Actuarial Outpost Another question: ASM Problem 14.3
 Register Blogs Wiki FAQ Calendar Search Today's Posts Mark Forums Read
 FlashChat Actuarial Discussion Preliminary Exams CAS/SOA Exams Cyberchat Around the World Suggestions

 Long-Term Actuarial Math Old Exam MLC Forum

#1
06-13-2009, 04:21 PM
 (/iropracy Member CAS Join Date: Nov 2008 Location: East College: Post Graduate Posts: 374
Another question: ASM Problem 14.3

Given that ${_{20}}p_{40} = 0.9$ and $i = 0.02$ for end of year insurance (EOYI), in the solution we have

${_{20}}E_{40} = v^{20}{_{20}}p_{40} = \frac{0.9}{1.02^{20}}$.

For EOYI the pure endowment is given by

${_n}E_{x} = \sum_{k=0}^n {_k}p_{x}q_{x+k}v^n = v^{n}\sum_{k=0}^n {_k}p_{x}q_{x+k}$ This is correct isn't it?

Now for EOYI, is it true in general that the pure endowment is given by ${_n}E_{x} = v^{n}{_{n}}p_{x}$.

I cannot figure out how $v^{n}{_{n}}p_{x} = v^{n}\sum_{k=0}^n {_k}p_{x}q_{x+k}$? Can anyone help me with this? Am I missing something? Thanks!
#2
06-13-2009, 06:20 PM
 jraven Member Join Date: Aug 2007 Location: New Hampshire Studying for nothing! College: Penn State Posts: 1,314

Quote:
 Originally Posted by (/iropracy Now for EOYI, is it true in general that the pure endowment is given by ${_n}E_{x} = v^{n}{_{n}}p_{x}$.
Yes. A (unit) pure endowment pays $1 at the end of the period if (and only if) you're alive at that time. So ${_{n}E_{x}} = {_{n}p_{x}} \,\cdot\, \nu^n$, i.e. the product of the probability that you survive to receive the payment times the present value of that payment. Quote:  I cannot figure out how $v^{n}{_{n}}p_{x} = v^{n}\sum_{k=0}^n {_k}p_{x}q_{x+k}$? Can anyone help me with this? Am I missing something? Thanks! They're not equal. $v^{n}\sum_{k=0}^{n-1} {_k}p_{x}q_{x+k}$ is the value of a policy that pays$1 at the end of n years if you died sometime over the course of those n years** (note: I had to change the upper limit of the sum to n-1 for what I just said to be correct) -- it's sort of the logical opposite of a pure endowment, which pays if you've survived those n years.

** This is easier to understand if you first rewrite the sum as
$\sum_{k=0}^{n-1} {_k}p_{x}q_{x+k} = {_{n}q_{x}}$
__________________
The Poisson distribution wasn't named after a fish -- it was named after a man ... who was named after a fish.
#3
06-13-2009, 11:55 PM
 Abraham Weishaus Member SOA AAA Join Date: Oct 2001 Posts: 7,253

No equality of this type appears in the question or solution to ASM 14.3.
#4
06-15-2009, 02:04 PM
 (/iropracy Member CAS Join Date: Nov 2008 Location: East College: Post Graduate Posts: 374

Ok your right that the equality is not given anywhere in 14.3.... but the implication (so I thought) was.

So to change the question ..... is the following correct for an end of year pure endowment..

${_n}E_{x} = \sum_{k=0}^n {_k}p_{x}q_{x+k}v^n = v^{n}\sum_{k=0}^n {_k}p_{x}q_{x+k}$

In the continuous case we sum with $v^n$ but on eoy we don't... is that correct?
#5
06-15-2009, 02:14 PM
 Abraham Weishaus Member SOA AAA Join Date: Oct 2001 Posts: 7,253

As Jeff Raven said, no. You're thinking of a term insurance, not a pure endowment, and the range of indices isn't right.
#6
06-15-2009, 02:17 PM
 (/iropracy Member CAS Join Date: Nov 2008 Location: East College: Post Graduate Posts: 374

That's what I thought. Yep, still learning.