How to Solve Insurance Risk Variables — Insurance — expected payment with deductible / policy limit Questions on Exam P
Sample Practice Problem
ID: #606EZ
A cargo unit is insured for 12 (in thousands) with a deductible of 1. Partial damage occurs with probability 3% and total loss with probability 2%. Conditional on partial damage, let loss \(X\) has truncated exponential density proportional to \(e^{-x/10}\) on \((0,12)\). Calculate the unconditional expected payment, in thousands.
📖 Worked Solution & Strategy
Loss \(X\) given partial damage has truncated exponential density \(f_X(x) = \frac{0.1 e^{-0.1x}}{1 - e^{-1.2}}\) on \((0, 12)\). With deductible \(1\), payment for partial damage is \((x-1)_+\). Evaluating \(\int_1^{12} (x-1) e^{-0.1x}\,dx\) by parts with \(u = x-1, dv = e^{-0.1x}\,dx\) yields \(\left[-10(x-1)e^{-0.1x}\right]_1^{12} + 10\int_1^{12} e^{-0.1x}\,dx = 100 e^{-0.1} - 210 e^{-1.2}\). Multiplying by the normalizing constant \(\frac{0.1}{1 - e^{-1.2}}\) gives mean partial payment \(m_P = \int_1^{12} (x-1)f_X(x)\,dx = 3.8971\). A total loss (probability \(0.02\)) pays policy limit minus deductible: \(12 - 1 = 11\). The unconditional expected payment combines partial and total loss payments weighted by occurrence probabilities: \(\operatorname{E}[Y] = \frac{3}{100} m_P + 0.02(11) = 0.3369\). The evaluated result is \(0.3369\), which matches the correct option.