2008年春季中国精算师资格考试04寿险精算数学答案详解

∫0
35
exp(−δ t ) ⋅ exp(− µ30+t t ) ⋅ µ 30+t d + 35E 30
5 0
∫0
8
exp(−δ t ) ⋅ exp(− µ30+t t ) ⋅ µ 30+t dt + ∫ exp(−δ t ) ⋅ exp(− µ30+t t ) ⋅ µ 30+t dt ]
µ 30 + t µ 30 + t + δ
.. a x:n = 1 − Ax:n = 5.20208 d
Ax:n 0.804 1000 P( Ax:n ) = 1000 .. = 1000 × = 155 5.20208 a x:n
19.
_
_
设该保险的均衡纯保费为P .. Ax 1 − d a x 1 − (1 − 0.9) × 5 Px = .. = .. = = 0.1 5 ax ax
..
L (π ) = 20000v 39+1 − π π ≥ 121.92
1 − v 39+1 = 1944.443754 − 15.94907468π ≤ 0 d
1 1 14. 令 P35:20 =x , P35:20 =y
1 1 A35:20 + A35:20 ⋅ A55 A35 1 1 = = P35:20 + P35:20 ⋅ A55 .. 20 P 35 = .. a35:20 a35:20
l
o
l10+t =40- t ,由均匀分布的性质可知 fT (10) (t ) =
E[T (10)] = ∫
ω −10
0
1 1 = ω − 10 40
t fT (10) (t )dt =20
E[T 2 (10)] = ∫
ω −10
0
t
2
fT (10) (t )dt =533
Var[T (10)] = E[T 2 (10)] − {E[T (10)]}2 =133
× 4 p71 =
3
p70
2
p
− µ × e ∫71 x dx =0.89
75
71
2. 由死亡服从 UDD 假设，可得 µ
x+
1 = 2
µx+ qx ，所以 q = x 1 1− 1 q 1 − 2 µ x+ 2 x
1 2
1 2
不难求出， q80 =0.02, q81 =0.04, q82 =0.06 故 80.5 岁的人在两年之内死亡的概率 2 q80.5 = 1 −
0 0
+∞
+∞
a 65 = ∫ v t ⋅ t p65 dt = ∫ e −0.04t ⋅ e −0.02t dt =
0 0
_
+∞
+∞
1 50 = 0.04 + 0.02 3
APV 均衡保费π = _ = 60 a 65
2
V = ∫ b2+t ⋅ v t ⋅ t p67 µ67 ( t ) dt − π a 67
2
18.
_ _
Ax:n
1 Ax :n =
_
1
_
1 Ax:n = Ax:n + n Ex ⇒ Ax :n = 0.804 − 0.6 = 0.204
i 1 1 A x:n ⇒ A x = 0.204 × 0.0392 ÷ 0.04 = 0.19992 :n δ 1 = Ax + n Ex = 0.19992 + 0.6 = 0.79992 :n
x+y= P =0.042 35:20
x+0.6099y=0.0299 解得：x=0.011 y=0.031
15. 令该保险的均衡纯保费为 P，由题意得：
1 1 P ⋅ a 30:10 = A30:10 + A30:10 ⋅10 P 1 A30:10 1 A30:10 − A30:10
..
P=
..
=
+∞
12. E (Y ) =
+∞
∑a
k =0
2
..
k +1
⋅ k | q95 =0.28×1+0.33×( 1 + v )+0.39×( 1 + v + v 2 )=2.0263
2
E (Y
2
) = ∑Y
k =0
⋅ k | q95 =0.28×1+0.33× (1 + v ) +0.39× (1 + v + v 2 ) =4.6573
1
故
ζ 0.9
= exp(−9.5δ ) =0.5655
5. 由题意可知，该保险相当于保额 1000 元的 35 年期两全保险+1000 元保额的 8 年期定期 保险（5-8 年内被保险人只有一个孩子小于 11 岁）+1000 元保额的 5 年期定期保险（5 年内 两个孩子都小于 11 岁） 此保单的趸交保险费=1000（ A30:35 + A30:8 + A30:5 ）= 1000[ +
(12 ) (12 )
p
35
=3.81
.. .. 1 1 9. a = a 50 - 12 = a 50 ⋅ α (12) + β (12) - 12
50
1 =13.5 × 1-0.4665- 12 =12.95
(注：课本上公式为 + β (12) ，但这里 β (12) 为正值，故改为 − β (12) )
3
2
Var (Y ) = E (Y 2 ) − ⎡ ⎣ E (Y ) ⎤ ⎦ =0.55
13.
2
Pr ⎡ ⎣ L (π ) > 0 ⎤ ⎦ < 0.5ak +1 Pr(20000v k +1 − π a k +1 > 0) < 0.5 由于 39 q40 = 0.4939及 40 q40 = 0.5109 并且L (π ) = 20000v k +1 − π 1 − v k +1 π π = (20000 + )v k +1 − d d d 是k的减函数,意味着L(π )取满足条件的最高值时,k必须取39,故
4. 令 h =
ln ξ 0.9 , v = exp( −δ ) < 1 ,则 ln v
+∞
Pr ( Z ≤ ζ 0.9 ) = Pr (vT ≤ ζ 0.9 ) = Pr (T ≥ h) = ∫ =∫
解之得：
h = 9.5 ,即 ln ζ 0.9
+∞
h
fT (t )dt
h
1 dt 95
1 = 95 (95 − h) = 0.9 = 9.5ln v
50
50
E(Z2)=
∫ (1 + 0.1t )
p
−2
Var(Z)= E(Z2)-( E(Z))2=0.04
1 1 8. A35:1 = A35:1 + A35:1 =v⋅
35
+ v ⋅ q35 = v =0.9439
(IA)35-A35=1E35 × (IA)36= v ⋅
p
35
× (IA)36
2
(IA)36=[(IA)35-A35]/ v ⋅
b1(k =1) {10 −b1(k =2)
Pr[ K (30) = 1] = q30 =0.1 Pr[ K (30) = 2] =
p q
30
31
=(1-0.1) × 0.6=0.54
E(Z)= b1 × 0.1 + (10- b1 ) × 0.54 E(Z2)=
b1 × 0.1 + (10 − b1)2 × 0.54 b1 -6.048 b1 (常数项省略)
2
k=50000 ÷ 12.95 ÷ 12=322 10. a x =
1 =10 µ +δ ⎛ 1 − vt ⎞ ln 0.6 ⎞ ⎛ > 10 ⎟ = Pr ( v t < 0.6 ) = Pr ⎜ T > Pr (a T ≥ a x ) = Pr ⎜ ⎟ −δ ⎠ ⎝ ⎝ δ ⎠
= Pr (T > 12.7706 ) = =0.4648
17.
_
4
2 ⎛ P⎞ 由Var ( L ) = 0.1 ⇒ ⎜1 + ⎟ ⎡ 2 A49 − ( A49 ) ⎤ = 0.1 ⎣ ⎦ ⎝ d⎠ P ⇒ = 0.772598818 d ⎛ ⎛ P⎞ P⎞ ⎛ P⎞ P E ( L ) = E ⎜ V K +1 ⎜1 + ⎟ − ⎟ = A49 ⎜ 1 + ⎟ − = −0.25 ⎝ d⎠ d⎠ ⎝ d⎠ d ⎝
10 x
20.
10
A40 _ V ( A40 ) = A50 − P ( A40 ) ⋅ a 50 = A50 − _ a 50 a 40
_ _
_
_ _
_
_
_
1 − A50 _ × = A50 − A40 × δ 1 − A40
_
_
_
δ
_ _ 1 − A50 _ = A50 − A40 _
1 − A40
.. Ax − vqx 1 − d a x − vqx 1 − (1 − 0.9) × 5 − 0.9 × 0.05 P = .. = = = 0.091 .. 5 ax ax
1 − Ax +10 .. .. V = Ax +10 − Px ⋅ a x +10 = Ax +10 − Px ⋅ ⇒ Ax +10 = 0.6, a x +10 = 4 d .. 10Vx = 5000( Ax +10 − P ⋅ a x +10 ) = 5000 × (0.6 − 4 × 0.091) = 1180
x
k = 95 − x
95 0 100 0.28 1
96 1 72 0.33
97 2 39 0.39
ቤተ መጻሕፍቲ ባይዱ
98 3 0 0
lx
k | 95
q
Y = ak +1 Y2
..
1+ v
1 + v + v2
2 2 2
1 + v + v 2 + v3
2
1
(1 + v )
(1 + v + v ) (1 + v + v
+ v3 )
1
a 30:10 − 10 A30:10
16.
1 − A30:10
= 0.039
1
d
− 10 A30:10
保险人面临正损失的概率即Pr( L > 0) = Pr(1000vT − 10 a T > 0) 1 − vT 1 > 0) = Pr(1200vT − 200 > 0) = Pr(vT > ) δ 6 2 35.8352 35.8352 ln 6 t = Pr(T < ) = Pr(T < 35.8352) = ∫ fT ( t )dt = = 0.51 ⏐ 0 0 δ 2500 = Pr(1000vT − 10
2008 年春季中国精算师资格考试 04 寿险精算数学答案详解
(申明：本解答由人大经济论坛保险精算与风险管理版 leon_yxw 所创 版权所有 侵权必究)
1. 3 p70
S (73)
=
=0.95
2
p71
S (73)
=
=0.96
S (70)
S (71)
p70 = 5
1 p70 × 4 p71 =
S (71) S (70)
=1000 （
{1 − exp[−35( µ
30 + t
+ δ
)]} + 35E 30 +
µ 30 + t µ 30 + t + δ
{1 − exp[−8( µ
30 + t
+ δ
)]} +
µ 30 + t µ 30 + t + δ
{1 − exp[−5( µ
30 + t
+ δ
)]}）
=796 6. 由题意得 Z= bk +1 ivk +1 =
∫12.77 exp(−µt ) ⋅ µ d
+∞
11.
ax:4 = ∑ ak ⋅ k −1| qx + 4 px ⋅ a4
k =1
..
4
..
..
= 1.00 × 0.33 + 1.93 × 0.24 + 2.80 × 0.16 + 3.62 × 0.11 + (1 − 0.33 − 0.24 − 0.16 − 0.11) × 3.62 = 2.2
5
_
A50 = ∫ vt t px µ x +t dt = ∫ v t ⋅
0 0
+∞
50
1 1 _ dt = a 50 = 0.3742 100 − 50 50
同理 A40 =
21.
_
1 _ a = 0.3233 ， 60 60
我们可以求出该险种的现值 :
APV = ∫ bt ⋅ vt ⋅ t p65 µ65 ( t ) dt = ∫ 1000e0.04t ⋅ e −0.04t ⋅ 0.02e −0.02t dt = 1000
2
2
Var(Z)= E(Z2)-( E(Z))2 =0.4464
当 b1 =6.048/(2 × 0.4464)=6.8 时，Var(Z)最小 7. 给付现值函数 Z = bt ⋅ vt = (1 + 0.1t ) E(Z)=
−1
∫ (1 + 0.1t )
0 50 0
50
−1
⋅ t P50 ⋅ µ 50+t dt =0.02 × 10 × ln(1 + 0.1t ) =0.35835189 ⏐ 0 ⋅ t P50 ⋅ µ 50+t dt =0.02 × 10 × (-1) (1 + 0.1t )−1 =0.16666667 ⏐ 0
l82.5 l80.5
= 1−
l80 ⋅
p80 ⋅ p81 (1 − 0.5q82 ) l80 (1 − 0.5q80 )
=0.0782
=1 − 3.
0.98 × 0.96 × (1 − 0.5 × 0.06 )
1 − 0.5 × 0.02
由 x = ω − x ,可知 x 服从均匀分布，由 e0 =25，可知 ω =50