TO SOLUTIONS ASSIGNMENT 5 ACTSC 431/831, FALL 2008 1. (a) We have E (S ) = E (N )E (X ) = 10 and Pr{S = 1} = Pr{S = 3} = 0, Pr{S = 0} = Pr{N = 0} = 0.1, Pr{S = 2} = Pr{N = 1, X1 = 2} = 0.04, Pr{S = 4} = Pr{N = 1, X1 = 4} + Pr{N = 2, X1 = 2, X2 = 2} = 0.128. Thus, we have E [(S 5)+ ] = E (S ) E (S 5)
4 4
= 10
x=1
x Pr{S = x} 5 1
x=0
Pr{S = x} = 5.748.
(b) We have E [(S 4.6)+ ] = E (S ) E (S 4.6)
4 4
= 10
x=1
x Pr{S = x} 4.6 1
x=0
Pr{S = x} = 6.041.
(c) We have E [(S 4)+ ] = E (S ) E (S 4)
3 3
= 10
x=1
x Pr{S = x} 4 1
x=0
Pr{S = x} = 6.48,
V ar(S ) = E (N )V ar(X ) + V ar(N )(E (X ))2 = 40, and
4
E [(S 4)2 ] = +
x=5
(x 4)2 Pr{S = x} = E [(S 4)2 ]
x=0 4
(x 4)2 Pr{S = x}
= V ar(S ) + (E (S ))2 8E (S ) + 16
x=0
(x 4)2 Pr{S = x} = 74.24.
Hence, V ar[(S 4)+ ] = E [(S 4)2 ] (E [(S 4)+ ])2 = 32.25. + 2. (a) We have E (N ) = 15 and
25
E [(Xi 25)+ ] = E (Xi ) E (Xi 25) = 20
ex/20 dx = 20e5/4 = 5.7301.
0
Thus, the net reinsurance premium for the excess-of-loss reinsurance is E (N )E [(Xi 25)+ ] = 85.95. 1
(b) The survival function of S is 1 FS (x) = 15 e 320 , x 0. Thus, the net reinsur16 ance premium for the stop-loss reinsurance is E [(S 150)+ ] = E (S ) E (S 150)
150
x
= E (N )E (Xi ) (c) We have
0
15 x/320 e dx = 300e15/32 = 187.735. 16
E [(S 150)+ ] = 300e15/32 , E (S 150) = 300(1 e15/32 ), E [(S 150)((S 150)+ )] = 150E [(S 150)+ ] = 150 300e15/32 . Thus, Cov ((S 150), (S 150)+ ) = 300e15/32 (150 300(1 e15/32 )) = 7084.226. Furthermore, we have V ar(S 150) = E [(S 150)2 ] (E (S 150))2 150 15 = 2x ex/320 dx 3002 (1 e15/32 )2 = 2925.525. 16 0 V ar[(S 150)+ ] = E [(S 150)2 ] (E [(S 150)+ ])2 + 15 ex/320 dx (300e15/32 )2 = 84906.023. (x 150)2 = 16 320 150 Thus, the correlation coecient of the retained loss S 150 and the ceded loss (S 150)+ for the insurer in the stop-loss reinsurance is Cov ((S 150), (S 150)+ ) V ar(S 150)V ar((S 150)+ ) 3. (a) The probability is Pr{N 3 = 3, N 5 = 5} = Pr{N 3 = 3, N 5 N 3 = 2}
12 12 12 12 12 12
= 0.4495.
= Pr{N 3 = 3} Pr{N 2 = 2} = 0.0278.
12
(b) The probability is Pr{N 1 = 2|N 1 = 6} =
12 2
Pr{N 1 = 2, N 1 = 6}
12 2
Pr{N 1 = 6}
2
=
Pr{N 1 = 2, N 1 N 1 = 4}
12 2 12
Pr{N 1 = 6}
2
=
Pr{N 1 = 2} Pr{N 5 = 4}
12 12
Pr{N 1 = 6}
2
= 0.20094.
(c) Let be the relative security loading in Ut . Then 220 = (1 + ) 10 20 and thus = 0.1 The adjustment coecient R for the surplus process Ut is the smallest positive solution to equation 1 + (1 + )E (X1 )t = 1 + 11t = MX1 (t) = 1 , 1 10t
which gives that R = 0.0091. Thus, the Lundberg upper bound b(u) = e0.0091u , u 0. 2
(d) Let T1 be the time of the rst claim. We have
1 (u) = Pr{X1 > 220T1 + u} =
0
Pr{X1 > 220t + u} 20e20t dt 10 u/10 e , u 0. 21
=
0
e
(220t+u)/10
20e
20t
dt =
(e) For any u 0, we have
u+220t 0 u+220t 0
2 (u) = =
0
1 (u + 220t x)
0
1 x/10 e dx + e(u+220t)/10 20e20t dt 10 10 (u+220tx)/10 1 x/10 e e dx + e(u+220t)/10 20e20t dt 21 10
1 (5510 210u)eu/10 + . = 9261 (f) Note that in the surplus process Ut , the premium rate c = 22022.5E (Xi 25)+ = 220 22.5(E (X ) E (X 25)) = 201.53. i. Let T1 be the time of the rst claim. For u 25,
1 (u) = Pr{X1 25 > cT1 + u} = 0.
Furthermore, for 0 u < 25,
1 (u) = Pr{X1 25 > cT1 + u} = (25u)/c 0
Pr{X1 25 > ct + u}20e20t dt
=
0
e(ct+u)/10 20e20t dt
= (0.498095 0.00342e0.19924u )eu/10 = (a beku )eu/10 , where a = 0.498095, b = 0.00342, and k = 0.19924. It is easy to show or to verify by Maple or Mathematica that (0.498095 0.00342e0.19924u ) 10/21 < 0 for 0 u < 25 and thus 1 (u) < 1 (u) for all u 0. ii. Let be the relative security loading for the surplus process Ut . Then c = (1 + ) 20E (Xi 25) = 201.53. The adjustment coecient R for the surplus process Ut is the smallest positive solution to equation 1 + (1 + )E (Xi 25) = 1 + 10.0765t = MX1 25 (t) 1 1 10te2.5+25t = e(x25)t ex/10 dx = . 10 1 10t 0 Solving the above equation by Maple or Mathematica, we have that R = 0.01166. Thus, the Lundberg upper bound b (u) = e0.01166u , u 0. Since R = 0.01166 > R = 0.009, it holds that b (u) = eR u < eRu = b(u) for all u > 0. 3
iii. (Bonus) Note that if T1 = t and u + ct 50 or t (50 u)/c, ruin will neither occur at the rst claim nor at the second claim; if T1 = t and 25 u + ct < 50 or (25 u)/c t < (50 u)/c, ruin will not occur at the rst claim and but may occur at the second claim; and if 0 u < 25, ruin may occur at the rst claim or at the second claim. Furthermore, the maximum drop size of the surplus Ut at a claim is 25. Thus, by conditioning on the time and amount of the rst claim, we have that if 0 u < 25,
2 (u)
=
0
25u c
+ =
0
50u c 25u c 25u c
+ =
0
50u c 25u c 25u c
+
50u c 25u c
1 x/10 e dx + e(u+ct)/10 20e20t dt 10 0 25 1 1 (u + c t x) ex/10 dx + 1 (u + c t 25)e25/10 20e20t dt 10 0 u+ct 1 1 (u + c t x) ex/10 dx + e(u+ct)/10 20e20t dt 10 0 25 1 1 (u + c t x) ex/10 dx + 1 (u + c t 25)e25/10 20e20t dt 10 u+ct25 u+ct 1 x/10 [a bek(u+c tx) ]e(u+c tx)/10 e dx + e(u+ct)/10 20e20t dt 10 0 25 1 x/10 [a bek(u+c tx) ]e(u+c tx)/10 e dx 10 u+ct25
1 (u + c t x)
u+ct
+ [a bek(u+c t25) ]e(u+c t25)/10 e25/10 20e20t dt = e0.1u (0.623472 210.875e0.19924u + 210.862e0.199241u + 0.02481u); if 25 u < 50,
2 (u)
=
1 x/10 e dx + 1 (u + c t 25)e25/10 20e20t dt 10 0 0 50u 25 1 c = 1 (u + c t x) ex/10 dx + 1 (u + c t 25)e25/10 20e20t dt 10 0 u+ct25 50u 25 1 x/10 c [a bek(u+c tx) ]e(u+c tx)/10 = e 10 0 u+ct25 + [a bek(u+c t25) ]e(u+c t25)/10 e25/10 20e20t dt.
1 (u + c t x)
(50u)/c
25
The expression can be obtained from this integral. 2 (u) = 0.
Clearly, if u 50,
It is easy to verify by Maple or Mathematics that there exists 0 < u0 < 10 such that 2 (u) > 2 (u) if 0 u < u0 and 2 (u) < 2 (u) if u0 < u 25. Also, it is easy to verify by Maple or Mathematics that if 25 u < 50, then 2 (u) < 2 (u). Obviously, 2 (u) < 2 (u) if u 50.
4