Clark

Question 1

Key elements of a statistical loss reserving model (2)

Answer 1

1. expected amount of loss to emerge (point estimate)

2. distribution of actual emergence around expected value (range of possible outcomes)

Question 2

Loglogistic growth curve

Answer 2

G(x) = x^omega / (x^omega + theta^omega)

Question 3

Weibull growth curve

Answer 3

G(x) = 1 - e^ -(x / theta)^omega

*shorter tail compared to Loglogistic

Question 4

Advantages to using parameterized curves to describe loss emergence pattern (3)

Answer 4

1. simple estimation
2. ability to use data from triangles w/o evenly spaced evaluation dates
3. creates a smooth curve

Question 5

Number of parameters in the LDF method (*Clark)

Answer 5

n + 2

• n AYs
• omega
• theta

Question 6

Number of parameters in the CC method (*Clark)

Answer 6

3

• omega
• theta
• ELR

Question 7

Reasons the CC method has a lower parameter and total variance (2)

Answer 7

1. reduced number of parameters

2. additional info included in the exposure base

Question 8

Difference b/w LDF and CC methods

Clark

Answer 8

LDF - assumes AYs are independent

CC - assumes a known relationship b/w ultimate losses across AYs&raquo_space; described by ELR

Question 9

Tests for constant ELR assumption (2)

Clark

Answer 9

1. plot ultimate LRs by AY

2. plot normalized residuals against expected incremental losses and look for a random scatter around the 0 line

Question 10

Variance / mean ratio (sigma^2)

Answer 10

= 1 / (n - p) * sum[ (actual - expected)^2 / expected)

> calculate chi-squared triangle and then sum

Question 11

Advantages of using the ODP distribution (2)

Clark

Answer 11

1. high flexibility - scaling factor, sigma^2, allows matching 1st and 2nd moments of any distribution
2. results presented in a familiar format (produces LDF and CC estimates)

Question 12

Advantage of MLE

Answer 12

works with negative or 0 incremental loss amounts

Question 13

Key assumptions from Clark’s model (3)

Answer 13

1. incremental losses are i.i.d.
2. var/mean scale parameter is fixed and known
3. variance is based on approx. to Rao-Cramer lower bound (minimized)

Question 14

Residual plots to test model assumptions (what to look for and 4 types of tests)

Answer 14

want: random scatter around zero line

can plot against:

1. increment age (how well loss emergence curve fits incremental losses at different dev. Periods)
2. expected loss (var/mean ratio is constant)
3. AY
4. CY (diagonal effects)

Question 15

Handling truncation (3)

Answer 15

1. ELR is always calculated before truncation
2. LDF method truncates LDF (fitted / truncated fitted)
3. CC method truncates % reported (difference)

Question 16

Which has the smallest variance and why: discounted or un-discounted reserves?

Answer 16

discounted reserves - the tail has the largest parameter variance but also receives the deepest discount

Question 17

How to handle partial exposure periods

Answer 17

scale by % earned

Question 18

Discounted reserves when decomposing AY reserves into CY components

Answer 18

discounted reserves = ultimate * change in growth curves / (1 + i) ^ (k - 1/2)

k = # years b/w current and avg age

Question 19

Process variance when decomposing AY reserves into CY components

Answer 19

= sigma^2 * ultimate * change in growth curves / (1 + i) ^ (2k - 1)

Question 20

Normalized residuals (Clark)

Answer 20

(Actual incremental loss - fitted incremental loss) / sqrt( sigma^2 *fitted incremental loss)

Question 21

Parameter variance calculation

Clark

Answer 21

Parameter variance = var(ELR) * premium^2

Question 22

Why is parameter variance generally greater than process variance?
(Clark)

Answer 22

Few data points in the triangle means that most of the uncertainty comes from parameter estimation (vs. randomness)

Question 23

MLE term

Clark

Answer 23

MLE term = actual incremental loss * ln(expected incremental loss) - expected incremental loss

(Needs to not be in $000’s)