Fall 2002 - Course 3 Solutions
Fall 2002 - Course 3 Solutions
Fall 2002 - Course 3 Solutions
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Question #36<br />
Answer: E<br />
k k+<br />
1<br />
p k<br />
β = mean = 4; = β / 1+<br />
β<br />
n P N = n<br />
b g<br />
b g x f<br />
b1<br />
g bxg f<br />
b2<br />
g bxg f<br />
b3<br />
g bxg<br />
0 0.2 0 0 0 0<br />
1 0.16 1 0.25 0 0<br />
2 0.128 2 0.25 0.0625 0<br />
3 0.1024 3 0.25 0.125 0.0156<br />
f<br />
b k g b x g = probability that, given exactly k claims occur, that the aggregate amount is x.<br />
b1<br />
g b g<br />
bkg b g<br />
b g<br />
x<br />
k−1<br />
e b gj b g<br />
f x = f x ; the claim amount distribution for a single claim<br />
b g<br />
∑<br />
f x = f j x f x − j<br />
j=<br />
0<br />
x<br />
sb g ∑ b g bkgb g<br />
k = 0<br />
k<br />
f x = P N = k × f x ; upper limit of sum is really ∞ , but here with smallest possible<br />
claim size 1, f<br />
f s 0 02<br />
b g b g = 0 for k<br />
x<br />
> x<br />
g = .<br />
g = . * . = .<br />
= . * . + . * . = .<br />
b g = + + =<br />
b g = . + . + . + . = .<br />
f s 1 016 0 25 0 04<br />
f s 2 016 0 25 0128 00625 0048<br />
f s 3 016 . * 025 . 0128 . * 0125 . 01024 . * 00156 . 0.<br />
0576<br />
F s 3 02 004 0048 0 0576 0346
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