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Q&Abank 3.12

N

Neetu Verma

Member
Q&A bank part 3 ,Q-no 3.12
in this question ,Assurance is payable immediately on the death of life aged 40 dies after life aged 45 & after his 50th birthday & before 55th birthday i.e between 10 & 15 years. the first integral value of the solution is ok but why & how to introduce second intergal however life ages 40 died within 15 year.
thanks,
 
Two integrals

Let's see like this...

We are modeling death of x (40) at time t. In order for benefit to be paid-
At time t,
1. x must die
2. y (45) should have died within 15 years before t. i.e. between time (t-15) and t.
3. x should be aged 50+, i.e 10 years hence from now (so our t can start values from 10 ...)

This would give us:
v^t : discounting factor
tpx. mu x+t: factor for x's death at time t
t-15py - tpy: probability that y dies between t-15 and t.

Now, let's put limits to integral for all these three factors:
We need t to start from 10 and go to infinity. (see 3 above why t should start from 10).

This should do everything for us..but a small hiccup here... for values of t between 10 and 15, t-15py has no meaning...

So, lets make the limits from 15 to infinity and keep the three factors...it would give your second integral term in solution..

now, for the duration of t=10 through15:
if x dies between 10 and 15 years from now (i.e when x age is 50 through 55), the benefit is paid if y is dead by then. Because, we know y is alive today - and if he dies between 10 and 15 years from now, it would automatically satisfy our point (2) above.

So, our factors would be:
v^t
tpx . mu x+t
tqy (which is also 1-tpy).
Limits for integrals here are 10 to 15.
This is your first integral term in solution.

This is a bit lengthy..and not sure if I could explain it clearly...please let me know if I am not clear..

Thanks,
Raj
 
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