Example 11: the reserve, with a distribution¶

A ten-year paid triangle, one reserve number every actuary reports, and the question that number cannot answer on its own — how wrong could it be? This page carries the Taylor–Ashe triangle from the chain-ladder point estimate to the full predictive distribution of the unpaid claims, checks the model has the right to run, reconciles the two error models, and turns the reserve into capital in risksim. Development and the deterministic methods come from actuarialpy (re-exported through reservingmodels); the distribution and diagnostics are reservingmodels; the capital is risksim. Every number on this page is the output of this exact fixed-seed run, pinned by a regression test in the reservingmodels suite.

The triangle, and the number everyone reports¶

import reservingmodels as rv

triangle = rv.datasets.taylor_ashe()
cl = rv.ChainLadder.fit(triangle)
cl.mack_standard_errors(triangle).loc["Total"]

reserve	Mack SE	CV
18,680,856	2,447,095	13.1%

The chain ladder develops the triangle to an 18.7m reserve; Mack (1993) puts a distribution-free standard error on it without simulating anything. That is already more than a point estimate — but a single number and its SD do not tell you the shape of the tail, and the tail is where a reserve fails.

The whole distribution, not just its width¶

boot = rv.BootstrapODP.fit(triangle)
dist = boot.reserve_distribution(size=100_000, rng=42)

boot.dispersion        # 52,601
dist.point_reserve     # 18,680,856  — reproduces the chain ladder exactly
dist.mean()            # 18,909,342  — the bootstrap mean, +1.2% (ratio bias)
dist.prediction_error()  # 2,996,849

The bootstrap fits the over-dispersed Poisson whose mean surface is the chain ladder, so point_reserve lands on 18,680,856 to the dollar; the estimated dispersion \(\phi = 52{,}601\) matches the published fit. Report the point estimate, not the bootstrap mean — the 1.2% gap is the known upward bias of a product of ratio estimators, not signal. What the bootstrap adds is the quantiles:

for q in (0.50, 0.75, 0.95, 0.99, 0.995):
    dist.quantile(q)
dist.tvar(0.99)

quantile	reserve
50% (median)	18,713,157
75%	20,770,994
95%	24,138,798
99%	26,886,153
99.5%	27,949,893
TVaR 99%	28,458,966

The 99th percentile sits 8.2m — a third — above the booked reserve, and TVaR carries the average of the worst 1%. A 13% CV on the point estimate sounds mild; the 99.5% quantile is what a capital calculation actually consumes.

Per-origin, where the risk actually sits¶

dist.to_frame()

origin	reserve	se	q95	q99
…
7	2,194,440	496,466	3,063,438	3,685,963
8	3,952,518	795,231	5,343,556	6,297,848
9	4,322,826	1,058,224	6,182,766	7,605,661
10	4,708,405	2,035,330	8,239,562	11,659,798
Total	18,909,342	2,996,849	24,138,798	27,949,893

The uncertainty is not spread evenly. The youngest origin (10) has the second largest reserve but by far the largest relative error — its q99 is nearly 2.5× its mean — because it is developed off a single observation and the bootstrap refits the whole pattern each replicate. The total SE (2,996,849) is far below the sum of the origin SEs: the origins are estimated through one shared development pattern, so their errors are correlated, and the bootstrap captures that automatically.

Is the bootstrap even entitled to run?¶

rv.residual_summary(triangle)
rv.calendar_year_effects(triangle)

n	mean	std	min	max
55	0.60	187.3	−403.8	533.2

The 55 Pearson residuals are centred near zero, and their spread sits in the neighbourhood of \(\sqrt{\phi}\approx 229\) — the ODP variance assumption is credible. calendar_year_effects shows no monotone drift across diagonals: the largest-magnitude diagonal means fall on the one-, two-, and three-cell diagonals, where a single residual dominates, not a systematic calendar-year (inflation) trend that would break the single-pattern assumption. When that check fails, the bootstrap’s tail is not trustworthy no matter how tidy the quantiles look — which is the point of running it.

Mack vs the bootstrap¶

mack = cl.mack_standard_errors(triangle).loc["Total"]
proc = (boot.dispersion * boot.point_reserve) ** 0.5     # process SD

method	reserve	SE	CV
chain ladder / Mack	18,680,856	2,447,095	13.1%
ODP bootstrap	18,909,342	2,996,849	15.9%

The bootstrap SE runs 22% above Mack’s, and that is a feature, not a discrepancy: Mack assumes \(\mathrm{Var}(C_{i,k+1}\mid C_{i,k})\propto C_{i,k}\), the ODP assumes \(\mathrm{Var}\propto\) mean, and the two answer the same question under different variance laws. The bootstrap SE decomposes cleanly — estimation error 2,828,156 and process error \(\sqrt{\phi\cdot R}=991{,}281\) combine as \(\sqrt{2{,}828{,}156^2 + 991{,}281^2} = 2{,}996{,}849\) — so nothing is double-counted; the width is exactly estimation risk plus process risk. Having both numbers is the honest position: they bracket the reserve risk under two defensible models.

The reserve as capital¶

A fitted bootstrap is a risksim component — one .sample draw is one simulated unpaid total — so the reserve joins a capital model with no dependency between the packages:

import risksim as rs

port = rs.Portfolio([rs.PortfolioItem("reserve_risk", boot)])
sim = port.simulate(100_000, rng=7)
rs.metrics.tvar(sim.gross_losses, 0.99)      # 28,318,860

booked reserve	TVaR(99%)	capital margin
18,680,856	28,318,860	9,638,004

That last row is the whole point of the exercise stated in currency: you book 18.7m, but holding to a 99% tail-value standard needs roughly 9.6m more. The portfolio TVaR reproduces the standalone distribution’s tail because the seam is transparent — risksim simply consumed the reserve model — and from here a second reserving segment aggregates in with a rank correlation exactly as Example 9: two lines, one tail does it. The reserve stops being a number and becomes a distribution you can hold capital against.