4.3 Confidence Intervals

Can have different assumptions on the distribution of the unpaid

Normal Estimation

\[\begin{equation} \hat{R}_i \pm Z_\alpha \: s.e.(\hat{R}_i) \tag{4.7} \end{equation}\]

Remark.

  • Under CLT we can assume that \(\hat{R}_i\) is normally distributed given that the outstanding claims are large

  • We can get the 95% CI with \(Z_{0.975} = 1.96\)

Log-Normal Estimation

\[\begin{equation} e^{\mu_i + Z_{\alpha} \sigma_i} = \hat{R}_i \times \exp\left \{ -\dfrac{\sigma_i^2}{2} \pm \: Z_\alpha \sigma_i \right \} \tag{4.8} \end{equation}\]

\[\sigma_i^2 = \mathrm{ln} \left [ 1 + \left ( \dfrac{s.e.(\hat{R}_i)}{\hat{R}_i} \right)^2 \right]\]

\[\mu_i = \mathrm{ln}(\hat{R}_i) - \dfrac{\sigma_i^2}{2}\]

Remark.

  • Use log-normal when the true distribution of \(R_i\) is skewed

  • Especially true when \(s.e.(\hat{R}_i) > \dfrac{R_i}{2}\)

    This would give us negative value for the bottom end of the CI if we use normal distribution

4.3.1 CI for All Years Reserves

Mean is easy: \(\hat{R} = \sum \limits_{i} \hat{R}_i\)

But since \(\hat{R}_{i}\) rely on the same LDFs they are not independent and we need to include a correlation factor for the \(MSE(\hat{R})\)

\[\begin{equation} [s.e.({\hat{R}})]^2 = \sum \limits_{i=2}^I \left\{ [s.e.(\hat{R}_i)]^2 + \hat{c}_{i,I} \left( \sum \limits_{j = i+1}^{I} \hat{c}_{j,I} \right) \left( \sum \limits_{k = I + 1 - i}^{I - 1} \dfrac{2 {\hat{\alpha_k}}^2 \big/ {\hat{f_k}}^2}{\sum_{n=1}^{I-k}c_{n,k}}\right) \right\} \tag{4.9} \end{equation}\]

If we want to simplify things we can use the square root rule to sum up the different AYs if we assume independence

4.3.2 CI Application & Range

Total CI

Consider the ratio \(\dfrac{s.e.(\hat{R}_i)}{\hat{R}_i}\)

  • Can be high for older years since reserve is small but absolute s.e. is also small \(\therefore\) not important

  • Good to calculate the ratio since we need it for \(\sigma_i^2\)

  • Tend to be high for most recent AY as well

    Driven by the large uncertainty around the development from age 12 - 24

  • Recall from previous section we mentioned that we should use log-normal distribution for the CI if \(s.e.(\hat{R}_i) > \dfrac{R_i}{2}\), this is equivalent to \(\dfrac{s.e.(\hat{R}_i)}{\hat{R}_i} > 50\%\)

  • Then calculate the total reserve CI

  • Total ratio should be greater than the simple sum because the LDFs are positively correlated

Allocate CI to each AY

After calculating the all year CI, we can allocate the upperbound the total CI to each AYs

  • Through trial & error to figure out the \(Z_{\alpha}\) for each AY that would yield the total upperbound CI

Empirical Limits

Use the max/min historical LDFs for each age \(k\) to get the upper and lower limit of the confidence interval

  • Caveat: CI is too small for older ages as there are only a few LDFs (and too large for younger ages)

    \(\therefore\) Not very useful