GLM Loss Reserving

GLM loss reserving fits a generalized linear model to the cells of a claims triangle. Each cell is treated as an independent observation from an exponential dispersion family distribution whose mean is linked to accident-period and development-period covariates, so one fitted model gives the reserve, its prediction error and the diagnostics that test it. Taylor and McGuire (2016) show that the chain ladder is itself such a model.

h(\mu_{kj}) = x_{kj}^{T}\beta

\mathrm{Var}[Y_{kj}] = \frac{\phi\,V(\mu_{kj})}{w_{kj}}

• Y_{kj} is the incremental amount of accident period k in development period j, with mean \mu_{kj}. h is the link function, x_{kj} the covariates, \beta the parameters, V the variance function (V(\mu) = \mu^p for a Tweedie member, p = 1 for the over-dispersed Poisson), \phi the dispersion parameter and w_{kj} a known weight.
• Two GLMs reproduce the chain ladder. The ODP cross-classified model has Y_{kj} \sim \mathrm{ODP}(\alpha_k\beta_j, \phi) with \sum_j \beta_j = 1: a log link, with accident and development year as categorical covariates. The ODP Mack model (Mack's model given an ODP distribution) takes the response \hat f_{kj} - 1 = Y_{k,j+1}/X_{kj}, an identity link, development year as the only covariate, and weight X_{kj}. Both give the chain ladder's forecasts, which Taylor (2011) showed are minimum variance unbiased under these models (for the cross-classified model once corrected for bias). They are different models, though, and give different prediction errors.
• Marginal sums. The cross-classified model's likelihood equations set every row total and every column total of fitted values equal to the observed total. That is why it reproduces the chain ladder, and why actual-versus-expected by accident or development year is always exactly 100%. A missing effect has to be found in the calendar direction, the cell-by-cell heat map and the residuals.
• Choosing the model means choosing the error distribution (a larger Tweedie p for a heavier tail), the covariates, the link (a multiplicative structure implies a log link) and the weights. A zero weight deletes a cell, so using only the latest diagonals or dropping an outlier is still a GLM fit. Weights inversely proportional to the residual variance correct heteroscedasticity.
• Prediction error. Its mean square (MSEP) is the expected squared parameter error plus the expected squared process error, the two being independent. It is estimated by the delta method from the fitted parameters' covariance matrix, or by a semi-parametric or parametric bootstrap, which gives the whole distribution. Model error lies outside the model and can be the largest part.
• Diagnostics use standardized deviance residuals plotted against accident, development and calendar period, spread plots, actual-and-expected plots, a 2-D heat map, a P-P plot and a histogram. The monograph's order is to fit a sensible link and distribution, fit the main effects, check for gross violations, improve the fit of the cell means, then review the distribution in detail.
• Extending the model is the point of the framework. When the diagnostics reject the chain ladder, the GLM is extended rather than abandoned. The monograph's Chapter 7, beyond the Exam 7 assignment, replaces the accident and development parameters with parsimonious curves and adds calendar-period (superimposed inflation) and interaction terms, comparing the models by AIC, BIC and prediction error.

The Chain Ladder as an ODP Cross-Classified GLM Incremental paid losses ($000):

AY	Dev 1	Dev 2	Dev 3
1	100	60	20
2	120	66
3	130

(a) Compute the chain ladder reserve with volume-weighted factors. (b) Fit the ODP cross-classified model by solving its marginal-sum equations with \beta_1 + \beta_2 + \beta_3 = 1, and forecast the same reserve.

Answer (a) Chain ladder. The cumulative rows are 100, 160, 180; 120, 186; and 130.

\begin{align*}
\hat f_1 &= \frac{160 + 186}{100 + 120} \\
&= 1.5727 \\
\hat f_2 &= \frac{180}{160} \\
&= 1.125 \\
\hat R_2 &= 186 \times (1.125 - 1) \\
&= 23.25 \\
\hat R_3 &= 130 \times (1.5727 \times 1.125 - 1) \\
&= 100.01
\end{align*}

The total reserve is 123.26.

(b) Marginal sums. Solve row 1, column 3, row 2, column 2, then row 3, using the constraint each time:

\begin{align*}
\hat\alpha_1 &= 100 + 60 + 20 \\
&= 180 \\
\hat\beta_3 &= \frac{20}{180} \\
&= 0.11111 \\
\hat\alpha_2 &= \frac{120 + 66}{1 - 0.11111} \\
&= 209.25 \\
\hat\beta_2 &= \frac{60 + 66}{180 + 209.25} \\
&= 0.32370 \\
\hat\beta_1 &= 1 - 0.32370 - 0.11111 \\
&= 0.56519 \\
\hat\alpha_3 &= \frac{130}{0.56519} \\
&= 230.01
\end{align*}

The forecasts are \hat Y_{23} = 209.25 \times 0.11111 = 23.25, \hat Y_{32} = 230.01 \times 0.32370 = 74.45 and \hat Y_{33} = 230.01 \times 0.11111 = 25.56. So \hat R_2 = 23.25 and \hat R_3 = 100.01, exactly the chain ladder's.

With \sum_j\beta_j = 1, each \hat\alpha_k is the chain ladder ultimate (180, 209.25, 230.01) and \hat\beta_j is the incremental emergence pattern. The factors are recovered as \hat f_1 = (\hat\beta_1 + \hat\beta_2)/\hat\beta_1 = 0.88889/0.56519 = 1.5727.

Fitting Only the Latest Two Diagonals with Zero Weights Cumulative paid losses ($000):

AY	Dev 1	Dev 2	Dev 3	Dev 4
1	1{,}000	1{,}500	1{,}650	1{,}700
2	1{,}100	1{,}700	1{,}850
3	1{,}200	1{,}900
4	1{,}300

Fit the ODP Mack GLM using only the latest m = 2 experience years, with weights w_{kj} = X_{kj}\,I(3 \le k + j \le 4). Compare AY 4's reserve with the all-year fit.

Answer Each response is \hat f_{kj} - 1. The GLM estimate for each development year is the weighted mean of its responses, and weight times response is just Y_{k,j+1}:

Response	k + j	Value	Weight
AY 1, dev 1	2	0.5000	0
AY 2, dev 1	3	0.5455	1{,}100
AY 3, dev 1	4	0.5833	1{,}200
AY 1, dev 2	3	0.1000	1{,}500
AY 2, dev 2	4	0.0882	1{,}700
AY 1, dev 3	4	0.0303	1{,}650

\begin{align*}
\hat f_1 &= 1 + \frac{600 + 700}{1{,}100 + 1{,}200} \\
&= 1.56522 \\
\hat f_2 &= 1 + \frac{150 + 150}{1{,}500 + 1{,}700} \\
&= 1.09375 \\
\hat f_3 &= 1 + \frac{50}{1{,}650} \\
&= 1.03030 \\
\hat R_4 &= 1{,}300 \times (1.56522 \times 1.09375 \times 1.03030 - 1) \\
&= 993.0
\end{align*}

The all-year fit gives \hat f_1 = 5{,}100/3{,}300 = 1.54545 and \hat R_4 = 964.0. The zero weight on AY 1's 1.500 link raises AY 4's reserve by 29.0 (3.0\%), because the first link has been rising. The later factors are unchanged here, since every cell they use lies on the last two diagonals.

What the Marginal Sums Hide An ODP cross-classified model is fitted to a triangle of incremental losses ($000). The actual values and the fitted values are:

AY	Actual 1	Actual 2	Actual 3	Actual 4
1	500	300	150	58
2	500	300	173
3	500	345
4	575

AY	Fitted 1	Fitted 2	Fitted 3	Fitted 4
1	484.9	305.5	159.6	58.0
2	496.7	312.9	163.4
3	518.4	326.6
4	575.0

Compute actual/expected by accident year, development year and calendar period, and say what the model is missing.

Answer By accident year and development year every ratio is 100\%. Row 2, for example, is 973/973.0 and column 2 is 945/945.0. The likelihood equations force this, so these plots say nothing.

By calendar period (the diagonals k + j - 1):

\begin{align*}
\text{CY 1} &= \frac{500}{484.9} \\
&= 103.1\% \\
\text{CY 2} &= \frac{300 + 500}{305.5 + 496.7} \\
&= 99.7\% \\
\text{CY 3} &= \frac{150 + 300 + 500}{159.6 + 312.9 + 518.4} \\
&= 95.9\% \\
\text{CY 4} &= \frac{58 + 173 + 345 + 575}{58.0 + 163.4 + 326.6 + 575.0} \\
&= 102.5\%
\end{align*}

The heat map tells the same story cell by cell. Every cell of calendar period 3 is below 100\% (94.0\%, 95.9\%, 96.5\%). The two interior cells of period 4 are above it (105.9\%, 105.6\%); its corner cells are fitted exactly because each is alone in its row or column.

The colours run along whole diagonals rather than clustering in accident and development groups, so the missing effect is a calendar-period one: here a jump in the latest period, such as a court precedent. The chain ladder has smeared it across the row and column parameters. The remedy is a calendar term with a simple structure, not a free parameter per diagonal (see Superimposed Inflation).