Estimation, Evaluation, and Selection of Actuarial Models

105
Exercise 25 The equations to solve are
µ
ln 18.25 − µ
0.2 = F (18.25) = Φ
σ
µ
ln 35.8 − µ
0.8 = F (35.8) = Φ
.
σ
The 20th and 80th percentiles of the normal distribution are −0.842 and 0.842 respectively. The
equations become
−0.842 = 2.904 − µ
σ
0.842 = 3.578 − µ .
σ
Dividing the first equation by the second yields
−1 = 2.904 − µ
3.578 − µ .
The solution is ˆµ =3.241 and substituting in either equation yields ˆσ =0.4. The probability of
exceeding 30 is
Pr(X >30) =
µ
ln 30 − 3.241
1 − F (30) = 1 − Φ
=1− Φ(0.4)
0.4
= 1− Φ(0.4) = 1 − 0.6554 = 0.3446.
Exercise 26 For a mixture, the mean and second moment are a combination of the individual
moments. The first two moments are
E(X) = p(1) + (1 − p)(10) = 10 − 9p
E(X 2 ) = p(2) + (1 − p)(200) = 200 − 198p
Var(X) = 200 − 198p − (10 − 9p) 2 = 100 − 18p − 81p 2 =4.
The only positive root of the quadratic equation is ˆp =0.983.
Exercise 27 For the inverse exponential distribution,
l(θ) =
nX
(ln θ − θx −1
j
− 2lnx j )=n ln θ − nyθ − 2
j=1
l 0 (θ) = nθ −1 − ny, ˆθ = y −1 ,wherey = 1 n
nX
j=1
1
x j
.
nX
ln x j
For Data Set B, we have ˆθ =197.72 and the loglikelihood value is −159.78. Because the mean
does not exist for the inverse exponential distribution, there is no traditional method of moments
estimate available. However, it is possible to obtain a method of moments estimate using the
j=1