X Assignment X4.14

Hi,

If i'm reading this correctly you are multiplying together two expressions, the value of a term assurance that pays immediately on death to life x:

integral from 0 to 20 [exp(-0.05t)exp(-0.005t)0.005 dt] = 1.99

and the probability that life x dies second in the 20 year term which can be calculated as:

integral from 0 to 20 [(1-exp(-0.005t))*exp(-0.005t)*0.005 dt]

or, given they are subject to the same force of mortality:

0.5*20q(xybar) = 0.004528

Now if we combined all terms together into one integral we would have:

integral from 0 to 20 [ discount factor * prob x survives to time t * force of mortality for x at time t * probability y has died before time t * prob x survives to time t * force of mortality for x at time t]

If we put it this way we can see that we have double counted x's survival probability and their force of mortality (the bolded parts).

It's not the easiest one to articulate in text this one so please let me know if not clear and i'll attempt to explain in another way

Hope this helps
Joe

 

Hi,

So when we set up these EPV expressions we are looking for payment x discount factor x probability.

Payment = 20
Discount factor = exp(-0.05t)

The probability we need here is that life x dies at exact time t and life y has already died which is tpx * mu_x+t * tqy.

The probability that y has already died by time t (tqy) is 1-tpy = 1 - exp(-0.005t)

So here we need an integral from 0 to 20 [20,000 * exp(-0.05t) * exp(-0.005t) * (1 - exp(-0.005t)* 0.005]. If you solve that you'll get the answer given in the solutions.

On your point about E[Kx], I can appreciate why you think you've made a mistake but what you've calculated is actually fine. The reason it's coming out so high is because we're using a constant force of mortality that's very low (and not consistent with human mortality across the full range of ages, maybe more like a bowhead whale...). You could contextualise this by looking in the AM92 Tables and seeing that this force of mortality corresponds with someone who is 56/57 years old and eg 100px = exp(-0.5)=0.6065 which is very high!

Joe

 

Hi,

Quite right! Thanks for pointing that out, I have edited my response above.

We never perform a calculation like v^[E[kx]+1]. This is because the expectation of a whole life assurance paying out at the end of the year (let's say) is not simply the benefit discounted back from the expected time of death. Instead we work out the expectation of v^Kx+1 so multiplying the discount factor that would apply in each year multiplied by the probability that the life dies in that particular year, summed up across all years ie

v*P[Kx=0] + v^2 * P[Kx=1] + ...

If you can I'd try to stick to my approach and thinking above.

If it's any consolation this is an unusual and challenging question. These types of questions will pop up time to time and typically come with a comment in the examiners report to say "this question was poorly answered". It may be one that you'd like to build up to by trying some simpler integral expression questions. Of course you may be intending to submit an assignment and you have a deadline to meet so in that case you just have to give it your best shot!

Joe