Clark

Question 1

growth function G(x)

Answer 1

loss emergence pattern

growth function as of time x, x is avg acc date to evaluation date

G(x)=1/CDF=p_k = cumulative % of loss reported or paid

can be described by Loglogistic and Weibull

Question 2

Weibull and loglogistic

Answer 2

Weibull: G(x) = 1-exp(-(x/theta)^w)

loglogistic: G(x)=x^w/(theta^w+x^w)

Question 3

average accident date to evaluation date

Answer 3

AvgAge(t) = t/2 for t< 12 and t-6 for t>12

Question 4

variance of actual loss emergence

Answer 4

total variance = process variance + parameter variance

Question 5

process variance

Answer 5

process variance = σ² * reserves

Question 6

why is σ² larger for LDF?

Answer 6

LDF requires more parameters

-LDF requires parameters for each AY Ult loss and parameters in G

µ = ULT_AY*[G(y)-G(x)]

-CC requires ELR parameter and parameters in G

µ = Prem_AY*[G(y)-G(x)]

Question 7

Why is CC perferred in general?

Answer 7

• CC has smaller parameter var since add info and fewer parameters
• process var can be higher or lower than LDF
• CC in general produces lower total var than LDF

Question 8

LDF method

Answer 8

Question 9

CC Method

Answer 9

Question 10

CoV and why CC’s is lower

Answer 10

CoV = std dev/estimated reserves

-CoV for CC is reduced from LDF because relying on more info like premium and this allows to make better estimate of reserve

Question 11

normalized residuals

Answer 11

Question 12

plot of residuals

Answer 12

• Plot of increment age vs normalized residuals should be randomly scattered around 0 -> if not, growth curve not appropriate
• plot can show CY effect by having string of negative residuals for one CY and positive residuals for another
• positive residuals = underestimates losses
• negative = overestimates losses

Question 13

variance of prospective loss

Answer 13

uses CC, if have estimate of future prem, can calc estimate of expected loss which would be estimated reserves, process var calc as usual

Question 14

CY development

Answer 14

rather than calc IBNR for each AY, estimate development for next CY period beyond latest diag -> take difference in growth fct @ 2 evaluation ages and mult by estimated ult loss

Question 15

benefit of estimated CY development to help validate model

Answer 15

12 month development estimate is testable within short time period compared to estimate of total unpaid loss reserves

within 1 yr, can see whether actual CY development falls within range of estimated CY development (based on expected and std dev of 12 month devel)

if development is within forecast range, indicates model may be reasonable

Question 16

Discounted reserves

Answer 16

• start with age of AY and age of truncation
• breakdown this diff in 12 month portions
• set up table with age, avg age, etc
• CY reserve = Ult*(growth i - growth i+1)
• then discount
• to do process variance you can do the above but discounting is squared and you need to multiply by sigma^2

Question 17

CY discount

Answer 17

1/(1+i)^(avg age i - age of AY)

Question 18

which period doesn’t have discounted reserve

Answer 18

AY age

Question 19

to calc expected incremental

Answer 19

incremental emergence % = G(y)-G(x)

even with truncation

Question 20

total variance for CC

Answer 20

total var = process var + parameter var

=σ²R+Var(ELR)*Prem²

Question 21

3 assumptions of Clark

Answer 21

1. incremental losses are i.i.d
   - one reserving period doesn’t affect surrounding periods
   - assumes emergence pattern is same for all AYs
2. var/mean scale parameter is fixed and known
3. variance estimates are based on approx to rao-cramer lower bound

Question 22

advantages of using parameterized curves to describe expected loss emergence pattern

Answer 22

• simplifies problem of estimating expected loss emergence because few parameters are needed
• can use data that’s not in triangle with evenly spaced eval dates
• indicated pattern is smooth curve, doesnt follow noise in historical age-to-age factors

Question 23

why is paramater var greater than process var?

Answer 23

only 6 data points but LDF model uses 5 parameters -> model is over-parameterized and overfits noise in data

there are few data points in loss reserve triangle so most of uncertainty in reserve estimate is from parameter estimate needed to estimate expected reserve, not random events

Question 24

advantage of tabular form

Answer 24

with tabular form, can use data with irregular evaluation periods

Question 25

tabular form

Answer 25

AY, From, To, Actual Incr Loss (c), Expected Incr Loss (u), MLE term

Question 26

MLE term

Answer 26

MLE term = c*ln(u)-u

Question 27

LDF method assumes

Answer 27

loss amount in each AY is independent from all other years (std CL)

Question 28

CC method assumes

Answer 28

there is known relationship between expected ultimate losses across AYs where relationship is identified by exposure base