Question

5. An insurance company's losses on a particular policy are normally distributed with a mean of $150 million and a standard deviation of $50 million. (Assume that the risks taken on by the insurance company are entirely non-systematic.) The one-year risk-free rate is 5% per annum with annual compounding. Estimate the cost of the following: (a) A contract that will pay, in one-year's time, 60% of the insurance company's costs. (b) A contract that pays $ 100 million in one-year's time if losses exceed $200 million (and zero otherwise). Hints: • In (a), the expected cost to the insurance company depends only on the mean losses; the expected cost should then be discounted to the present. • In (b), the expected cost to the company is $100 million times the probability that the losses exceed $200 (you must calculate this probability from the mean and standard deviation given in the problem). Once you have computed the expected cost, you must discount it to the present. • A discount rate of 5% with annual compounding implies a one-year PV factor of 1/1.05.

Answer 1

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Answer(a):

Given that the losses by the insurance company are normally distributed.

Also,

Mean of the distribution =$ 150 Mn

Std. Dev of the distribution = $50 Mn

Risk free rate = 5%

We describe the probability function of the above situation using $\Phi$

=> The losses can be described using normal distribution as $\Phi$ (150 , 50²)

where 150 Mn is the mean and 50 is the standard deviation.

Now, payout will become $\Phi$ (90 , 50²) as we have been given only 60% of the losses can be recovered

Again, for finding the cost of reinssurance from the payout formulation, we discount it at the risk free rate given

=> The Cost of the contract becomes 90e^{-0.05 * 1} = 85.61 Mn (We are using the expected payoff $90 Mn to find the cost and discounting back at 5% to get the cost)

Answer(b):

We have to find the probability that losses will be greater than $200 Mn.

In this case, std dev is $50 Mn. So, we have to find the probability that the normal distribution is one std dev above mean (Since $200 - $150 Mn = $50 Mn which is 1 standard deviation)

So, the probability for being exactly above 1 standard deviation = 0.159 ( From the below graph, prob of being greater than 1 standard deviation = (100-68.2)/2 = 15.9%)

Standard Deviations 68% -3 -2 -1 +1 +2 +3

=> So, the expected payoff is therefore = 15.9e^{-0.05 * 1} = $15.10 Mn