Probability of Fire The probability that a fire will occur is $0.001$. If there is a fire, the amount of damage, $X$, will have a Pareto distribution given by $P(X>x) = \left(\frac{2(10)^6}{2(10)^6+x} \right)^2$. An insurance will pay the excess of the loss over a deductible of $100,000$. For this coverage the one-time insurance premium will be $110 \%$ of the expected payment. Calculate the premium. 
So the expected payment is $E[W]$ where $W$ denotes the payment. Then $E[W] = E[W| \text{fire}]P(\text{fire})+E[W| \text{no fire}]P(\text{no fire})$. To calculate $E[W| \text{fire}]$, we could use $\int_{0.1}^{\infty} [1-F(x)] \ dx$? This would be: $\int_{0.1}^{\infty} 1-\left[1-\left(\frac{2(10)^6}{2(10)^6+x} \right)^2\right] \ dx$ which equals $\int_{0.1}^{\infty} \left(\frac{2(10)^6}{2(10)^6+x} \right)^2 \ dx$?

So $E[W| \text{fire}] = \int_{0.1}^{\infty} [1-F(x)] \ dx = 4 \int_{0.1}^{\infty} (2+x)^{-2} \ dx = 4/2.1$. Thus $E[W] = (4/2.1)(0.001)+0 = 0.00190476$. So the premium is $(0.0019047)(10^6)(1.1) =2,095$. 
