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  #31  
Old 10-27-2017, 10:02 PM
hunkal hunkal is offline
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Originally Posted by mel.fel View Post
yeah sick or healthy
so i did 1-p(dead). since it seemed easier
Did you get .1335 or phi(-1.101)?

Tha wasn't an answer. I must have calculated my probabilities wrong. What was the answer you got if you remember?
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  #32  
Old 10-27-2017, 10:04 PM
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ngxxx081 ngxxx081 is offline
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i did prob of being sick at the beg of year 3 (so two time periods) then 1000*prob and variance is 1000*(1-prob)*prob and then did the normal approx.
it wasn't four because i struggled with four.. and i remember totally guessing that.
It was #3 - one thing I didn't catch at first was that you wanted the probability of being alive (healthy OR sick, but NOT dead) at the BEGINNING of t=3 which is the end of t=2. I was calculating probability of being healthy only which and combining sick with dead. After realizing the syntax error I corrected my calculations and got something that made more sense and one of the answer choices.
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  #33  
Old 10-27-2017, 10:05 PM
hunkal hunkal is offline
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How did you guys calculate the variance of the 10 pay annuity at time 5 where they give you premium is 166.5 or something? This was one of the written answer, later they ask for covariance of the whole life insurance and term annuity which I also couldn't get

I also couldn't get the premium for the common shock model. Or the pv of the benefit when the guy dies after the girl

Also, what is the Kolmogorov forward differential equation for tpx in state j? Why only one state?

Last edited by hunkal; 10-27-2017 at 10:18 PM..
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  #34  
Old 10-27-2017, 10:20 PM
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ngxxx081 ngxxx081 is offline
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Originally Posted by hunkal View Post
How did you guys calculate the variance of the 10 pay annuity at time 5 where they give you premium is 166.5 or something? This was one of the written answer, later they ask for covariance of the whole life insurance and term annuity which I also couldn't get

I also couldn't get the premium for the common shock model. Or the pv of the benefit when the guy dies after the girl
The "simpler" approach to calculating variance of an annuity is to look at it in terms of insurance so a_x = [1-A_x]/d; taking the variance of this expression and with a little algebra you get Var((P/d)(1-v^Kx+1)) which can be re-written as Var((P/d)-(P/d)v^Kx+)). Then taking the variance of each of the two items you get:

(P/d)^2*Var(v^Kx+1) (P/d is a constant and variance of a constant is zero).

You will then recognize Var(v^Kx+1) as variance for an insurance item (this was a simplified approach, use appropriate corresponding annuity/insurance relation).

I couldn't prove the premium for the common shock model question - at this point I had lost steam and still had to get to the last question.
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  #35  
Old 10-27-2017, 10:23 PM
mel.fel mel.fel is offline
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Originally Posted by ngxxx081 View Post
It was #3 - one thing I didn't catch at first was that you wanted the probability of being alive (healthy OR sick, but NOT dead) at the BEGINNING of t=3 which is the end of t=2. I was calculating probability of being healthy only which and combining sick with dead. After realizing the syntax error I corrected my calculations and got something that made more sense and one of the answer choices.
yeah i caught that too but my brain wasn't working so i ended up going 1-dead, whicha lso worked.
C /#3 sounds right to me,
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  #36  
Old 10-27-2017, 10:24 PM
mel.fel mel.fel is offline
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The "simpler" approach to calculating variance of an annuity is to look at it in terms of insurance so a_x = [1-A_x]/d; taking the variance of this expression and with a little algebra you get Var((P/d)(1-v^Kx+1)) which can be re-written as Var((P/d)-(P/d)v^Kx+)). Then taking the variance of each of the two items you get:

(P/d)^2*Var(v^Kx+1) (P/d is a constant and variance of a constant is zero).

You will then recognize Var(v^Kx+1) as variance for an insurance item (this was a simplified approach, use appropriate corresponding annuity/insurance relation).

I couldn't prove the premium for the common shock model question - at this point I had lost steam and still had to get to the last question.
ha im the opposite. i got the premium but not the variance.
again, just ran out of time. lmao.
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  #37  
Old 10-27-2017, 10:28 PM
Demo99 Demo99 is offline
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ha im the opposite. i got the premium but not the variance.
again, just ran out of time. lmao.
I ended up just writing formulas for this entire question with all the parts, without solving for a single value. Hoping to get at least a few points on this whole thing.
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  #38  
Old 10-27-2017, 10:33 PM
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they asked for ADB.
I got an answer, did ADB as 1.5AV6 then foudn AV6 and x1.5 to get an answer
This is correct and the answer was E. The AV without the CF was larger then the AV with the CF which means the CF is in play. The question asked for the ADB which would be 1.5AV. If you did not use the CF and just did ADB = FA - AV you would have got choice A which is not correct.
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  #39  
Old 10-27-2017, 10:40 PM
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Originally Posted by noone View Post
This is correct and the answer was E. The AV without the CF was larger then the AV with the CF which means the CF is in play. The question asked for the ADB which would be 1.5AV. If you did not use the CF and just did ADB = FA - AV you would have got choice A which is not correct.
I thought for these types of problems you went with the method that yielded the higher cost of insurance which for this question I determined it to be without corridor factor, which is it might seem just as many chose FA-AV as those who did AV*(corridor-1).

Of course this was one of last few concepts I started understanding better in the last month so maybe my calculations could have been off.
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  #40  
Old 10-27-2017, 10:41 PM
cixelsidmaikool cixelsidmaikool is offline
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yeah, that sounds right (unfortunately). For some reason I keep mixing up AV and COI. higher COI/lower AV = corridor in play, but I flipped it.
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