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radrach
04-11-2005, 07:30 PM
2001 exam, Q41:

41. (2 points)
Based on Kelley, “Homeowners Insurance to Value - An Update,” and the following
information, calculate the reduction in loss per policy in 2001 which results from non-renewing
the in-force policies with unrepaired roof damages. Show all work.
Expected 2001 statewide average loss cost per policy,
if all policies are renewed 300
Percentage of total policies with unrepaired roofs 10%
Expected percentage of total claims on policies with unrepaired
roof damages resulting from unrepaired roofs 30%

Expected ratio of severity on claims resulting from unrepaired roofs
to severity for all other claims 40%.

This last part I do not understand. The solution shows that the unrepaired roof claims are have a severity 40% higher than all claims. IE, the ratio would be 140%. When I set up this problem, I had E[ severity unrepaired roof claim ]/E[ severity all other ] = 0.4 . Which implies the severity is less than all other claims.

Where is my logic wrong?

dumples
04-12-2005, 10:01 AM
Because you are also providing coverage on the rest of their house too. Just because you have a crappy roof, that doesn't mean that you are less likely to have a pipe in your basement burst and ruin your foundation.

Rice
04-12-2005, 10:31 AM
I agree with radrach. The question should have stated 140% rather than 40%.

Colymbosathon ecplecticos
04-12-2005, 11:29 AM
You can do the problem as stated. Some claims (like major fires) are worse than the roof claims. You are told that the average roof claim is 40% of an average non-roof claim. Also these are extra claims (30% higher frequency) than a normal house. Additionally, 10% of the houses have bad roofs.

Now it's just a little algebra.

dumples
04-13-2005, 12:48 PM
You can do the problem as stated. Some claims (like major fires) are worse than the roof claims. You are told that the average roof claim is 40% of an average non-roof claim. Also these are extra claims (30% higher frequency) than a normal house. Additionally, 10% of the houses have bad roofs.

Now it's just a little algebra.

30% of total frequency is not the same as 30% higher frequency

Colymbosathon ecplecticos
04-13-2005, 01:44 PM
Yes, that's right, there is some algebra to do, like I said. The point is that the 30% roof claims are due to the unrepaired roofs. The problem, as stated, is doable.

dumples
04-13-2005, 03:40 PM
When you compare the question that was asked to the question in the paper, the solutions are the same (except for the 150 and 300 which don't matter), but the questions asked appear very much different. This leads me to agree with people that this isn't correct. However, I do not thing that changing the .4 on the exam to 1.4 is the solution. Regardless, I've already wasted more time on this than I should have.

New question: Is the example in the book correct?

radrach
04-13-2005, 09:49 PM
The example in the book is the way we are supposed to do it on the exam. Regardless of whether we agree with the logic or not. Mahler's notes explain that he would solve it differently (diff. soln.); however, we still have to replicate the example from the text.

tbug
02-08-2007, 12:52 PM
*bump*

I'm having trouble with this question. All10 calculates as follows:

Loss Cost Per Policy with Unrepaired Roof = (s/w avg loss cost)*(%total claims on policies with unrepaired roof damage)*(Claim Severity Factor from unrepaired roofs) + (s/w avg loss cost)*(% total claims on policies w/out roof damage)*(Claim Severity Factor on policies w/out roof damage)

I don't get the logic behind this equation. The loss cost per policy with unrepaired roof is some sort of weighted average of severities on repaired and unrepaired roofs?

carrytheCrřss
02-08-2007, 01:11 PM
*bump*

I'm having trouble with this question. All10 calculates as follows:

Loss Cost Per Policy with Unrepaired Roof = (s/w avg loss cost)*(%total claims on policies with unrepaired roof damage)*(Claim Severity Factor from unrepaired roofs) + (s/w avg loss cost)*(% total claims on policies w/out roof damage)*(Claim Severity Factor on policies w/out roof damage)

I don't get the logic behind this equation. The loss cost per policy with unrepaired roof is some sort of weighted average of severities on repaired and unrepaired roofs?
Per Mahler, I believe you can think of:

1) your first term as the expected losses affected when there is unrepaired roof damage and

2) your second term as the expected losses unaffected by whether or not there is roof damage.

We have to add to 2), the expected losses per policy that will occur regardless of whether roof damage is present, 1), the expected losses per policy affected when there is roof damage. Once I thought of it like this, it made much more sense, and hopefully this will clarify. And if I messed anything up, please help out.

tbug
02-08-2007, 01:28 PM
I think I'm mostly getting it now; thanks!

Aces Wild
02-08-2007, 05:21 PM
2001 exam, Q41:

41. (2 points)
Based on Kelley, “Homeowners Insurance to Value - An Update,” and the following
information, calculate the reduction in loss per policy in 2001 which results from non-renewing
the in-force policies with unrepaired roof damages. Show all work.
Expected 2001 statewide average loss cost per policy,
if all policies are renewed 300
Percentage of total policies with unrepaired roofs 10%
Expected percentage of total claims on policies with unrepaired
roof damages resulting from unrepaired roofs 30%

Expected ratio of severity on claims resulting from unrepaired roofs
to severity for all other claims 40%.

This last part I do not understand. The solution shows that the unrepaired roof claims are have a severity 40% higher than all claims. IE, the ratio would be 140%. When I set up this problem, I had E[ severity unrepaired roof claim ]/E[ severity all other ] = 0.4 . Which implies the severity is less than all other claims.

Where is my logic wrong?

I think that this is a mistake on the part of whomever wrote the question. It is pretty much identical to the example presented in the paper on page 546, where they say "The claims would be 40% higher". So, the question should read "Expected ratio of ...... is 1.40. Stating that the ratio is 0.40 means that bad roof homes are a better risk which does not at all Jive with the paper.

Aces Wild
02-08-2007, 05:27 PM
*bump*

I'm having trouble with this question. All10 calculates as follows:

Loss Cost Per Policy with Unrepaired Roof = (s/w avg loss cost)*(%total claims on policies with unrepaired roof damage)*(Claim Severity Factor from unrepaired roofs) + (s/w avg loss cost)*(% total claims on policies w/out roof damage)*(Claim Severity Factor on policies w/out roof damage)

I don't get the logic behind this equation. The loss cost per policy with unrepaired roof is some sort of weighted average of severities on repaired and unrepaired roofs?

Yeah, I totally see shat you are saying. The first, second, and..... time I read the example on pg. 546 of the paper, I said to mysef "this makes no sense". What does make sense to me is

statewide average loss per policy = (percent of bad roof homes)*X*1.4 +(percent of good roof homes)*X

where X is the average severity for good roof homes.

Clearly I am not getting something here, so time to go talk to my boss who is much much smarter than me!

Aces Wild
02-08-2007, 05:50 PM
Yeah, I totally see shat you are saying. The first, second, and..... time I read the example on pg. 546 of the paper, I said to mysef "this makes no sense". What does make sense to me is

statewide average loss per policy = (percent of bad roof homes)*X*1.4 +(percent of good roof homes)*X

where X is the average severity for good roof homes.

Clearly I am not getting something here, so time to go talk to my boss who is much much smarter than me!

Indeed my boss is much smarter than I am. I was interpreting (as I'm sure many others have) the 40% to be the cost difference between a good and bad roof, but what the author is actually saying is that whatever the severity for bad roofs was last year, it will be 40% higher next year.

lxj034000
02-15-2007, 10:29 AM
Indeed my boss is much smarter than I am. I was interpreting (as I'm sure many others have) the 40% to be the cost difference between a good and bad roof, but what the author is actually saying is that whatever the severity for bad roofs was last year, it will be 40% higher next year.

If this is true, then wouldn't the Statewide Average PP increase for next year? But the last step of calculation in the paper is:
150 = 10% * Loss Cost Per Policy with Unrepaired Roof + 90% *Loss Cost Per Policy without Unrepaired Roof
The "150" is the Statewide Average PP of last year.

Actually, IF you interpret 1.4 as "Expected ratio of severity on claims resulting from unrepaired roofs to severity for all other claims is 140%", then

Loss Cost Per Policy with Unrepaired Roof = X * (%total claims on policies related with unrepaired roof damage)*(Claim Severity Factor from unrepaired roofs) +X * (% total claims on policies unrelated withroof damage)

But the author meant that "Expected ratio of severity on claims resulting from unrepaired roofs to severity for all claims is 140%", so

Loss Cost Per Policy with Unrepaired Roof = (s/w avg loss cost)*(%total claims on policies related with unrepaired roof damage)*(Claim Severity Factor from unrepaired roofs) + (s/w avg loss cost)*(% total claims on policies unrelated with roof damage)

NewTubaBoy
02-16-2007, 08:14 AM
Isn't this almost word for word the exaple right out of the paper?

phillyactuary33
02-20-2007, 10:13 PM
i think it's funny when someone says 'All10 solved it this way'..

if by solved it you mean copied and pasted it directly from the paper, then yes.. they solved it alright..

and in that same respect, they also did a wonderful job of taking a 20 page paper, shrinking the font, making bullet points and little notes saying 'review calculations on page xxx' and making it into 15 pages of notes..

three cheers for the TOTAL lack of effort...

tbug
02-21-2007, 08:32 AM
Yeah, I'm definitely wondering what the heck I bought the thing for. I'm glad i have Infinite Actuary as well, so Ken can actually EXPLAIN the papers instead of just restating them!

NewTubaBoy
02-21-2007, 08:58 PM
I completely agree. All 10 is lame. I guess the online exams are helpful though... if you purchase that part.

badger
03-01-2007, 03:50 PM
Anyone want to take another shot at explaining this to me.

edit:
I don't understand why they use the average loss cost for all houses.

It seems to me the equation to solve would be

.1 (.3*1.4X +.7X) +.9X=300
where X is the expected loss costs of the regular roofs <300

So the term in () is the average loss costs of the leaky roofed houses, which differs from the textbook because X is the expected loss of regular roofs instead of the expected loss of all houses.

It seems like if their equation for calculating the loss costs per policies with unrepaired roofs was correct it would work when applied to the subset of houses with repaired roofs:
0%*1.4*300+100%*1.0*300=300 <> 296

tdconrad
03-09-2008, 04:03 PM
I agree with badger-

The solution seems to be treating $300 as the lost cost for policies w/o unrepaired roof damage. But loss costs for policies without unrepaired roof damage will be less than $300, since the statewide average loss cost has been inflated by the presence of houses with damaged roofs.