actu loss model Essay example

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Actu 457/4 and Mast 724/4 (Winter 2011)
Solutions to Assignment 3
No.1 Exercise 9.12 (Exercise 6.11, 2004 Edition).
If Nm denotes the number of males out of the 8 employees, then 8 − Nm is the number of females, where Nm has a binomial(8, 0.4) distribution. Hence, using Table 6.6, the
N
8−N number S = 1 m Xm,i + 1 m Xf,j of cigarettes these employees smoke in a day
(where Xm,i and Xf,j are indeppendent) has a conditional mean of
E(S | Nm ) = Nm (6) + (8 − Nm )(3) = 3Nm + 24 , which gives
E(S) = E E(S | Nm ) = 3E(Nm ) + 24 = 3(8)(.04) + 24 = 33.6 .
Similarly for the variance
V(S) = V E(S | Nm ) + E V(S | Nm ) = V(3 Nm + 24) + E Nm (64) + (8 − Nm )(31)
= 9 V(Nm ) + (64 − 31) E(Nm ) + 248 = 9(8)(0.4)(0.6) + 33 (8) (0.4) + 248
= 370.88 .
√
The standard deviation is thus 370.88 = 19.26.
No.2 Exercise 9.50 (Exercise 6.43, 2004 Edition).
Here the total length of care for all patients, S = SA +SC is the sum of two independent compound Poisson sums, SA = NA Xk with NA ∼Poisson(3), and SC = NC Yk k=1 k=1 with NC ∼Poisson(2) and severity distribution f(x) = 3/5 fX (x) + 2/5 fY (x). Hence
S is also compound Poisson with parameter λ = 3 + 2 = 5. We need the probability
FS (4) = P{S ≤ 4} as the cost would then be ≤ 4 × $200 = $800: fS (0) = P{S = 0} = e−5 ,
1

fS (1) =

y λ fS (0) f(1) = 5 e−5 [3 (0.4) + 2 (0.9)]/5 = 3 e−5 ,
1
y=1

fS (2) = 5 (1/2) 3e−5 [3 (0.4) + 2 (0.9)]/5 + (2/2) e−5 [3 (0.6) + 2 (0.1)]/5
= (13/2) e−5 , fS (3) = 5 (1/3) (13/2)e−5 [3 (0.4) + 2 (0.9)]/5 + (2/3) 3e−5 [3 (0.6) + 2 (0.1)]/5
= (21/2) e−5 , fS (4) = 5 (1/4) (21/2)e−5 [3 (0.4) + 2 (0.9)]/5 + (2/4) (13/2)e−5 [3 (0.6) + 2 (0.1)]/5
= (115/8) e−5 .
Hence FS (4) =

4
1

fs (k) = [1 + 3 +

13
2

+
1

21
2

+

115 −5
]e
8

=

283
8

e−5 = 0.2384.

No.3 No need here to use recursive or convolution formulas; the that aggregate claims can be 600 only in 2 ways, with 2 claims of 500 and 100, or with 6 claims of 100. Since
N ∼ Poisson(λ = 5) and fX (100) = 0.8, while fX (500) = 0.16, then
P{N = 6, X1 = X2 = . . . = X6 = 100} =
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