Fall 2002 - Course 3 Solutions
Fall 2002 - Course 3 Solutions
Fall 2002 - Course 3 Solutions
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Question #23<br />
Answer: C<br />
Time of first claim is T 1 = − 1/ 3log 05 . = 023 .<br />
0.<br />
23<br />
b g . Size of claim = =<br />
Time of second claim is T 2 = 0. 23− 1/ 3log 0. 2 = 0.<br />
77<br />
Time of third claim is T 3 = 077 . − 1/ 3log 01 . = 1.<br />
54<br />
10 1.<br />
7<br />
0.<br />
77<br />
b g . Size of claim = =<br />
1.<br />
54<br />
b g . Size of claim = =<br />
Since initial surplus = 5> first claim, the first claim does not determine c<br />
0 z.<br />
77<br />
4<br />
Test at T 2 : Cumulative assets = 5 + ct dt = 5 + 0 . 054c<br />
Cumulative claims = 1.7 + 5.9 = 7.6<br />
Assets ≥ claims for c ≥ 4815 .<br />
Test at T 3 : Cumulative claims = 1.7 + 5.9 + 34.7 = 42.3<br />
Cumulative assets = 5+ ct dt = 5+<br />
1732 . c<br />
0<br />
z. 154<br />
0<br />
Assets ≥ claims for c ≥ 2154 .<br />
If c < 4815 . , insolvent at T 2 .<br />
If c > 4815 . , solvent throughout.<br />
49 is smallest choice > 48.15.<br />
4<br />
10 5.<br />
9<br />
10 34.<br />
7<br />
Question #24<br />
Answer: C<br />
µ xy = µ x + µ y = 014 .<br />
µ<br />
Ax<br />
= Ay<br />
=<br />
+ = 007 .<br />
= 05833 .<br />
µ δ 007 . + 005 .<br />
µ xy<br />
Axy<br />
=<br />
axy<br />
+ = 0.<br />
14 014 .<br />
1<br />
= = 0 7368 =<br />
µ δ 0 14 + 0 05 019<br />
µ + δ<br />
= 1<br />
. and<br />
. . .<br />
014 . + 0.<br />
05<br />
xy<br />
P<br />
A xy A + A − A<br />
= =<br />
a a<br />
xy<br />
x y xy<br />
xy<br />
b g<br />
2 0. 5833 −0.<br />
7368<br />
=<br />
= 0.<br />
0817<br />
52632 .<br />
xy<br />
= 5.<br />
2632