CS 70 FALL 2006 — DISCUSSION #10D. GARMIRE, L. ORECCHIA & B. RUBINSTEIN1. Administrivia(1) Course Information• Midterm #2 will be held in-class Friday• No homework due this week2. Life InsuranceAs an extended example of probability, we analyze a simple life insurance system.A real system would be too cumbersome to look at, so we make many simplificationshere.Here are the basic rules for our system:1. You pay b dollars to the insurance company when you are born. You never haveto pay again.2. If you die before age c, the company pays your beneficiaries d dollars.3. The insurance company is non-profit, so just wants to break even.Given these rules, what should the insurance company set as the values of b andd, in terms of c? Let X be the age at which a person dies. The fraction of itscustomers the insurance pays is then the fraction of those that die before age c, orPr (X < c). Then b and d are related by b = d · Pr (X < c).Lets do a detailed example, where c = 60 and d = $1, 000, 000. We need tocompute Pr (X < 60).2.1. Distribution of Death. Before we can calculate Pr (X < 60), we need toknow what the distribution of X looks like. First, lets assume that nobody lives past100. Now we cant just take the distribution to be uniform in the range {1, . . . , 100},since a person is more likely to die as they get older. So let’s assume a lineardistribution of the form Pr (X = k) =kNfor k ∈ {1, . . . , 100}.Exercise 1. Calculate the constant N in order to ensure the probabilities sum to1.2.2. Life Expectancy. The first thing we should calculate is the expected age atwhich a person dies.Exercise 2. Calculate E [X]. Use the identityPni=1i2=n(n+1)(2n+1)6.Date: November 1, 2006.The authors gratefully acknowledge Chris Crutchfield and Amir Kamil for the use of theirprevious notes, which form part of the basis for this handout.12 D. GARMIRE, L. ORECCHIA & B. RUBINSTEINKnowing just the expectation is not enough to calculate Pr (X < 60). Considerthe two distributions (A) where Pr (X = 67) = 1 and (B) where Pr (X = 55) =Pr (X = 79) = 0.5. In (A), Pr (X < 60) = 0, whereas in (B) Pr (X < 60) = 0.5.Notice that in both cases E [X] = 67.The variance is what makes the difference in the above distributions. It is vari-ance that makes insurance useful. If there were no variance, everyone would knowwhen they would die and thus no one would need or provide life insurance.2.3. Variance and Chebyshevs Inequality. We proceed by calculating the vari-ance of the age at which a person dies.Exercise 3. Calculate Var (X), using the identityPni=1i3= (Pni=1i)2.Now recall Chebyshevs inequalityPr (|X − E [X]| > r) ≤Var (X)r2.Exercise 4. Now use Chebyshev’s inequality to upper-bound Pr (X < 60). Whatis the problem with this bound?Exercise 5. When does Chebyshev’s give a bound less than 1?Even in general, Chebyshevs still gives us a weak bound. Its usefulness is due tothe fact that it is easy to compute and only requires knowledge of the expectationand variance of a random variable.2.4. Exact Solution. In this case, since the distribution is so simple, we cancompute Pr (X < 60) directly.Exercise 6. Compute this probability directly, and thus determine the appropriateb that the insurance company should charge clients at
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