2008 年春季中国精算师资格考试04 寿险精算数学答案详解

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
∫12.77 exp(−μt ) ⋅ μ d
.. 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 = = 0.1 Px = .. = .. 5 ax ax
-3-
Provided by leon_yxw Edited by clzu@
12. E (Y ) =
+∞
∑a
..
k =0
+∞
k +1
⋅ k | q95 =0.28×1+0.33×( 1 + v )+0.39×( 1 + v + v 2 )=2.0263
2 2
E (Y 2 ) = ∑ Y 2 ⋅ k | q95 =0.28×1+0.33× (1 + v ) +0.39× (1 + v + v 2 ) =4.6573
2
Var(Z)= E(Z2)-( E(Z))2 =0.4464
b1 -6.048 b1 (常数项省略)
−1
2
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 + 0.1t )
0 50 0
Provided by leon_yxw Edited by clzu@
2008 年春季中国精算师资格考试 04 寿险精算数学答案详解
1. 3 p70
=
S (73) S (70)
=0.95
2
p71
=
S (73) S (71)
=0.96
p70 = 5
1 p70 × 4 p71 =
S (71) S (70)
_
_ _ 1 − A50 _ = A50 − A40 _ _
1 − A40
50
A50 = ∫ v t t px μ x +t dt = ∫ v t ⋅
0 0
+∞
1 1 _ dt = a 50 = 0.3742 100 − 50 50
同理 A40 =
21.
_
1 _ a = 0.3233 ， 60 60
18.
Ax:n = A x:n + n Ex ⇒ Ax:n = 0.804 − 0.6 = 0.204
1 1
_ _
_
i
_
Ax:n =
1
δ
1 1 Ax ⇒ Ax = 0.204 × 0.0392 ÷ 0.04 = 0.19992 :n :n
1 Ax:n = A x + n Ex = 0.19992 + 0.6 = 0.79992 :n
1 1 = ω − 10 40
ω −10
l10+t =40- t ,由均匀分布的性质可知 fT (10) (t ) =
E[T (10)] = ∫
0
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
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
-2-
Provided by leon_yxw Edited by clzu@
(IA)35-A35=1E35 × (IA)36= v ⋅ (IA)36=[(IA)35-A35]/ v ⋅
× 4 p71 =
3
p70
2
p
− μ × e ∫71 x dx =0.89
75
71
μx+ qx 2. 由死亡服从 UDD 假设，可得 μ = ，所以 qx = 1 1 x+ 1 − q 1 − 2 μ x+ 2 x
1 2
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 − =1 − 3. 由 x=
(12 )
(12 )
p p
35
× (IA)36
=3.81
35
.. .. 1 1 9. a = a 50 - 12 = a 50 ⋅ α (12) + β (12) - 12
50
x
k = 95 − x
95 0 100 0.28 1
96 1 72 0.33
97 2 39 0.39
98 3 0 0
lx
k | 95
8
∫0
35
exp(−δ t ) ⋅ exp(− μ30+t t ) ⋅ μ 30+t d + 35E 30
30 + t
∫0 exp(−δ t ) ⋅ exp(−μ
μ 30 + t μ 30 + t + δ
t ) ⋅ μ 30+t dt + ∫ exp(−δ t ) ⋅ exp(− μ30+t t ) ⋅ μ 30+t dt ]
.. Ax − vqx 1 − d a x − vqx 1 − (1 − 0.9) × 5 − 0.9 × 0.05 = = = 0.091 P = .. .. 5 ax ax
1 − Ax +10 .. .. ⇒ Ax +10 = 0.6, a x +10 = 4 V = Ax +10 − Px ⋅ a x +10 = Ax +10 − Px ⋅ d .. 10Vx = 5000( Ax +10 − P ⋅ a x +10 ) = 5000 × (0.6 − 4 × 0.091) = 1180
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
50
50
E(Z2)=
∫ (1 + 0.1t )
10 x
-5-
Provided by leon_yxw Edited by clzu@
20.
10
A40 _ V ( A40 ) = A50 − P( A40 ) ⋅ a 50 = A50 − _ a 50 a 40
_ _
_
_ _
_
_
_
= A50 − A40 ×
_
_
1 − A50 δ _ × δ 1 − A40
= Pr(1000vT − 10
1 − vT
17.
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 ⎝ 2
b1(k =1) {10 −b1( k =2)
Pr[ K (30) = 1] = q30 =0.1 Pr[ K (30) = 2] =
p30 q31 =(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
h = 9.5 ,即 ln ζ 0.9
-1-
Provided by leon_yxw Edited by clzu@
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[ +
k =0
Var (Y ) = E (Y 2 ) − ⎡ ⎣ E (Y ) ⎤ ⎦ =0.55
2
13.
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 L (π ) = 20000v 39+1 − π 1 − v 39+1 = 1944.443754 − 15.94907468π ≤ 0 d