SOA Exam ASTAM Spring 2025, Question 4
Study guide: Exam ASTAM
An actuary is analyzing the reported losses from an auto insurance with an ordinary deductible of d. Let X denote the ground-up loss random variable, which is assumed to have a Pareto distribution with known \theta and unknown \alpha.
A random sample of 20 reported ground-up losses was collected. Note that losses below the deductible are not reported to the insurer, so only losses above the deductible were sampled.
Part a (3 points)
(i) Determine the log-likelihood function for the sample in terms of \theta, \alpha, d, and the sample data, x_1, x_2, \cdots, x_{20}.
(ii) You are given that \theta = 5{,}000, d = 1{,}000, and that \displaystyle\sum_{i=1}^{20} \ln(\theta + x_i) = 181.66. Show that the maximum likelihood estimate of \alpha, denoted \hat{\alpha}, is 2.6 to the nearest 0.1. You should calculate the value to the nearest 0.001.
Part b (1 point)
Estimate a 95% confidence interval for \alpha using Fisher's information.
Part c (2 points)
Let
Y = \begin{cases} 0 & \text{if } X \le d \\ X - d & \text{if } X > d \end{cases}
Calculate the maximum likelihood estimate of E[Y].
Part d (2 points)
For X with a Pareto distribution with \theta = 5000 and \alpha = \hat{\alpha}, show that the loss elimination ratio for an ordinary deductible of d = 1000 is 0.25 to the nearest 0.05. You should calculate the value to the nearest 0.001.
Part e (1 point)
The actuary's colleague notes that an ordinary deductible of d = 1000 produces a similar reduction in expected payment as does a 75% coinsurance.
Explain why an insurer may prefer to impose an ordinary deductible of 1000 over 75% coinsurance.
Answer this question and see the worked solution
Concepts tested
- Maximum Likelihood Estimation
- Truncation
- Fisher Information
- Approximate Confidence Interval
- Loss Elimination Ratio
- Coinsurance