ST3 S07 Q2 (ii)

[mario]

Hi

Please can someone tell me whether my answer to the question is reasonable?

Assumptions:

- each policy sold is identical (apart from when it incepts)
- accidents are incurred uniformly over each policy year
- there are no cancellations
- there is no event delay (so the date of each accident is certain)

My method was to consider the mid point of the accidents that occur in 2006, which is:

for Sept: 1 Nov
for Oct: 15 Nov
for Nov: 1 Dec
for Dec: 15 Dec

(e.g. there are two months of exposure for the November policies for 2006, so given that the accident must occur in 2006 to count, the mid-point is 1 Dec)

Then we can just take a weighted average of the accident dates for each month of policies written:

[1000(10/12) + 1500(10.5/12) + 2000(11/12) + 2500(11.5/12)]/[1000+1500+2000+2500] = 0.9107

Then multiplying that by 12 we get 10.928, so we know that the average date is in November (between 10 and 11 months), and to get the date we just do 30 x 0.928 = 27.84

So the answer I get is 28 November. Is the approach above reasonable?

I don't really understand the method in the examiner's report (but it seems a lot shorter than mine so would like to understand it if possible!)

 

[Ian Senator]

Looks reasonable to me! The examiners said they'd accept any reasonable method as long as the assumptions make sense. It all comes down to how you interpret 'average' eg is it the mean or median? Is it weighted or not?

Another possible method:
1,000 policies were sold in September, giving 1,000 units of exposure in each of September, October, November and December.
1,500 policies were sold in October, giving 1,500 units of exposure in each of October, November and December.
2,000 policies were sold in November, giving 2,000 units of exposure in November and December.
2,500 policies were sold in December, giving 2,500 units of exposure in December.
So, in total, we have 1,000 units of exposure in September, 2,500 in October, 4,500 in November and 7,000 in December. This gives a total of 15,000 units of exposure in 2006.
If we take the accident dates in each month to be on average half-way through each month, then we can weight the exposure to find the mean accident date. For example, September’s 1,000 exposure units happen 0.5 months after 1 September, October’s 2,500 exposure units happen 1.5 months after 1 September, etc, to give an average of:
(1000*.5)+(2500*1.5)+(4500*2.5)+(7000*3.5)/15000=2.66 ie 20th November.

An alternative, using the median:
We know we have 15,000 exposure units in total for 2006. So the median exposure unit will be the 7,500th one. September has 1,000, October has 4,500, and so the 7,500th one will be 4,000 units through November’s 4,500 units, ie 8/9ths of the way through November (a few days later than the mean).

 

[mario]

Thanks Ian - the alternatives all make sense too