Question Bank Part 5

Has anyone been up to the stage of Q&A Bank Part 5 yet?

First is part i) from Q 5.5 regarding transition matrix of no claim discount system.
I understand that the transition matrix is 5*5 and I understand how to derive the first two rows and the last row, just I dont know why the 3rd and 4th rows become:

(0.3 0 0 0.7 0 )
(0.1 0.2 0 0 0.7)

As if there is exactly one claim was made at the end of the year, a p/holder drops back two levels, but the 3rd low above looks like dropping back three levels and the probability is the sum of 0.1 and 0.2. Can anyone explain this?

The second is part ii) and iii) from Q 5.11.

ii) When calculating K-M estimate, "Durations at 01.01.1998 "are also included, but the period for this mortality investigation period is from 01.01.1998 to 31.12.2003. So why should we include the durations before 01.01.98 of the first two lives?

iii) Estimating the force of matality, how exactly to calculate the central exposed to risk from age 67 to 68 for this particular question? I see the duration line, but I dont know how to get 6 from duration =0 to duration =3m?

Thank you very very much if someone can help. I have thought about them for long, just no clue at all.

Cheers!

 

Avviey,

the third row gives the transition probabilities out of state 3 (40% discount). If one claim is made (with a probability of 0.2) you drop back 2 levels, to state 1 (no discount). If more than one claim is made, you go back to state 1 (no discount) with probability 0.1.

The same thing happens if you're in state 4 (50% discount) - 20% probability of dropping 2 levels to state 2, and a 10% probability of going to state 1.

As for 5.11, the durations at 1.1.1998 correspond to the incremements in the table at the bottom of page 18 in the answers. The durations at 1.1.1998 aren't being included in the exposed to risk, but excluded. For example, the first life retired on 1.4.1995 but we only started observing him on 1.1.1998 so he is only included in the exposed to risk from age 67yrs 9months.

For part iii), the 6 is the number of lives we observe between ages 67yrs and 67yrs 3months. We only observe members 2, 3, 4, 5, 6 & 7 over this period - member 1 doesn't enter observation until he's 67yrs 9m old, while we stop observing member 8 at the end of the investigation on 31 December 2003 when he's only 67yrs 0months old. Since we observe 6 lives over this 3 month period of their lives, this contributes 18 months to the exposed to risk. Similarly we observe 5 lives over the range 67yrs 3m to 67yr 7m (as member 4 died aged 67yr 3m), contributing another 20 months, etc.

Hope that helps,

Giles

 

Q&A part 5

Hi Giles,

Thank you very much for your explanation.

Yes, I got them for Q5 and part iii) of Q11. But there is still something not clear for part ii) of Q11.

If you look at the solution, ok, when j=0, Nj=6, which means the first two lives has been removed, but when duration = 5 months, the 2nd life was added into the exposure. However, this starting point of this duration is on 1 Aug 1997 (his date of retirement). But the starting point of the investigation is 1 Jan 1998. So when looking at tj ( the 2nd column of the last table on page 18), it listed durations with different starting points, this is what confused me as I thought durations all starting from the the same dates, at least this is the first I have encountered so far. Looking at the answer, looks like my thought was incorrect. But I'm not convinced yet.

I do agree the first two lives should be included in the exposure, at least the first one should as the death occurred during the investigation period.

What do you think?

Cheers!

 

Has anyone or any tutor got the chance to have a look at this?
Exam is approaching soon.

Thanks alot.

 

Let's have a look at how to crack a question like this.

I always draw a timeline. Let's see what my timeline looks like.

Remember we are measuring time from the 65th birthday. We're not interested in when (in absolute time) any of these events occurred. Also, I'm going to work in base 12 (so time 0.8 refers to age 65 and 8/12ths).

At time 0, 6 lives enter (lives 3-8, who all became 65 after the start of the investigation, and so are observed from their 65th birthdays onwards).

At time 0.5 another life enters (Life 2, who is 65.5 at the start of the investigation). We now have 7 lives under observation.

At time 2.0, one life is censored (Life 8, who is 67 at the end of the investigation). We now have 6 lives under observation.

At time 2.3, a life dies (Life 4, who is aged 67.3 at death). We now have 5 lives under observation.

At time 2.7, another life dies (Life 7, who dies age 67.7). We now have 4 lives under observation.

At time 2.9, another life enters (Life 1, who is aged 67.9 at the start of the investigation). We now have 5 lives under observation.

At time 2.10, another life is censored (Life 6, who is aged 67.10 at the end of the investigation). We now have 4 lives under onservation.

At time 4.5, there is a death (Life 5, 3 left). At time 5.11 there is a censor (Life 3, 2 left). At time 6.5 there is another censor (Life 2, 1 left). At time 7.1 there is a death (Life 1, all gone).

If you draw out this time line and look carefully at the proportion of lives dying at each death time, you should get the survival function given in the solution.

 

Question Bank Part 1

Thank you very much for the explanation, Michael.

I just have another quick question here about survival probability tPx.

Its the definition that tPx = exp(∫Ux+s ds), integrating from 0 to t.

But for Q 1.19 from Q&A Bank part 1, when calculating the probability of a 60 year old surviving to age 80, it integrates from 60 to 80 rather than 0 to 20.
Can anyone tell me why?? Many thanks.

 

I think this has now been answered in another thread.