Course 3 detailed solutions

[thing]

#28

Rewrite F(x) = .8(1-e^(-.02x))+.2(1-e^(-.001x)), so this is a mixture of exponential distributions.

For theta=50, E[X ^ 1000] = 50(1-e^(-20)) = 49.99
For theta = 1000, E[X ^ 1000] = 1000(1-e^(-1)) = 632.12

.8(49.99) + .2(635.12) = 166.42, C

[thing]

#22

Set up a transition matrix, and you get the equations:
.2R + .3S = R
.8R + .7S = S
R + S = 1
Either of the first two equations gives you 8R = 3S, and so S = 8/11, R = 3/11

If in R, 1/5 of trips end in R, so probability of a trip from R to R = 3/55
If in S, 7/10 of trips end is S, so probability of a trip from S to S = 28/55
That leaves 24/55 probability of a trip between the towns.

(3/55)(1)+(28/55)(1.2)+(24/55)(2) = 1.5381...
B

[thing]

#27

Poisson-Gamma distribution = negative binomial distribution, with alpha = r, theta = beta.

Mean of gamma = 2 = theta*alpha
Variance of gamma = 4 = theta^2(alpha)(alpha + 1) - (theta*alpha)^2 = (theta^2)alpha
So theta = 2 = beta,
alpha = 1 = r.

For the negative binomial distribution,
p_1 = r*beta/(1+beta)^(r+1) = 2/3^2 = .222...
A

(These are some of the ones I'm relatively proud of myself for getting... Please jump in! There's 20 or so I didn't get and so far I only know why for 2 of them...)

[MathGuy]

#1

e_25:25 = (# of years lived in first 15 years) + Prob(survive 15 years)*(# of years lived in next 10 years)

= int(0,15, e^-.04x) + e^-.04*15*int(0,10, e^-.05x)

= (1/.04)*(1-e^-.6) + e^-.6*(1/.05)(1-e^-.5)

= 11.2797 + 4.3188

= 15.6 (E)

<font size=-1>[ This Message was edited by: MathGuy on 2001-11-08 12:59 ]</font>

[MathGuy]

#2

₅p_[60]+1 = (1 - q_[60]+1)*(1 - q_[60]+2)*(1 - q₆₃)*(1 - q₆₄)*(1 - q₆₅)

₅p_[60]+1 = (1-.11)(1-.13)(1-.15)(1-.16)(1-.17)

₅p_[60]+1 = (.89)(.87)(.85)(.84)(.83)

₅p_[60]+1 = .4589 (C)

[MathGuy]

#3

a_x = 1/(μ+δ) = 1/(2μ)
2μ = 1/12.5
μ = .04 = δ

A_x = μ/(μ+δ) = 1/2
²A_x = μ/(μ+2δ) = 1/3

Var(a_T(x)) = δ^-2(²A_x - A_x²) = .04^-2(.33333 - .5²) = 625(.08333) = 52.08333

Standard Deviation = 52.08333^.5 = 7.217 (E)

<font size=-1>[ This Message was edited by: MathGuy on 2001-11-08 13:21 ]</font>

[MathGuy]

#5

To guarantee a 0% chance of loss, you must collect enough to pay in the event of death. This is because Loss = P.V. of paid out - collected. So, if the premium is 0.95, and the amount paid out is 1 (at the end of one year), and the loss is zero, we can find v:

0 = v - .95
v = .95

Now, to abuse some symbols,

A_x = (1-p_x)v + p_x(1-p_x+1)v² = (.25)(.95) + (.75)(.20)(.95)² = .372875

Now, to calculate ²A_x, we use the square of the discount factor, so

²A_x = (1-p_x)v² + p_x(1-p_x+1)v⁴ = (.25)(.95)² + (.75)(.20)(.95)⁴ =
.3478

Var(Z) = ²A_x - A_x² = .3478 - .372875² = .2088 (D)

[MathGuy]

#6

This one is easy. We expect 300 claims. After severity rises, we expect 25% to be 60, 25% to be 120, 25% to be 180 and 25% to be 200. After deductible, the expected claim size is .25(0)+.25(20)+.25(80)+.25(200) = 75. With 300 expected claims, the total claim payment should be 75*300 = 22500. (D)

[MathGuy]

#7

E[N] = 50
Var[N] = 100
E[X] = 200
Var[X] = 400

E[S] = E[N]*E[X] = 10000
Var[S] = E[X]²*Var[N] + E[N]*Var[X] = 4,020,000
SD[S] = 2005

P(Z < 8000) = P((Z-10000)/2005 < (8000-10000)/2005) = P((Z-10000)/2005 < -.9975) = 1 - &Phi;(1) = .16 (A)

[MathGuy]

#8

A_x:5 = e^-5(μ+δ)μ/(μ+δ) = .2426

²A_x:5 = e^-5(μ+2δ)μ/(μ+2δ) = .11233

Variance = .11233 - .2426² = .053475

Aggregate Expected Loss = 100*10*.2426 = 242.6
Aggregate Variance = 100*10²*.053475 = 534.75
Aggregate Standard Deviation = 534.75^.5 = 23.12

To ensure 95% probability of sufficient funds, F must be:

242.6 + 1.645(23.12) = 280.64 (A)

<font size=-1>[ This Message was edited by: MathGuy on 2001-11-08 13:54 ]</font>

<font size=-1>[ This Message was edited by: MathGuy on 2001-11-08 13:54 ]</font>

[jaegar]

[jaegar]

What is the solution for #25? I got something different than the answer key, but I am unable to see where I went wrong, and how they derived their answer.

[thing]

#21

_.7q'₄₀⁽¹⁾ = .07

(1-.07)(.125) = .11625, C

(MathGuy, you're so systematic! But you skipped #4...)

<font size=-1>[ This Message was edited by: thing on 2001-11-08 14:51 ]</font>

<font size=-1>[ This Message was edited by: thing on 2001-11-08 17:16 ]</font>

[WQN]

Many of us were convinced that #25 should be zero however it is 83. I first calculated 83 then second guessed myself and calculated 0. I'm unsure of how I got the 83 since I already convinced myself I was wrong. I wish I hadn't gone back on that one. It would've given me 24 instead of 23. Ugh.

[jaegar]

WQN:

Using the idea that the reserve is the expectation of the loss function, I got the APV of future benefits to be a constant, since the accumulation of interest on the benefit offsets witht discounting. Additionally, since you have constants forces, annuities are equivalent for all ages. Thus, the APV of future premiums is a constant as well. V = APV(B) - APV(P) = 0

Using Facklers Theorem, I get the terminal reserve of year 2 to be 88, however discounting (interest and mortality) this back to the beginning of year 1, i get 82.88, which isnt the end of year 2, nor the reserve.

Where is my thinking flawed? Any thoughts?

[Anonymous]

#25
p65=p66=e(-.02)=.98
q65=q66=.02
(1+i)=e(.04)
APV of Ins. = Int[1000e(.04)e(-.04).02e(-.02)] = 1000
APV of Annuity = 1/(.02+.04) = 16.667
P = (APV of Ins.)/ (APV of Annuity) = 60

b_1 =1000e(.04)= 1040.81, benefit at t=1
b_2 =1000e(.08)=1083.29

You have
(0_V + P)(1+i)= 1_V * p65 + b_1 * q65
which 0_V = 0 as the benefit reserve at t=0
=> 60e(.04) = 1_V * .98 + 1040.81 *.02
=> 1_V = 42.48

(1_V + P)(1+i)= 2_V * p66 + b_1 * q66
=>(42.48 + 60)e(.04) = 2_V *.98 + 1083.29 *.02
=> 2_V = 86.73

Therefore, I got (E)

[WQN]

How do you justify using Fackler's Recursive formula for a fully continuous insurance, that is for discrete only? I got the same results, a premium of 60 then I realized the insurance is worth 1083 at time 2 and got lost in the shuffle. I can't add much more. The solutions will help (in January).

[jaegar]

WQN: I agree with you as far as facklers theorem. There definitely seems to be some inconsistencies in the answer of 83.

[MathGuy]

Yes, I am quite systematic. I just don't know how to do #4! :smile:

Anyway, on #25, you guys are making it way too hard.

At age (65), the insurance has an APV of 1000. This is because the benefit increase and force of interest cancel each other out. Thus, assuming he survives to age 67, the APV of the insurance will be:

1000e^.04*2 = 1083.29.

Because forces of interest and mortality are constant, the value of the annuity is:

a_x = (&mu;+&delta;)^-1 = 1/.06 = 16.666667,

for all x. So, the level benefit premium is 1000/16.666667 = 60, and the benefit reserve after 2 years is:

1083.29 - 60*16.66667 = 83.29. (E)

[WQN]

Mathguy that is how I did it the first time on the exam and then thought that I should second guess myself (just so I could lose a point). If I don't pass this it is not because I didn't study enough, its because I am an awful test taker. Ugh again.

[NY_Ranger]

#4
I think 4 goes something like this.
You treat the insurance (when calculating the premium) as a one year deferred whole life insurance.

1|Ax= Ax-Ax:n
= .5-.045
=.455

10Vx = 1-(ax+10/ax)
ax+10 = 4

10V = Ax+10 - (1|Ax/ax) * ax+10
= .6 - (.455/5) * 4
= .236
Times 5000 = 1180

[thing]

#23
Funds at the end of year n = funds at end of year n-1 + 2 - claims - dividend

Table of funds at end of year and probability of being at that fund level:

Year 1:
3 .8
1 .2

Year 2:
3 .8*.8+.2*.15 = .67
2 .2*.25 = .05
1 .2*.4+.8*.2 = .24
0 .2*.2 = .04

Year 3:
3
2 | do we really care?
1 /
0 .04+.05*.2+.24*.2 = .098

1-.098 = .902, D

[thing]

#26

a-umlaut_x = .72/(.05/1.05)=15.12
a_x = 15.12 - 1 = 14.12

a-umlaut_x+20 = .6/(.05/1.05) = 12.6
a_x+20 = 12.6 - 1 = 11.6

a_x:n-| = 14.12 - .25*11.6 = 11.22

B

I missed this one on the actual test -- just convinced myself there must have been an umlaut... :sad:

[jaegar]

MathGuy:Thanks for the insight. I actually wrote out a long response of why I didnt think that 1083 was the APV and why 1000 is. However, it finally dawned on me. The intergration of the APV of the future benefit from time t=2 on is intergrate from 2 to infinity, not 0 to infinity, since the death benefit is accumulated from time t=0. Therefore, the APV of future benefits is infact 1083. That one was driving me nutty.

[thing]

#9 (almost systematic...)

₃p_Keith = p₄₈₊₂*p₄₉₊₂*p₅₀₊₂=.9713*.9698*.9682 = .912 (or so -- store in calculator)
₃p_Clive = p_[50]*p_[50]+1*p_[50]+2=.9849*.9819*.9682 = .936 (again, store in calculator for greater accuracy...)

Prob(both alive) = .912...*.936.... = .8539...

Prob(one alive) = ₃p_Keith + ₃p_Clive -2*Prob(both alive) = .14046..., D



<font size=-1>[ This Message was edited by: thing on 2001-11-08 18:20 ]</font>

[Anonymous]

On 2001-11-08 12:04, thing wrote:
#28

Rewrite F(x) = .8(1-e^(-.02x))+.2(1-e^(-.001x)), so this is a mixture of exponential distributions.

For theta=50, E[X ^ 1000] = 50(1-e^(-20)) = 49.99
For theta = 1000, E[X ^ 1000] = 1000(1-e^(-1)) = 632.12

.8(49.99) + .2(635.12) = 166.42, C




#28 A more efficient approach: Simply integrate 1-F(x) from 0 to 1000. It is a simple integration.

[thing]

idikoko: but then you won't feel clever for noticing that it's a mixture of exponentials... :smile:

[thing]

#10 (took me a while...)

Probability that T. Rex dies is
probability that he eats no scientists day one +
probability that he eats only one scientist day one and none day two
(eating more than that, T's got calorie reserves to make it to time 2.5)

So Pr(T dies by 2.5) = e^-1 + (e^-1)(e^-1) = .50321..., C

[MathGuy]

#11

Expected Calories in Day 1: 10,000
Expected Calories in Day 2: 10,000*Prob(Survive Day 1) = 10,000*(1-e^-1) = 6321
Expected Calories in Day 2.5 = 5,000*Prob(Surviving Days 1 and 2) = 5,000*(1-Prob(eat 0 on day 1)-Prob(eat 1 on day 1, 0 on day 2) = 5,000*(1-e^-1-e^-2) = 2484

E[Calories] = 10,000 + 6321 + 2484 = 18,805 (B)

[thing]

#12

q₇₀ = .03318
_1|q₇₀ = (6396609-6164663)/66161555 = .03506
₂p₇₀ = 1 - .03318 - .03506 = .93176

P = .03318(10)/1.08 + .03506(10)/1.08^2 + .93176(P)/1.08^2

P = 3.021411..., C

[Anonymous]

On #25 you can integrate from 0 to infinity but realize that the benefits at t=2 are (1+i)^(2+t) or e^(.08)*(1+i)^t.

[thing]

#18

E[(S-2)₊] = E[(S-1)₊] -[1 - F(1)]
= E[S] - [1 - F(0)] - [1 - F(1)]

E[S] = E[N]E[X] = 2*2 = 4

F(0) = prob of zero claims = e^-2 = .1353
F(1) = prob of zero or one claims = e^-2 + e^-2/3 = .2255

E[(S-2)₊] = 4 - 1 + .1353 - 1 + .2255 = 2.3606..., B

another one I think I missed on the actual test...

[thing]

#20

₅p^(t)₅₀*₅q^(t)₅₅
= (45/50)(e^(-.05*5))*(1-((40/45)(e^-.05*5)))
= .215696..., A


<font size=-1>[ This Message was edited by: thing on 2001-11-09 14:51 ]</font>

[jaegar]

JASA:
You can integrate from 0 to infinity. However, the benefit is measured from time t=0. Therefore, the setup is from 2 to infinity. After a change of variables, in order to get the limits of integration being from 0 to finity, you have the constant of 1000*e^(.04*2) time the new integral. So yeah. Dumb me. should have been a slam dunk.

[Anonymous]

For question 19:
With the information given, one can determine that E((type II severity)^2)= 1000.
The probability of a claim being type II is 2/3. Since this is Poisson, Then Type II claims occur according to a Poisson process with mean and variance equal to 3000*(2/3)= 2000.
Being that this is Poisson the Aggregate variance is simply 2000*E(severity^2)=2,000*1,000=2,000,000

[Anonymous]

#13 I think this is how it works
Prob. of success for binomial is the prob. of person age 70 dying(I quess you can call it success) in one year which = q70. Set binomial parameter q=q70=.03318 and n=100.
From recursion:
f(0)=(1-q)^100=.0342423
f(1)=100(q/(1-q))f(0)=.11387
f(2)=(99/2)(q/(1-q))f(1)=.1934396

We're looking for F(x-1) < u < F(x) and x is the number of claims.
Since u=.18 then
F(1)=.1481123 < .18 < F(2)=.3415519
gives us 2 claims at $10 a piece equals $20 Answer C.
I actually answered B because I picked the left end point. I hate it when that happens.

<font size=-1>[ This Message was edited by: jasa on 2001-11-09 14:56 ]</font>

[thing]

On 2001-11-09 14:46, Tenacious R wrote:
For question 19:
With the information given, one can determine that E((type II severity)^2)= 1000.


Show me how you derived that, and I'll believe the rest! :smile:

[jaegar]

On #19, I am not sure how the wonderboy derived 1000. However, I solved as such,

Poisson with rate = 3000
Prob I = 1/3 Severity constant at 10
Prob II = 2/3 Severity is unknown
Aggregate Variance is 2.1M

Using Compound Poisson Aggregate Loss Variance Formula,
Var(S) = lamda*E(X^2)
Thus, E(X^2) = 700

Therefore E(X^2) attributable to I is 100/3. Even more so, the variance attributable to I is 100,000...[(100/3)*3000]. This means the variance attributable to II, which is what we are solving for, is 2M.

[thing]

#24
pi + ₂p₈₀pi/1.06^2 = 1000A₈₀ + q₈₀(pi/2)/1.06 + _2|q₈₀(pi/2)/1.06^3

₂p₈₀ = 3284542/3914365
_2|q₈₀ = (3600038-3284542)/3914365
Others come directly from table.

Regrouping, you get

1.675pi = 665.75,
pi = 397.46...., C

[thing]

#29

row 1 of Matrix^2 = [.44 .16 .4 0]

which is enough to give you the upper left entry of Matrix^3 = .468

500v^3(.468)= 170.586, E

[thing]

#30

E[N] = 50
Var(N) = 50
E[X] = 2.5
E[X^2] = 7.2
Var(X) = .95

E[S] = E[N]E[X] = 125
Var(S) = E[N]Var(X) + Var(N)E[X]^2 = 360

E[S_1 + ... + S_13] = 13E[S] = 1625
Var(S_1 + ... + S_13) = 13Var(S) = 4680
sigma = 68.41....

90th percentile = 1625 + 1.282(68.41) = 1712.702...

B

[thing]

#32

T(x) = (100-x)R₁ is certainly a legitimate way to generate the future lifetime of x, given our mortality assumption. T(xy) = min(T(x),T(y)) is just the definition of joint-life. So Method 1 must be valid.

For Method 2, the only issue is that the probability of x dying before y depends on their relative age. If x = (2) and y = (98), it's absurd to think they are equally likely to die next. But if they're the same age, subject to the same mortality law, they are equally likely to die. So Method 2 is valid for x = y but not otherwise.

E

[phdmom]

thing,

I put E on #32 also (same reasoning as you). But the key says D...any thoughts?

[phdmom]

#14

The waiting time between claims is exponential with mean 1/2 (since it's a Poisson process with mean 2). The distribution is then F(t)=1-e^(-2t). Solve for t (I put in u for F(t)): t=-1/2 ln(1-u). So u=.83 gives t=.886 (i.e. first claim at time t=.886) and u=.54 gives t=.388 (i.e. second claim at .886+.388). The second claim happens after t=1, so at t=1 there is one claim.

For severity, F(x)=1-(1000/(x+1000))^2, or x=1000/sqrt(1-u) -1000. So u=.89 gives x=2015.

We also need to find c: lambda*(1+theta)*E[X]=c; plug in 2*1.1*1000=2200=c.

surplus=2000+2200*1-2015=2185.

I nearly jumped up and screamed when I got that in the test! :grin:

[thing]

#32 again. Well, I think their answer is wrong, then! :smile:

Okay. Maybe because Method 2 doesn't really generate both lifespans? To work, perhaps we'd have to generate three random variables: use R₁ to decide which of x and y gets the lesser of R₂ and R₃

That's not quite right.

How about this: joint life doesn't care who dies first, so the bit about assigning the first death to x or to y effectively means we ignore the first random number. Then we're assigning a joint-life value when we've really only generated a single life value.

I could kind of see that. Weirdly, I was misreading the answer key, and thought I had gotten that one right. My score just dropped another point, I guess... :sad:

[thing]

#33

At age 60, the force of mortality for men is 1/(w - x) = 1/15
So the force of mortality for women at age 60 is (3/5)(1/15) = 1/25, which implies w = 85.

The expected life of a man at age 65 is (w-x)/2 = 5
For a woman at age 60, (85-60)/2 = 12.5

The expected future joint life is the integral from 0 to 10 of (1-x/10)(1-x/25), which works out to 4.33333...

So the expected time for the last survivor = 5 + 12.5 - 4.3333... = 13.1666666...

Or 13 and 1/6 if you like fractions

[thing]

#34

A = mu/(mu + delta) = 1/3
²A = mu/(mu + 2delta) = 1/5
a = (1 - A)/delta = 25/3

P = A/a = 1/25

Var(L) = (1 + P/delta)^2(²A - A^2)
= (1 + (1/25)(2/25))^2(1/5-1/9)
= (9/4)(4/45) = 1/5

B (and yes, on the test I actually solved it with fractions. Goodness knows why...)

[thing]

#35

Mean excess loss = (E[X] - E[X ^ d])/(1 - F(d))
=(331 - 91)/(1 - .2) = 300

Why is E[X] = E[X ^ 1000]? Because F(1000) = 1.

B (another one I missed on the actual test)

[thing]

#37

From the tables, E[X] = 4alpha/(alpha -1) = 8, which solves to alpha = 2.

The probability of living 6 years =

1 - F(6) =

1 - (1 - (4/6)^2) = 4/9 = .4444...

A

okay I'll stop for tonight... :smile:

[jerkweed]

Regarding #32, method 2 is incorrect since if the person with the larger w were chosen with the first simulation there is a possibility that there joint lifetime status would survive after the person with the smaller w is supposed to be dead.

[HOSS]

On 2001-11-09 21:40, phdmom wrote:
thing,

I put E on #32 also (same reasoning as you). But the key says D...any thoughts?

Any answers to #31??? Why can't d be right?

[thing]

To HOSS re: 31

Under the stated conditions, ruin is possible in year 1 if there are twice as many claims as usual in year 1. That's possible, so ruin is possible.

Suppose the initial surplus was 60 instead of 6. It would take 10 consecutive years of having twice as many claims as usual, but ruin could still occur.

Set the initial surplus as high as you like, and I can tell you how many consecutive years of twice-usual claims to ruin. You mentioned (in another thread) that the surplus could be set to "near infinite", but the time frame is actually infinite. You're thinking just because if we had a billion surplus that ruin wouldn't happen before the sun goes nova, that ruin is impossible. :smile:

[thing]

#39

The increasing annuity cancels inflation, so a whole life annuity-due would be worth 1000e-dot_x = 11050. Take away 3000 for the deferal period to get the present value of the deferred annuity is 8050.

We need P such that P + .99P/1.04 + (.99)(.98)P/1.04^2 = 8050. This solves to P = 2825.6247..., B

[HOSS2000]

Will it be possible to argue that #31 can be D considered the near infinity to supernova situation??

[thing]

by my count, the only 4 questions which have not yet been addressed (in this thread or on another) are 15, 16, 17, and 38. But personally, the obsessive need to know I can actually solve these problems (given enough time) is beginning to burn off...

[dinosaur]

#15

This is a (P-P)/P problem. (Batten helps)

(Px - P^(1)x:n)/Px:n^(1) = kVx
(WL - Term)/Endowment = kVx

so (.09-Term)/.00864 = .563
therefore Term = .0851

[dinosaur]

#16

Note: All continuous, not discrete
A50 - P40 * a50
or
1-a60 / a50
among others

Since De Moivre's Law
A40 = a(w-x)/(w-x) => a60/60
A40=.3233
A50=.3742
so
a40=13.869
a50=12.826

Answer = .0752

No need to integrate!

[dinosaur]

#38

use formula p'(j)x=p(t)x^[q(j)/q(t)x]

p(t)64=(1-.025)*(1-.035)*(1-.2)=.7527

q^(1)64=.02204

2q^(1)64=q^(1)64+p^(t)64*q^(1)65

2q^(1)64=.02204+.7527*(.02716)=.04248



<font size=-1>[ This Message was edited by: dinosaur on 2001-11-13 09:57 ]</font>

[MathGuy]

Drum roll please...

17.

Just use weighted averages to determine A_x and ²A_x:

A_x = .3*A_x;Smokers + .7*A_{x;Non-Smokers}
= .3*(.06/(.06+.08)) + .7*(.03/(.03+.08))
= .31948

²A_x = .3*²A_x;Smokers + .7*²A_{x;Non-Smokers}
= .3*(.06/(.06+2*.08)) + .7*(.03/(.03+2*.08))
= .192344

Var(a_T(x)) = &delta;^-2(²A_x - A_x²)
= .08^-2(.192344 - .31948²)
= 14.105 (D)

And that's that

[Macroman]

Well done all, especially you Math Guy.

[MathGuy]

I just wish I could've gotten them right during the exam. It's looking tight for me. I can't believe I didn't get either dinosaur problem... :sad:

[dinosaur]

I actually think I did alright on the exam. I verified my answers through the SOA & I have 28 of 40 plus any right guesses so I'm thinking I'll be safe.

Next in line, C6 since I passed 4 on an earlier sitting.

[thing]

MathGuy: I feel your pain. No, wait -- actually I have that same pain. Sigh...

[Anonymous]

I have 22 right after checking the SOA key. This seems better than I've done before and I've received 5's on my previous tries. This is the first time I remember;-) my actual answers so I don't know exactly what I had previously. I do remember seeing something about needing 43 points on last November's exam. I'm kicking my self for not getting some more...but thats what these dag gone exams do to you. Good luck to all.

[Smash Puny Human]

Anyone have a solution for #36? Thanks in advance.

[thing]

On 2001-11-19 15:05, Smash Puny Human wrote:
Anyone have a solution for #36? Thanks in advance.

http://www.actuary.ca/phpBB/viewtopic.php?topic=853&forum=18&10

[trinity]

Can anybody tell me how to do #4? I couldn't understand the previous discrip0tion in this thread.

[rabbit]

Here goes: First, using the annuity formula for reserves ---> 0.2 = 10_V_x = 1 - (add_x+10)/(add_x). This tells us that add_x+10 = 4.

Now, let P be the premium for the insurance in the problem. Then
P = (5000A_x - 5000vq_x)/add_x = (5000A_x - 225)/5.

We can find A_x = 1 - d add_x = 1 - (1-v) add_x = 0.5. So P = 455. Similarly,
A_x+10 = 1 - (0.1)add_x+10 = 0.6.

Finally, 10_V = 5000 A_x+10 - P add_x+10 = 1180.

So the key is D. Hope that helps!

Robin Cunningham
Arch Solutions

<font size=-1>[ This Message was edited by: rabbit on 2001-12-04 21:51 ]</font>