About
January 14th, 2008
This blog will be used to communicate with students registered for The Infinite Actuary’s Exam MLC Online Seminar.
My name is James Washer. You can find out more information about me here.
This blog will be used to communicate with students registered for The Infinite Actuary’s Exam MLC Online Seminar.
My name is James Washer. You can find out more information about me here.
I am currently working on on calculating premiums using the equivalence method. Although I feel i have a good grasp of the material, my issue with these kinds of problems is that they involve a lot of calculator work and it winds up taking me a long time just to input the data. Do you have any calculator tips for these kinds of problems.
Other than for DML (use financial keys to calculate insurance factors) I don’t have any shortcuts. The best way to get faster is to work a ton of problems. Many students prefer the TI-30 XS Multiview for these types of problems.
When i solve problems requiring a transfer from the discrete to the continuous using the “i over omega” formula, if the interest rate is 6%, my study manual tells me that these values are available in the tables given on the exam. I am unable to find this. Can you tell me where this is? Also,in reference your last post, what does DML stand for?
I tried solving #25 for Fall 2001 of the CAS/SOA exam. I tried using the reserve formula tVx= (Ax+t-Ax)/(1-Ax). I got terribly stuck and could not solve the problem that way. Where did I go wrong? Is this formula only valid with a constant benefit? Thanks.
DML – DeMoivre’s Law
Do you mean “i over delta”? delta = ln(1+i). You don’t need a table for that, but I think the table you are asking about is on the last page of the MLC notes.
https://img.skitch.com/20111004-pwitx2a4kepshxd16ep7axejcc.jpg
Correct. For non-level benefit insurance or non-level premium insurance you can only use the prospective and retrospective method.
I was doing Fall 2004 #20 SOA and I was unsure about the wording of these kinds of problems. When they say the probability that x survives are they referring to the probability without the common shock or with it? Is there a general rule for how to determine this based on the question’s wording?
In order for x to survive he has to survive the common shock too.
Did you watch the lesson on common shock? I detail how I always solve these problems.
I will have to watch it. I have come across problems that state that mortality follows a “linear model”. This looks a lot like DML.Is it?
Sounds like it. Could be UDD if just linear between integral ages.
For Fall 2000 #7 SOA/CAS you are given the three associated decrements and told that withdrawals don’t happen until the end of the year. So, I got that.I understood that you ignore the withdrawals until the end of the year and treat it like a two decrement model. What I am not sure of is: after I calculate the total survival until year end, do I use the associated decrement for withdrawal or do I use or I need to find the true q60^(3).
The true probability would already account for surviving the other two decrements. I think in the lesson I work an exact example like this one. If not exactly like it then one very similar.
I tried the Fall 2004 #10 SOA and I tried to use the retrospective method to accumulate the premiums without having to deal with the premium change after the 4th year. My answer was none of the choices.What did I do wrong?
The retrospective method only works for benefit reserves. This problem asks for a contract reserve. So you have to use the prospective method.
What’s the difference between contract reserve and benefit reserve?
Contract reserve is the present value of future benefits (and maybe expenses depending on the wording of the problem) minus the present value of future contract premiums. For contract reserves you can only use the prospective method.
Benefit reserves are what we are talking about 95% of the time in MLC.
I just worked on a problem that referred to “the net amount at risk for policy year h.” What does this term mean?
See lesson B.5.3 page 1.
When calculating the variance of the loss random variable, how do you know which formula to use and what verbal clues will the problem give? Thank You
Jacob, it just depends on the given information. Can you give me some specific examples?
This is a problem from my review manual: For fully continuous whole life insurance
1) Ax(bar)=1/3
2) delta=.1
3)L is loss at issue random variable using the fully continuous level annual benefit premium based on the equivalence principle.
4) Var(L)=1/5
5)L’ is the loss at issue random variable using the continuous level premium of .2.
Calculate Var (L’).
So, I understood that you need to use the variance formula given in step 4) to find the value of 2^A(bar) and then use that with the new premium to calculate the answer. By trial and error I used the (2^Ax(bar)-Ax(bar)^2)/(delta*ax(bar)for the original calculation and (b+(P/delta)(2^Ax(bar)-Ax(bar)^2) for the second step. It is notclear to me why though.
That should be (2^Ax(bar)-Ax(bar)^2)/(delta*ax(bar)^2
Jacob, when posting a 3rd party problem please give me the suggested answer (not the solution just the value) so I can be sure I didn’t make a careless mistake. Anytime you have the variance for one premium and the variance for another premium you should look at the ratio (like I covered in the lesson). See this solution. Does that help?
http://dl.dropbox.com/u/426149/mlc/mlc_blog_solution.pdf
Yes,I think I understand. I was working on the Spring 2007 Exam and I was totally lost on number 21. I knew to integrate, but could not get rid of the alpha term.
Also, on that same exam #24 I somehow went wrong. I recognized the integral to be the exponent of the survival function, then i added the annuities up. A certain 1 + (e^-.01)/1.11 + (e^-.01*2^1.2)/(1.11)^2. Apparently, that is wrong. Please explain. Thank you.
Also, is there any way to be prepared for a question like #14 on that exam.
You don’t have to integrate for #21.
http://dl.dropbox.com/u/426149/mlc/mlc_Spring07_21.pdf
For #24, you must be making a mistake somewhere.
http://dl.dropbox.com/u/426149/mlc/mlc_spring07_24.pdf
For #14, yes. See lesson B.6.5.
When i tried #26 0n the 2007 exam i got very mixed up. I played around with the numbers enough and got the right answer, though. Can you explain the solution please.
Also, for #22 on that exam. Thank you.
#26 – http://www.youtube.com/watch?v=ZnXSETYvLQ4
#22 – http://www.youtube.com/watch?v=aWtYzEey00s
For SOA sample problem 151: I know how to solve it, I just am not used to the notation. What does p^01,p^02, etc mean?
Jacob, you must be looking at the new 282 (for the Spring 2012 sitting). You can find the SOA problems for the Fall 2011 sitting here:
Questions: http://www.soa.org/files/pdf/edu-2008-spring-mlc-questions.pdf
Solutions: http://www.soa.org/files/pdf/edu-mlc-09-08-solutions.pdf
Could you show me how to do those types of problems where you get the benefit reserve in addition to the death benefit?
See lesson B.5.3 pages 3-5.
Can you explain SOA #213
We want the expected time until the first train arrives. From the poisson lesson this is the first inter arrival time. Inter arrival times are distributed exponentially with mean 1/lambda. So for the first 45 minutes we have an exponential distribution with mean = 30. And thereafter we have an exponential distribution with mean = 15/2.
What mortality assumption do you think of when you are told exponential? Constant force!! Under constant force future lifetime is distributed exponential with mean 1/mu. So what we have here is the time until the first train is constant force with mu = 1/30 for the first 45 minutes and 1/(15/2) = 2/15 thereafter.
That is the SOA’s second approach and the method I would use.
A questions on contignet probabilities – prob of (x) dies before (y) – the formula on the top of page 9/23 of section I.2.1 is inconsistent with the formula on the bottom of page 2/23 of the same section. I think the formula on page 2 should be the correct answer. Further, i think it is coincidence that A, B & C on page 9 add up to 1 because the prob of (y) living forever is 0.
@Ting-Ting Ting, the symbols on page 2 and page 9 are equivalent. If I have two lives x and y, if x dies first then y dies second. If x dies second then y dies first. So the probability that x dies first is equal to the probability that y dies second.
Yes, the probability of (y) living forever is 0, but what we have added up is the probability that (x) dies before the end of time which is 1. We broke up this probability into three mutually exclusive exhaustive cases: (A) before y, (B) more than n-years after the death of y and (C) within n-years after the death of y. (x) has to die during one of those cases, so the sum of the three probabilities must equal 1.