Chapter 2: Sec 2

[jensen]

In this section where they rewrite a_x in terms of sum of the survival probabilities multiplied by discount, I don't understand why the sum of k|q_x equals to j+1_p_x.

My understanding is that k|q_x = k_p_x . q_x+k, so how come the sum becomes j+1_p_x?

Can someone please shed some light?

Thanks.

 

[DevonMatthews]

In this section where they rewrite a_x in terms of sum of the survival probabilities multiplied by discount, I don't understand why the sum of k|q_x equals to j+1_p_x.

My understanding is that k|q_x = k_p_x . q_x+k, so how come the sum becomes j+1_p_x?

Can someone please shed some light?

Thanks.

I havnt got CT5 notes with at the moment and have not looked at the subject for ages so don't quote me on this but i think if the sum starts at j+1 (check this) and goes to infinity then the sum of k|q_x is just the probability of the life dying any time after j+1 so its equivalent to surviving up to j+1 given alive at x, ie j+1_p_x.

 

[jensen]

I havnt got CT5 notes with at the moment and have not looked at the subject for ages so don't quote me on this but i think if the sum starts at j+1 (check this) and goes to infinity then the sum of k|q_x is just the probability of the life dying any time after j+1 so its equivalent to surviving up to j+1 given alive at x, ie j+1_p_x.

Thanks Devon

You're right; the sum does start from j+1.

I can see where the p is coming from, but i cant get over the missing q. You're saying that we're certain that the life survived from age x up to x+j+1 but we don't know when he/she is going to die after that, hence the sum is just equal to j+1_p_x. Is my understanding correct?

 

[didster]

Two ways:
1) Intuitive (as Devon said): sum of k|qx from k=j+1 to infinity (or omega if you prefer) is the sum of the probabilities of dying after j+1. To die this late you must have survived to at least j+1.

2) Mathematical :
Let S(j) = sum of k|qx from k= j to ω;

Using induction (albeit backwards) to cut down on writing notation here.

Inductive step:
Assume S(j+1) = (j+1)p(x)

S(j) = j|q(x) + S(j+1)
= j|q(x) + (j+1)p(x)
= jp(x) * q(x+j) + jp(x) * p(x+j)
= jp(x) {q(x+j) + p(x+j)}
= jp(x)

I'll leave proof of the base case to you.

 

[jensen]

...
S(j) = j|q(x) + S(j+1)
= j|q(x) + (j+1)|p(x)
= j|p(x) * q(x+j) + j|p(x) * p(x+j)
= j|p(x) {q(x+j) + p(x+j)}
= j|p(x)

Hi didster

May I know what is this "(j+1)|p(x)" term here actually, please?

I tried expanding the sum (sorry for the notation overload):

Sum[k|q_x; j+1 to inf] = (j+1)_p_x . q_(x+j+1) + (j+2)_p_x . q_(x+j+2) + ...

= (j+1)_p_x . {q_(x+j+1) + p_(x+j+1).q_(x+j+2) + ...}

so if the {} equals to 1 (because that's the prob of a live dying eventually), I have my answer. Is this correct?

 

[didster]

Sorry I put the | in to separate the t in tpx. (removed now)
You're expansion is correct but it doesn't help that much in terms of a rigorous mathematical proof.

The trick is to look at the last two terms, factor out a tpx leaving a p(x+t) and a q(x+t) which sum to 1 - repeat as necessary until you get rid of all the terms at the end which you didn't want.

However, your reasoning relating to the certainty of death is correct and probably sufficient.