Actuarial Notes

Over-Dispersed Poisson Model

Study guide: Exam ASTAM

Over-Dispersed Poisson Model (ODP) is the stochastic model for a triangle of incremental losses in which each cell q(w,d) is an independent over-dispersed Poisson variable. Its mean is an accident-year level times a development-period share, and its variance is a constant scale parameter \phi times the mean. Fitted as a GLM with a log link, it reproduces the volume-weighted chain ladder, which is why it underlies the ODP bootstrap.

E[q(w,d)] = m_{w,d} = x_w\, y_d

\mathrm{Var}[q(w,d)] = \phi\, m_{w,d}

\ln m_{w,d} = c + \alpha_w + \beta_d

  • Two ways of writing it. x_w is the expected ultimate loss of accident year w, and y_d is the proportion of ultimate that emerges in development period d, with \sum_d y_d = 1. That is Verrall's form. In the log-link form of Shapland's formula 3.7, \alpha_1 = \beta_1 = 0; the \alpha_w adjust the level and the \beta_d the development trend after the first period. The reserve estimates are the same either way.
  • What "over-dispersed" means. If X is Poisson with mean \mu, then Y = \phi X has E(Y) = \phi\mu and \mathrm{Var}(Y) = \phi^2\mu = \phi\,E(Y). A quasi-likelihood approach means the losses need not be integers. \phi is usually above 1 but need not be, and Shapland notes that "quasi-Poisson" is the more general name. Clark (2003) writes the loss dollars as a Poisson count times a scale factor \sigma^2.
  • Why it is the chain ladder. Its maximum likelihood equations equate each row sum and each column sum of fitted values with the actual ones (Taylor and McGuire). The solution is the volume-weighted chain ladder, found by dividing the latest diagonal backwards through the factors and differencing. That needs a parameter for every accident and every development period, a common \phi, and positive column sums; a negative column total makes the GLM fail. Clark finds likewise that the ODP likelihood reproduces the LDF and Cape Cod ultimates.
  • The scale parameter is estimated from the Pearson residuals r_{w,d} = (q - m)/\sqrt{m} as \phi = \sum r^2/(N - p). N is the number of cells and p = 2n - 1 the number of parameters of an n \times n triangle. (Shapland's errata corrects the printed 2(n-1).) It is the dispersion parameter of a GLM whose variance is \phi\,m^{z} with z = 1; z = 0, 2 and 3 give the Normal, gamma and inverse Gaussian. Taylor and McGuire call the ODP the Tweedie case p = 1.
  • Where it is used. Because the cells are independent, the process variance of a reserve R is \phi R. Shapland resamples its residuals in the ODP bootstrap. Verrall's over-dispersed negative binomial model gives the same predictive distribution, and his Bayesian Bornhuetter-Ferguson model puts priors on its row parameters x_w.

Row and Column Parameters from the Chain Ladder Incremental paid losses ($000s):

AY 1 2 3
1 2{,}000 1{,}200 400
2 2{,}400 1{,}500
3 2{,}700

Fit the ODP model. Find x_w, y_d, the fitted values and the reserve.

Answer The cumulative losses are 2{,}000, 3{,}200, 3{,}600 for AY 1, 2{,}400, 3{,}900 for AY 2 and 2{,}700 for AY 3. The volume-weighted factors are:

\begin{align*}
\hat f_1 &= \frac{3{,}200 + 3{,}900}{2{,}000 + 2{,}400} \\
&= 1.61364 \\
\hat f_2 &= \frac{3{,}600}{3{,}200} \\
&= 1.125
\end{align*}

Column parameters, the share of ultimate emerging in each period:

\begin{align*}
y_1 &= \frac{1}{1.61364 \times 1.125} \\
&= 0.5509 \\
y_2 &= \frac{1}{1.125} - 0.5509 \\
&= 0.3380 \\
y_3 &= 1 - \frac{1}{1.125} \\
&= 0.1111
\end{align*}

Row parameters, each year's chain ladder ultimate: x_1 = 3{,}600, x_2 = 3{,}900 \times 1.125 = 4{,}387.5 and x_3 = 2{,}700 \times 1.81534 = 4{,}901.4.

Fitted values m_{w,d} = x_w y_d, with the future cells in brackets:

AY 1 2 3
1 1{,}983.1 1{,}216.9 400.0
2 2{,}416.9 1{,}483.1 (487.5)
3 2{,}700.0 (1{,}656.8) (544.6)

The fitted row sums over the observed cells are 3{,}600 and 3{,}900, and the column sums are 7{,}100 and 2{,}700. Each equals its actual total, which is the marginal-sum property behind the equivalence. The two corner cells are fitted exactly because each has a parameter of its own.

\begin{align*}
\hat R &= 487.5 + 1{,}656.8 + 544.6 \\
&= 2{,}688.9
\end{align*}

That is the chain ladder reserve: 3{,}900 \times 0.125 = 487.5 for AY 2 and 2{,}700 \times 0.81534 = 2{,}201.4 for AY 3.

The Scale Parameter and the Process Variance of the Reserve Incremental losses ($000s), and the ODP fitted values from the chain ladder (factors 1.58788, 1.17059 and 1.05263):

AY 1 2 3 4
1 1{,}000 600 300 100
2 1{,}100 700 280
3 1{,}200 640
4 1{,}300
Fitted 1 2 3 4
1 1{,}022.2 600.9 276.9 100.0
2 1{,}119.0 657.9 303.1
3 1{,}158.8 681.2
4 1{,}300.0

The chain ladder reserve is 1{,}780.3. Estimate \phi and the process standard deviation of the reserve. What would the misprinted p = 2(n-1) have given?

Answer Residuals, r = (q - m)/\sqrt{m}. For example, r_{1,1} = (1{,}000 - 1{,}022.2)/\sqrt{1{,}022.2} = -0.694.

AY 1 2 3 4
1 -0.694 -0.038 1.389 0
2 -0.569 1.643 -1.328
3 1.211 -1.579
4 0

The squares sum to 11.160. There are N = 10 cells and p = 2(4) - 1 = 7 parameters: four accident-year levels and three development trends.

\begin{align*}
\hat\phi &= \frac{11.160}{10 - 7} \\
&= 3.720 \\
\mathrm{Var}(R) &= 3.720 \times 1{,}780.3 \\
&= 6{,}623 \\
\mathrm{SD}(R) &= 81.4
\end{align*}

The process CoV is 81.4/1{,}780.3 = 4.6\%.

With the misprint, p = 6 gives \hat\phi = 11.160/4 = 2.790 and a standard deviation of \sqrt{2.790 \times 1{,}780.3} = 70.5. That is 13\% too low, because the degrees of freedom were overstated by one. Either way this is process variance only. The parameter variance is what the ODP bootstrap adds by resampling the residuals.