11. Let L be a random variable that takes the value $500 with probability 999/10

Question

11. Let L be a random variable that takes the value $500 with probability 999/1000; otherwise, L $200,000. Suppose that L represents a loss that you are facing. You are considering buying insurance. (a) Compute E[L]. (b) Compute E[(L-1000)+]. (c) If you buy full coverage insurance, the premium charged by the in- surance company is twice the expected loss. How much would full coverage cost? (d) If you "self-insure," what is your expected cost? (e) To lower your average cost, you consider buying insurance with a $1,000 deductible. With a deductible d, you incur the first d 2 0 of any loss and the insurance company incurs any loss above d· That is, the loss L is split into two parts: L Ad and (L -d)t. Show that these two pieces add up to L. (f) If d = $1,000 and the insurance company charges you double the expected loss that they will incur, what is the premium? (8) What is your total expected cost, which would be the premium plus the expected cost of the loss you incur assuming a $1,000 deductible? (h) Under what circumstances would a $1,000 deductible be better than either full coverage or self-insuring? By the way, even though the numbers here are made up, the idea of reducing your expected cost but also avoiding a catastrophic loss through a large deductible is often a wise strategy.

Explanation / Answer

Ans:

a) E[L] = 500*.999 + 200000*0.001 = 699.5

b) E[(L-1000)+] : This term on the left denotes the Payment per loss, i.e

Suppose a loss equal to L happens.

In this question there are only two cases:

1) L = 500,
Here, since the loss is below 1000, the whole amount will be borne by us only and the insurance company pays nothing

2) L = 200000

Here, the insurance company will pay the amount of loss above 1000 i.e. 200000-1000 = 199000
The expected value will be 199000*0.001 = 199

So, the total expected loss will be the sum of cases a) and b) = 0+199 =199

c) The full coverage cost will be twice the expected loss i.e.
2 * E[L] = 2* 500*.999 + 200000*0.001 = 2*699.5 = 1399

d) If we self insure then we need to set aside an amount equal to the loss we are expecting:
Cost of self insurance = E[L] = 500*.999 + 200000*0.001 = 699.5

e) The loss is divided into two part:
1) Payment per loss = (L-d)+ This term represents that part of the loss which is borne by the insurance company
2) Limited Payment per loss: L^d This term represents that part of the loss which is borne by the insured party

We need to prove that the total value of both these parts is equal to L.
Let us prove this by dividing the situation into two cases

f) The premium will be the double of the loss that the insurance company is expecting to bear:
Premium = 2* E[(L-d)] = 2*199000*0.001 = 199*2 = 398

g) Our total expected cost will be the sum of
1) Premium paid to the company = 398
2) Expected cost of the loss, we have two cases of losses

Total cost = 398 + 500.5 = 898.5

h) Let the probabilities of loss of 500 be p and 200000 be 1-p
Total expected cost in the deductible case = 500p + 1000(1-p) + 2*199000*(1-p)

Total expected cost for self -insurance case: 500p + 200000(1-p)

Total expected cost for full coverage case: 2*[500p+200000(1-p)]

Deductible is better than full coverage case when: 500p + 1000(1-p) + 2*199000*(1-p) <  2*[500p+200000(1-p)]
Solving, we get p<0.5

Deductible is better than self-insurance case when: 500p + 1000(1-p) + 2*199000*(1-p) < 500p+200000(1-p)
Solving we get p>1, which implies never. So, under no circumstance will deductible be better than self-insurance.

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