Help: Batten's Ch.3&4 Practice Problems

[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!!

[macchow]

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

[macchow]

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!

[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!

[macchow]

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)

[Sunny]

Originally Posted by macchow

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 _np_x, which is 1/2 = _x+ns₀/_xs₀ = ℓ_x+n/ℓ_x = (1 + x)/(1 + x + n), which becomes n = 1 + x, therefore median [y] = 1 + y.

[Sunny]

Does anyone still need answers to other problems? I have the solutions.

[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

also can you provide solution to #76

thanks

[Sunny]

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_x. Namely, for UDD,
f = _.5│q₆₀ = _.5p₆₀ (1 - _.5p_60.5 _.5p₆₁) = (1 - .5 * q₆₀) [1 - (1 - .5 * q₆₀/(1 - .5 * q₆₀)) (1 - .5 * q₆₁)] = 0.29,

Similarly, for Balducci,
g = _.5│q₆₀ = _.5p₆₀ (1 - _.5p_60.5 _.5p₆₁) = p₆₀/(1 - .5 * q₆₀) [1 - (1 - .5 * q₆₀) (1 - .5 * q₆₁/(1 - .5 * q₆₁))] = 0.298529412,

Therefore, 1000(g - f) = 85.29411765.

The second method employs the number of life as opposed to the q_x. Namely, for UDD,
f = _.5│q₆₀ = (ℓ_60.5 - ℓ_61.5)/ℓ₆₀, make ℓ₆₀ = 100 hypothetically, then

ℓ_60.5 = ℓ₆₀ (1 - .5 * q₆₀) = 100 * (1 - .5 * .3) = 85 and

ℓ_61.5 = ℓ₆₀ (1 - q₆₀) (1 - .5 * q₆₁) = 100 * (1 - .3) * (1 - .5 * .4) = 56,

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

Similarly, for Balducci,
g = (ℓ_60.5 - ℓ_61.5)/ℓ₆₀, using the relationship 1/ℓ_60.5 = (1-t)/ℓ₆₀ + t/ℓ₆₁,

After some algebraic manipulation,

ℓ_60.5 = 2ℓ₆₀ℓ₆₁/(ℓ₆₀ + ℓ₆₁) = 2(100)(70)/170 = 82.35294118

ℓ_61.5 = 2ℓ₆₁ℓ₆₂/(ℓ₆₁ + ℓ₆₂) = 2(70)(42)/112 = 52.5

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

Again 10000(g - f) = 85.29411765.

Ans. B

[Sunny]

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_x = (ω - x)^r and

^°e_x = (ω - x)/(r + 1),

here r = 4 and ω = 100, so we have

µ(x) = 4/(100 - x),

ℓ_x = (100 - x)⁴ and

^°e_x = (100 - x)/5 = T_x/ℓ_x,

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

T_x = ^°e_x ℓ_x = (100 - x)⁵/5

₁₀p₃₀ ^°e_40:20┐ = ℓ₄₀/ℓ₃₀ * (T₄₀ - T₆₀)/ℓ₄₀ = (T₄₀ - T₆₀)/ℓ₃₀ = 5.624323199

Ans. C