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, the insurer payment is \(Y_{i}=(R_{i}-1)_{+}\), so \(\operatorname{E}[Y_{i}]=0(0.4)+0(0.3)+1(0.2)+2(0.1)=0.4\). Because both factories have the same coverage, apply the \(5\%\) loading to each expected payment and add the two premiums: \(P=2(1+\frac{1}{20})(0.4)=\frac{21}{25}\). The retained cost is \(M_{i}=\min(R_{i},1)\), so \(M_{i}=0\) with probability \(0.4\) and \(M_{i}=1\) with probability \(0.6\). Independence gives \(\Pr(M=0)=0.4^2=0.16\), \(\Pr(M=1)=2(0.4)(0.6)=0.48\), and \(\Pr(M=2)=0.6^2=0.36\), where \(M=M_{1}+M_{2}\). The amount available before retained costs is \(3-0.45-P=\frac{171}{100}\). Thus \(D=\frac{171}{100}\) when \(M=0\), \(D=\frac{171}{100}-1\) when \(M=1\), and \(D=0\) when \(M=2\). Therefore \(\operatorname{E}[D]=0.16(\frac{171}{100})+0.48(\frac{171}{100}-1)=\frac{384}{625}\).