Data Diagnostic Analysis

Study guides: Exam 5, Exam 7, Exam GI-101

Data diagnostic analysis is the examination of claims and exposure data before a reserving method is applied. The data are arranged as triangles of ratios and averages, or as residuals from a fitted model, to detect changes the method would otherwise absorb silently: case reserving, settlement speed, mix, inflation and data errors. Its purpose is to test the method's assumptions and to decide how the data must be adjusted.

$\bar O_{w,d} = \frac{L^{\text{rep}}_{w,d} - L^{\text{paid}}_{w,d}}{N^{\text{open}}_{w,d}}$

$r_{w,d} = \frac{q_{w,d} - m_{w,d}}{\sqrt{m_{w,d}}}$

The first block is the average case outstanding for accident year $w$ at age $d$ : reported less paid losses, per open claim. The second is the unscaled Pearson residual of a fitted stochastic model, with actual incremental $q_{w,d}$ and fitted mean $m_{w,d}$ .
Triangle diagnostics (Exam 5, Friedland). The standard triangles are:
- reported, closed and open claim counts, and closed-to-reported ratios (closure rates);
- paid, reported and average-case-outstanding severities;
- paid-to-reported loss ratios;
- losses to premium (claim ratios) and frequencies.
Reading down a column compares years at the same age, which shows trends and mix. Reading along a diagonal isolates a calendar period, which shows changes in claims handling.
What the common changes look like.
- Case strengthening: average case outstanding rises faster than trend on the latest diagonals, and paid-to-reported falls.
- Faster settlement: closure rates and paid-to-reported rise together.
- A mix change: shows up down the columns, not along the diagonals.
Each of these breaks a chain ladder assumption. The usual fixes are a Berquist-Sherman restatement, or separating large and catastrophe losses.
Model diagnostics (Exam 7, Shapland).
- Plot residuals against development period, accident period, calendar period and fitted value. They should be patternless and have a common variance.
- Test normality by plotting the residuals against the normal inverse, with a P-value, $R^2$ , AIC and BIC. The ODP bootstrap does not need normal residuals, but the test compares parameter sets and gauges skewness.
- Find outliers with a box-whisker plot.
- Where residual variance differs by development group, rescale with a hetero-adjustment factor $h_i = \mathrm{sd}(\text{all } r)/\mathrm{sd}(r \in \text{group } i)$ .
Data issues to adjust for (Data Issues): negative incremental values, missing cells, outliers, a partial first or last diagonal, exposure changes and tail factors. Venter's tests for the age-to-age factor assumptions add linearity, stability over accident years, correlation between columns, and high or low diagonals.

Spotting Case Strengthening Average case outstanding ($) at year-end 2024, with severity trend of about $5\%$ a year:

AY 12 24 36

2021 $7{,}600$ $11{,}400$ $15{,}000$

2022 $8{,}000$ $12{,}000$ $16{,}550$

2023 $8{,}400$ $13{,}250$

2024 $9{,}300$

Closure rates at each age have been stable. What is happening?
Answer Read down each column, year over year:
 $\begin{align*} \text{12 mo: } 8{,}400/8{,}000 &= 1.050 \\ 9{,}300/8{,}400 &= 1.107 \\ \text{24 mo: } 12{,}000/11{,}400 &= 1.053 \\ 13{,}250/12{,}000 &= 1.104 \\ \text{36 mo: } 16{,}550/15{,}000 &= 1.103 \end{align*}$ 
Every cell on the 2024 diagonal is about $10\%$ above the prior year's, against roughly $5\%$ in earlier years. A calendar-year jump across all ages, with stable closure rates, points to case strengthening in 2024, not a change in settlement or a severity shock in one accident year.

The consequence is that the reported chain ladder will apply historical factors, earned on weaker reserves, to a stronger diagonal and overstate ultimates. The fix is to restate the earlier diagonals' average case outstanding to the 2024 adequacy level, detrending the 2024 values at $5\%$ , and rebuild the reported triangle before projecting.

AY	12	24	36
2021	$7{,}600$	$11{,}400$	$15{,}000$
2022	$8{,}000$	$12{,}000$	$16{,}550$
2023	$8{,}400$	$13{,}250$
2024	$9{,}300$

Residuals with Unequal Variance An ODP bootstrap's residuals have standard deviation $1.60$ in development periods 1–3 and $0.80$ in periods 4–9. The standard deviation of all residuals is $1.20$ .

Compute the hetero-adjustment factors and describe how a resampled residual is used.
Answer
 $\begin{align*} h_{1\text{-}3} &= 1.20/1.60 \\ &= 0.75 \\ h_{4\text{-}9} &= 1.20/0.80 \\ &= 1.50 \end{align*}$ 
Each residual is multiplied by its group's $h$ , so all groups share a standard deviation of $1.20$ and can be pooled for resampling.

When a pooled residual $r^*$ is placed back in a cell, it is divided by that cell's factor: $q^* = m + (r^*/h_i)\sqrt{m}$ . Drawn into period 2, a residual of $1.2$ becomes $1.2/0.75 = 1.6$ standard units. Drawn into period 6, it becomes $1.2/1.5 = 0.8$ . The pseudo-triangles then reproduce the variances actually seen in the data.

The two factors are extra parameters, so they reduce the degrees of freedom used for the scale parameter. The same factors must also scale the process variance of future cells.