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ShortTerm Actuarial Math Old Exam C Forum 

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#41




I feel myself struggling with level 6 Adapt exams, I can pass level 5 with no issue, but there are problems I just get stuck on with my level 6 exams and they take so long kinda panicking now. I'm wondering if I should just continue hitting concepts with level 5 exams..
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#43




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#44




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#45




Which "note" are you referring to? The Supplement to Chapter 3 note or the Individual Health Insurance note? Or both?
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#47




SOA Sample Q108
For the SOA sample problem #108, the solution stated that the distribution is (a,b,0) but since I did not see any definition of p(0) I assumed (a,b,1).
They had the answer being 0.09 (C) but my answer is 0.10 (D). Kindly clarify which answer is correct.
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#48




The definition given is the definition of (a,b,0) (or equivalent to other definitions). (a,b,0) means the relationship between p(0) and p(1) can be expressed in the same form as the relationship between p(1) and p(2), and between p(k) and p(k+1). That same form applies whether you are defining p(k) in terms of p(k1) (as here), for k=1,2,... or whether you choose to define p(k) in terms of p(k+1), for k=0, 1, 2,..., or p(k1) in terms of p(k) for k=1,2,...

#49




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So to clarify, if they had said k=2,3,... instead then either this is Truncated (assume p(0)=0) or Modified (p(0)=some p). And because k starts at 1, which would make the provided recursion contain p(0) which would imply that p(0) can be determined in the regular form of whatever distribution imply (a,b,0). Is that it?
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#50




Yes. But I was focusing on the limits of the expression. The form does have to be right (as it is here).
So if it said (for example) that p(k) = ap(k1)^2 + b, k=1,2,3,... that would not be (a,b,0) even though all the relationships would have the same form and the limits are OK. The form of the relationship would be wrong. 
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