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D.W. Simpson 

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#453




Hey guys, I may have a stupid question, but I don't understand something on ASM Practice Exam 1, #11. After you recognize that this is a beta distribution, to find e_x:[5], why can't you just use the recursion formula by finding e_0, 5P0, and e_5?
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#454




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Solving implicitly for E(X^5) gives you 2.91667 which ASM also gets by integrating over (0,5) tp0*dt. In general, E(X^n) = integral (0,n) of f(x)*x*dx + n*np0 So for all values beyond the warranty period, 5, the value 5 is being assigned which we need to discard... So the author subtracts 5*5p0 from 2.91667, then divides by Pr(X<5) because we're conditioning on it. With either approach, direct or by backing it out from the recursive formula, you still need to account for that. Last edited by dfunkhou; 04282012 at 10:11 PM.. 
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#456




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the complete expectation from e from x5 tpx...notice that the last piece is 5tp0 (the probability of surviving to time 5...meaning lasting beyond the warranty....meaning that this current expectation that you calculated includes the 5*the probability of not getting the warranty.... If your still confused...the other way to doing a "partial complete expectation" is from x5 tpx(ux) + n*tpx (the n*tpx is not in the integral)....so that very last piece is part of the complete expectation when normally calculating and what it essentially states is surviving through the time period....BUT... what the questions asks is the complete expectation from those that die within the warranty...so when you think about it like that...YOU DONT WANT THE LAST PIECE! (even though its in the regular complete expectation calc). but your not done yet.... you have to condition on all those pieces not leaving the time period.....so divide by that probability... Last edited by billyMcStuds4life; 04292012 at 11:30 AM.. 
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E(X^5) = E(XX<5)*F(5) + E(5X>=5)*S(5) = E(XX<5)*F(5) + 5*S(5) => E(X^5)  5*S(5) = E(XX<5)*F(5) => E(XX<5) = [E(X^5)  5*S(5)]/F(5) Last edited by dfunkhou; 04292012 at 01:44 PM.. 
#458




In ASM example 53D, instead of using reversionary annuities, shouldn't you be able to add 5 times the ten year temporary life annuity for 70 and 7 times the ten year temporary life annuity for 65, then subtract two times the the joint ten year temporary life annuity and then add back in ten times the ten year deferred joint life annuity? I'm coming up with a much different answer that way.
Edit: now that I'm really digging into this, it seems that the solution is assuming that payments to 65 and 70 when their counterpart is dead do not begin until ten years after today, when the question states that those payments end ten years after today. Is anyone else with me? Edit 2: ok found my mistake. This shit is confusing as hell.
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Last edited by Practicing; 04292012 at 02:23 PM.. Reason: Figured it out 
#459




In fact, all of the terminology seems mixed for reversionary annuities... Do they pay after the later of the certain time period and the death of the other life or do they pay after the death of the other life but stop at the death of the certain period?
Edit: I think I figured it out. It's not one or the other and that's the trouble. So confusing.
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Last edited by Practicing; 04292012 at 02:24 PM.. 
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