Question

The model for the amount of damage to a particular property during a one-month period is as follows: there is a .99 probability of no damage, there is a.01 probability that damage will occur, and if damage does occur, it is uniformly distributed between 1000 and 2000. An insurance policy pays the amount of damage up to a policy limit of 1500. It is later found that the original model for damage when damage does occur was incorrect, and should have been uniformly distributed between 1000 and 5000. Find the amount by which the insurer's expected payment was understated when comparing the original model with the corrected model A 116 15 ) E8

Answer 1

Here in first case, where x is the amount of damage

P(x) = 0.99, x = 0

= 0.01, x E U(1000, 2000)

but the maximum payment here is 1500

so here

E[X] =
rp(x)
= 0.99 * 0 + 0.01 *
1500 r/(2000 -1000)dr 1000
+ 0.01 * 1500 * $\int_{1500}^{2000} 1/(2000 - 1000)dx$

= 0 + 0.01 * (1/1000) (1500² -1000²)/2 + 0.01 * 1500 * 500 /1000

= 13.75

Now, in the second case,

P(x) = 0.99, x = 0

= 0.01, x E (1000, 5000)

so here

E[X] =0.99 * 0 + 0.01 * $\int_{1000}^{1500} x/(5000 - 1000)dx$ + 0.01 * 1500 * $\int_{1500}^{5000} 1/(5000 - 1000)dx$

= 0 + 0.01 * (1/4000) * (1500² -1000²)/2 + 0.01 * 1500 * 3500/4000

= 1.5625 + 13.125

= 14.6875

so here the understated difference in expected amount = 14.6875 - 13.75 = 0.9375 = 15/16

Option C is correct here