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#1
12-07-2001, 09:12 AM
 Anonymous Guest Posts: n/a

Hi Again

Okay - I am just gonna start asking my questions here so I don't start tons of topics. Please help me if you can - I really want to get this exam and I need some help to do it - Thanks in advance!

okay - this question, I thought I knew how to do it but I am not getting the answer.

Death benefits payable at the end of year of death of (94) are 3, 2, 1, 1, 2, 3 for a 6 year term policy. If mu(x) = 2/(100-x) for 0<=x<=100 and i = 0 find the variance of the present value random variable. (ans 2/3)

I started by noting this was a modified demoivre law - so I know that w = 100 and
tPx=((100-x-t)/(100-x))^2
with i=0 we know that v=1

SO - I am just trying to do your typical
E(Z) and E(Z^2) but I can't get the number - I am getting a negative variance. :???: Does this mean that I can't do it this way?

Thanks again!
#2
12-07-2001, 10:06 AM
 rabbit Member Join Date: Sep 2001 Posts: 37

I think you are doing it right -- just like an old Course 1 problem.

E[Z] = 2, and E[Z^2] = 168/36, so Var[Z] = 24/36.
#3
12-07-2001, 10:39 AM
 Anonymous Guest Posts: n/a

OOPS!

that was silly of me! I need to learn how to add fractions!

Thanks
#4
12-07-2001, 02:41 PM
 Anonymous Guest Posts: n/a

Okay, I have another one - now please don't even tell me I am doing another stupid mistake!

A special fully discrete 3 year endowment insurnace to (60) has a death protection of 300 in each of the first 2 years and 200 in the third. The pure endowment benefit at age 63 is 200. Assuming demoivres law with
w=80 and i=0, find the standard deviation of the present value random variable for the policy (ans 30)

okay so what I am doing is since i=0 we get:
E(Z)= 200*3q60 + 100*2q60 +200*3P60
(where 2q60 = 1/10)
so E(Z) = 210

So shouldn't E(Z^2)= 200^2 + 100^2*2q60
but this can't be since I once again get a negative variance!

Any ideas?

Thanks!
#5
12-07-2001, 03:01 PM
 MathGuy Official Speelchecker CAS AAA Join Date: Oct 2001 Location: Boston-ish Favorite beer: Panda Posts: 2,038 Blog Entries: 5

Since (x) = (60) and w = 80, the probability of (x) dying in a particular year is 1/20. Since i = 0, Z = 300 if death occurs in the first two years (p = 2/20 = .1). Otherwise, because this is an endowment insurance, 200 is received (p = 18/20 = .9). So,

E[Z] = .1(300) + .9(200) = 210
E[Z^2] = .1(300^2) + .9(200^2) = 45000
Var[Z] = 45000 - 210^2 = 900
SD[Z] = 900^.5 = 30.
#6
12-07-2001, 03:06 PM
 Anonymous Guest Posts: n/a

okay, so I can't break it up the way I was?

I was just trying to rearrange it and since i=0 I thought that makes my v=1 and therefore timing doesn't really matter?

I see what you are doing - thanks - I just don't know why my solution won't work

appreciate it! THANKS!
#7
12-07-2001, 04:08 PM
 MathGuy Official Speelchecker CAS AAA Join Date: Oct 2001 Location: Boston-ish Favorite beer: Panda Posts: 2,038 Blog Entries: 5

The problem is that you are breaking up the payouts. I'm not sure why this is wrong, but it is. Suppose there is a 10% chance of a payment of 4. By breaking up the loss, you are saying that

E[Z] = .1(4) = .1(2) + .1(2) = .4,

which is okay, but

E[Z^2] is not .1(2^2) + .1(2^2) = .8.

In fact it is

E[Z^2] = .1(4^2) = 1.6

I would say that by breaking up the payments, you saying that the probability exists that (60) will receive the first 200 in the first 2 years, but not the other 100, which is in fact not possible.
#8
12-10-2001, 08:29 AM
 Anonymous Guest Posts: n/a

Ah, I see - that makes more sense! Thanks so much! I need to watch that - I do that kinda thing whenever I can.

Yes, it makes sense since the way I say it is that there is a chance to get something completely different!

Thanks Again!
#9
12-11-2001, 02:45 PM
 Anonymous Guest Posts: n/a

Hi Again

Can anyone help me with the following:
I don't really understand how the book is setting this up.

A single premium whole life policy is issued to (20). The 1000 face amount is paid at the end of year of death. In addition, the single benefit premium is paid

a) with interest at the val. rate
b) without interest

if death occurs prior to age 35. Find the single benefit premium given
A_20=0.5,
a(due)_20=13
15_E_20=0.34705 and
A_20:angle15 = 0.54705.

(ans a) 800, b) 625)

Thanks for any help!
#10
12-11-2001, 07:35 PM
 Anonymous Guest Posts: n/a

Part b) is easier. The premium must equal the present value of the 1000 benefit plus the present value of getting the premium back without interest or ----->

P = 1000A_20 + P*A_{20^1:angl(15)}
(where A_{20^1:angl(15)} is the APV of a 15 year term policy)

A_{20^1:angl(15)} = A_20:angle15 - A_20angle15)^1 = 0.54705 - 0.34705 = 0.2

So P = (500)/(1 - 0.2) = 625.

Part A) Again the Premium P must equal the present value of the benefit. The 1000 death benefit part is the same. For the return of premium with interest, the present value today of the benefit (if you get it) is just P since it is accumulating with interest. You only get this benefit if you die in the next 15 years. All this can be summed up by saying

P = 1000A_20 + P*15_q_20

To find 15_q_20, note that a(dd)_20 = (1 - A_20)/d ----> d = 0.038462. This implies that v = 0.961538.

Now use 15_E_20 = v^15 * 15_p_x to get that
15_p_20 = 0.625017 -----> 15_q_x = 0.374983.

So P = 500/(1 - 15_q_20) = 800.

Hope this helps....

Robin Cunningham
Arch Solutions

<font size=-1>[ This Message was edited by: Arch on 2001-12-11 19:36 ]</font>

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