| Author |
Comment/Response |
Student0002012
|
 12/22/12 10:15pm
Hi again,
Im pretty sure i found the solution for my problem.
For my problem i used the mathematica function Nest[] that is really usefull for nested functions.
I would be really happy if someone that has the "knowledge" would confirm that my solutions is a possible way to solve this problem.
pension = 1000000; Risky = 0.6; Riskless = 0.4; riskfree = 0.04; mu = \
0.12; sigma = 0.3; draw = 150000;
value[pension_] :=
pension*(Risky*
Exp[mu + sigma*RandomReal[NormalDistribution[0, 1]]] +
Riskless*Exp[riskfree]) - draw;
Count[{Table[Nest[value, pension, 10], {100}]}, _?NonNegative, 2]/
100 // N
Attachment: basic-solved.nb, URL: Right?, |
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