How to Solve Insurance Risk Variables — Insurance — variance/sd of the payment random variable Questions on Exam P
Sample Practice Problem
ID: #8VR7Q
Two factories have independent repair costs \(R_1,R_2\), each taking values \(0,1,2,3\) with probabilities \(0.4,0.3,0.2,0.1\). Insurance pays \((R_i-1)_+\) and charges expected claim cost plus \(5\%\). Combined revenue is \(3\), other expense is \(0.45\), and dividend \(D\) equals any positive remaining profit. Let \(P\) be total premium. Calculate \(\operatorname{E}[D]\).
📖 Worked Solution & Strategy
For one factory, insurer payment \(Y_i = (R_i-1)_+\) has expected value \(\operatorname{E}[Y_i] = 0(0.4)+0(0.3)+1(0.2)+2(0.1) = 0.4\). With \(5\%\) loading, total premium for two factories is \(P = 2(1+\frac{1}{20})(0.4) = \frac{21}{25}\). The retained cost per factory \(M_i = \min(R_i,1)\) is Bernoulli: \(M_i=0\) with probability \(0.4\) and \(M_i=1\) with probability \(0.6\). For two independent factories, total retained cost \(M = M_1+M_2\) takes values \(\{0,1,2\}\) with probabilities \(0.16, 0.48, 0.36\). Insurer profit before retention is \(\text{Profit}_0 = 3 - 0.45 - P = \frac{171}{100}\). Profit after retention is \(\frac{171}{100}-M\). Positive profit occurs for \(M=0\) (dividend \(\frac{171}{100}\)) and \(M=1\) (dividend \(\frac{171}{100}-1\)). Taking expected values yields \(\operatorname{E}[D] = 0.16\left(\frac{171}{100}\right) + 0.48\left(\frac{171}{100}-1\right) = \frac{384}{625}\). This evaluated result matches the correct option.