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#451
04-27-2012, 08:07 PM
 mistykz Member SOA Join Date: Mar 2008 Posts: 214

Definitely multiview, I can't stand not seeing exactly what was entered on the BA II+ (in exam conditions, at least). Just wish the keys were a bit more responsive!
#452
04-28-2012, 11:46 AM
 Practicing Member SOA Join Date: Sep 2010 Location: NYC Posts: 872

I don't know how I passed MFE with the BAII Plus.
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Quote:
 Originally Posted by Abraham Weishaus I thought the actual exam was very similar to one of the last 2 ASM practice exams, but perhaps those underestimate the difficulty of the exam.
#453
04-28-2012, 08:34 PM
 AdmiralWen Member CAS Join Date: Sep 2010 Posts: 283

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
04-28-2012, 09:55 PM
 dfunkhou Member Join Date: Aug 2010 Posts: 191

Quote:
 Originally Posted by AdmiralWen 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?
You can, but you still need to condition on breaking down in the warranty period.

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; 04-28-2012 at 10:11 PM..
#455
04-29-2012, 01:13 AM
 psp-fifa-fan Member Join Date: Jul 2010 Posts: 661

Quote:
 Originally Posted by dfunkhou You can, but you still need to condition on breaking down in the warranty period. 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.
I still don't get this problem... If we're solving the life expectancy within the warranty, why can't E(X^5) be the answer?
#456
04-29-2012, 11:25 AM
 billyMcStuds4life Member Join Date: Jun 2010 Posts: 350

Quote:
 Originally Posted by psp-fifa-fan I still don't get this problem... If we're solving the life expectancy within the warranty, why can't E(X^5) be the answer?

the complete expectation from e from x---5 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 x---5 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; 04-29-2012 at 11:30 AM..
#457
04-29-2012, 01:35 PM
 dfunkhou Member Join Date: Aug 2010 Posts: 191

Quote:
 Originally Posted by psp-fifa-fan I still don't get this problem... If we're solving the life expectancy within the warranty, why can't E(X^5) be the answer?
"Calculate the expected number of years to breakdown for those automobiles that break down during the warranty period [between 0 and 5]." So you want E(X|X<5). This is how I worked out the problem, but maybe I'm off? I summed up all Partial Expectations, breaking them up in two PE's - one before 5 and one after 5, and writing them as Conditional Expectations times probability of condition. Then just solved for E(X|X<5). I believe you can get E(X^5) by either evaluating it directly or by backing it out of the recursive formula.

E(X^5) = E(X|X<5)*F(5) + E(5|X>=5)*S(5)
= E(X|X<5)*F(5) + 5*S(5)

=> E(X^5) - 5*S(5) = E(X|X<5)*F(5)

=> E(X|X<5) = [E(X^5) - 5*S(5)]/F(5)

Last edited by dfunkhou; 04-29-2012 at 01:44 PM..
#458
04-29-2012, 01:57 PM
 Practicing Member SOA Join Date: Sep 2010 Location: NYC Posts: 872

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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Quote:
 Originally Posted by Abraham Weishaus I thought the actual exam was very similar to one of the last 2 ASM practice exams, but perhaps those underestimate the difficulty of the exam.

Last edited by Practicing; 04-29-2012 at 02:23 PM.. Reason: Figured it out
#459
04-29-2012, 02:05 PM
 Practicing Member SOA Join Date: Sep 2010 Location: NYC Posts: 872

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.
__________________
P FM MFE MLC C VEE IA FA

Quote:
 Originally Posted by Abraham Weishaus I thought the actual exam was very similar to one of the last 2 ASM practice exams, but perhaps those underestimate the difficulty of the exam.

Last edited by Practicing; 04-29-2012 at 02:24 PM..
#460
04-29-2012, 02:39 PM
 liquidz Member SOA Join Date: May 2011 Posts: 484

Quote:
 Originally Posted by Practicing 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.
do you have TIA? James does a good job explaining it. ASM tends to complicates simple subjects.

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