Consider an industrial risk where hazards can lead to a loss, x, of $1000 with p
ID: 3040320 • Letter: C
Question
Consider an industrial risk where hazards can lead to a loss, x, of $1000 with probability 0.3 and $500 with a probability of 0.7. Suppose the loss significance of the $1000 loss, s(1000), is equal to 1, and that for the $500 loss, s(500) is 0.4.
For this industrial risk, what is;
i) the expected value of the loss and
ii) the expected loss significance.
The owners of this industry are risk averse and are considering purchasing insurance as an alternative to assuming this risk. Assume they correctly perceive the true values of the loss significances s(1000) and s(500), and have a perceived quadratic loss function of the form s(x) = a + bx2 , where x is equal to the amount of the loss, and a and b are constants in the quadratic function.
What is the highest insurance premium that this company will be willing to pay and why? How does this change if the probability of a $1000 loss changes to 0.7?
Explanation / Answer
1. Expected value of loss = 1000*0.3 + 500*0.7 = 650
2. Expected loss significance = 1*0.3 + 0.7*0.4 = 0.58
s(x) = a + bx^2
s(1000) = 1
a+b*1000000 = 1
s(500) = 0.4
a + b*250000 = 0.4
Solving, 750000*b = 0.6
b = 0.6/750000 = 8*10^(-7)
a = 0.2
s(x) = 0.2 + 8*10^(-7)*x^2
The company would be ready to pay an insurance x such that the loss significance then is equal to the expected loss significance now
So, 0.2 + 8*10^(-7)*x^2 = 0.58
x = 689.2
If probabilities are interchanged, expected loss significance = 0.7*1+0.3*0.4 = 0.82
0.2 + 8*10^(-7)*x^2 = 0.82
x = 880.34