Mack 1994 - Chain Ladder

Question 1

3 benefits of confidence intervals

Answer 1

1. Estimated ultimate losses are not an exact forecast of the true ultimate losses
2. They allow the inclusion of business policy by using a specific confidence probability
3. They allow comparison between the chain-ladder and other reserving procedures

Question 2

Mack Chain Ladder Assumption 1

Answer 2

Expected losses in the next development period are proportional to losses-to-date.

E(Ci_k+1 given Ci_1 to Ci_k) = Ci_k * LDF

The chain ladder method uses the same LDF for each accident year

Uses most recent losses-to-date to project losses, ignoring losses as of earlier development periods

Question 3

Briefly describe a major consequence of Assumption 1

Answer 3

It implies that subsequent development factors are uncorrelated.

We should NOT apply the chain-ladder method to a book of business where we usually observe a smaller-than-average increase between development periods k and k+1 following a larger-than-average increase between k-1 and k.

Question 4

Mack Chain Ladder Assumption 2

Answer 4

Losses are independent between accident years.

Question 5

Briefly describe a major consequence of Assumption 2

Answer 5

It cannot be used for triangles where calendar year effects, such as a change in claims handling or case reserving practices, affect several accident years in the same way.

Question 6

Mack Chain Ladder Assumption 3

Answer 6

Variance of losses in the next development period is proportional to losses-to-date with proportionality constant (a_k)^2 that varies by age.

Var(Ci_k+1 given Ci_1 to Ci_k) = Ci_k * (a_k)^2

Question 7

Summarize the 3 mack chain ladder assumptions

Answer 7

1. Expected losses in the next development period are proportional to losses-to-date
2. Losses are independent between accident years
3. Variance of losses in the next development period is proportional to losses-to-date with proportionality constant (a_k)^2 that varies by age

Question 8

Calculate the MSE of an accident year’s ultimate loss estimate.

Answer 8

MSE(Ult) = sum over dev periods of Ult^2 * (a^2_k/LDF^2) * (1/Chat_k + 1/sum over AYs before curr. diagonal C_k)

Question 9

Calculate a^2

Answer 9

a^2_k = 1/(I-k-1) * sum over AYs of C_k*(C_k+1/C_k - LDF_k)^2

Question 10

Name an issue from the a^2 formula

Answer 10

It does not provide an estimator for a^2_I-1

Question 11

Calcualte a^2_I-1 for the final development period

Answer 11

If last LDF = 1.000, then a^2_I-1 = 0

If you can extrapolate nicely, a^2_I-1 = (a^2_I-2)^2 / a^2_I-3

Otherwise, a^2_I-1 = min((a^2_I-2)^2/a^2_I-3, min(a^2_I-2, a^2_I-3))

Question 12

Calculate the standard error of the estimated reserves for an accident year

Answer 12

s.e.(Chat_iI) = MSE(Chat_iI) = s.e.(Rhat_i)

Question 13

Calculate the confidence interval for the reserve estimate

Answer 13

CI = R +- z*s.e.(R)

z is the z-score at (1-a/2) percentile of the std normal distr.

If lognormal distr.:
s^2 = ln(1+(s.e.(R)/R)^2)
CI = R * exp(+- z*s-s^2/2)

If min of CI is negative OR s.e.(R) greater than 50% of R, use lognormal.

Question 14

Briefly describe 2 potential problems when using the normal distribution as an approximation to the true distribution of R

Answer 14

1. If data is skewed, it is a poor approximation (s.e.(Ri)/Ri greater than 50%)
2. The CI interval can have negative limits, even if a negative reserve is not possible

Question 15

Briefly explain why reserves estimate by accident year are dependent.

Answer 15

The estimators Rhat_i are all influenced by same age-to-age factors, fhat_k, resulting in positive correlation between accident reserve estimates.

This causes the overall MSE to be greater than the sum of individual accident year MSEs.

Question 16

If we assume Var(C_k+1) = 1
a) which weights should we use for LDF calculation?
b) what are the weighted residuals?

Answer 16

a) C^2_i,k weighted

b) e= C_i,K=1 - C_i,k*f_k,0

Question 17

If we assume Var(C_k+1) = C_i,k
a) which weights should we use for LDF calculation?
b) what are the weighted residuals?

Answer 17

a) C_i,k (volume weighted)

b) e= (C_i,K+1 - C_i,k*f_k,1) / (C_i,k)^0.5

Question 18

If we assume Var(C_k+1) = C^2_i,k
a) which weights should we use for LDF calculation?
b) what are the weighted residuals?

Answer 18

a) 1 (simple average)
b) e= (C_i,k+1 - C_i,k*f_k,1) / (Ci,k)^0.5

Question 19

Calculate CI(C_i,I) using empirical limits

Answer 19

1. The lower empirical limit results from applying the min age-to-age factors for each dev period to the losses-to-date
2. The upper empirical limit results from applying the max age-to-age factors for each dev period to the losses-to-date

Question 20

Calculate the total s.e.(R)

Answer 20

s.e.(Rhat) ^2 = sum over I between 2 and I of (s.e.(R_i)^2 + chat_i,I(sum over I of Chat_I)sum over k between I+1-i and I-1 of (2*a^2_k/(f^2_k * sum between 1 and I-k of C_nk))

Unlikely would need to execute this formula on exam

Question 21

Briefly describe Mack regression plot to test assumption 1

Answer 21

Plot cumulative losses at k+1 (y-axis) against losses at k (x-axis).

Fit a straight line through the origin with a slope of the calculated LDF.

The line should fit the data reasonably well.

If not, then the chain-ladder assumption is violated.

Question 22

Briefly describe Mack residual plot to test assumption 3

Answer 22

Plot the weighted residuals (y-axis) against cumulative losses at the prior development period (x-axis).

Residuals should be random around zero without trends or patterns.

If not, then the assumption 3 is violated.

Question 23

Calculate the spearman’s test of correlation of adjacent development factors.

Answer 23

Used to test correlations between subsequent development factors

S_k = sum of (Rank_k - Rank_prior)^2

T_k = 1 - S_k / n*(n^2-1)/6

T = sum of weights(k)*T_k / sum of weights(k)

Var(T) = 1 / (#AYs - 2)*(#AYs - 3)/2

CI = 0 +- z*sqrt(Var(T))

If T lies outside of the CI, we reject the null hypothesis that subsequent development factors are uncorrelated.

If T lies inside of the confidence interval, we fail to reject the null hypothesis that subsequent development factors are uncorrelated.

Question 24

Give 2 reasons why we use T as Test statistic for correlation between dev factors.

Answer 24

1. Some correlation will occur in subsequent LDFs purely due to random chance
2. It’s more important to know whether correlations prevail globally than to find a small part of the triangle with correlations.

Question 25

Explain the Calendar Year Effects Test for assumption 2

Answer 25

Zj = min(Sj, Lj)
nj = #S + #L in diagonal
mj = (nj-1)/2
E(Zj) = n/2 - (n-1 choose m)n/2^n
Var(Zj) = n(n-1)/4 - (n-1 choose m)n(n-1)/2^n + E(Zj) - E^2(Zj)
CI = E(X) +- z*sqrt(Var(Z))

Test is sum of Zj is within the CI

If Z lies outside of the CI, we reject the null hypothesis that losses are independent between AYs.

If Z lies inside of the CI, we fail to reject the null hypothesis that losses are independent between AYs.

Question 26

Name 3 examples of CY influences

Answer 26

1. Reserve strengthening or weakening
2. Changes in payment processes
3. Changes in inflation