Q&A Bank Q1.9

Hi

I had a different approach to this question, and I wondered if it's acceptable because the answers are slightly different.

I used:

2_q_65 = 2 (d_65 + d_66) / l_65

instead of:

2_q_65 = 1 - 2_p_65

where 2_p_65 = (1 - 2 . q_65) . (1 - 2 . q_66)

 

Question in the current CT5 notes is completley different, But why do you have 2*(d65+d66)? wouldnt d65+d66 / l65 give the korrect answer???? ??? ? ? ?

2_q_65 = 1q65 + 1p65 *1q66

= d65/l65 + (l66/l65)*(d66/l66)

= (d65 + d66) / l65

 

G'day Devon

Apologies, but the question asked for probability of a life aged 65 to die before age 67, if the annual initial rate of mortality is twice that of AM92 Ultimate over the next two years.

So is my method correct, if I doubled the deaths?

 

sounds like a legit approach, how much is it out by?

 

Nice idea but I think it may be wrong
Reasoning:
It the initial rate which is doubled, not the deaths (a very subtle difference).
Since we have more deaths at age 65 (last birthday), there are less people to die at age 66. So the number of deaths is slightly less than 2 times the standard number of deaths at 66.

Actually if we extend this beyond two years, it exasperates the error. If we multiply all the numbers of deaths after age 65 up to omega, we'll have twice as many deaths as people that we started with. Clearly this is a flaw, and any proportions based on this will in turn be flawed.

 

Thats a nice piece of reasoning there, over an age range of 1 year though doubbling the deaths would be a perfect aprox to doubbling the initial rate of mortality since 2*qx = (2*dx)/lx, so over two years over this part of the age range the aprox should be resonably valid since lx doesn't really change by that much (proportionally), it would be a terrible aprox. between say x=98-100

 

The difference is about 1%, which can be insignificant but I'm just wondering why the two methods produced slightly different numbers, when doubling the initial rates is the same as doubling the deaths.

 

To answer your question, doubling the deaths is not the same as doubling the initial rates. This is because if you increase the death rates, you leave less people to die at later ages. I said this before and am not sure how I can say it differently so I'll try an example

Say you have 1000000 people aged 65 exact and the base mortality rates are 1%.
At age 66 you are left with 990000 people, ie 10000 deaths
At age 67 you are left with 980100 people, ie 9900 deaths

If you double the mortality rates to 2%.
At age 66 you are left with 980000 people, ie 20000 deaths
At age 67 you are left with 960400 people, ie 19600 deaths

The latter number is not twice as the number under standard rates.

Put another way, you are increasing the chance of dying now, which means that you are decreasing the chance of dying later (as chance of dying at some time sums to 1). Ie, you can't simply increases deaths at all ages across the board.

Devon, you seem to understand the finer issues. It is exactly right for one period, and I agree that it is a reasonably close approximation (turns out to be 1% in this case). However, you should never make approximations, regardless of how good they are, without realising that you are making approximations and appreciate the impact such approximations have. In exams (and real life) you should always note what approximations you make and there are often marks for this.

 

Oh ok. I see the difference now. It'll work for a single period (hence the 20,000 deaths) but for later years the error starts to show.

Cheers didster.