
#1




SOA Sample Question 196
This is not a particularly difficult sample problem, but it is an extremely tedious one. While doing timed practice problems I came across this one, and after spending a few minutes setting up the solution I thought to myself "F**k this, the algebra here is going to kill me!" and I moved on. However, after finishing a few other problems I went back to #196 (after looking at the answer to ensure that there was no special trick I was missing), and decided to time how long it would take me to complete it. Knowing exactly how to solve the problem, and being confident in that method, it took me 9:42 to come up with the correct solution: 3.089. Worse still, the solution is very close to being in between the ranges of the solutions given, which would cause me to panic in an exam setting (I like not having to worry about rounding errors, etc). For those of you who have taken the exam, how many questions are there that require over 9 minutes to come up with the answer (assuming you know exactly how to solve it)? Alternatively, should I be able to solve this problem faster than I did?
__________________
FAP APC 
#3




I'm confused on why the losses with nonzero deductibles are calculated conditioned on those losses exceeding the deductible. What part of the question implies truncation of losses below the deductibles?
__________________
DPGH ADVGH ERM Modules 
#5




I did this question in about 34 minutes
The trick is this... The 3 losses of 750 have a deductible of 200. With deductibles in MLE's, you have to condition on the loss being above the deductible. (also, you can drop the "+1" from the (x+theta)^(a+1), since (x+theta)^1 is just a constant So, for the losses of 750 the likelihood function is (a^3)*(10000^3a)*(10200^3a) divided by (10750^3a)*(10000^3a). So, the (10000^3a) part will cancel out, and you're left with (a^3)*(10200^3a) divided by (10750^3a). But, you can cancel that further once you see that there are 3 losses of 200, which its likelihood function of (a^3)*(10000)^3a divided by (10200^3a). So, the (10200^3a) will cancel out between the 750 losses and the 300 losses. So, combined, their likelihood function is (a^6)*(10000)^3a divided by (10750)^(3a) The same thing will happen with the 4 losses of 400 with a 300 deductible, and the 4 losses of 300. So, you'll have (a^8)*(10000)^4a divided by (10400)^4a for those 8 losses. Now, all you're left with is the 6 censored observations, which have a likelihood function of (10000^6a) divided by (20000^6a). Combined, the likelihood is a^14 * (10000)^13a divided by (20000^6a) * (10400^4a) * (10750^3a) log and take derivative, you have (14 / a) + 13 (ln 10000)  6 (ln 20000)  4 (ln 10400)  3 (ln 10750). So, you get 3.089.
__________________
Last edited by francois; 06032013 at 11:29 AM.. 
#6




Quote:
I understand the math just fine, it's the setup I'm having an issue with. This is the only question I've seen where you are supposed to infer truncation rather than it being a given assumption. In the ASM Chapter 30 practice problems, the only time the formula adjusts for truncation is when there is an explicit assumption to do so. In this question there is no assumption that would prevent an occurrence of a recorded loss being less than the deductible, because there is nothing indicating that losses below the deductible have been truncated.
__________________
DPGH ADVGH ERM Modules 
#7




Quote:
And, specifically in this question, treating it as truncated causes the simplifications I mentioned. Thus, you can infer that the writers intended you to condition on the loss being above the deductible.
__________________

Thread Tools  
Display Modes  

