View Full Version : 1993 #57: Skurnick Table L question
09-29-2005, 05:27 PM
I tried to solve this buy using the double sum method while the All 10 manual did not. And we don't agree. The table of of loss ratios does not seem to be in synch with the other parts of parts of the problem. Is this the reason my solution did not work. Or, have I missed some fundamental concept here?
I see that the Casualty Study Manuals and the All 10 Manual solve this problem differently and come up with different answers.
The CSM uses a method similar to that found in the Skurnick paper by
a. treating the given "Expected Unlimited Loss Ratio to Standard Premium" in the problem as Skurnick's "Permissible LR" or E<hat>:
b. building a table including adjusting the entry ratios for the difference between the expected unlimited LR (E<hat>) and the actual unlimited LR (E or E(A)): (r=R*(1-k)*S1,0/S2,0);
c. using the *actual* unlimited LR (E or E(A)) when calculating the excess charge as (k+(1-k)*S2,i/S2,0);
d. using the *actual* unlimited LR (E or E(A)) in the calculation of the net insurance charge (c*(XrG-SrH)*E(A))
The All 10 solution, on the other hand,
- Treats the given "Expected Unlimited Loss Ratio to Standard Premium" in the problem as Skurnick's E or E(A);
- does not build an actual table or make the adjustment to r in b., above;
- does not seem to be using k at all in the calculation of XrH in c., above, and
- uses the *expected* unlimited LR in c. & d., above.
I can't be CERTAIN that who, if any, is wrong, but right now, to me, it SEEMS that one of them is.
Here are some questions:
1. Does anyone have access to the CAS official solution for this problem? It would be nice to see if they, for some reason, accepted both methods.
2. Could you please be more specific about how YOU solved the problem, Keith, and the solution that you got? It would make it easier to analyze it then.
09-30-2005, 03:22 PM
The solution will be easier to read in a table format. How do that in this tread?
09-30-2005, 04:05 PM
Note that the mean loss ratio from the data is 67% which is greater than the given unlimited loss ratio of 60%.
Firstly, as you write, the loss ratio of the data is 67%. But the data is LIMITED losses. The denominator in Table L is the UNLIMITED Losses, so you need to divide the 67% by (1-k) = (1-0.0467) giving you 70.6%.
Secondly, Skurnick adjusts his R's to r's (adjusting for the difference between the original expected unlimited LR of 60% and the actual expected unlimited LR of 70.6%).
(see attached txt file)
09-30-2005, 06:24 PM
Thanks for correcting my solution. Somehow I have totally missed performing the adjustment. I'll go back and review the article again.
As to the correctness of one or the other, I'm not sure either. They both seem plausible to me.
I wondered if the 1 risk at a 0% LR might be an issue. I am generally unclear how 0% LR accounts affect Table M (or L) in general. Should they be excluded from the analysis?
To me, the loss at 0% LR means exactly that -- if you were to draw a Lee-style graph, that loss would have a height of 0. Follow the rest of the steps from there.
Effectively, you will get a Lee curve where the curve does NOT intersect the axes at the origin (or even at r>0 (on Y-axis)) rather the curve intersects the X-axis at F(r)>0).
[Referring back to the original question at the top of this thread and my first response...]
FWIW, I just saw the CAS model solution to this question (there are 2 of them, but they are variations of the same solution) and surprisingly (at least to me) they are the same as the All-10 solution!
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