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#1
04-03-2012, 10:02 PM
 Justin.M Guest Posts: n/a
SoA #263

Liquidz and I were wondering if either Abe or Jim Daniel could shed some light on the SoA's solution to SoA #263.

We fully understand the initial set up:

Integate over t=0 to t=.25, the probability that (y)=40.5 is already dead times the density function of (x)=30.5

Clearly, the probability that y is already dead is (t*.6) / (1-.5*.6), by our UDD assumption. Now we just need our density function for x. This should be tPx * mu(x+t).

However, it appears the solution just has mu(x+t). Where are we going wrong?

http://www.soa.org/files/pdf/edu-201...-solutions.pdf

Thanks.
#2
04-03-2012, 10:41 PM
 liquidz Member SOA Join Date: May 2011 Posts: 472

Thanks for looking into this Justin.
#3
04-04-2012, 10:16 AM
 Gandalf Site Supporter Site Supporter SOA Join Date: Nov 2001 Location: Middle Earth Posts: 26,563

Quote:
 Originally Posted by Justin.M Clearly, the probability that y is already dead is (t*.6) / (1-.5*.6), by our UDD assumption. Now we just need our density function for x. This should be tPx * mu(x+t). However, it appears the solution just has mu(x+t). Where are we going wrong? http://www.soa.org/files/pdf/edu-201...-solutions.pdf Thanks.
mu(x+t) is a function of t. Where do you see mu(x+t) in that link?
#4
04-04-2012, 10:32 AM
 Justin.M Guest Posts: n/a

The way I see it, the expression .4 / (1 - .5 * .4) is mu(30.5). But the way the problem is worked, the SoA sees this expression as being the density function.

I'm suspecting the solution has something to do with this fact of UDD: I know that under UDD, q(x) is our density function, tPx * mu(x + t), so would q(x + s), where s<1, be our density f(x) at some age 30< 30+s<31?

This doesn't seem to work either, since q(30.5) = (.5 * .4)/ (1-.5*.4), correct?

Last edited by Justin.M; 04-04-2012 at 10:43 AM..
#5
04-04-2012, 12:39 PM
 Gandalf Site Supporter Site Supporter SOA Join Date: Nov 2001 Location: Middle Earth Posts: 26,563

You seem to agree that you want $_tp_{30.5} \, \mu(30.5+t)$

You seem to agree that the expression is right for the specific instant that t=0, since $_0p_{30.5}=1$

Try it for t=0.1. What are $_{0.1}0p_{30.5}$ and $\mu(30.6)$, and thus what is the product? If .4 / (1 - .5 * .4), that's a highly encouraging sign; worth at least trying something like t=0.2; maybe worth jumping to the general case. If not .4 / (1 - .5 * .4), then the SOA formula might happen be right, but only by coincidence.
Quote:
 Originally Posted by Justin.M Since q(30.5) = (.5 * .4)/ (1-.5*.4), correct?
Actually, not right, since the symbol q(30.5) is the probability that 30.5 dies between 30.5 and 31.5, and you have the probability of dying only in the next 6 months.
#6
04-04-2012, 02:17 PM
 liquidz Member SOA Join Date: May 2011 Posts: 472

I think we are making it harder than what it needs to be.

simply put, since it is UDD, and since they both have to die in the next .25 years. Half the time it will be 30.5 dying the other half it will be 40.5 dying.

=.5*.25q30.5*.25q40.5
#7
04-04-2012, 02:42 PM
 Justin.M Guest Posts: n/a

Alright Liquidz, here we go...

.1 P 30.5 * mu(30.6)::
[1-(.1Qx)/(1-.5Qx)] * (Qx)/(1-.6Qx) = .5 (after calculating)

.2 P 30.5 * mu(30.7)::
[1-(.2Qx)/(1-.5Qx)] * (Qx)/(1-.7Qx) = .5

General Case:

(t-.5) P 30.5 * mu(30 + t)

= [1- [(t-.5)Qx]/[1-.5Qx]] * (Qx)/(1-tQx)

= [1- (tQx-.5Qx)/(1-.5Qx)] * (Qx)/(1-tQx)

= [(1-.5Qx)/(1-.5Qx) - (tQx-.5Qx)/(1-.5Qx)] * (Qx)/(1-tQx)

= (1-tQx)/(1-.5Qx) * (Qx)/(1-tQx)

= Qx/(1-.5Qx)

Wow, wasn't it nice of the SoA to carry out some algebraic steps to show us where they got .4/(1-.5 Qx) from ?!? Geez, I would never want to do this on an exam.

@Gandalf
Is there some sort of property that I'm missing that would allow me to automatically know that our density is Qx/(1-.5Qx)??
#8
04-04-2012, 03:00 PM
 Justin.M Guest Posts: n/a

Quote:
 Originally Posted by liquidz I think we are making it harder than what it needs to be. simply put, since it is UDD, and since they both have to die in the next .25 years. Half the time it will be 30.5 dying the other half it will be 40.5 dying. =.5*.25q30.5*.25q40.5
I don't know why I'm making this so hard. I follow your logic, and it obviously yields the correct answer, but I'm missing a piece of the puzzle here.

More specifically, how do we know that half the time 40.5 dies second, and the other half of the time, 30.5 dies second. Since 40.5 has a higher Qx, wouldn't it make sense that 40.5 dies first more than half of the time?

Super confused right now.
#9
04-04-2012, 05:44 PM
 Gandalf Site Supporter Site Supporter SOA Join Date: Nov 2001 Location: Middle Earth Posts: 26,563

Quote:
 Originally Posted by Justin.M I don't know why I'm making this so hard. I follow your logic, and it obviously yields the correct answer, but I'm missing a piece of the puzzle here. More specifically, how do we know that half the time 40.5 dies second, and the other half of the time, 30.5 dies second. Since 40.5 has a higher Qx, wouldn't it make sense that 40.5 dies first more than half of the time? Super confused right now.
These are conditional probabilities. As liquidz said, you require that both of them did die. Given that 30.5 died, and UDD, he is equally likely to have died at any moment of the interval. Similarly for 40.5.
#10
04-04-2012, 05:58 PM
 Gandalf Site Supporter Site Supporter SOA Join Date: Nov 2001 Location: Middle Earth Posts: 26,563

Quote:
 Originally Posted by Justin.M @Gandalf Is there some sort of property that I'm missing that would allow me to automatically know that our density is Qx/(1-.5Qx)??
Had you studied from the old text Actuarial Mathematics, that formula would be there. Some other formulas you need for the new material wouldn't be.

Generally, you could realize that since deaths are UDD over the period (30,31) the density must be constant over (30.5,31). That constant must integrate to the probability that 30.5 dies within .5.

Integral of a constant over an interval of length .5 is just .5*the constant

Note: I said "you could realize" not "you should realize".

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