Mack Chain Ladder Model

Study guide: Exam GI-301

Mack Chain Ladder Model is the distribution-free stochastic model that Mack (1994) shows the chain ladder method implicitly assumes, together with the standard error of chain ladder reserves that follows from it. For cumulative claims C_{ik} of accident year i at development year k in an I \times I triangle, with accident years independent:

E(C_{i,k+1} \mid C_{i1}, \dots, C_{ik}) = C_{ik}\,f_k

\text{Var}(C_{i,k+1} \mid C_{i1}, \dots, C_{ik}) = C_{ik}\,\alpha_k^2

• Where the three assumptions come from. The chain ladder uses one factor per age for every accident year and projects from the latest amount alone (the mean assumption); it ignores any dependence between accident years (independence); and it weights the individual factors by C_{jk}, which gives the minimum-variance unbiased average only if the variance is proportional to C_{jk}. None holds for every triangle — see Chain Ladder Assumptions, Calendar Year Effect and Development Factor Correlation Test.
• Estimators. \hat f_k = \sum_j C_{j,k+1} / \sum_j C_{jk} is unbiased. For k \le I-2, \alpha_k^2 is estimated without bias by the first formula below. The last one, \hat\alpha_{I-1}^2, is 0 if development is finished; otherwise Mack extrapolates it, for instance by the second formula:

\hat\alpha_k^2 = \frac{1}{I-k-1}\sum_{j=1}^{I-k} C_{jk}\left(\frac{C_{j,k+1}}{C_{jk}} - \hat f_k\right)^2

\hat\alpha_{I-1}^2 = \min\left(\frac{\hat\alpha_{I-2}^4}{\hat\alpha_{I-3}^2},\ \hat\alpha_{I-3}^2,\ \hat\alpha_{I-2}^2\right)

• Standard error of one year's reserve. The mean squared error of \hat R_i = \hat C_{iI} - C_{i,I+1-i} is taken conditional on the observed triangle. Its square root, estimated below, is the same for the reserve as for the ultimate. In the bracket, the 1/\hat C_{ik} term estimates the future random variation (process) and the 1/\sum C_{jk} term the error in the \hat f_k (estimation). \hat C_{ik} is the triangle completed by the chain ladder, with \hat C_{i,I+1-i} = C_{i,I+1-i}.

\begin{aligned}
\text{s.e.}(\hat R_i)^2 &= \hat C_{iI}^2 \sum_{k=I+1-i}^{I-1} \frac{\hat\alpha_k^2}{\hat f_k^2} \\
&\quad \times \left(\frac{1}{\hat C_{ik}} + \frac{1}{\sum_{j=1}^{I-k} C_{jk}}\right)
\end{aligned}

• Standard error of the total. The \hat R_i are positively correlated because they share the \hat f_k, so their squared standard errors do not simply add:

\begin{aligned}
\text{s.e.}(\hat R)^2 &= \sum_{i=2}^{I}\Bigg\{\text{s.e.}(\hat R_i)^2 + \hat C_{iI}\sum_{j=i+1}^{I}\hat C_{jI} \\
&\qquad \times \sum_{k=I+1-i}^{I-1}\frac{2\hat\alpha_k^2/\hat f_k^2}{\sum_{n=1}^{I-k} C_{nk}}\Bigg\}
\end{aligned}

• Confidence limits. When the volume is large, a Normal approximation gives \hat R_i \pm z\,\text{s.e.}. When \text{s.e.}(\hat R_i) exceeds about 50\% of \hat R_i, Mack recommends a Lognormal with the same mean and variance. That fit has \sigma_i^2 = \ln\!\big(1 + \text{s.e.}^2/\hat R_i^2\big), and its z-percentile is \hat R_i\exp(z\sigma_i - \sigma_i^2/2). Adding C_{i,I+1-i} turns reserve limits into limits for the ultimate.
• Allocating a total percentile. To give the total a chosen percentile, Mack sets that percentile of \hat R and finds the one z at which the accident years' Lognormal percentiles add up to it. Every year then sits at the same confidence level.

Estimating the Factors and Their Variances From a Paid Triangle Cumulative paid claims ($000):

AY	k=1	k=2	k=3	k=4
1	1{,}000	2{,}500	3{,}000	3{,}150
2	1{,}500	2{,}400	3{,}120
3	800	2{,}400
4	1{,}200

Development is not believed finished at k = 4, but no tail is projected. Estimate \hat f_k and \hat\alpha_k^2 for k = 1, 2, 3, and the chain ladder reserves.

Answer The individual factors are 2.5, 1.6, 3.0 for k=1; 1.2, 1.3 for k=2; and 1.05 for k=3.

\begin{align*}
\hat f_1 &= \frac{2{,}500 + 2{,}400 + 2{,}400}{1{,}000 + 1{,}500 + 800} \\
&= \frac{7{,}300}{3{,}300} \\
&= 2.21212 \\
\hat f_2 &= \frac{3{,}000 + 3{,}120}{2{,}500 + 2{,}400} \\
&= 1.24898 \\
\hat f_3 &= \frac{3{,}150}{3{,}000} \\
&= 1.05000
\end{align*}

\hat\alpha_1^2 has I - k - 1 = 2 degrees of freedom and \hat\alpha_2^2 has 1:

\begin{align*}
\hat\alpha_1^2 &= \tfrac{1}{2}\big[1{,}000(0.28788)^2 + 1{,}500(0.61212)^2 \\
&\qquad + 800(0.78788)^2\big] \\
&= \tfrac{1}{2}(82.87 + 562.04 + 496.60) \\
&= 570.76 \\
\hat\alpha_2^2 &= 2{,}500(0.04898)^2 + 2{,}400(0.05102)^2 \\
&= 5.997 + 6.247 \\
&= 12.245
\end{align*}

\hat\alpha_3^2 cannot be estimated from one factor, so it is extrapolated:

\begin{align*}
\hat\alpha_3^2 &= \min\left(\frac{12.245^2}{570.76},\ 570.76,\ 12.245\right) \\
&= 0.2627
\end{align*}

Completing the triangle gives ultimates 3{,}276.0 for AY 2, 3{,}147.4 for AY 3 and 3{,}481.2 for AY 4. The reserves are \hat R_2 = 156.0, \hat R_3 = 747.4 and \hat R_4 = 2{,}281.2, a total of 3{,}184.7.

\hat\alpha_k^2 falls fast with k because the early factors scatter widely (1.6 to 3.0) around \hat f_1 and the later ones barely at all.

Standard Error of One Year and of the Total Continue the triangle above. Compute \text{s.e.}(\hat R_4), split it into its process and estimation parts, and compute \text{s.e.}(\hat R) for all years combined. Take \text{s.e.}(\hat R_2)^2 = 1{,}672 and \text{s.e.}(\hat R_3)^2 = 49{,}844 as given.

Answer AY 4 is at k = 1, so the sum runs over k = 1, 2, 3, with \hat C_{41} = 1{,}200, \hat C_{42} = 2{,}654.5 and \hat C_{43} = 3{,}315.5:

k	\hat\alpha_k^2/\hat f_k^2	1/\hat C_{4k}	1/\sum_j C_{jk}	Term
1	116.64	1/1{,}200	1/3{,}300	0.132541
2	7.850	1/2{,}654.5	1/4{,}900	0.004559
3	0.2383	1/3{,}315.5	1/3{,}000	0.000151
Sum				0.137252

\begin{align*}
\text{s.e.}(\hat R_4)^2 &= 3{,}481.25^2 \times 0.137252 \\
&\approx 1{,}663{,}360 \\
\text{s.e.}(\hat R_4) &= 1{,}289.7
\end{align*}

That is 56.5\% of \hat R_4. Summing only the 1/\hat C_{4k} parts gives the process standard deviation, 1{,}102; the 1/\sum C_{jk} parts give the estimation standard deviation, 670 (1{,}102^2 + 670^2 \approx 1{,}289.7^2). Almost all of it arises at k = 1, the widely scattered first factor applied to one year's 1{,}200.

The total. The cross terms use 2\hat\alpha_k^2/\hat f_k^2 \div \sum_n C_{nk}, which is 0.0032039 for k=2 and 0.00015885 for k=3:

\begin{align*}
i = 2&: \ 3{,}276.0 \times 6{,}628.68 \times 0.00015885 \\
&= 3{,}450 \\
i = 3&: \ 3{,}147.43 \times 3{,}481.25 \times 0.0033628 \\
&= 36{,}846
\end{align*}

\begin{align*}
\text{s.e.}(\hat R)^2 &= 1{,}672 + 49{,}844 + 1{,}663{,}360 \\
&\quad + 3{,}450 + 36{,}846 \\
&= 1{,}714{,}876 + 40{,}296 \\
&= 1{,}755{,}172 \\
\text{s.e.}(\hat R) &= 1{,}324.8
\end{align*}

That is 41.6\% of the 3{,}184.7 total. Treating the years as independent would have given \sqrt{1{,}714{,}876} = 1{,}309.5; the shared factors add the covariance.

Normal Versus Lognormal Confidence Limits From the example above, \hat R_4 = 2{,}281.2 with \text{s.e.} = 1{,}289.7 and C_{41} = 1{,}200. Find the 10th and 90th percentiles of R_4 under a Normal and under a Lognormal approximation (z_{0.90} = 1.28). Then give the lognormal interval for the ultimate.

Answer Normal:

\begin{align*}
2{,}281.2 \pm 1.28 \times 1{,}289.7 &= 2{,}281.2 \pm 1{,}650.8 \\
&= (630.4,\ 3{,}932.0)
\end{align*}

Lognormal: the coefficient of variation is 1{,}289.7/2{,}281.2 = 0.5654.

\begin{align*}
\sigma^2 &= \ln(1 + 0.5654^2) \\
&= 0.2774 \\
\sigma &= 0.5267 \\
\text{90th} &= 2{,}281.2\,e^{1.28(0.5267) - 0.2774/2} \\
&= 2{,}281.2\,e^{0.5355} \\
&\approx 3{,}897 \\
\text{10th} &= 2{,}281.2\,e^{-1.28(0.5267) - 0.2774/2} \\
&= 2{,}281.2\,e^{-0.8129} \\
&\approx 1{,}012
\end{align*}

The Lognormal interval (1{,}012,\ 3{,}897) sits higher at the bottom than the Normal one. At the top it is slightly lower, which is Mack's point that neither approximation's 90th percentile is always the larger one.

With the standard error above 50\% of the reserve, Mack's advice is the Lognormal. A symmetric Normal 95% interval, \hat R_4 \pm 2\,\text{s.e.}, would already have a negative lower limit, -298.

Shifting by the 1{,}200 already paid gives an 80% interval for AY 4's ultimate of (2{,}212,\ 5{,}097).