ASM PE5 Q12

[boxbox122]

Survival time for each of (x) and (y) follows deMoivre's law with omega = 100.

Calculate 10|10 q(60:70) with 70 dies first.

I use 10p(60:70)*integral(tp70*tq80*mu(70))from 0 to 10. Did I miss something?

Thanks!!

[levenzaha]

don't you need to double integrate and have y go from 0 to x

[actuary_aspire]

Originally Posted by boxbox122

Survival time for each of (x) and (y) follows deMoivre's law with omega = 100.

Calculate 10|10 q(60:70) with 70 dies first.

I use 10p(60:70)*integral(tp70*tq80*mu(70))from 0 to 10. Did I miss something?

Thanks!!

Should be tp80*mu(80+t)*tp70.

[Math]

actuary aspire is right.

Also you can do:

int from 10 to 20 of:

tP60 tP70 mu_70+t

At least i think you can, maybe someone can confirm.

[actuary_aspire]

Math,

Yeah, that works too.

[RMDoolabh]

To follow up to Math's post.

int ( t_q_70 * t_p_60*u_60+t, t, 10,20 ) + 10|10_q_70 * 20_p_60

Shouldn't this also work? When I do the work out I get 7/24 -- not 5/24?

Am I missing something here? This has been bugging me for quite some time.

[Bballry1234]

Originally Posted by Math

actuary aspire is right.

Also you can do:

int from 10 to 20 of:

tP60 tP70 mu_70+t

At least i think you can, maybe someone can confirm.

I'm a big fan of this way of integrating t|s q x problems...it takes the thinking out of it!

[boxbox122]

Originally Posted by actuary_aspire

Should be tp80*mu(80+t)*tp70.

if 10q(70:80)with 80 dies first is equal to integral[tp80*mu(80+t)*tp70] from 0 to 10, dose 10q(70:80)with 80 dies first mean (80) dies first within 10 years but (70) is alive during this 10 years? what's the meaning for 10p(70:80) with 80 dies first?

Thanks!

[actuary_aspire]

Originally Posted by boxbox122

if 10q(70:80)with 80 dies first is equal to integral[tp80*mu(80+t)*tp70] from 0 to 10, dose 10q(70:80)with 80 dies first mean (80) dies first within 10 years but (70) is alive during this 10 years? what's the meaning for 10p(70:80) with 80 dies first?

Thanks!

It means that (80) dies within the next 10 years and (70) is still alive at the time of death of (80).

[pizzabox23]

For these type of problems, it's often De Moivre, for which case I like to use the "draw a picture" method described in ASM.

Draw the total area (a 30 x 40 rectangle). (height=30, width=40)
Draw a diagonal line from bottom-left corner to point (30,30). Above the line (triangle) is the probability of (60) dying first, and below the line (the trapezoid) is the probability of (70) dying first.
Then to account for the part that they live 10 years, and die within the next 10 years, you need the "slice" of the trapezoid from age 80 to 90.

Here are the coordinates of the region I am describing (since I don't think I'm doing a great job describing it): (10,10), (40,10), (40,20), (20,20)

That is a trapezoid with area 250. Divide that by the area of the rectangle, 1200. That leaves 5/24.

Remember that this only works for De Moivre, because the deaths are uniformly distributed throughout the rectangle.