Actuarial Notes

Over-Dispersed Negative Binomial Model

Over-Dispersed Negative Binomial Model (ODNB) is Verrall's (2000) recursive form of the stochastic chain ladder. Each incremental loss is modeled given the cumulative loss to date: its mean is (\lambda_j - 1)D_{i,j-1}, the chain ladder's own projection, and its variance is proportional to that mean. It gives the same predictive distribution as the over-dispersed Poisson model, but it is written in terms of the development factors \lambda_j. That is what lets Verrall (2007) put expert-opinion priors directly on the factors.

E[C_{ij} \mid D_{i,j-1}] = (\lambda_j - 1)\,D_{i,j-1}

\mathrm{Var}[C_{ij} \mid D_{i,j-1}] = \phi\,\lambda_j(\lambda_j - 1)\,D_{i,j-1}

  • C_{ij} is the incremental and D_{i,j-1} the previous cumulative loss of accident year i, \lambda_j the development factor into year j, and \phi the over-dispersion parameter. Adding D_{i,j-1} gives the cumulative form: D_{ij} has mean \lambda_j D_{i,j-1} and the same variance. Written that way, the connection with the chain ladder is immediately apparent from the mean. "Over-dispersed" again means Y = \phi X for a negative binomial X, with a quasi-likelihood so that losses need not be integers.
  • Same answers as the ODP. The reserve estimates equal the chain ladder's, and the predictive distribution is exactly that of the ODP model, whether it is written for incremental or cumulative losses. The same positivity constraint applies: every column sum must be positive, because \lambda_j < 1 would make the variance negative. For that case Verrall notes a model close to Mack's that keeps the mean but uses a normal distribution with variance \phi_j D_{i,j-1}, at the cost of another set of variance parameters.
  • Priors on the factors (Verrall's Section 4, after Verrall and England 2005). Each row gets its own factor \lambda_{i,j} with a prior distribution:
    • Reproduce the chain ladder. Set \lambda_{i,j} = \lambda_j within each column and give the \lambda_j vague priors. This yields prediction errors and a predictive distribution for the chain ladder itself.
    • Intervene in some rows. If a change in settlement makes an older factor wrong for recent rows, those rows share a separate factor. Its prior mean is the expert's factor, and its variance W is set to reflect the strength of the opinion.
    • Use only recent years. Give the most recent rows factors \lambda_j and the earlier rows factors \lambda^*_j, both with large-variance priors. That is a stochastic version of a latest-years volume-weighted average.
  • Why it matters. Changing a factor by hand abandons the stochastic model's estimates, and it is not clear what estimation error an inserted parameter should carry. A Bayesian prior answers both, and MCMC simulation in winBUGS produces the posterior. The prior's form (gamma, lognormal) is chosen so the numerical procedures work well. The same recursive form, running down the rows instead of across the columns, carries the Bayesian Bornhuetter-Ferguson model.

Mean and Variance of the Next Development An accident year has cumulative losses D_{i,2} = 4{,}000 ($000s). The next factor is \lambda_3 = 1.25 and \phi = 60. Find the mean, variance and CoV of C_{i3} and the mean of D_{i3}. What goes wrong if \lambda_3 = 0.98?

Answer

\begin{align*}
E[C_{i3}] &= (1.25 - 1)(4{,}000) \\
&= 1{,}000 \\
\mathrm{Var}[C_{i3}] &= 60(1.25)(0.25)(4{,}000) \\
&= 75{,}000 \\
\mathrm{SD}[C_{i3}] &= 273.9
\end{align*}

The CoV is 273.9/1{,}000 = 27.4\%. D_{i3} has mean 1.25 \times 4{,}000 = 5{,}000 and the same variance, since D_{i2} is given.

With \lambda_3 = 0.98 the variance would be 60(0.98)(-0.02)(4{,}000) = -4{,}704, which is impossible. A column of negative development cannot be modeled this way. The normal version keeps the mean, -80, and uses \phi_3 D_{i2}. To match this example's variance at \lambda_3 = 1.25 it would need \phi_3 = 60 \times 1.25 \times 0.25 = 18.75.

An Expert Prior on One Development Factor In Verrall's Taylor and Ashe illustration, information implies that the second development factor, from column 2 to column 3, should be 1.5 for rows 7 to 10. The chain ladder estimate is 1.7473.

(a) His winBUGS program puts a gamma prior on \lambda - 1. Find its shape and rate for a prior mean of 1.5 and a standard deviation of 0.1, and describe the alternative \mathrm{Gamma}(0.005, 0.01).

(b) The fitted factor for rows 7 to 10 is 1.971 under the vague prior and 1.673 under the tight one. Accident year 9 has cumulative losses of 1{,}363{,}294 at development year 2, and its later factors multiply to 2.36877. Approximate its reserve in each case.

Answer (a) For \lambda - 1 the mean is 0.5 and the standard deviation 0.1:

\begin{align*}
\text{shape} &= (0.5/0.1)^2 \\
&= 25 \\
\text{rate} &= 0.5/0.1^2 \\
&= 50
\end{align*}

That is the program's dgamma(25, 50). The alternative has mean 0.005/0.01 = 0.5 but standard deviation \sqrt{0.005}/0.01 = 7.07. It is effectively no opinion, so the data decide.

(b) \hat R_9 = 1{,}363{,}294\,(\lambda \times 2.36877 - 1):

Second factor \hat R_9 Verrall's reserve
1.7473 (chain ladder) 4{,}279{,}323 4{,}278{,}972 (Table 2)
1.971 (vague prior) 5{,}001{,}726 4{,}998{,}000 (Table 6)
1.673 (prior sd 0.1) 4{,}039{,}384 4{,}044{,}000 (Table 6)

Left to their own data, rows 7 and 8, whose individual factors are 1.878 and 2.016, pull the separate factor up to 1.971, and the reserve rises by 17\%. The tight prior pulls it toward 1.5, and the reserve falls by 6\%. As Verrall notes, the intervention changes the reserve and its prediction error considerably, while the prediction error as a percentage barely moves: Table 6 gives 24\%, 27\% and 25\%.