Fall 2002 - Course 3 Solutions
Fall 2002 - Course 3 Solutions
Fall 2002 - Course 3 Solutions
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Question #34<br />
Answer: A<br />
b g b g = + =<br />
µ τ x t<br />
0100 . 0004 . 0104 .<br />
t<br />
p<br />
b g =<br />
-0.<br />
104<br />
τ<br />
x<br />
e<br />
t<br />
Actuarial present value (APV) = APV for cause 1 + APV for cause 2.<br />
z5<br />
0<br />
z<br />
−0. 04 t −0 . 104 t 5<br />
−0. 04 t −0.<br />
104 t<br />
0<br />
z5<br />
−0.<br />
144t<br />
. , .<br />
0<br />
b g b g<br />
2000 e e 0100 . dt + 500, 000 e e 0400 . dt<br />
c b g b gh e dt<br />
= 2000 010 + 500 000 0 004<br />
e<br />
b gj<br />
2200 −<br />
= − =<br />
0 144 1 0.<br />
144 5<br />
e<br />
.<br />
7841<br />
Question #35<br />
Answer: A<br />
R = 1 − p = q<br />
x<br />
x<br />
S = 1− p × e since e<br />
0<br />
= e<br />
x<br />
b g<br />
z<br />
1<br />
1<br />
1<br />
−k − µ x + − µ x −<br />
0<br />
0<br />
So S = 0. 75R ⇒ 1− p × e = 0.<br />
75q<br />
x<br />
e<br />
e<br />
−k<br />
k<br />
−k<br />
t k dt t dt k dt<br />
= e<br />
−<br />
z<br />
z<br />
d b g i b g<br />
x<br />
q<br />
= 1−<br />
0.<br />
75<br />
p<br />
x<br />
x<br />
1<br />
0<br />
µ<br />
x<br />
b g<br />
t dt<br />
px<br />
1−<br />
qx<br />
= =<br />
1 − 0. 75q<br />
1−<br />
0.<br />
75q<br />
L<br />
NM<br />
1−qx<br />
k = ln<br />
1−0.<br />
75q<br />
x<br />
x<br />
O<br />
QP<br />
x<br />
e<br />
z<br />
−z<br />
1<br />
0<br />
k dt