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Question 12, Spring 2007 Exam 4/C May 27, 2007

Posted by Peter in Exam 4/C.

12. For 200 auto accident claims you are given:

  1. Claims are submitted t months after the accident occurs, t = 0, 1, 2, ….
  2. There are no censored observations.
  3. \hat{S}(t) is calculated using the Kaplan-Meier product limit estimator.
  4. \displaystyle c_S^2(t) = \frac{\widehat{\rm Var}[\hat{S}(t)]}{\hat{S}^2(t)} , where \widehat{\rm Var}[\hat{S}(t)] is calculated using Greenwood’s approximation.
  5. \hat{S}(8) = 0.22, \; \hat{S}(9) = 0.16, \; c_S^2(9) = 0.02625, \; c_S^2(10) = 0.04045  .

    Determine the number of claims that were submitted to the company 10 months after an accident occurred.

    Solution. There are two key observations we need to make. The first is that we are given the risk set at time t = 0, namely r_0 = 200 . The second observation is that because no observations are censored, the Kaplan-Meier estimator of the survival time takes on a particularly simple form. This is because in the absence of censoring, r_{j+1} = r_j - s_j ; that is, the risk set at time y_{j+1} is simply the risk set at time y_j minus those who died in the meantime. Therefore

    \displaystyle \hat{S}(y_n) = \prod_{j=0}^n \frac{r_j - s_j}{r_j} = \prod_{j=0}^n \frac{r_{j+1}}{r_j} = \frac{r_{n+1}}{r_0} = \frac{r_{n+1}}{200}.

    So with this in mind, we have r_{10} = 200\hat{S}(9) = 32 . Recalling Greenwood’s approximation,

    \displaystyle c_S^2(t) = \frac{\widehat{\rm Var}[\hat{S}(t)]}{\hat{S}^2(t)} = \sum_{j=1}^t \frac{s_j}{r_j(r_j-s_j)},

    so \displaystyle c_S^2(10) - c_S^2(9) =  \frac{s_{10}}{r_{10}(r_{10}-s_{10})} = 0.0142.

    Substituting and solving, we obtain s_{10} = 9.9978 \approx 10.



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