9. For the change of variable, the prior density of p is f(p) = 100p, 0 < p < .1
f(p) = 20 - 100p, .1 < p < .2
The likelihood of observing a 1 and then a 0 is p(1-p) and so the posterior density is proportional to
g(p) = 100(p^2)(1-p)=100p^2 - 100p^3, 0<p<.1
g(p) = 20p - 120p^2 + 100p^3, .1<p<.2
Integrating this function yields .03083 for the first part and .0575 for the second part. This the actual posterior density is the above g(p) dividied by .08833. The posterior probability of being below .1 is the integral from 0 to .1 or .03083/.08833 = .34903. The answer is (B)
10. The model variance is p(1-p) and the expected process variance is the integral of p(1-p) times the density f(p) above. The result is the .08833 because it happens to be the same calculation used in 9. The model mean is p and for the variance of the hypothetical means we first need E(p) which is the integral of pf(p) which is .1 and E(p^2) which is the integral of p^2f(p) which is .0116667 and so the variance is .0016667 and k = 53. then Z = 2/55 and the estimate is .5(2/55) + .1(53/55) = 63/550 for answer (D)
11. The distribution function is F(x) = 1 - [t/(t + x)]^a using t for lambda. For Y we have as a distribution function:
F(y) = Pr(Y<=y) = Pr[ln(1 + X/t) <= y]
= Pr[1 + X/t <= exp(y)]
= Pr[X <= t*{exp(y)-1}]
= 1 - [t^a]/[t + t*{exp(y)-1}]^a.
Note that the t's cancel, leaving
= 1 - [1 + exp(y) - 1]^(-a) = 1 - exp(-ay)
which is the exponential distribution. answer (B)
12. The grand mean is the same for both. Under Bayesian estimates it is [5*3 + 2*4.5 + 5*6]/12 = 4.5. So under Buhlmann we have [5x + 2*3.8 + 5*6.1]/12 = 4.5 or 5x = 15.9 for x = 3.18, for answer (C).
21. We want the sum of x(x-1)...(x-9)exp(-t)*(t^x)/x! from x=10 to infinity (the previous terms are all zero). Note the cancellation, so the summand becomes
exp(-t)*(t^x)/(x-10)!. Now do a change of variable, letting y = x-10 and so the sum on y starts at zero. The summand is
exp(-t)*(t^10)*(t^y)/y!. The first, third, and fourth terms are Poisson probabilities and so sum to one. That leaves t^10, answer (D).
26. Let Y be the 1996 losses. The mean for 1997 is exp(mu + lnk + s^2/2). Thus
1 - p = Pr(Y < kexp(mu + lnk + s^2/2))
= Pr(Z < [mu + lnk + s^2/2 - mu]/s)
= Pr(Z < lnk/s + s/2)
Thefore, zp = -lnk/s - s/2. Multiply by 2s to get
2zps = -2lnk - s^2 or 0 = s^2 + 2zps + 2lnk.
So s = -2zp +- sqrt[4zp^2 - 8*lnk] all divided by 2.
This reduces to answer (B).
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Stuart Klugman, FSA
Principal Financial Group Professor of Actuarial Science
Drake University
2507 University Avenue
Des Moines, IA 50311 USA
ph: 515-271-4097
e-mail: Stuart.Klugman@drake.edu