Mack 1994 - Chain Ladder Flashcards
3 benefits of confidence intervals
- Estimated ultimate losses are not an exact forecast of the true ultimate losses
- They allow the inclusion of business policy by using a specific confidence probability
- They allow comparison between the chain-ladder and other reserving procedures
Mack Chain Ladder Assumption 1
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
Briefly describe a major consequence of Assumption 1
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.
Mack Chain Ladder Assumption 2
Losses are independent between accident years.
Briefly describe a major consequence of Assumption 2
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.
Mack Chain Ladder Assumption 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.
Var(Ci_k+1 given Ci_1 to Ci_k) = Ci_k * (a_k)^2
Summarize the 3 mack chain ladder assumptions
- Expected losses in the next development period are proportional to losses-to-date
- Losses are independent between accident years
- Variance of losses in the next development period is proportional to losses-to-date with proportionality constant (a_k)^2 that varies by age
Calculate the MSE of an accident year’s ultimate loss estimate.
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)
Calculate a^2
a^2_k = 1/(I-k-1) * sum over AYs of C_k*(C_k+1/C_k - LDF_k)^2
Name an issue from the a^2 formula
It does not provide an estimator for a^2_I-1
Calcualte a^2_I-1 for the final development period
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))
Calculate the standard error of the estimated reserves for an accident year
s.e.(Chat_iI) = MSE(Chat_iI) = s.e.(Rhat_i)
Calculate the confidence interval for the reserve estimate
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.
Briefly describe 2 potential problems when using the normal distribution as an approximation to the true distribution of R
- If data is skewed, it is a poor approximation (s.e.(Ri)/Ri greater than 50%)
- The CI interval can have negative limits, even if a negative reserve is not possible
Briefly explain why reserves estimate by accident year are dependent.
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.
If we assume Var(C_k+1) = 1 (constant var)
a) which weights should we use for LDF calculation?
b) what are the weighted residuals?
a) C^2_i,k weighted
b) e= C_i,K=1 - C_i,k*f_k,0
If we assume Var(C_k+1) = C_i,k (proportional var)
a) which weights should we use for LDF calculation?
b) what are the weighted residuals?
a) C_i,k (volume weighted)
b) e= (C_i,K+1 - C_i,k*f_k,1) / (C_i,k)^0.5
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?
a) 1 (simple average)
b) e= (C_i,k+1 - C_i,k*f_k,1) / (Ci,k)^0.5
Calculate CI(C_i,I) using empirical limits
- The lower empirical limit results from applying the min age-to-age factors for each dev period to the losses-to-date
- The upper empirical limit results from applying the max age-to-age factors for each dev period to the losses-to-date
Calculate the total s.e.(R)
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
Briefly describe Mack regression plot to test assumption 1
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.
Briefly describe Mack residual plot to test assumption 3
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.
Calculate the spearman’s test of correlation of adjacent development factors.
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.
Give 2 reasons why we use T as Test statistic for correlation between dev factors.
- Some correlation will occur in subsequent LDFs purely due to random chance
- It’s more important to know whether correlations prevail globally than to find a small part of the triangle with correlations.
Explain the Calendar Year Effects Test for assumption 2
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.
Name 3 examples of CY influences
- Reserve strengthening or weakening
- Changes in payment processes
- Changes in inflation
