Actuarial Outpost Course 3 May 2001 #8
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 Long-Term Actuarial Math Old Exam MLC Forum

#1
11-04-2001, 01:08 PM
 phdmom Member SOA AAA Join Date: Nov 2001 Posts: 295

This is how I would work this problem...

P(t)=100e^(.01 + .02Y(t)) where Y(t) is standard Brownian motion. Y(1) is N(0,1), so the simulated value .1587 gives Y(1)=-1 and P(1) = 99. Y(2) is N(0,2), so the simulated value .9332 gives Y(2)=1.5*sqrt(2) and P(2)=104, for a difference of 5. Wrong answer.

Can anyone tell me what's wrong with the above reasoning?
#2
11-05-2001, 02:34 AM
 Kiitos Member CAS AAA Join Date: Nov 2001 College: BYU Favorite beer: Root Posts: 37

The problem in your logic comes when you get the second simulated value. Y(2) is not N(0,2), but N(0,1). The reason for this comes that we assume that after the first simulated number, we start over again. Hope that helps.
#3
11-05-2001, 08:28 AM
 boognish Member Join Date: Oct 2001 Posts: 80

PhdMom,

I think the key is when they state in the problem that "simulation projects the stock price in steps of time 1"