This dataset, discussed in Gesmann & Morris (2020), contains cumulative insurance loss payments over the course of ten years.
Format
A data frame of 55 observations containing information on the following 4 variables.
- AY
Origin year of the insurance (1991 to 2000)
- dev
Deviation from the origin year in months
- cum
Cumulative loss payments
- premium
Achieved premiums for the given origin year
Source
Gesmann M. & Morris J. (2020). Hierarchical Compartmental Reserving Models. CAS Research Papers.
Examples
# \dontrun{
# non-linear model to predict cumulative loss payments
fit_loss <- brm(
bf(cum ~ ult * (1 - exp(-(dev/theta)^omega)),
ult ~ 1 + (1|AY), omega ~ 1, theta ~ 1,
nl = TRUE),
data = loss, family = gaussian(),
prior = c(
prior(normal(5000, 1000), nlpar = "ult"),
prior(normal(1, 2), nlpar = "omega"),
prior(normal(45, 10), nlpar = "theta")
),
control = list(adapt_delta = 0.9)
)
#> Compiling Stan program...
#> Start sampling
#>
#> SAMPLING FOR MODEL 'anon_model' NOW (CHAIN 1).
#> Chain 1:
#> Chain 1: Gradient evaluation took 3.3e-05 seconds
#> Chain 1: 1000 transitions using 10 leapfrog steps per transition would take 0.33 seconds.
#> Chain 1: Adjust your expectations accordingly!
#> Chain 1:
#> Chain 1:
#> Chain 1: Iteration: 1 / 2000 [ 0%] (Warmup)
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#> Chain 1:
#> Chain 1: Elapsed Time: 2.05 seconds (Warm-up)
#> Chain 1: 0.986 seconds (Sampling)
#> Chain 1: 3.036 seconds (Total)
#> Chain 1:
#>
#> SAMPLING FOR MODEL 'anon_model' NOW (CHAIN 2).
#> Chain 2: Rejecting initial value:
#> Chain 2: Error evaluating the log probability at the initial value.
#> Chain 2: Exception: normal_lpdf: Location parameter[1] is nan, but must be finite! (in 'anon_model', line 68, column 4 to column 41)
#> Chain 2:
#> Chain 2: Gradient evaluation took 2.1e-05 seconds
#> Chain 2: 1000 transitions using 10 leapfrog steps per transition would take 0.21 seconds.
#> Chain 2: Adjust your expectations accordingly!
#> Chain 2:
#> Chain 2:
#> Chain 2: Iteration: 1 / 2000 [ 0%] (Warmup)
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#> Chain 2:
#> Chain 2: Elapsed Time: 2.095 seconds (Warm-up)
#> Chain 2: 1.143 seconds (Sampling)
#> Chain 2: 3.238 seconds (Total)
#> Chain 2:
#>
#> SAMPLING FOR MODEL 'anon_model' NOW (CHAIN 3).
#> Chain 3: Rejecting initial value:
#> Chain 3: Error evaluating the log probability at the initial value.
#> Chain 3: Exception: normal_lpdf: Location parameter[1] is nan, but must be finite! (in 'anon_model', line 68, column 4 to column 41)
#> Chain 3: Rejecting initial value:
#> Chain 3: Error evaluating the log probability at the initial value.
#> Chain 3: Exception: normal_lpdf: Location parameter[1] is nan, but must be finite! (in 'anon_model', line 68, column 4 to column 41)
#> Chain 3:
#> Chain 3: Gradient evaluation took 2.1e-05 seconds
#> Chain 3: 1000 transitions using 10 leapfrog steps per transition would take 0.21 seconds.
#> Chain 3: Adjust your expectations accordingly!
#> Chain 3:
#> Chain 3:
#> Chain 3: Iteration: 1 / 2000 [ 0%] (Warmup)
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#> Chain 3:
#> Chain 3: Elapsed Time: 1.954 seconds (Warm-up)
#> Chain 3: 0.936 seconds (Sampling)
#> Chain 3: 2.89 seconds (Total)
#> Chain 3:
#>
#> SAMPLING FOR MODEL 'anon_model' NOW (CHAIN 4).
#> Chain 4: Rejecting initial value:
#> Chain 4: Error evaluating the log probability at the initial value.
#> Chain 4: Exception: normal_lpdf: Location parameter[1] is nan, but must be finite! (in 'anon_model', line 68, column 4 to column 41)
#> Chain 4: Rejecting initial value:
#> Chain 4: Error evaluating the log probability at the initial value.
#> Chain 4: Exception: normal_lpdf: Location parameter[1] is nan, but must be finite! (in 'anon_model', line 68, column 4 to column 41)
#> Chain 4:
#> Chain 4: Gradient evaluation took 2e-05 seconds
#> Chain 4: 1000 transitions using 10 leapfrog steps per transition would take 0.2 seconds.
#> Chain 4: Adjust your expectations accordingly!
#> Chain 4:
#> Chain 4:
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#> Chain 4:
#> Chain 4: Elapsed Time: 2.186 seconds (Warm-up)
#> Chain 4: 1.213 seconds (Sampling)
#> Chain 4: 3.399 seconds (Total)
#> Chain 4:
# basic summaries
summary(fit_loss)
#> Family: gaussian
#> Links: mu = identity
#> Formula: cum ~ ult * (1 - exp(-(dev/theta)^omega))
#> ult ~ 1 + (1 | AY)
#> omega ~ 1
#> theta ~ 1
#> Data: loss (Number of observations: 55)
#> Draws: 4 chains, each with iter = 2000; warmup = 1000; thin = 1;
#> total post-warmup draws = 4000
#>
#> Multilevel Hyperparameters:
#> ~AY (Number of levels: 10)
#> Estimate Est.Error l-95% CI u-95% CI Rhat Bulk_ESS Tail_ESS
#> sd(ult_Intercept) 732.58 222.78 433.87 1284.77 1.00 1162 2052
#>
#> Regression Coefficients:
#> Estimate Est.Error l-95% CI u-95% CI Rhat Bulk_ESS Tail_ESS
#> ult_Intercept 5300.74 294.32 4742.62 5905.04 1.00 1062 1757
#> omega_Intercept 1.34 0.05 1.24 1.44 1.00 2324 2280
#> theta_Intercept 46.24 2.18 42.39 50.92 1.00 2083 2139
#>
#> Further Distributional Parameters:
#> Estimate Est.Error l-95% CI u-95% CI Rhat Bulk_ESS Tail_ESS
#> sigma 140.41 15.94 113.81 176.04 1.00 2569 2446
#>
#> Draws were sampled using sampling(NUTS). For each parameter, Bulk_ESS
#> and Tail_ESS are effective sample size measures, and Rhat is the potential
#> scale reduction factor on split chains (at convergence, Rhat = 1).
conditional_effects(fit_loss)
# plot predictions per origin year
conditions <- data.frame(AY = unique(loss$AY))
rownames(conditions) <- unique(loss$AY)
me_loss <- conditional_effects(
fit_loss, conditions = conditions,
re_formula = NULL, method = "predict"
)
plot(me_loss, ncol = 5, points = TRUE)
# }