Actuarial Outpost Help: Batten's Ch.3&4 Practice Problems
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#1
03-28-2004, 11:11 PM
 Pennykei Join Date: Dec 2003 Posts: 21
Help: Batten's Ch.3&4 Practice Problems

How do you get the answers?

Ch.3 #86
A mortality table has a force of martality u(x+t) and mortality rate qx.
A second mortality table has a force of mortality u*(x+t) and mortality rate q*x.
You are given u*(x+t) = 0.5u(x+t) for 0&lt;=t&lt;=1. Calculate qx*.

Ch.3 #87
You are given s(x)=1/(1+x). Determine the median future lifetime of (y).

Ch.3 #89
You are given F(x)=1-(1/(x+1))
Which of the following are true?
(i) xp0 = 1/(1+x)
(ii) u(49) = 0.02
(iii) 10p39 = 0.8

Ch.3 #91
You are given
(i) qx = 0.04
(ii) u(x+t) = 0.04+0.001644t, 0&lt;=t&lt;=1
(iii) u(y+t) = 0.08+0.003288t, 0&lt;=t&lt;=1
Calculate qy

Ch.4 #80
For a select and ultimate mortaliy table with a one-year select period ,
q[x] = 0.5qx
Determine Ax-A[x]

Ch.4 #84
You are given:
(i) A60 = 0.58896
(ii) A60:1 (pure endowment ins.) = 0.9506
(iii) A61 = 0.60122
Calculate q60.

Ch.4 #87
Z is the PV random variable for an n-year- term insurance payable at the moment of death of (x) wih bt=(1+i)^t. Determine Var(Z)

THANKS!!
#2
03-29-2004, 03:43 AM
 macchow Member Join Date: Apr 2003 Posts: 232
Re: Help: Batten's Ch.3&4 Practice Problems

Quote:
 Originally Posted by Pennykei How do you get the answers? Ch.3 #86 A mortality table has a force of martality u(x+t) and mortality rate qx. A second mortality table has a force of mortality u*(x+t) and mortality rate q*x. You are given u*(x+t) = 0.5u(x+t) for 0<=t<=1. Calculate qx*. (answer: 1-(1-qx)^1/2) Ch.3 #87 You are given s(x)=1/(1+x). Determine the median future lifetime of (y). (answer: y+1) Ch.3 #89 You are given F(x)=1-(1/(x+1)) Which of the following are true? (i) xp0 = 1/(1+x) (ii) u(49) = 0.02 (iii) 10p39 = 0.8 (answer: all of the above) Ch.3 #91 You are given (i) qx = 0.04 (ii) u(x+t) = 0.04+0.001644t, 0<=t<=1 (iii) u(y+t) = 0.08+0.003288t, 0<=t<=1 Calculate qy (answer: 0.0784) Ch.4 #80 For a select and ultimate mortaliy table with a one-year select period , q[x] = 0.5qx Determine Ax-A[x] (Answer: A[x]:1*(1-Ax+1)) Ch.4 #84 You are given: (i) A60 = 0.58896 (ii) A60:1 (pure endowment ins.) = 0.9506 (iii) A61 = 0.60122 Calculate q60. (Answer: 0.018) Ch.4 #87 Z is the PV random variable for an n-year- term insurance payable at the moment of death of (x) wih bt=(1+i)^t. Determine Var(Z) (Answer: nqx*npx) THANKS!!
Oh..I cannot get the answer of Ch.4 #84, my answer is 0.046
#3
03-29-2004, 07:21 AM
 macchow Member Join Date: Apr 2003 Posts: 232
Re: Help: Batten's Ch.3&4 Practice Problems

Quote:
 Originally Posted by Pennykei How do you get the answers? Ch.3 #86 A mortality table has a force of martality u(x+t) and mortality rate qx. A second mortality table has a force of mortality u*(x+t) and mortality rate q*x. You are given u*(x+t) = 0.5u(x+t) for 0<=t<=1. Calculate qx*. (answer: 1-(1-qx)^1/2) Ch.3 #87 You are given s(x)=1/(1+x). Determine the median future lifetime of (y). (answer: y+1) Ch.3 #89 You are given F(x)=1-(1/(x+1)) Which of the following are true? (i) xp0 = 1/(1+x) (ii) u(49) = 0.02 (iii) 10p39 = 0.8 (answer: all of the above) Ch.3 #91 You are given (i) qx = 0.04 (ii) u(x+t) = 0.04+0.001644t, 0<=t<=1 (iii) u(y+t) = 0.08+0.003288t, 0<=t<=1 Calculate qy (answer: 0.0784) Ch.4 #80 For a select and ultimate mortaliy table with a one-year select period , q[x] = 0.5qx Determine Ax-A[x] (Answer: A[x]:1*(1-Ax+1)) Ch.4 #84 You are given: (i) A60 = 0.58896 (ii) A60:1 (pure endowment ins.) = 0.9506 (iii) A61 = 0.60122 Calculate q60. (Answer: 0.018) Ch.4 #87 Z is the PV random variable for an n-year- term insurance payable at the moment of death of (x) wih bt=(1+i)^t. Determine Var(Z) (Answer: nqx*npx) THANKS!!
Sorry, I made something wrong, I got it!
#4
03-31-2004, 03:30 PM
 calla Member Join Date: Aug 2003 Posts: 34

I have correct solutions to two of the problems:

Ch. 3, #89

0px = s(x) = 1-F(x) = 1/(x+1) -- (i) is true

u(x) = -s'(x)/s(x) = [1/(x+1)^2]/[1/(x+1)] = 1/(x+1)
u(49) = 1/(50) = 0.02 -- (ii) is true

10p39 = s(49)/s(39) = (1/50)/(1/40) = 0.8 -- (iii) is true

Ch. 4 #87

First notice that the n-year term insurance in this question has PV = (1+i)^t * v^t * nqx = nqx

V(Z) = ^2A_x^1:n - (A_x^1:n)^2
= nqx - (nqx)^2
= nqx - nqx(1-npx)
= nqx * npx

I have incorrect solutions to ch. 3 #87 and ch.4 #84, so I'd appreciate it if someone could tell me where I went wrong!

Ch.3 #87

Let m be the median future lifetime, so that mpy = 0.5
mp0 = yp0 * mpy
1/(1+m) = 1/(1+y) * 0.5
m = 2y +1

Ch. 4 #84

First notice that for a 1-yr endowment insurance, A_60:1 = v
A_60:1 = A_60^1:1 + A_60:1^1
v = (A_60 - v*p60*A_61) + v*p60
Substituting in the given values and using 0.9506 for v results in p60 = 0.95399 --> q = 0.046 (answer should be 0.018)

#5
04-01-2004, 02:20 AM
 macchow Member Join Date: Apr 2003 Posts: 232

Quote:
 Originally Posted by calla I have correct solutions to two of the problems: Ch. 3, #89 0px = s(x) = 1-F(x) = 1/(x+1) -- (i) is true u(x) = -s'(x)/s(x) = [1/(x+1)^2]/[1/(x+1)] = 1/(x+1) u(49) = 1/(50) = 0.02 -- (ii) is true 10p39 = s(49)/s(39) = (1/50)/(1/40) = 0.8 -- (iii) is true Ch. 4 #87 First notice that the n-year term insurance in this question has PV = (1+i)^t * v^t * nqx = nqx V(Z) = ^2A_x^1:n - (A_x^1:n)^2 = nqx - (nqx)^2 = nqx - nqx(1-npx) = nqx * npx I have incorrect solutions to ch. 3 #87 and ch.4 #84, so I'd appreciate it if someone could tell me where I went wrong! Ch.3 #87 Let m be the median future lifetime, so that mpy = 0.5 mp0 = yp0 * mpy 1/(1+m) = 1/(1+y) * 0.5 m = 2y +1 (answer should be y+1) Ch. 4 #84 First notice that for a 1-yr endowment insurance, A_60:1 = v A_60:1 = A_60^1:1 + A_60:1^1 v = (A_60 - v*p60*A_61) + v*p60 Substituting in the given values and using 0.9506 for v results in p60 = 0.95399 --> q = 0.046 (answer should be 0.018) Thanks in advance!
A_60:1 is pure endowment, not 1-yr endowment, his question shows that A_60:1 (pure endowment)
#6
04-01-2004, 06:23 AM
 Sunny Member Join Date: Nov 2003 Posts: 3,940

Quote:
Originally Posted by macchow
Quote:
 Originally Posted by calla I have correct solutions to two of the problems: Ch. 3, #89 0px = s(x) = 1-F(x) = 1/(x+1) -- (i) is true u(x) = -s'(x)/s(x) = [1/(x+1)^2]/[1/(x+1)] = 1/(x+1) u(49) = 1/(50) = 0.02 -- (ii) is true 10p39 = s(49)/s(39) = (1/50)/(1/40) = 0.8 -- (iii) is true Ch. 4 #87 First notice that the n-year term insurance in this question has PV = (1+i)^t * v^t * nqx = nqx V(Z) = ^2A_x^1:n - (A_x^1:n)^2 = nqx - (nqx)^2 = nqx - nqx(1-npx) = nqx * npx I have incorrect solutions to ch. 3 #87 and ch.4 #84, so I'd appreciate it if someone could tell me where I went wrong! Ch.3 #87 Let m be the median future lifetime, so that mpy = 0.5 mp0 = yp0 * mpy 1/(1+m) = 1/(1+y) * 0.5 m = 2y +1 (answer should be y+1) Ch. 4 #84 First notice that for a 1-yr endowment insurance, A_60:1 = v A_60:1 = A_60^1:1 + A_60:1^1 v = (A_60 - v*p60*A_61) + v*p60 Substituting in the given values and using 0.9506 for v results in p60 = 0.95399 --> q = 0.046 (answer should be 0.018) Thanks in advance!
A_60:1 is pure endowment, not 1-yr endowment, his question shows that A_60:1 (pure endowment)
And for Chap 3 #87 your setup is incorrect. .5 should equal to <sub>n</sub>p<sub>x</sub>, which is 1/2 = <sub>x+n</sub>s<sub>0</sub>/<sub>x</sub>s<sub>0</sub> = ℓ<sub>x+n</sub>/ℓ<sub>x</sub> = (1 + x)/(1 + x + n), which becomes n = 1 + x, therefore median [y] = 1 + y.
#7
04-01-2004, 06:27 AM
 Sunny Member Join Date: Nov 2003 Posts: 3,940

Does anyone still need answers to other problems? I have the solutions.
#8
04-02-2004, 11:08 AM
 testtaker Member Join Date: Nov 2001 Posts: 67
Chp 3 #73

help me on #73

f: is UDD - person age 60 dies between 60.5 and 61.5
g: is Bald - person age 60 dies between 60.5 and 61.5

what is 10000* (f-g)

i did (.5p60)*(1q60.5) for each and used the fractional formulas

the result i got was 851. the answer is 85, not sure why i am off by factor of 10

also can you provide solution to #76

thanks
#9
04-04-2004, 03:48 AM
 Sunny Member Join Date: Nov 2003 Posts: 3,940
Re: Chp 3 #73

Quote:
 Originally Posted by testtaker help me on #73 f: is UDD - person age 60 dies between 60.5 and 61.5 g: is Bald - person age 60 dies between 60.5 and 61.5 what is 10000* (f-g) i did (.5p60)*(1q60.5) for each and used the fractional formulas the result i got was 851. the answer is 85, not sure why i am off by factor of 10
#73
There are two ways to do this problem. I prefer the second method.

First, one can do this the way of q<sub>x</sub>. Namely, for UDD,
f = <sub>.5│</sub>q<sub>60</sub> = <sub>.5</sub>p<sub>60</sub> (1 - <sub>.5</sub>p<sub>60.5</sub> <sub>.5</sub>p<sub>61</sub>) = (1 - .5 * q<sub>60</sub>) [1 - (1 - .5 * q<sub>60</sub>/(1 - .5 * q<sub>60</sub>)) (1 - .5 * q<sub>61</sub>)] = 0.29,

Similarly, for Balducci,
g = <sub>.5│</sub>q<sub>60</sub> = <sub>.5</sub>p<sub>60</sub> (1 - <sub>.5</sub>p<sub>60.5</sub> <sub>.5</sub>p<sub>61</sub>) = p<sub>60</sub>/(1 - .5 * q<sub>60</sub>) [1 - (1 - .5 * q<sub>60</sub>) (1 - .5 * q<sub>61</sub>/(1 - .5 * q<sub>61</sub>))] = 0.298529412,

Therefore, 1000(g - f) = 85.29411765.

The second method employs the number of life as opposed to the q<sub>x</sub>. Namely, for UDD,
f = <sub>.5│</sub>q<sub>60</sub> = (ℓ<sub>60.5</sub> - ℓ<sub>61.5</sub>)/ℓ<sub>60</sub>, make ℓ<sub>60</sub> = 100 hypothetically, then

ℓ<sub>60.5</sub> = ℓ<sub>60</sub> (1 - .5 * q<sub>60</sub>) = 100 * (1 - .5 * .3) = 85 and

ℓ<sub>61.5</sub> = ℓ<sub>60</sub> (1 - q<sub>60</sub>) (1 - .5 * q<sub>61</sub>) = 100 * (1 - .3) * (1 - .5 * .4) = 56,

therefore f = (85 - 56)/100 = 0.29,

Similarly, for Balducci,
g = (ℓ<sub>60.5</sub> - ℓ<sub>61.5</sub>)/ℓ<sub>60</sub>, using the relationship 1/ℓ<sub>60.5</sub> = (1-t)/ℓ<sub>60</sub> + t/ℓ<sub>61</sub>,

After some algebraic manipulation,

ℓ<sub>60.5</sub> = 2ℓ<sub>60</sub>ℓ<sub>61</sub>/(ℓ<sub>60</sub> + ℓ<sub>61</sub>) = 2(100)(70)/170 = 82.35294118

ℓ<sub>61.5</sub> = 2ℓ<sub>61</sub>ℓ<sub>62</sub>/(ℓ<sub>61</sub> + ℓ<sub>62</sub>) = 2(70)(42)/112 = 52.5

Thus g = (82.35294118 - 52.5)/100 = 0.298529412.

Again 10000(g - f) = 85.29411765.

Ans. B
#10
04-04-2004, 04:43 AM
 Sunny Member Join Date: Nov 2003 Posts: 3,940
Re: Chp 3 #73

Quote:
 Originally Posted by testtaker also can you provide solution to #76 thanks
#76

From the Generalized/Modified DeMoivre's Law,

µ(x) = r/(ω - x),

ℓ<sub>x</sub> = (ω - x)<sup>r</sup> and

<sup>°</sup>e<sub>x</sub> = (ω - x)/(r + 1),

here r = 4 and ω = 100, so we have

µ(x) = 4/(100 - x),

ℓ<sub>x</sub> = (100 - x)<sup>4</sup> and

<sup>°</sup>e<sub>x</sub> = (100 - x)/5 = T<sub>x</sub>/ℓ<sub>x</sub>,

Let the average number of years lived by (x) be

T<sub>x</sub> = <sup>°</sup>e<sub>x</sub> ℓ<sub>x</sub> = (100 - x)<sup>5</sup>/5

<sub>10</sub>p<sub>30</sub> <sup>°</sup>e<sub>40:20</sub>┐ = ℓ<sub>40</sub>/ℓ<sub>30</sub> * (T<sub>40</sub> - T<sub>60</sub>)/ℓ<sub>40</sub> = (T<sub>40</sub> - T<sub>60</sub>)/ℓ<sub>30</sub> = 5.624323199

Ans. C

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