RESIDUAL
-1
-2
-3
-4
-5
-6
-7
CUMULATIVE LOSS
4,000
5,000
6,000
Chain Ladder Model Residual -4
CHAI N LADDER M ODEL RESI DUALS OVER T I M E
( ACROSS ACCI DENT YEARS)
5.0
4.0
STD RESIDUALS
3.0
2.0
1.0
0.0
1998
1999
2000
- 1.0
2001
Age 1 t o 2
Age 2 t o 3
Age 3 t o 4
- 2.0
ACCIDENT YEAR
Chain Ladder Model Residual -5
CHAI N LADDER M ODEL RESI DUALS OVER T I M E
( ACROSS DEVELOPM ENT PERI ODS)
5.0
4.0
STD RESIDUALS
3.0
2.0
1.0
0.0
2
3
4
5
- 1.0
1998
1999
2000
- 2.0
AGE
Additional Models
• What to do if tests of assumptions
underlying chain ladder fail?
• Alternative Models
–
–
–
–
Linear with Constant
Additive Model
Bornhuetter-Ferguson (BF)
Cape Cod-Special Case of BF (CC)
Approach to CC and BF
• Estimate Ultimate Claim Counts
• Parameterize incremental severity by functional
form f(d)*h(w), where f(d) varies by development
age and h(w) varies by accident year
• Fitting accomplished by an iterative approach with
constraints (Solver add-in to Excel)
• Cape Cod assumes h(w) constant over all accident
years (reduces parameters)
Claim Count
AY
1998
1999
2000
2001
2002
AY
1998
1999
2000
2001
2002
1
CUMULATIVE COUNT
2
3
4
5
228
11
215
226
227
133
598
684
688
632
2,731
3,004
877
3,577
665
1
11
133
632
877
665
INCREMENTAL COUNT
1 to 2
2 to 3
204
11
465
86
2,099
273
2,700
3 to 4
1
4
4 to 5
1
Chain Ladder Method-Claim
Count
Wgtd Avg
f(d)
% Reported
AY
1998
1999
2000
2001
2002
1 to 2
4.3079
3.3079
0.2081
AGE TO AGE FACTORS
2 to 3
3 to 4
1.1044
1.0055
0.1044
0.0055
0.6884
0.0936
Chain Ladder Method Projected Count
1
1 to 2
2 to 3
11
215
226
133
598
684
632
2,731
3,004
877
3,577
3,950
665
2,865
3,164
4 to 5
1.0044
0.0044
0.0054
5 to Ult
3 to 4
227
688
3,021
3,972
3,181
4 to 5
228
691
3,034
3,990
3,195
0.0044
Claim Count Incremental vs Cumulative-1
AGE 1 T O 2 I NCREM ENT AL VERSUS AGE 1 CUM ULAT I VE
3,000
2,500
INCREMENTAL COUNT
2,000
1,500
1,000
500
0
0
100
200
300
400
500
600
CUMULATIVE COUNT
700
800
900
1,000
Claim Count Incremental vs Cumulative-2
AGE 2 T O 3 I NCREM ENT AL VERSUS AGE 2 CUM ULAT I VE
300
250
INCREMENTAL COUNT
200
150
100
50
0
0
500
1,000
1,500
CUMULATIVE COUNT
2,000
2,500
3,000
Claim Count Incremental vs Cumulative-3
AGE 3 TO 4 I NCREMENTAL VERSUS AGE 3 CUMULATI VE
5
INCREMENTAL COUNT
4
3
2
1
0
0
100
200
300
400
CUMULATIVE COUNT
500
600
700
800
Claim Count Residual - 1
CLAI M COUNT RESI DUAL PLOT OVER T I M E
( ACROSS ACCI DENT YEARS)
2.0
STD RESIDUALS
1.0
0.0
1998
1999
2000
2001
- 1.0
Age 1 t o 2
Age 2 t o 3
Age 3 t o 4
- 2.0
ACCIDENT YEAR
Claim Count Residual - 2
CLAI M COUNT RESI DUAL PLOT OVER T I M E
( ACROSS DEVELOPM ENT PERI ODS)
2.0
STD RESIDUALS
1.0
0.0
1
2
3
4
5
- 1.0
1998
1999
2000
- 2.0
AGE
Significance of Intercept-Claim
Count
Intercept:
Standard Error
T Statistic
Degrees of Freedom
Student t probability
Conclusion
R Squared:
1 to 2
136.181
66
2.079
2.000
0.173
Not Significant
From Zero
0.997
2 to 3
6.901
22
0.307
1.000
0.810
Not Significant
From Zero
0.981
3 to 4
-0.480
0
N/A
0.000
N/A
Not Significant
From Zero
1.000
Linear Regression Results-Claim
Count
Slope:
Standard Error
T Statistic
Degrees of Freedom
Student t probability
Conclusion
R Squared:
SEy
Selected f(d):
1 to 2
3.168
0.114
27.858
3.000
0.000
Significant
From Zero
0.990
123.868
2 to 3
0.102
0.007
14.484
2.000
0.005
Significant
From Zero
0.979
19.682
3 to 4
0.006
0.000
13.464
1.000
0.047
Significant
From Zero
0.979
0.305
1 to 2
3.168
2 to 3
0.102
3 to 4
0.006
Linear Regression Projections
Cumulative Count + Modeled Incremental Count
1
1 to 2
2 to 3
11
215
226
133
598
684
632
2,731
3,004
877
3,577
3,941
665
2,772
3,054
AY
1998
1999
2000
2001
2002
AY
1998
1999
2000
2001
2002
Modeled Incremental Count
1
1 to 2
11
35
133
421
632
2,002
877
2,778
665
2,107
2 to 3
22
61
278
364
282
3 to 4
227
688
3,021
3,963
3,071
4 to 5
228
688
3,021
3,963
3,071
3 to 4
1
4
17
22
17
Observed Severity
AY
1998
1999
2000
2001
2002
AY
1998
1999
2000
2001
2002
1
152
1,480
1,377
1,029
1,112
1
152
1,480
1,377
1,029
1,112
Severity Per File
2
6,152
6,588
5,407
5,272
3
6,775
7,360
5,980
Incremental Severity
1 to 2
2 to 3
6,000
623
5,109
772
4,030
573
4,243
4
6,787
7,403
5
6,798
3 to 4
11
42
4 to 5
11
Initial Seed for f(d) and h(w)
Cape Cod
H(w) BASED ON THE ULTIMATES FROM THE CHAIN
LADDER MODELS (LOSS DIVIDED BY CLAIM COUNTS)
5,943 FOR ALL YEARS
F(d) BASED ON THE PAYMENT PATTERN FROM THE
CHAIN LADDER MODEL
CL Model
Final Selections:
Cumulative
Incremental
Starting Values
% Reported
1
1.000
1.000
1 to 2
3.512
4.512
3.512
2 to 3
0.107
4.994
0.482
3 to 4
0.005
5.021
0.027
4 to 5
0.000
5.021
0.000
19.92%
69.95%
9.59%
0.54%
0.00%
Iterative Process Results-CC
F(d)
Fitted
h(w)
6,507
6,507
6,507
6,507
6,507
15.86%
AY
1998
1999
2000
2001
2002
74.50%
9.65%
FITTED INCREMENTAL SEVERITY
1
1 to 2
2 to 3
1,032
4,848
628
1,032
4,848
628
1,032
4,848
628
1,032
4,848
628
1,032
4,848
628
0.00%
0.00%
3 to 4
-
4 to 5
-
Fitted Ultimates and Projected Development-CC
AY
1998
1999
2000
2001
2002
AY
1998
1999
2000
2001
2002
1
1,032
1,032
1,032
1,032
1,032
FITTED CUMULATIVE SEVERITY
2
3
5,880
6,507
5,880
6,507
5,880
6,507
5,880
6,507
5,880
6,507
ULT CLAIMS
4
6,507
6,507
6,507
6,507
6,507
5
6,507
6,507
6,507
6,507
6,507
228
688
3,021
3,963
3,071
ULT LOSS
1,483,663
4,477,018
19,659,486
25,789,508
19,984,231
PROJECTED LOSS DEVELOPMENT
1
2
3
4
5
34,691
1,402,761
1,544,751
1,547,344
1,549,844
1,018,092
4,532,796
5,063,937
5,093,082
5,093,082
4,161,397
16,336,148
18,067,843
18,067,843 18,067,843
4,077,332
20,892,524
23,380,324
23,380,324 23,380,324
3,415,880
18,303,297
20,231,087
20,231,087 20,231,087
Initial Seed for f(d) and h(w)
Bornhuetter-Ferguson
H(w) BASED ON THE ULTIMATES FROM THE CHAIN
LADDER MODELS (LOSS DIVIDED BY CLAIM COUNTS)
1998
6,798
1999
7,403
2000
6,013
2001
5,866
2002
5,585
F(d) BASED ON FINAL PATTERN FROM CAPE COD MODEL
Starting Values
% Reported
1
1 to 2
2 to 3
3 to 4
4 to 5
15.86%
74.50%
9.65%
0.00%
0.00%
Iterative Process Results-BF
F(d)
Fitted
h(w)
7,666
6,895
5,484
5,663
7,570
AY
1998
1999
2000
2001
2002
14.69%
1
1,126
1,013
806
832
1,112
75.60%
9.71%
FITTED INCREMENTAL SEVERITY
1 to 2
2 to 3
5,795
744
5,212
669
4,146
532
4,281
550
5,723
735
0.00%
0.00%
3 to 4
-
4 to 5
-
Fitted Ultimates and Projected Development-BF
AY
1998
1999
2000
2001
2002
1
1,126
1,013
806
832
1,112
AY
1998
1999
2000
2001
2002
FITTED CUMULATIVE SEVERITY
2
3
6,922
7,666
6,225
6,895
4,952
5,484
5,113
5,663
6,835
7,570
1
34,691
1,018,092
4,161,397
4,077,332
3,415,880
4
7,666
6,895
5,484
5,663
7,570
ULT CLAIMS
ULT LOSS
228
688
3,021
3,963
3,071
1,747,831
4,743,672
16,569,195
22,443,025
23,249,242
5
7,666
6,895
5,484
5,663
7,570
PROJECTED LOSS DEVELOPMENT
2
3
4
1,402,761
1,544,751
1,547,344
4,532,796
5,063,937
5,093,082
16,336,148
18,067,843
18,067,843
20,892,524
23,071,364
23,071,364
20,992,140
23,249,250
23,249,250
5
1,549,844
5,093,082
18,067,843
23,071,364
23,249,250
Summary of Projected Ultimates
AY
1998
1999
2000
2001
2002
Total
Dev
1
2
3
4
5
CL Method
1,549,844
5,101,311
18,183,945
23,297,180
17,695,084
65,827,364
CL Method
19.30%
70.37%
9.68%
0.48%
0.00%
Ultimate Losses (Age 5)
CL Model
1,549,844
5,093,082
18,165,559
23,248,164
17,150,490
65,207,139
Cape Cod
1,549,844
5,093,082
18,067,843
23,380,324
20,231,087
68,322,181
FH
1,549,844
5,093,082
18,067,843
23,071,364
23,249,250
71,031,383
% Reported (incremental)
CL Model
Cape Cod
19.92%
15.86%
69.95%
74.50%
9.59%
9.65%
0.54%
0.00%
0.00%
0.00%
FH
14.69%
75.60%
9.71%
0.00%
0.00%
Cape Cod and FH Model are driven by AY 1998, Dev 1 loss amount
Comparison of Prediction of
Incremental Loss
SSE (000,000's)
DF
MSE (000,000's)
CL Model
13,741,273
11
1,249,207
Cape Cod
13,213,917
11
1,201,265
BF
2,818,248
7
402,607