Process Risk

Process risk is the variability of actual outcomes around their expected value that would remain even if the model and its parameters were known exactly — the "random" part of a reserve's variability. Together with parameter risk it makes up the prediction error. In Clark (2003), the process variance of a reserve is its expected value times a constant variance-to-mean ratio $\sigma^2$ .

$\mathrm{Var}_{\text{process}}(R) = \sigma^2 \sum \mu_{AY;x,y}$

$\sigma^2 = \frac{1}{n-p}\sum_{AY,t}\frac{(c_{AY,t} - \hat\mu_{AY,t})^2}{\hat\mu_{AY,t}}$

$R$ is the reserve for a period or group of periods, $\mu_{AY;x,y}$ the expected incremental loss of each accident year between ages $x$ and $y$ , and $c$ the actual increments. There are $n$ increments in the data and $p$ fitted parameters: $3$ for Clark's Cape Cod method, or one ultimate per accident year plus $\omega$ and $\theta$ for his LDF method. The formula for $\sigma^2$ has the form of a chi-square statistic.
Where the ratio comes from. Clark assumes each increment is over-dispersed Poisson: a Poisson count times $\sigma^2$ , so $E[c] = \mu$ and $\mathrm{Var}(c) = \sigma^2\mu$ . The same $\sigma^2$ scales the normalized residuals $(c - \hat\mu)/\sqrt{\sigma^2\hat\mu}$ used to check the fit. The model treats $\sigma^2$ as fixed and known, which ignores the variance of the variance.
It adds across periods; parameter risk does not. Increments are independent in the model, so the process variance of any group of periods is $\sigma^2$ times its expected total. The process CV of a reserve, $\sqrt{\sigma^2/R}$ , therefore shrinks as the reserve grows. Parameter variance, $(\partial R)'\,\Sigma\,(\partial R)$ , comes from parameters shared by every year, and does not diversify this way.
Total variance. Clark adds the two, treating them as independent. In his example, parameter risk dominates: on the LDF method's reserve of $28{,}987{,}633$ , the process standard deviation is $1{,}372{,}966$ against a parameter standard deviation of $4{,}688{,}826$ . He notes that the Cape Cod method may carry somewhat more process variance than the LDF method but much less estimation error. The same split appears elsewhere: in Mack's standard error the $1/\hat C_{i,k}$ term is process variance (see Prediction Error), and the ODP bootstrap adds process noise to every future cell after resampling for parameter risk (see Stochastic Reserving).
Not model risk. Changes in mix of business or in how claims are settled lie outside a model's assumptions. Clark calls these "model variance" (model risk); neither process nor parameter variance measures them.

Process Variance by Year and in Total A Clark Cape Cod fit gives reserves of $1{,}200$ , $2{,}500$ and $4{,}300$ for three accident years ($000s), with $\sigma^2 = 50$ . The parameter standard deviation of the total reserve is $900$ .

Find the process standard deviation and CV for each year and for the total, and the total standard deviation.
Answer Each year's process variance is $\sigma^2$ times its reserve:

AY Reserve Process SD Process CV

1 $1{,}200$ $\sqrt{50 \times 1{,}200} = 244.9$ $20.4\%$

2 $2{,}500$ $\sqrt{50 \times 2{,}500} = 353.6$ $14.1\%$

3 $4{,}300$ $\sqrt{50 \times 4{,}300} = 463.7$ $10.8\%$

For the total, the process variances add:
 $\begin{align*} \mathrm{Var}_{\text{process}} &= 50 \times 8{,}000 \\ &= 400{,}000 \\ \mathrm{SD}_{\text{process}} &= 632.5 \\ \mathrm{SD}_{\text{total}} &= \sqrt{400{,}000 + 900^2} \\ &= 1{,}100.0 \end{align*}$ 
The total's process CV is $632.5/8{,}000 = 7.9\%$ , below every single year's. Its process standard deviation is also well below the sum of the three, $1{,}062.2$ , because independent increments partly offset one another. The total CV is $13.75\%$ , and parameter variance is $810{,}000/1{,}210{,}000 = 67\%$ of the total variance. That is the pattern Clark reports: at the level of the whole reserve, parameter risk is the larger part.

AY	Reserve	Process SD	Process CV
1	$1{,}200$	$\sqrt{50 \times 1{,}200} = 244.9$	$20.4\%$
2	$2{,}500$	$\sqrt{50 \times 2{,}500} = 353.6$	$14.1\%$
3	$4{,}300$	$\sqrt{50 \times 4{,}300} = 463.7$	$10.8\%$

Testing Next Year's Development Against Its Process Range A loglogistic growth curve $G(x) = x^{1.5}/(x^{1.5} + 40^{1.5})$ is fitted by Clark's LDF method, with $\sigma^2 = 30$ . At year-end, two accident years are $24$ and $36$ months old, with estimated ultimates of $5{,}000$ and $6{,}000$ .

Forecast the development over the next calendar year and its process standard deviation. One year later, actual development is $1{,}400$ . Is that surprising?
Answer The average ages move from $18$ to $30$ and from $30$ to $42$ . The curve gives $G(18) = 0.2319$ , $G(30) = 0.3938$ and $G(42) = 0.5183$ :
 $\begin{align*} \mu_1 &= 5{,}000 \times (0.3938 - 0.2319) \\ &= 809.5 \\ \mu_2 &= 6{,}000 \times (0.5183 - 0.3938) \\ &= 747.0 \\ \mu &= 1{,}556.5 \\ \mathrm{SD}_{\text{process}} &= \sqrt{30 \times 1{,}556.5} \\ &= 216.1 \end{align*}$ 
The process CV is $13.9\%$ . The actual development is $1{,}400$ :
 $\begin{align*} r &= \frac{1{,}400 - 1{,}556.5}{216.1} \\ &= -0.72 \end{align*}$ 
That is within one process standard deviation, so it is not surprising. Adding parameter variance would widen the range further. This is why Clark forecasts calendar year development: unlike the ultimate reserve, it can be checked against actual emergence within a year.